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S T U D E N T S O L U T I O N S M A N U A LF O R M E N D E N H A L L , B E A V E R , A N D B E A V E R ’ S
I N T R O D U C T I O N T O
P R O B A B I L I T Y
S T A T I S T I C S
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Duxbury
Student Solutions Manual
for
Mendenhall, Beaver, and Beaver's
tro d u ctio n to P ro b a b ility and Statistics
Twelfth Edition
Barbara M. BeaverUn iversity o f California, Riverside
T H O M S O N
* — -------
B R O O K S / C O L E
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Table of Contents
1: Describing Da ta with Graphs
2: De scribing Data with Nu merical M easures
3: Describing Biva riate Data
4: Prob ability and Probability Distributions
5: Several Usefu l Discrete Distributions
6: The Norm al Probability Distribution
7: Sam pling Distributions
8: Large-S am ple Estimation
9: Large-S am ple Tests of Hypotheses10: Inference from Small Sam ples
11: Th e Analysis o f Variance
12: Linear Regression and Correlation
13: M últiple Regression Analysis
14: Analysis o f Categ orical Data
15: N on parametric Statistics
1
9
23
31
43
53
65
71
81
91
107
119
135
141
151
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1: Describing Data with Graphs
a The experime ntal unit , the individual or object on which a v ariable is measured,
is the student.
b The ex perimental u nit on w hich the nu m ber of errors is me asured is the exam.
c The experimental unit is the patient.
d The experimental unit is the azalea plant.
e The experime ntal unit is the car.
The popu lat ion of interest consists o f voter opinions (for or against the candidate) at
the time of the election for all perso ns voting in the election. No te that when a sample
is taken (at some t im e p rior or the elect ion), we are not actual ly sam pling from the popu la tion o f in te re st. A s tim e passes, vote r opin io ns change. H enee, the popula tion
o f voter opinions chan ges w ith t ime, and the sample may not be representative of the
popu la tion o f in te re st.
a The variable “reading score ” is a quanti tat ive variable, which is probably
integer-valued and henee discrete.
b Th e individual on wh ich the variable is m easured is the student.
c Th e pop ulation is hypo thetical - i t does not exist in fact - but consists of the
reading scores for all students who could possibly be taught by this method.
a The percentages given in the exercise only add to 94% . W e should add another
category called “Other”, which will account for the other 6 % o f the responses.
b Eithe r type of cha rt is appropriate. Since the data is already p resented as
perc enta ges o f th e w hole g ro up. we choose to use a p ie chart, sh ow n in the fi gure
belo w .
c Answers will vary.
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1.15 a Th e total percentage of responses given in the table is only
(40 + 34 + 19)% = 93% . Henee there are 7% o f the opinions not record ed. w hichshould go into a category cal led “O ther” or “More than a few days” .
b Yes. The bars are very cióse to the correct proportions.
c Similar to previou s exercises. The pie chart is shown below. The ba r cha rt is
1.21 a Since the variable of interest can only take the valúes 0, 1, or 2, the classe s can
be cho sen as th e in te ger valú es 0 , 1, and 2. The ta ble belo w show s th e c la sses, th eir
corresp ond ing frequencies and their relative frequencies. Th e relative frequeney
histogram is shown below. ___________ _______________________
V alué Frequeney R elative Frequeney
0 5 .25
1 9 .452 6 .30
0.5
b Using the table in part a, the proport ion of measurements greater then 1 is the
same as the p roport ion o f “2”s, or 0.30.
2
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c Th e propo rt ion o f m easurem ents less than 2 is the same as the proport ion of “0”s
and “ l ”s , or 0 .25 + 0 .45 = .70 .
d The proba bil ity of select ing a “2” in a random select ion from these twenty
m easurem ents is 6 / 2 0 = 30 .
e Th ere are no outl iers in this relat ively sym m etric, m oun d-shaped distr ibution.
a Th e test scores are graphed u sing a stem and leaf plot generated by Mi ni t ab .
Stem-and-Leaf Display: ScoresStem-and-leaf of Scores N = 2 0Leaf Unit = 1.0
2 5 575 6 1238 6 5789 7 2(2) 7 56
9 8 247 8 66793 9 134
b-c The d istribution is not m ound -shaped , but is rather bimoda l w ith two peaks
centere d around the scores 65 and 85. Th is might indícate that the students are
divided into two group s - those w ho unde rstand the ma terial and do well on exams,
and those who do not have a thorough com m and o f the m ater ial.
a The data ranges from .2 to 5.2, or 5.0 units. Since the num ber of class intervals
should be betw een five and tw enty, we cho ose to use eleven class intervals, with each
class interval having length 0.50 (5.0/11 = .45 , w hich, round ed to the nearest
con venien t fract ion, is .50). W e m ust now select interval bou ndaries such that nom easurem ent can fal l on a bou nda ry point . Th e subintervals .1 to < .6 , .6 to < 1.1,
and so on, are convenient and a tally i s constructed. ^ ^ _______________________
Class
i
Class
Boundaries
Tally í Relative frequency,
f j n
1 0.1 to < 0.6 11111 11111 10 .167
2 0.6 t o < 1.1 11111 11111
11111
15 .250
3 1.1 to < 1.6 11111 11111
11111
15 .250
4 1.6 to < 2.1 11111 11111 10 .1675 2.1 to < 2.6 1111 4 .067
6 2.6 to < 3.1 1 1 .017
7 3.1 to < 3.6 11 2 .033
8 3.6 to < 4.1 1 1 .017
9 4.1 to < 4 .6 1 1 .017
10 4.6 to < 5.1 0 .0 0 0
11 5.1 to < 5.6 1 1 .017
The relat ive frequency histogram is shown below.
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15/60
10/60
IV
I 5/60
0
a The distribution is skewed to the right, with several unusually large
observations.
b Fo r som e reason, one person had to wait 5.2 minutes. Perhaps the superm arket
was un derstaffed that day, or there may have been an unu sually large num ber of
custom ers in the store.
c The two graphs convey the same information. The stem and leaf plot allows us
to actually recreate the actual data set, while the histogram does not.
1.35 a H istograms will vary from student to student. A typical histogram, generated by
Mi n i í ab is show n on the next page. ^
b Since 2 o f the 20 players have averages abov e 0.400, the chance is 2 out of 20 or
2 / 2 0 = 0 . 1 .
1.39 To d etermine w hether a distr ibution is likely to be skewed, look for the likelihood of
observing extremely large or extrem ely small v alúes o f the variable of interest.
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a The distr ibution of non-secured loan sizes might be skewed (a few extremely
large loans are possible) .
b The distr ibution o f secured loan sizes is not l ikely to contain unu sually large or
small valúes.
c N ot likely to be sk ew ed.
d N ot likely to be sk ew ed.e If a packag e is drop ped , it is likely that allthe shells will be bro ken. H enee, a
few large n um ber of broken sh ells is po ssible.The distribution will be sk ew ed.
f If an animal has one t ick, he is likely to have m ore than one. There will be some
“0”s w ith uninfected rabbits, and then a larger num ber of large valúes. The
distribution will not be symmetric.
1.43 a Stem and leaf displays may va ry from student to student. Th e most obvious
choice is to use the tens digit as the stem and the ones digit as the leaf.
7 | 8 9
8 | 0 1 79 | 0 1 2 4 4 5 6 6 6 8 8
1 0 1 1 7 9
11 | 2
Th e display is fair ly mo und-sha ped, w ith a large pe ak in the m iddle.
1.47 A nsw ers wil l vary from student to student . The students should notice that the
distr ibution is skewed to the r ight w ith a few presiden ts (Truman , Cleveland, and F.D.
Roosevelt) east ing an unusually large number of vetoes. _________________
Vetoes
1.51 a The popu lar vote w ithin each state should vary depen ding on the size of the
state. Since there are several v ery large States (in po pulation ) in the U nited States, the
distr ibution should be skew ed to the r ight .
b-c Histogram s w ill vary from stud ent to student , bul should resem ble the histogram
generated by Mi n i t ab in the f igure below . Th e distribution is indeed skewed to the
right , with two o utl iers - Ca lifornia and New York.
5
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O 1000 2000 3000 4000 5000Popubr Vote
1.55 a-b An swers w ill vary from student to student . The l ine chart should loo k sim ilar to
the one show n below.
c Th e percentage of people who were not w orried was r ising at a slow rate until
Sep tem ber 11, 2001 , at wh ich time the percentages reversed them selves dram atical ly.
d The horizontal axis on the www.g a l lup . co m chart is not an actual time line, so
tha t the t ime f rame in which these changes occur m ay be dis tor ted.
1.59 a-b An sw ers wil l vary. A typical histogram is show n below. N otice the gaps and the
b im odal natu re o f th e h is to gram , p robab ly due to th e fa ct th at th e sam ple s were
collected at different locat ions.
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1.63 a-b The M ini t ab s tem and le af plot is show n below. The d istr ibution is sl ightly
skew ed to the left .
Stem-and-Leaf Display: PercentStem-and-leaf of Percent N = 5 1
Leaf Unit = 1.0
1 0 72 0 83 1 04 1 36 1 4512 1 66677720 1 88888999
(11) 2 0000000111120 2 2222233312 2 444445554 2 6771 2 9
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c G eorgia (7.5) and Arkan sas (8.0) have gasoline taxes that are som ew hat sm aller
than m ost, but they m ay no t be “outliers” in the sense that they lie far away from therest o f the m easurem ents in the data set.
1.67 a-b The distr ibution is approximately mound-shaped. with one unusual
m easurem ent, in the class w ith midpoint at 100.8°. Perhaps the person w hose
temperature w as 100 .8 has som e sort of i l lness coming on ?
c Th e valué 98.6° is slightly to the r ight of center.
1.69
a The distr ibution is somewhat mound-shaped (as much as a small set can be);
there are no outliers.
b 2/10 = 0. 2
1.73 a Th ere are a few extreme ly large num bers, indicating that the distr ibution is
p ro bably skew ed to th e right.
b-c The d istribution is indeed skewe d r ight with three possible outliers - Y ahoo!,
Time Warner and MSN-Microsof t .
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2: Describing Data with Numerical Measures
2.1 a The dotplot shown below plots the f ive measurements along the horizontal axis.
Since there are two “ 1 ”s, the correspon ding d ots are placed one abo ve the other. The
approxim ate center of the data appears to be around 1 .
b The mean i s the sum o f the m easurements d ivided by the num ber of
measurements , or
2.5
_ = ^ = 0 + 5 + 1+ U 3 = m = 2 x ~ n 5 ~ 5 ~
To ca lcúlate the m edian, the observations are f i rst ranked from sm allest to largest : 0,
1, 1, 3, 5. Then since
n = 5 ,the positiono f the m edian is 0.5(n + l) = 3, and the m edian is the 3rd ranked
measurement , orm = 1 .The mode is the measurement occurring most frequently, or
mode = 1 .
c The three measures in part b are located on the dotplot . Since the m edian and
m ode are to the left of the mean, w e conclude that the me asurem ents are skewed to
the right.
a Although there may be a few households who own more than one DVD player,
the m ajori ty should own e i ther 0 or 1. Th e distribution should be sl ightly skew ed to
the right.b Since most households wil l have only one DVD player, we guess that the mode
is 1.
c The m ean is
Z jL = 1 + 0 + - + , a 27
n 25 25
To calcúlate the m edian, the ob servations are f i rst ranked from sm allest to largest:
Th ere are six Os, thír teen ls, four 2s, and two 3s. The n since n = 25 , the position o f
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the me dian is 0.5(n +1) = 13, w hich is the 13th ranked m easurem ent, or m = 1. The
mode is the measurement occurring most frequently, ormo d e = 1 .
d The relat ive frequency histogram is shown below, with the three measures
superim pose d. N otice that the m ean fal ls sl ightly to the right of the median an d
m ode, indicat ing that the m easurem ents are slightly sk ewe d to the r ight .
0 1 2 3 VCRsmedian meanmode.
2.9 The distr ibution o f sports salaries wil l be skew ed to the r ight , becau se o f the very high
salaries of som e sports figures. Henee , the me dian salary w ould be a bet ter measureo f cen ter than the mean.
n 5
b Create a table of differences, ( x ¡ - x ) and their squares, (*. - T ) 2 .
X¡ x¡ - x ( * / - * )
2 - 0 .4 0.161 - 1 .4 1.96
1 -1 .4 1.96
3 0 .6 0.36
5 2 .6 6.76
Total 0 1 1 .2 0
Then^
j3 _ ! ( * , - ? ) • _ ( 2 - 2 A ) 2 -i------H 5 - 2 .4 ): . 11.20 . „
n - 1 4 4
c The sam ple standard deviat ion is the posi t ive square root of the variance or
s = V 7 = V l 8 =1 .673
Calcúlate ' Z x f = 2 2 + 1 2 +- - - + 52 = 4 0 . T he n
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2.17
2.19
2.21
n _ 1 - = — = 2.8 a nd s = y f ? = - J l .8 = 1.673 .4n - 1 4
Th e results o f parts a and b are identical.
a The range is R = 2 .39 -1 .28 = 1 .11 .
b Calcúlate I *,2 = 1,282 + 2 .3 9 2 + • •• + 1 .5 l2 = 15.415 . Th en
^ , ( I x ) 2 (8 .56 )2 Z * ? - - 15.451 —-— —1— 7 . n ? 8 — ---------" — = -------------------5 — = ^ 6 0 2 8 _ 1 9 0 0 7
n —1 4 4
and 5 = 7 7 = V. 1 90 07 = .4 36
c The range, R = 1.11, is 1.1 1/.436 = 2.5 standard d eviations.
The range of the data is R = 6 - 1 = 5 and the range approxima t ion wi th n = 10 is
s ~ — = 1.673
The s tandard d eviation o f the sample is
= V 7 =U -
& * , f n _
n - 1
130 — (32)
= 7 3 . 0 6 6 7 = 1.7 51
which is very cióse to the estimate for part a.
c-e From the dotplot on the next page, you can see that the data set is not mound-
shaped. Henee you can use T ch eb ysh eff s Theo rem, but not the Em pirical Rule to
describe the data.
a The interval from 40 to 60 represents f i ± g = 50 ± 10. S ince the distribu tion is
relatively m ound-shaped , the proportion o f m easurem ents between 40 and 60 is 6 8 %
according to the Empirical Rule and is shown on the next page
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b A gain, using the Em pirical Rule, the interval / / ± 2 o = 50 ± 2(10) or between 30
and 70 co nta ins approximately 95% o f the measurements .
c Re fer to the figure below .
Since approximately 6 8 % o f the measurements are between 40 and 60, the sym metry
o f the dis t ribut ion im plies that 34% o f the measurem ents are b etween 50 and 60.
Similarly , s ince 95% o f the measurements are betw een 3 0 and 70, approximately
47.5% are between 30 and 50. Thus , the propor t ion o f m easurements between 30 and60 is
0.34 + 0.475 = 0.815
d From the f igure in part a, the propor t ion o f the m easuremen ts between 50 and 60
is 0.34 and the proportion o f the mea surem ents w hich are greater than 50 is 0.50.
Th erefore, the propo rtion that are greater than 60 must be0 . 5 - 0 . 3 4 = 0 .1 6
2.25 A ccording to the Em pirical Rule, if a distr ibution o f m easurem ents is appro xim ately
mound- shaped .
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a approximately 6 8 % or 0 . 6 8 o f the m easu rem ents fall in the interval
H ± o = 12± 2.3 or 9.7 to 14.3
b app roxim ately 95% or 0.95 of the me asurem ents fall in the interval
/ / ± 2 a = 12 ± 4.6 o r 7.4 to 16.6
c app roxim ately 99.7% or 0.997 o f the measu reme nts fall in the interval
/y ±3<r = 12 ± 6.9 or5 .1 to 18.9
Th erefore, app roxim ately 0.3% or 0.003 wil l fal l outside this interval.
2.31 a W e choo se to use 12 classes of length 1.0. The tal ly and the relat ive frequency
-------------
C la ss i C la ss Bo un da rie s Tally fi Relat ive frequency , / / / !
1 2 to < 3 1 1 1/70
2 3 to < 4 1 1 1/70
3 4 to < 5 111 3 3/70
4 5 to < 6 11111 5 5/70
5 6 to < 7 11111 5 5/70
6 7 to < 8 1 1 11 1 11111 11 12 12/70
7 8 to < 9 11111 11111 11111 111 18 18/70
8 9 to < 10 11111 11111 11111 15 15/70
9 10 to < 11 11111 1 6 6/70
10 11 to < 12 111 3 3/70
11 12 to < 13 0 0
12 13 to < 14 1 1 1/70
20/70
« 10/70■E
L1
5 10TRH3
15
V* y 1b Calcúla te n = 70, X x = 5 4 1and X x f = 4453 . T hen x = —— = ------ = 7.729 is
n 70
an es t ím ate of / / .
c The sample standard deviat ion is
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U - í 541):J = V — = \ -------------- ^ — = V 3 .9 39 8 = 1.985
\ n - 1 \ 69
Th e th ree in te rv a ls , T ± ^ f o r k = 1,2,3 are calculated below. Th e table shows the
actual percentage of me asurem ents fal ling in a part icular interval as well as the
percenta ge pred ic te d by T cheby sh eff’s T heorem and th e E m pir ic a l R ule . N ote that
the Em pirical R ule should be fair ly accurate, as indicated by the m ound-sha pe of thehistogram in part a. ______________________________________________________________
k x ± k s Interval F raction in
Interval
Tchebyshef f Empir ical
Rule
1 7 .729 ±1 .985 5.744 to 9.714 50/70 = 0.71 at least 0 - 0 . 6 8
2 7 . 7 2 9 ± 3 . 9 7 0 3.759 to
11.699
67/70 = 0 .96 at least 0 .75 - 0 . 9 5
3 7 .7 2 9 ± 5 .9 5 5 1.774 to
13.6847 0 /7 0 = 1.00 at least 0 .89 - 0 .9 9 7
2.35 a Calcúlate R = 2 .3 9 -1 .28 = 1 .11 so that s - R / 2 .5 = 1.11/2.5 = .444 .
b In Exercise 2 .17, we calculated X*. =8.56 and
I *,2 = 1.2 82 + 2 .3 9 2 +--- + 1.512 = 15.41 5. Th en
,2■> ( Z * (8. 56)"
I xf-± 15.451 - - L.76028
= .19007n - 1 4 4
and 5 = V 7 = V-19007 = .436, which i s very cióse to our es tímate in par t a.
2.39 a T he data in this exercise have been arranged in a frequen cy table.
Xi 0 1 2 3 4 5 6 7 8 9 10
f i 10 5 3 2 1 1 1 0 0 1 1
Using the frequency table and the grouped formu las, calcúlate
' Lx i f = 0(10) +1(5) H------1-10(1) = 51
X * ,2/; = 0 2(10) + 12(5) h------1-102 (1) = 293
Then
I = I i A = 5 1 = 2 0 4
25n
U f -( Z x i í ) 2 2 9 3 -
s~ =
(51)1
^ - = 7.873 a ndn - 1 24
s = y ¡ l . 8 7 3 = 2 .8 0 6 .
b-c The three intervals x ± k s for k = 1,2,3 are calculated in the table a long with the
actual proport ion of m easurem ents fal ling in the intervals. T ch eb ys he ff s Th eorem is
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satisf icd and the approximation given by the Empirical Rule are fair ly cióse for
k = 2 and k = 3 .
k x ± k s Interval F raction in
Interval
T ch eb y sh eff E m pirical
Rule
1 2.04 ± 2 .8 06 -0 .7 6 6 to 4.846 21/25 = 0 .8 4 at least 0 = 0 . 6 8
2 2 .0 4 ± 5 .6 1 2 -3 .5 7 2 to 7.652 23/25 = 0 .92 at least 0.75 = 0.95
3 2.04 + 8.418 -6 .3 7 8 to
10.458
25/25 = 1.00 at least 0.89 - 0 .9 9 7
2.41 The data have already been sorted. Find the positions of the quartiles. and the
measurements that are just above and below those positions. Then f ind the quartiles
by in te rp o la ti on .
Sorted D ata Set P osition o f
Q.
Above
and below
Qi
1, 1.5, 2 ,2 , 2.2 .2 5 (6 )= 1.5 1 and 1.5 1.25
0, 1.7, 1.8, 3.1, 3.2,
7, 8 , 8 .8 , 8.9, 9, 10
.25(12) = 3 None 1.8
.23, .30, .35, .41,
.56, .58, .76, .80
.25(9) = 2.25 .30 and .35 .30 + .25(.05) = .3125
Posit ion o f Q 3 Above and below q3
.75(6) = 4.5 2 and 2 .2 2.1
.75(12) = 9 None 8.9
.75(9) = 6.75 .58 and .76 .58 + .75(.18) = .7150
2.45 The ordered data are:
2 , 3 , 4 , 5 , 6 , 6 , 6 ,7 ,8 , 9 , 9, 10, 22
Fo r « = 13, the p os i tion o f the median is 0 .5(« +1) = 0.5(13 +1) = 7 and m = 6 . The
p osit io ns o f th e q uartil es are 0.2 5(n + l) = 3.5 and 0.7 5(n + l) = 10 .5 , so th atQ¡ = 4.5, Qy = 9, and IQR = 9 - 4.5 = 4.5 .
The l ower and upper f ences are:
Qx - 1 . 5 I Q R = 4 .5 - 6 .75 = -2 .25
Q, + 1.5/0/? = 9 + 6.75 = 15.75
The valué x = 22 lies outside the upp er fence and is an outlier . The b ox plot is shown
belo w . T he low er w h is ker connects th e box to th e sm allest valué th at is not an
outlier , which happens to be the mínimum valué, x = 2. The upper whisker connects
the box to the largest valué that is not an outlier or x = 1 0 .
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2.49 a For n = 18 , the po sition of the m edian is 0.5(n +1) = 9.5 an d the p ositions o f the
qua rtiles are 0.25(n + 1) = 4.75 and 0.75(n + l) = 14.25 . Th e low er quartile is V a the
way betw een the 4 lh and 5111 measurements and the upper quartile is V a the way
betw een th e 14Ih an d 15,h m easure m ents . T he sorte d m easurem en ts are show n belo w .Favre: 10, 12, 13, 14, 15, 15, 18, 19, 21, 22, 22, 23, 23, 23, 23, 25, 25, 26
McNabb: 9, 10, 11, 15, 15, 16. 16, 17, 18, 18, 18, 18, 19, 21, 21, 23, 24, 27
For Brett Favre, m = ( 21 + 2 2 ) / 2 = 2 1 . 5 , 0 , = 1 4 + 0 . 7 5 ( 1 5 - 1 4 ) = 1 4.7 5 a nd
0 3 = 2 3 + 0 . 2 5 ( 2 3 - 2 3 ) = 2 3 .
For Donovan M cNabb, m = ( 18 + 1 8 ) / 2 = 1 8 , 0 , = 15 + 0.75 (15 -1 5) = 15 and
0 3 = 2 1 + 0 .2 5 ( 2 1 - 2 1 ) = 2 1 .
Then the f ive-num ber summ ar ies are
Min Q. Median 03 M ax
Favre 10 14.75 21.5 23 26M cN abb 9 15 18 21 27
b For B rett Favre, calcúlate IQR = 0 , - 0 , = 2 3 - 1 4 . 7 5 = 8 .2 5 . T h e n th e lower
a n d u p p er f en ees are:
0 , - 1 .5 IQR = 14.75 -1 2.3 75 = 2.375
0 , + 1 . 5 / 0 / ? = 2 3 + 1 2 . 3 7 5 = 3 5.3 75
For Donovan McNabb, ca lcúla te IQR = 0 , - 0 , = 2 1 -1 5 = 6 . Th en the l ower and
u p p er f en ces are:
0 , - 1 . 5 / 0 / ? = 1 5 - 9 = 6
0 , + 1 .5 IQ R = 21 + 9 = 30
Th ere are no outliers, and the box plots are shown on the next page.
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3
5
c Answers will vary. The Favre distribution is skewed left, while the Donovan
distribution is roughly symmetric , probably mound-shaped. The McNabb dis tribution
is s l ightly m ore variable; Favre has a higher m edian num ber of com pleted passes .
Answers will vary. The student should notice the outliers in the female group, that the
med ian female tem perature is higher than the median m ale temperature.
The ordered sets are shown below:
G eneric Sunm aid
24 25 25 25 26 2 2 24 24 24 24
26 26 26 26 27 25 25 27 28 2827 28 28 28 28 28 29 30
For n = 14, the position o f the m edian is0.5(/i + l) = 0.5(14 + 1) = 7.5 and the p ositions
o f the quarti les are 0.25 (n +1) = 3.75 and 0.75(n +1 ) = 11.25 , so that
Generic: m = 26, Qx= 25, Qy = 27.25,and IQR = 27.25 - 25 = 2.25
Sunmaid: m = 26, (2, = 2 4, <2, = 2 8, an d IQR = 28 - 24 = 4
Generic: L ow e r a nd u pp e r f e n c e s are:
<2, -1 5 I Q R = 25 - 3 .375 = 21.625
Q, +1 .5 IQ R = 27.25 + 3.375 = 30.625
Sunmaid: L ow e r an d up pe r f e n c e s are:
Q , - 1 . 5 I Q R = 2 4 - 6 = 18
Q3 + X.5 IQR = 28 + 6 = 34
T he box plots are shown on the next page. Th ere are no outliers.
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Sunmaid -
Generic
n ----------1----------1----------1----------1--------- 1---------- 1----------1----------1----------1-21 22 23 24 25 26 27 28 29 30
Raisins
d If the b oxes are not being und erfil led, the av erage size of the raisins is roughly
the same for the two brands. How ever, since the num ber of raisins is more va riable
for the Sun m aid brand, i t wou ld appear that some o f the Sun m aid raisins are large
wh ile others are small. Th e individual sizes of the gene ric raisins are not as variable.
2.59 The ordered data are shown below.0.2 2.0 4.3 8.2 14.7
0.2 2.1 4.4 8.3 16.7
0.3 2.4 5.6 8.7 18.00.4 2.4 5.8 9.0 18.0
1.0 2.7 6.1 9.6 18.4
1.2 3.3 6.6 9.9 19.2
1.3 3.5 6.9 11.4 23.1
1.4 3.7 7.4 12.6 24.01.6 3.9 7.4 13.5 26.7
1.6 4.1 8.2 14.1 32.3
Since n = 5 0 , the posi tion of the median is 0.5 (n +1) = 25.5 and the po si t ions of the
lower and upper qu art iles are 0.25(n +1) = 12.75 and 0.15(n +1) = 38.25 .
Then m = (6.1 + 6.6 ) /2 = 6.35, 0 , = 2 .1 + 0.7 5 (2.4 -2 .1) = 2.325 and<2, = 12 .6 + 0.25(13 .5-12 .6) = 12.825. Then 1QR = 12 .82 5-2 .325 = 10 .5 .
The l ower and upper f enees are:
0 , - 1 . 5 / 0 / ? = 2 . 3 2 5 - 1 5 . 7 5 = - 1 3 .4 2 5
- 0 , + 1 . 5 / 0 / ? = 1 2.8 25 + 1 5.7 5 = 2 8 .5 7 5
and the box plot is shown on the next page. There is one outlier, x = 32.3 . The
distribution is skewed to the right.
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2.63 The following inform ation is available:
n = 400, x = 600. s 2 = 4900
The s tandard dev ia t ion o f these scores is then 70, and the resul ts o f T chc by she f fsTheo rem fol low:
k x ± k s Interval T cheb ysheff
1 6 0 0 ± 7 0 530 to 670 at least 0
2 600 ± 140 4 6 0 to 740 at least 0.75
3 6 00 ± 2 1 0 390 to 810 at least 0.89
If the distr ibution o f scores is mou nd-shap ed, w e use the Em pirical Rule, and
conclude that approximately 6 8 % o f the scores w ould lie in the interval 530 to 670
(which is x ± 5 ). A pprox imately 95 % of the scores would lie in the interval 460 to
740.
2.69 If the distribution is m oun d-shap ed, then alm ost all of the measurem ents will fall in
the interval j u ± 3(7 , which is an interval 6(7 in length. Th at is, the range o f the
measurem ents should be approximately 6(7 . In this case , the range is8 0 0 - 2 0 0 = 6 0 0 , s o th at (7 = 6 0 0 / 6 = 1 0 0 .
2.73 Th e diameters o f the trees are approxim ately moun d-shaped with mean 14 and
standard deviation 2 .8 .
a The valué x = 8.4 lies two standard de viations below the m ean, while the valué x
= 22.4 is three standard deviations above the mean. Use the Empirical Rule. The
fraction o f trees with diam eters betw een 8.4 and 14 is half of 0.95 o r 0.475, while the
fraction o f trees with diameters betw een 14 and 22.4 is half of 0.997 or 0.4985. The
total fraction o f trees w ith diam eters betwee n 8.4 and 22 .4 is
0.475 + 0.4985 = .9735
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b The valué x = 16.8 l ies one standard deviat ion above the mean. Using the
Em pirical Rule, the fraction o f t rees with diam eters between 14 and 16.8 is half of0.68 or 0.34, and the fraction o f t rees with d iameters g reater than 16.8 is
0 .5 -0 .34 = .16
2.77
y x 4182.81 a Calcúlate n = 50, Z x = 418 , so that x = — — = ------ = 8 . 3 6 .
n 50
b Th e posi t ion o f the med ian is ,5(n +1) = 25.5 and m = (4 + 4)/2 = 4.
c Since the m ean is larger than the median, the distr ibution is skew ed to the r ight.d Since n = 5 0 , the posi tions of Q¡ and 0 3 are .25(51) = 12.75 and .75(51) =
38 .25 , r e spec tively Then 0 , = 0 + 0 .7 5(1 -0) = 12 .75 ,
0 3 =1 7 + .25 (19-1 7) = 17 .5 and I Q R = 17 .5- . 75 = 16.75 .
The l ow e r and uppe r f e ne e s are:
_ 0 , - 1 .5 /0 / ? = .7 5 -2 5 .1 2 5 = -2 4 .3 7 5
0 , + 1 .5 IQR = 17.5 + 25.125 = 42.625
and the box plot is show n on the next page. There are no outl iers and the data is
skew ed to the r ight.
a Th e percentage o f col leges that have betw een 145 and 205 teach ers correspo nds
to the fract ion of me asurem ents expected to l ie within two standard deviat ions o f the
m ean. T ch eb ys he ff s Theo rem States that this fract ion will be at least 3á or 75% .
b If the popu lat ion is normally distr ibuted, the Em pirical Rule is approp riate and
the desired fraction is calculated. Referring to the normal distr ibution sh ow n below,
the fract ion of area lying between 175 and 190 is 0.34, so that the fract ion o f col leges
m ore than 190 teachers is 0.5 - 0.34 = 0.16 .
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87 a-b As the valué of x gets smaller , so does the mean.
c The med ian does not chan ge until the green dot is smaller than x = 10, at which
poin t th e green do t becom es the m edia n.
d The largest and sm allest possible valúes for the median are 5 < m < 10.
91 Th e box plot show s a distribution that is skew ed to the left, but with one outlier to the
right of the other observations ( x = 5 2 0 ) .
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3: Describing Bivariate Data
3.1 a The side-by-side pie charts are constructed as in Chapter 1 for each o f the two
groups (men and women) and are displayed below using the percentages shown in thetable below.
Group 1 G roup 2 G roup 3 T otal
M en 23% 31% 46% 100%
W o m e n 8 % 57% 35% 100%
b-c The side-by-side and stacked bar charts in the next two f igures m easure the
frequency o f occu rrence for each o f the three groups. A sepárate bar (or portion o f a
bar) is used fo r m en and w om en. — - ■*>-., ----- ~~.y--'.------ . ... 7---------- i
'' “— —' ' >*
60
Gender Male Female Male Female Male FemaleGroup 1 Group 2 Group 3
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d The d ifferences in the proport ions of men and wom en in the three groups is mostgraphical ly port rayed by the pie char ts , s ince the unequal num ber of m en and wom en
tend to confuse the interpretation o f the bar charts. H ow ever, the bar charts are useful
in retaining the actual frequen cies of occurrence in each group , w hich is lost in the pie
chart.
3.5 a The popu lation o f interest is the populat ion of responses to the que st ion about
free time for all parents and children in the U nited States. Th e sam ple is the set of
responses generated for the 198 parents and 2 0 0 children in the survey.
b The data can be considered bivariate if , for each person interviewed, we record
the person’s relationship (Parent or Child) and their response to the question (just ther ight amount , not enough , too m uch, do n’t know). Since the mea surem ents are not
num erical in nature, the variables are qualitative.
c The entry in a cell represents the number of people who fell into that
relationship-opinion category.
d A pie chart is created for both the “parent” and the “childre n” categories. The
size of each sector angle is propo rtional to the fraction o f m easu rem ents falling into
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e Either stacked or com parative bar charts could be used, but since the height of
the bar rcpresents the frequen ey o f occu rrence (and henee is t ied to the sample size),
this type o f chart w ould be misleading . The com parative pie charts are the best
choice.
3.9 Follow the instructions in the M y Personal Tra iner section. T he correct answ ers are
shown in the table.
X y xy C alcúlate: C ovaria nce
1 6 6 n = 32 0 - ( 6 ) ( , 2 )
s *y 2 3 23 2 6
Sx = 1
2 4 8 sy = 2 Corre lat ion Coeff íc ient
I * - 6 Z y = 12 2 > = 2 0 r = _ 2 = —1
1 ( 2 )
3.13 a The scatterplot is shown below.
b There appears to be a negative relationship between x and y; that is, as x
increase, y d ecreases.
c Use your scientif ic calculator to calcúlate the sums, sums of squares and sum of
cross prod ucís for the pairs {xt, y , ) .
I * , = 2 1 ; I y . = 2 4 .3 ; I * 2 = 9 1 ; I y ,2 = 1 0 3 .9 9 ; I * , y . = 7 5 .3
Then the covar iance is
1 53J m ±
s „ = * ^---------
= -----------------
£-------
= - 1 . 9 5* n - 1 5
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3.17
and the samp le standard deviations are
s . = \ \ --------------^— = — = 1.8708 andn - 1 5
i r a no <24-3)26 _ s v = \ f — = ------------- =1.0559
l / i - l 5
The correlation coeff icient is
s —1 95r = — = ------------ '-------------= - 0 . 9 8 7
5I5> (1.8708X1.0559)
This valué of r indicates a strong negative relationship betw een x and y.
a- b The scatterplot is shown below. The re is a slight positive trend betw een pre-
and post- test scores, but the trend is not too pronoun ced.
c Calcúlate
n = 7; I x¡ = 677; I y. = 719; Z x f = 65 ,9 93 ; I y? = 74,585; I x i y i = 70,00 6 . Then
the covariance is
5 n = — = 7 8 .0 7 1 42 9* n - 1
The sample standard deviations are sx = 9.286447 and s y = 11.056134 so that
r = 0.7 60 . Th is is a relatively strong positive correlation, confirm ing the
interpretation of the scatterplot.
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23 a For each person interviewed in the survey, the following variables are recorded:
the service branch (qua litative), the age (quantitative con tinuo us) and the rank
(enlisted vs. officer, qualitative).
b The population of interest is the population of ages for all people in the military.
A lthough the source o f this data is not given, i t is probably b ased on censusinformation, in which case the data represents the entire population.
c A com parative (s ide-by-side) bar chart has been used. A n alternative
p resenta ti on can be ob ta in ed by usi ng com parative p ie chart s, w ith th e ages d iv id ed
into eight age groups, and com pared for the Army versus the M arine Corps.
d The enlisted men tend to be younger.
e The M arine Corps tends to have a much h igher percentage o f young er enlis ted
personnel.
29 a-c N o. T here se em s to be a la rge clu ste r o f poin ts in th e lo w er le ft hand córner
show ing no apparent relationship between the variables , while 7-10 d ata points fromtop left to bottom right show a negative linear relationship.b Th e pattern described in parts a and c would indícate a weak correlation:
r = ̂ -8 3 .5 3 0 = ^ Q32
sxs y ^ (712.603X9346.603)
d N um ber o f w aste si te s is only sl ig htly affecte d by th e si ze o f th e st at e. Som e
other possible ex planatory v ariables m ight be local environmental regulations,
p opu la ti on per square m ile, o r geographic al re gió n in th e U nit ed Sta te s.
.33 a The 1) num ber of home networks (quanti tat ive discrete) have been measured,along with the year (q uanti tative continuous) and the type of netwo rk (quali tat ive).
b-c Answers will vary. We choose to use a line chart for the two types of networks.
As the num ber of wireless netw orks increase, the num ber of wired netw orks
decreases .
.37 a-b T he scatterplot is show n on the next page. Th ere is a strong positive linear
relationship between x and y.
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3.38
3.43
Test 1 (x)
a Calcúlate
n = 8 ; I x, = 6 34 ; I y, = 386; X x2 = 52270;
I >’(2 = 1 9 8 7 6 ; I * , ? , = 3 2 1 3 6
Then the covariance is
sn = — ^ = 2 20 .7 85 71* n —1
The sample s tandard devia t ions are sx = 17 .010501 and $v = 13.37 107 75 so that
r = 0 . 9 7 1 .
b Since the co rrelation co eff icient is so cióse to 1, the strong correlation indicates
that the secon d and q uicker test could be used in place o f the long er test- interview.
a Calcúla te
,2 = 8 ; I * . = 4 5 1 ; L y , = 5 55 ; X x 2 = 2 9 , 6 1 9 ; X y f = 4 3 ,2 0 5 ; I x , ^ = 3 5 , 0 8 2 . T hen
the covariance is
= = 5 41 .9 82 1a; - 1
The sample standard deviations are sx = 24.4770 and s y = 25.9171 so that
r = 0 . 8 5 4 4 .
b-c The scatterplot should look like the one shown on the next page. The
correlation coe ff icient should be cióse to r = 0.85 . Th ere is a strong positive trend.
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4: Probability and Probability Distributions
4.1 a Th is experim ent involves tossing a s ingle die and observing the outcom e. The
sample space for this expe rimen t consis ts o f thefollowing s im ple eyents: E\. O bserve a 1 £ 4: O bservea 4
E 2: O bserve a 2 £ 5: Ob servea 5
£ 3: O bserve a 3 £ 6: O bservea 6
b E vents A th rough F are com pound even ts and ar e com posed in th e fo llow in g
manner:
A: (E2) D: (E2)
B: ( E 2, E 4 , E b) E: ( E 2, E ^ E 6)
C: ( £ 3, £4, £5, £ 6) F: contains no s imple events
c Since the s imple events £ ,- , /= 1, 2, 3, 6 are equally l ikely, £ (£ ,) = 1/6 .
d To find the probab il ity o f an event, we sum the proba bil i t ies assigned to the
simple events in that event. F or example,
/>( A) = /> (£ , ) = j6
Similarly, P(D) = \ /6: P (B ) = P ( E ) = £ ( £ 2) + £ ( £ 4) + £ ( £ 6) = - = - ; and6 2
4 2 P (C ) = — = — . S ince event £ conta ins no s imple events , P (£ ) = 0 .
4.5 a The exp er iment cons is ts of choos ing three coins a t random from four. The order
in which the coins are drawn is unimpo rtant. Henee, each s imp le event consis ts of a
triplet, indicating the three coins draw n. U sing the letters N, D, Q, and H to represen!
the nickel, dime, quarter, and half-dollar, respectively, the four possible simple events
are listed below.
£ , : (NDQ ) £ 2: (ND H) £ 3: (NQH ) £ 4: (DQH )
b T he event th at a half -doll ar is chosen is associa te d w ith th e sim ple even ts £ 2, £3,and £4. Henee ,
P lchoos e a ha lf -do ll ar ] = £ (£ , ) + P ( E , ) + P ( £ 4) = - + - + - = -4 4 4 4
since each simple event is equally likely.
c Th e s imple events along with their m onetary valúes follow:
E^ N D Q $0.40
e 2 N D H 0.6 5
E i N QH 0.80
£4 D Q H 0.85
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4.9
4.13
4.17
4.21
4.25
4.29
Hen ee, P[total am ount is $0.60 or more] = P ( E 2) + P ( E y) + P (E 4) = 3 /4 .
The four possible outeome s o f the experime nt, or simple events, are represented as
the c el ls o f a 2 x 2 table, and have proba bil i ties as given in the table.
a R[adult judg ed to need glasses] = .44 + .14 = .58
b P[adult needs glasses but does not use them] = . 14
c F[a du lt uses glasses] = .44 + .02 = .46
a E xper iment : A taster tastes and ranks three variet ies o f tea A, B, and C,
according to preference.
b Simple events in S are in triplet form. E l : (1,2,3) E 4 : (2,3,1)
E 2 : (1,3,2) £ 5 : (3,2,1)
E y : (2,1,3) £ 6 : (3,1,2)
Here 1 is assigned to the mos t desirab le, 2 to the next m ost desirable, and 3 to the
least desirable.
c De fine the even ts D: variety A is ranke d first
F: variety A is ranke d third
Then
P (D ) = />(£,) + P ( E 2) = 1/6 + 1/6 = 1/3
The probability that A is least desirable is
£ ( £ ) = P(E 5) + P ÍE , ) = 1 /6 +1 /6 = 1/3
Use the mn Rule. Th ere are 10(8) = 80 possible pairs.
8 'Since order is important, you use permutat ions and p \ = —j = 8(7)(6)(5)(4 ) = 672 0 .
Since order is unim portant, you use combinat ions a nd C í° = ^ = 120.3! 17! 3(2)(1)
a Each student has a choice of 52 cards, since the cards are replaced between
select ions. The mn R ule al low s you to f ind the total num ber o f configurat ions for
three students as 52(52)(52) = 140,60 8.
b N o w e ac h student m ust p ic k a d if fe re nt card . T hat is, the fir st stu dent has 52
choices, but the second and third students have only 51 and 5 0 choices, respectively.
The total num ber of configurat ions is found using the mn Rule or the rule for
perm uta tions:
mnt = 52(51)(50) = 132,600 or £ 5,2= — = 132,600 .49!
c Let A be the event of interest. Since there are 52 differen t card s in the deck,
there are 52 configurations in which all three students pick the same card (one for
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each card). Th at is , there are nA = 52 ways for the event A to occur, out of a total of
N =1 40 ,608 poss ible conf igurations f rom par t a. The probahil i ty of interest is
P (A ) = ^ - = — — — = .00037 N 140,608
d Ag ain, let A be the event o f interest . T here are nA =1 32 ,600 ways (f rom par t b)for the eve nt A to occu r, out of a total of N = 140,608 possible configurations froTn
part a, and the prob ahil i ty o f interest is
= ̂ = 1 3 2 , 6 0 0 ^943
N 140,608
33 N otice that a sam ple of 10 nurses will be the same no matter in w hich orde r they were
selected. Hen ee, order is unim portant and com binations are used. The num ber of
sam ples of 10 selected from a total of 90 is
9 0 i 2.07 590 76 Í1019) .Cm = - = -------------- = 5.720645 (10 )10 !80! 3.628 8(10 ) V ’
.37 Th e si tuation presented here is analog ous to draw ing 5 i tems from a ja r ( the f ive
m em bers v oting in favor of the plaint iff). If the jar con tains 5 red and 3 white i tems
(5 wo m en and 3 m en), wh at is the proba hility that all five items are red ? T ha t is, if
there is no sex bias, f ive o f the eight m em bers are random ly chosen to be those voting
for the plaint iff . W hat is the probah il ity that all five are wo m en? There are
8 ' N = C \ = — — = 56
5!3!
simple events in the experiment, only one of which results in choosing 5 women.
Henee,
P (fiv e w om en) = ̂ .
.41 Follow the instruct ions given in the My Personal Trainer section. The an swe rs are
given in the table.
P(A) P (B ) C o n ditio ns fo r e ve nts A a n d B P (A n B) P(A u B) P(A|B)
.3 .4 Mutually exclusive 0 .3 + .4 = .7 0
.3 .4 Independent .3(4 ) = .12 .3 + .4- ( .3) ( .4 ) = .58 .3
.1 .5 Independent .1 (5 ) = .05 , l + . 5 - ( . l ) ( .5 ) = .55 .1
.2 .5 Mutually exclusive 0 .2 + .5 = .7 0
.47 Refer to the solut ion to Exe rcise 4.1 where the six simple events in the exp erime nt are
given, w ith P{Ej ) = 1 /6 .
a 5 = { E l , E 2 , E i , E A , E 5 , E 6 } and P (S ) = 6 / 6 = 1
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4.48
4.49
4.55
4.59
b P ( A \ B ) = « A n B ) = y i = x
v ' P (B ) 1/3c B = { £ „ £ , } and P (B ) = 2 /6 = 1/3
d A n f i n C c o nta in s n o s im p le e ve n ts , a nd P ( A n B n C ) = 0
e P ( A n B ) = P (A \ B ) P ( B ) = 1(1/3) = 1/3
f A n C c on ta in s no sim p le ev en ts , an d P ( A n C ) = 0
g B n C c o n t a in s no sim p le e ve nts , a nd P ( f i n C ) = 0
h A k j C = S and P ( A u C ) = l
¡ 8 u C = { £ l, £ , , £ 4, £ s , £ j a nd P ( P u C ) = 5 /6
a From Exerc ise 4 .47 , P(A n B ) = 1/3 , P (A | B ) = 1, P(A ) = 1/2 ,
P ( A | B ) * P ( A ) , so t ha t A and B are no t i ndependen t. P (A n B )^ 0 , so t hat A and
B are not m utually exclusive.
b P ( A |C ) = P ( A n C ) /P ( C ) = 0 , P (A ) = 1 /2 , P ( A n C ) = 0 . Since
P(A | C ) * P (A ) , A and C are dependent . S ince P ( A r \ C ) = 0 , A and C are
m utually exclusive.
a Since A and B are independen t , P(A r \ B ) = P { A ) P { B ) = .4(.2) = .08 .
b P ( A u B ) = P ( A ) + P ( P ) - P ( A n f i ) = . 4 + .2 - (.4 )( .2 ) = . 52
Define the fol lowing events:
A: project is approved for funding
D: project is disappro ved for funding
For the first group, P { \ ) = .2 and P(D ,) = .8 . Fo r the second group,
P [same decisión as f i rst group] = .7 and P [reversal] = .3. Th at is,
P( A 2 \ A t ) = P (D 1 \ D t ) = . l and P ( \ \ D, ) = P (D 7 \ A ) = .3 .
a P ( A n 4 ) = P ( A ) í>(A2 |A ) = -2 (.7 ) = .14
b P(D¡ n D 2 ) = P ( D ¡)Pi ,D2 \ £),) = . 8(.7) = . 56
c
P(D, n A 2) + P ( A i n D 2) = P ( D , ) P ( A , | D , ) + P ( A , )P ( D 2 1A,) = ,8(.3) + ,2(.3) = .30
Fix the birth date o f the f irst person e ntering the room . Th en define the fol lowing
events:
A2: seco nd perso n’s birthda y differs from the first
A 3: third perso n’s birthda y differs from the first and sec ond
A 4: fourth pers on ’s birthday differs from al l preceding
An: nlh pe rso n’s birthda y differs from all prece ding
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' 364̂1̂363 p 65- / i + nv 365J ,365 J l 365 )
Then
P (A ) = P (A 2) P ( A ) - P ( A ii) =
since at each step, one less birth date is available for selection. Since event B is the
complem en t o f even t A ,
P (B ) = \ - P ( A )
a F o r n = 3 , P(A ) = (364)(363) = 99l8 and P (B ) = 1 - .99 18 = .0082(365)
b F or n = 4 , P (A ) = (364X363X362) = 9836 and P (B)= 1 - .9 8 3 6 = .0164(365)
63 D efine A: smo ke is detected by device A
B: smok e is detected by device BIf i t is given that P (A ) = .95, P (B ) = .98, and P ( A n B ) = .94 .
a P{A k j B ) = P{A) + P ( B ) - P( A n B ) = .95 + .98 - .94 = .99
b P ( A C n B c ) = l - P ( A k j B ) = 1- .99 = .01
69 a Use the Law of Total Probabil i ty, writing
P (A ) = P(5 , )P (A 15 ,) + P ( S 2 ) P ( A | S2) = ,7 (.2) + .3(.3) = .23
b U se th e resu lt s o f part a in th e fo rm o f B ayes’ Rule :
P ( S |M ) = -------------W , ) P ( » | 5 . )
P ^ P i A l S ^ + P i S ^ P i A l S , )
F o r i = l , P(S, | A) = ----------------- = — = .60871 ,7(.2) + .3(.3) .23
For í = 2 , P(5 , |A) = -------------- = ^ = .39132 ,7(.2 ) + .3(.3) .23
73 De fine A: m achine produces a defective i tem
B: wo rker follows instructions
Then P ( A \ B ) = . 0\ , P (B ) = . 90, P (A \ B c ) = .03, P ( B C) = .10 . The probabi l ity of
interest is
P (A ) = P ( A r ^ B ) + P ( A n B c )
= P{A | B ) P ( B ) + P ( A | B C) P ( B C)
= ,01(.90) + .03(.10) = .012
77 Th e pro bab ility o f interest is P (A \ H ) wh ich can be calculated using B ayes’ Rule
and the proba bil i t ies given in the exercise.
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P ( A \ H ) = ________________ P ( A ) P ( H \ A ) ________________
P ( A ) P ( H | A ) + P ( B ) P ( H | B ) + P ( C ) P ( H \ C)
__________ .01(.90) ____________ .009
.01(.90) + .005(.95) + .02( .75) ” .02875= .3130
4.83 a Since one of the requiremen ts of a probabi l ity dis t ribut ion is that ^ p ( x ) = 1 X
we need
p (3 ) = 1- (. 1+ .3 + .3 + . 1) = 1- .8 = .2
b The probability histogram is shown below.
c For the random var iable x given here,
H = E ( x ) = Z xp ( x ) = 0( . 1) + 1(.3) H— + 4(. 1) = 1.9
The var iance of x is defined as
( j 1 ^ E ^ x - p f Y ^ x - p ) 1 p (x ) = (0 - 1.9)2(.1) + (1 -1 .9 )2(.3) + --- + ( 4 - 1.9)2(.l) = 1.29
a n d í t = VT 2 9 = 1 . 1 3 6 .d Using the table form o f the probab ility distr ibution given in the exercise,
P ( x > 2 ) = .2 + . l = . 3 .
e P ( x < 3) = 1 - P ( x = 4) = 1- . 1 = .9 .
4.87 a-b On the f irst try, the probab ility o f selecting the p rope r key is 1/4. I f the key is
not found oír the f irst try, the proba bility chan ges on the se cond try. Let F denote a
failure to f ind the key and S deno te a success. Th e random variable is x, the number
o f keys tr ied before the correct key is found. Th e four associated sim ple events are
shown below.
E , : S ( x = \) E3: FFS (x = 3)
E 2:F S (^ = 2) E4: FFFS (* = 4)
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c-d Then
P ( l ) = P ( x = 1) = P ( S ) = 1/4
p{ 2 ) = P ( jc = 2) = P (F 5) = P ( F ) P ( S ) = (3 /4 ) (1/3) = 1/4
p ( 3 ) = P (* = 3) = P (F FS ) = P ( F ) P ( F ) P ( S ) = (3 /4) (2/3) (1/2) = 1/4
p (4 ) = F(jc = 4 ) = P ( F F F S ) = P ( F ) P ( F ) P ( F ) P ( S ) = ( 3 / 4 ) ( 2 / 3 ) ( l / 2 ) ( l ) = 1 /4
The probability distr ibution and
JC 1 2 3 4
p( x) 1/4 1/4 1/4 1/4
pro babil ity h is to gra m fo llow .
1 Let x be the nu m ber o f drilling s until the first success (oil is struck). It is given that
the prob ability o f striking oil is P (O) = .1, so that the proba bility o f no oil is
P ( N ) = .9
a p{ 1) = F [o il struck on first drilling] = P ( 0 ) = .1
p ( 2) = P [o il s truck on second d rill ing ] . Th is is the probab ility that oil is not
found on the first drilling, but is found on the second drilling. Using the
M ult ipl ica tion Law ,
p (2) = P ( N O ) = (.9)(. 1) = .0 9.
Finally, p ( 3) = P ( N N O ) = (.9)(.9 )(. 1) = .081 .
b -c For th e fir st success to occur on tr ia l x, (jc - 1) failures must oc cur b efore the
first success. Thu s,
p ( x ) = P ( N N N ... N N O ) = (.9)* "' (. 1)
since there are (jc - 1) N ’s in the sequence. The pro bability histogram is show n on the
next page.
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1 3 5 7 9 11 13 15 17
x
4.95 The random variable G, total gain to the insurance company, will be D if there is no
theft, but D - 50,000 if there is a theft during a given year . The se two even ts will
occu r with probab ility .99 and .01, respectively. H enee, the prob ability distr ibution
for G is given below.
G _____________ p(G) The expe cted gain is
D .99 E{G ) = l G p ( G ) = . 99 D + . 0 1(D - 5 0 ,0 0 0 )
= D -5 0 ,0 0 0
D - 5 0,0 00 .01
In order that E (G ) = 1000 , it is necessary to h ave 1000 = D - 5 0 0 o r D = $ 1 5 0 0 .
4.97 a Similar to E xercise 4 .9 1. Fo r the fírst no n-b eliev er to be fo und on cali jc, (x - 1)
people w ho do believe in heaven mu st be called before the f irst non-believ er is found.
Thus , p ( x ) = P ( N N N . . . N N Y ) = ( .8 1)*"1(. 19)
b A s w ith o th er phone surv eys, th ere is a lw ay s a p rob lem o f no n-r espon se -
p eople who do not answ er th e tele phone or dec li n e to parti c ipa te in th e surv ey. A lso,
there is a problem of truthfulness of the response for a question such as this w hich
may be a sensitive subject for som e people.
4.101 Define the following events:
A: wo rker fails to repo rt fraud
B: w orke r suffers reprisal
It is given that P ( B | A c ) = .23 and P (A) = .69 . The p robability of interest is
P ( A C r \ B ) = P (B | Ac )P(A r ) = .23( .31) = .0713
4.105 Tw o systcms are selected from seven, three of w hich are defective. D enote the seven
system s as G 1( G 2, G 3, G 4, D(, D 2, D3 acco rding to w heth er they are go od o r defective.
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Each simple ev ent w ill represen t a particular pair of systems cho sen for testing, and
the sam ple space, consisting o f 21 pairs , is shown below.
G1G2 G jD | G2D3 G4D2 G1G3 G1G2 G3D1 G4D3
G,G4 G,D3 g ] d 2 D ,D2 G2G3 G 2D, G3D3 D ,D3
G2G4 G2D2 G4D1 D2D3 G3G4
N ote th at th e tw o syste m s are dra w n sim ulta neously and th at o rder is unim port ant in
identifying a sim ple event. H enee, the pairs G |G 2 and G 2G! are not considered to
represen t two d ifferent simple events. The eve nt A, “no defectives are selected”,
consis ts of the s imple events G |G 2, G1G3, G1G4, G 2G 3, G2G4, G3G4. Since the
systems are selected at rando m , any pa ir has an equal proba bility o f being selected.
He nee, the probab ility assigned to each simple event is 1/21 and P (A) = 6 /21 = 2 /7 .
4.109 a P ( c o l d ) =
Define:
49 + 43 + 34 126= .4565
276 276
F: person has four or f ive relationshipsS: person has six or more relationships
Then for the two people chosen from the total 276,
P ( o n e F and one S ) = P ( F n S ) + P ( S n F )
100
276
( 96+
r 96 ^ í k x T
275y ^ 2 7 6 , 1,275,= .2530
. P í t h r e e o r f e w e r n c o l d ) 4 9 / 2 7 6 4 9P Í T h r e e o r fe w e r c o id ) = = — = ----- = .
v 1 ’ P ( co ld) 126/276 1263889
4.113 a Define the following events:Bj: client buy s on f irst contact
B2: client buys on second contact
Since the client m ay b uy on e ither the f irst o f the second con tact, the desired
p rob abili ty is
P[cl ien t w ill buy] = P[cl ient b uys on f i rs t contact]
+ P [client doesn' t buy on f irst , but buys on second ]
= P ( P , ) + ( l - P ( f í 1) ) P ( f i 2) = .4 + ( l- .4 ) ( .5 5 )
= .73
b Th e pro bability that the client w ill not buy is one m inus the proba bility that the
client will buy, or 1- .73 = .27 .
4.117 Ea ch hall can be chosen from the set (4, 6) and there are three such balls. Henee,
there are a total of 2(2)(2 ) = 8 potential w inning numb ers.
4.121 a Co nside r a single tr ial which consists of tossing two coins. A m atch occurs
w hen either HH or TT is observed. Henee, the probability of a match on a single tr ial
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is P ( H H ) + P(T T ) = 1/4+ 1/4 = 1 /2. Let MM M denote the event “match on tr iá is 1,
2, and 3”. Then
P ( M M M ) = P ( M ) P ( M ) P ( M ) = (1 /2)3 = 1/8.b On a single tr ial the event A, “two trails are ob served ” has probability
P (A ) = P(T T ) = 1/4 . H enee , in thre e triáis
P ( A A A ) = P ( A ) P ( A ) P ( A ) = (1/4) ' = 1/64
c This low probability would not suggest collusion, since the probability of three
matches is low only if we assume that each student is merely guessing at each answer.
If the students have stud ied togethe r or if they b oth know the correct answ er, the
pro babil it y o f a m atc h on a si ng le tr ia l is no longer 1/2 , but is substa nti a ll y hig her.
Hen ee, the occurrence of three m atches is not unusual.
4.125 Define the events: A: the man waits f ive m inutes or longer
B: the wom an waits f ive minu tes or longer
The two events are independent, and P (A ) = P (B ) = .2 .
4.129 Since the f irst pooled test is positive, we a re interested in the p robab ility o f requir ing
five single tests to detec t the disease in the single affected pe rson. Th ere are
(5)(4)(3)(2)(1) w ays of ordering the f ive tests , and there are 4(3 )(2)(1) w ays of
ordering the tests so that the diseased person is given the f inal test . H enee, the desired
4! 1 pro babili ty is — = - .
I f two p eop le are diseased, six tests are need ed if the last two tests are given to the
diseased p eople. There are 3(2)(1) ways o f ordering the tests of the other three people
and 2(1) ways of ordering the tests of the two diseased peo ple. H enee, the probability
2!3! 1that six tests will be needed is ------ = — .
5! 10
4.133 a Define P: shopp er prefers Pepsi and C: shop per prefers Coke . Then if there is
actually no difference in the taste, P(P) = P(C) = 1/2 and
a P ( A r ) = l - / > M ) = .8
b P ( A r f lr ) = / > ( / l c ) /> ( £ c ) = (.8 )(.8 ) = .64
c P [at least one waits f ive minu tes or longer]
= 1- P [ne i the r wa i ts f ive minutes o r longer ] = 1- P ( A ( B c ) = 1 - . 6 4 = .3 6
b
P(exactly one prefers Pepsi) = P ( P C C C ) + P ( C P C C ) + P ( C C P C ) + P(C CC P)
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4.137 Refer to the Toss ing D ice applet, in which the simple events for this experiment are
displayed. Each sim ple event has a part icular valué of T associated with it , and by
sum m ing the p robabil i t ies o f all simple eve nts producing a part icular valué of T, the
fol lowing probah il ity distribution is obtained. The d istribution is mound -shaped.
a-b
T p( T) T p(T)2 1/36 8 5/36
3 2/36 9 4/36
4 3/36 10 3/36
5 4/36 11 2/36
6 5/36 12 1/36
7 6/36
2 4 6 8 10 12X
- ______________
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5: Several Useful Discrete Distributions
5.1 Follow the instructions in the Personal Trainer section. Th e answ ers are show n in thetables below.
k 0 1 2 3 4 5 6 7 8
P(x < k) .000 .001 .011 .058 .194 .448 .745 .942 1.000
T h e P r o b l e m Lis t the
V a lú e s o f x
W r i te t h e
p ro b a b ili ty
Rewr i te the
p ro b a b il it y
F ind the
p ro b a b il it y
Three or less 0 , 1 , 2 , 3 P (x < 3 ) .058Three or more 3, 4. 5, 6. 7, 8 P (x > 3) 1 - P(x < 2) 1—.011 = .989
M ore than three 4 , 5, 6, 7, 8 P (x > 3) 1 - P(x <, 3) 1 - . 058 = .942
Few er than three 0 , 1 ,2 P (x < 3) P (x < 2 ) .011
Between 3 and 5 (inclusive) 3 , 4 , 5 P<3 < jc < 5) P (x < 5) - P (x < 2) . 448 - .011 = .437
Exactly three 3 P fx = 3) P (x < 3 ) - P (x < 2) r -~
s I I ©
r o o
O
5.5 a
b
c
d
5.11 a
b
P ( x > 4 ) = 1- P ( x < 4) = 1- P ( x < 3).
These probabilit ies can be foun d individually using the binom ial formula, or
alternatively using the cumulative binomial tables in Appendix I .
P ( x = 0) = Co°( .4)° ( .6) '° = .006 P ( x = 1) = C !°( .4 )‘ (.6 )9 = .040
P ( x = 2 ) = C!.0(.4 )2 (.6 )8 = .121 P ( x = 3) = C 3°(-4)3 (.6 )? = .215
The sum of these probabilit ies gives P ( x < 3) = .382 an d P ( x > 4) = 1 - .382 = .618.
c Use the results o f parts a and b.
P ( x > 4 ) = 1- P ( x < 4) = 1 - ( .3 8 2 + .25 1) = .367
C ¡ ( -3) ; ( ,7 )6 = |Z 2 ( .0 9 ) ( . l 17649) = .2965
C S ( .0 5 ) °( .9 5 ) 4 = ( . 9 5 ) ‘ = . 8 1 4 5
C'J»(.5)S(.5)7= ^ ^ (.5 r= .1 17 2
Ci7(-2)‘ (.8)6 = 7 (.2 )(.8 )6 = .3670
For n = 10 and p = .4, P ( x = 4 ) = C 4°(-4)4 (.6)6 = .251 .
To calcúlate P ( x > 4 ) = p ( 4 ) + p ( 5) + ---+ p (10) it is easiest to write
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5.15
5.19
5.23
d From par t c , P { x < 4 ) = P ( x < 3) + P ( x = 4) = .382 + .251 = .63 3.
e n = np = 10( .4) = 4f o = yj ñpq = 7 l 0 ( . 4 ) (. 6 ) = y ¡2A = 1.549
a P f j t < 12] = P [ x < 1 1] = .74 8
b P [ x < 6] = .610
c />[jc > 4] = 1- ^ [ jc < 4 ] = 1- .6 3 3 = .367
d P [ x > 6 ] = 1 - P [ x ^ 5] = 1 —.034 = .966
e P [ 3 < J t < 7 ] = P [ x < 6 ] - P [ ; t < 3 ] = .8 2 8 - .1 7 2 = .656
a p (0 ) = C o°(-l )° (- 9 )20 = .1 215767 p (3 ) = C f ( - l ) ' (- 9)17 = .1 901199
M D = C?0(.1)1(.9)19 = .2701703 p(4 ) = C 4 ,( . l )4 ( .9 ) '6 = .0897788
^( 2 ) = C l°(-1)2 (-9) '8 = .2851798
so that P [ x < 4] = p ( 0) + p ( 1) + p { 2) + p ( 3) + p ( 4) = .9568255
b Using Table 1, Appendix I , P [ x < 4] is read directly as .957.
c A dding the entr ies for x = 0 ,1 ,2 ,3 ,4 , we have P [x < 4] = .956826 .
d p = np = 20(. 1) = 2 an d O = yjnpq = VE8 = 1.3416
e For ¿ = 1, / /± c r = 2 ± l .342 or .658 to 3 .342 so tha t
P [ .658 < jc < 3.342] = P [ 1 < x < 3] = .2702 + .2852 +. 1901 = .7455
For k = 2 , / j ± 2 ct = 2 ± 2 .683 or - .68 3 to 4.683 so that
P [ - . 6 8 3 < x < 4 .6 8 3 ] = P [ 0 < * < 4 ] = .9 56 9
For k =3 , p ± 2 o = 2 ± 4.025 or -2.02 5 to 6.025 so that
P [-2 .02 5 < jc < 6.025] = P [0 < x < 6] = .9977
f The resul ts are consistent with T ch eb ysh ef fs Theo rem and the Emp irical Rule .
Define x to be the num ber of alarm system s that are tr iggered. Then
p = P [alarm is tr iggered ] = .99 and n = 9 . Since there is a table availab le in
Appendix I for n = 9 and p = .99 , you sho uld use it rather than the binom ial formula
to calcúlate the necessary probabilit ies.
a P[at least one alarm is tr iggered] = P ( x > 1) = 1 - P ( x = 0 ) = 1- . 0 0 0 = 1 . 00 0 .
b P[more than seven] = P ( x > 7 ) = \ - P ( x < 7 ) = 1- .0 0 3 = .997
c P[e ight or fewer] = P ( x < 8) = .086
5.25 Define x to be the num ber of cars that are black. Th en p = P [bla ck ] = .1 and ti = 25.
Use Table 1 in App endix I.
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a />( jc> 5 ) = 1 - / >( jc< 4 ) = 1- .9 02 = . 098
b P ( x < 6 ) = .991
c P ( x > 4 ) = l - P ( x < 4 ) = 1 -.9 0 2 = .098
d P ( x = 4 ) = P ( x < 4 ) - P ( x < 3 ) = .9 0 2 -.7 6 4 = .138
e P ( 3 < jc < 5 ) = P ( x < 5 ) - P ( x < 2 ) = .9 6 7 - .5 3 7 = .4 30
f P ( m o r e th an 2 0 not b lack) = P ( less than 5 b lack) = P ( x < 4) = .902
5.29 De fine x to be the num ber of f ields infested with whitefly.
Then p = P [infec ted field] = .1 and n = 100.
a p = n p = 10 0(.l) = 10
b S in ce n is large, this binom ial distr ibution should be fair ly m ound -shaped , even
though p = .1 . Henee you would expect approximately 95% o f the measurem ents to
l ie w ithin two standa rd deviat ion of the m ean w ith o = y j npq = ^10 0(.1)( .9) = 3. The
lim its are calculated as
/i ± 2 ít => 10 ± 6 or from 4 to 16
c From par t b, a valué of x = 25 would be very u nlikely, assum ing that the
characterist ics of the binom ial experime nt are met and that p = . 1 . If this valué were
actually observed, it might be possible that the triáis (fields) are not independent.
This could e asi ly be the case, since an infestation in one field m ight quickly sprea d to
a neighboring field. This is evidenc e of contagión.
5.33 Def ine x to be the numb er of Am ericans who are “tas ters”. T hen, n = 20 and p = .7 .
Using the binom ial tables in Appe ndix I ,
a P ( x > 17 ) = 1 - P ( x < 16)= 1 - . 8 9 3 = .1 07
b P ( x < 15) = .762
5.35 Follow the instruct ions in the My Personal T rainer sect ion. The answ ersare shown in
the table below.
P robability F orm ula C alcula ted va lu é
P(x = 0) 2 . 5 V 250!
.0821
P ( * = l ) 2.51e~25
1!
.2052
P(* = 2) 2 . 5 V 25
2!
.2565
P(2 or fewer
successes)
P (x = 0 ) + P (x = 1) + P (x = 2) .5 438
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5.39
5.43
5.47
5.51
Using p ( x ) = p x e~^
x \2°e~2
X -22* e
X !
a p[ jc = 0l = ---------- = .1353351 J O!
2 V 2 b /> [* = l] = —— = .2 7067
c P [ x > 1] = 1- P [ x < 1] = 1 - .13 5 33 5 - .27 0 67 = .593994
9-V2d P [ * = 5] = --------- = .036089
1 J 5!
Let x be the num ber of misses during a given m onth. The n x has a Poisson
distribution with p = 5.
p ( 0) = e '5 = .00675 V 5
p( 5) = ---------= .17555!
c P [ x > 5] = 1 - />[* < 4] = 1- .4 4 0 = .560 f rom Table 2 .
The random var iable x, n um ber of bacteria, has a Poisson distr ibution w ith p = 2 .
The probab il ity of interest is
P [ x exceeds máx imum coun t ] = P [ x > 5]
Using the fact that p = 2 and < j = 1.414 from E xercise 5.47, m ost of the obse rvations
should fall within p ± 2cr or 0 to 4. H enee, it is unlikely that x will exc eed 5. In fact,
the exact Poisson probab il ity is P [ x > 5] = .017.
C4C"The formula for p {x ) is p ( x ) = x_ ¿ x for x = 0,1,2,3
p ( 0) =c 4c " 165
C j5 45 5= .36
^ = £ ^ = « - = .15Cj5 455
C 4r “ 2 2 0 p ( \ ) = = — — = .48
Cj5 455
C 4 C “ 4 P( 3) = - b r - = — - .0 1
c 15 455
b T h e p robabil it y h is to gram is sho wn belo w .
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5.55
5.61
c U sing the form ulas given in Section 5.4.
( í 4 ^
b i i
’ j j
, 1 5 ]
( N - M > Í N - n > o í 4 "l f 15 —4 ̂i * U - l y i , 5 ) I . 5 - 1 )
= .50286.
d Ca lcúlate the intervals
p ± 2 a = .8 ±2>/ .50286 = .8 + 1 .418 or - .6 1 8 to 2 .218
/ / ± 3 cr = . 8 ± 3-V-50286 = .8 ± 1 .418 or -1 .3 2 7 to 2 .927
Then,
P [- .6 18 < jc < 2.218] = p ( 0) + p ( 1) + p ( 2) = .99
P [ -1 .327 < x < 2 .927] = p ( 0) + p ( 1) + p {2) = .99
These results agree with Tchebysheff’s Theorem.
a Th e random variable x has a hype rgeom etric distr ibution with
N = 8,M = 5 and rc = 3. Th en
-.5 -̂3
p (x ) = * for x = 0,1,2 ,3C
b P ( x = 3) =C¡C¡ _ 10
C*= — = .1 78 6
56= — = . 0 17 86
56
r 5r ? r 5r 3 1 + 15
d P ( x < 1) = + - 4 - ^ - = I j l i l = .2857V ' C¡ C3* 56
Refer to E xercise 5 .60 and assume that p = . 1 instead o f p = .5 .
a P [ . t = 0 ] = p ( 0 ) = C o(- l )° ( -9) ' = .729
P [ j : = 1]= />(1) = C¡'(-1)‘(-9 )2 = .24 3
/>[.v = 2] = p(2 ) = C 5 (.l )2 (.9) ' = .027
P [ . i = 3] = p ( 3 ) = C ! ( . l ) ' ( . 9 ) “ = .0 0 1
b N o te th at th e p robab il it y d is tr ib ution is no lo nger sym m etr ic ; th at is , sin ce th e
p robabil ity o f observ in g a head is so sm all, th e p robabil it y o f observ in g a sm all
num ber of heads on three f lips is increased (see the f igure on the nex t page).
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0 1 2 3
c f j = n p = 3(. 1) =.3 and (T = yjn pq = ^ 3 ( . l ) ( .9 ) = . 52 0
d Th e desired intervals are
/ / + ít = . 3 ± . 5 2 0 o r - . 2 2 0 t o .8 20
/y ±2(7 = .3±1 .04 or - .7 4 0 to 1 .34
The only valué of x which falls in this first interval is jc = 0, and the fraction of
me asurem ents in this interval will be .729. Th e valúes of * = 0 and jc = 1 are enclosed
by th e second in te rv al, so th at .7 2 9 + .243 = .9 72 o f th e m easure m en ts fa ll w ithin tw o
standard devia t ions of the mean, cons is tent with both Tch eby shefPs Theorem and the
Empirical Rule.
5.65 Refer to Ex ercise 5.64. Red efine x to be the numb er of people who cho ose an inter ior
num ber in the samp le of n = 20. Then x has a binomial distr ibution with p = .3 .
a /> [jc > 8] = 1 - / > [ . * < 7 ] = 1 - 7 7 2 = .2 2 8
b Observing eight or more people choosing an inter ior number is not an unlikely
event, assum ing that the integers are all equally likely. Therefore, there is no
evidence to indícate that people are more likely to choose the inter ior numbers than
any others.
5.69 It is given that jc = nu m ber of pat ients with a psychosomatic problem, n = 25 , and
p = f ’ fpatient has psych osom atic pro ble m ]. A psyc hiatr ist w ishes to determine
whether or not p = .8 .
a A ssum ing that the psyc hiatr ist is correct ( that is, p = .8 ), the expected valué of
jc is E ( x ) = n p = 25(.8) = 20.
b ( f L = n pq = 25( .8) (.2) = 4
c G iven that p = .8 , P [ x < 14] = .006 from Tab le 1 in Appen dix I.
d Assum ing that the psych iatr ist is correct,the probab ility of observing jc = 14 or
the more u nlikely valúes, x - 0, 1, 2, . . . , 13 is very unlikely. H enee, one of twocon clusions can be drawn . Either we have observed a very unlikely even t, or the
p sychiatr is t is in corr ect and p is actually less than .8. W e would probab ly conc lude
that the psychiatr ist is incorrect. Th e prob ability that we have m ade an incorrect
decisión is
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P [ x < 14 given p = .8] = .006
which is quite small.
5.71
5.75
5.79
Define x to be the n um ber of students 30 years or older ,
with ti = 200 and p = ^ [stud en t is 30+ years] = .25.
a S ince x has a binom ial distribution, p = n p = 200(.25) = 50 and
G - y fñ p q - yj 2 0 0 (. 2 5 ) (. 1 5 ) = 6.124.
b The observed valué, x = 35 , lies
^ « = - 2 . 4 56.124
standard deviations below the mean. I t is unlikely that p = .25.
a The random variable x, the num ber of plants with red petáis , has a binomialdistr ibution with n = 10 and p = P[re d petáis] = .75.
b Since the valué p = .75 is not given in Table 1, you m ust use the binom ial
formula to calcúlate
P ( x > 9 ) = C¿° (.75 f ( .25)' + C¡¡¡ ( .75) '° ( .25)° = . 1877 + .0563 = .2440
c P ( x < 1) = C (.75)° (.25 )'° + C,10 (.75)' ( ,2 5)9 = .0000 296 .
d Refer to part c. The proba bility o f observing x = 1 or som ething even more
unlikely ( x = 0) is very small - .0000296. This is a highly unlikely event if in
fact p = .75 . Perhaps there has been a nonrandom choice o f seeds, or the 75% figureis not co rre d for this particular genetic cross.
a The distr ibution of x is actually hyp ergeom etr ic, w ith N = 1200, n = 20 and M =
num ber of defectives in the lot. How ever, s ince N is so large in com parison to n , the
distr ibution of * can be closely approxim ated by the binom ial distribution w ith
// = 20and p = P [defective].
b If p is small, with np < 7, the Poisson approxim ation can be used.
c If there are 10 defe ctive s in the lot, then p = 10/1200 = .008333 and p = .1667.
The probability that the lot is shipped is, ( .1 6 6 7 )% -1667
p ( x = 0 W ---------= .850 !
If there are 20 d efectives, p = 20/1200 and p = .3333. Then
x ( .3 3 3 3 Í V 3333/> (jc = 0 ) * - = .72V ’ 0!
If there are 30 d efectives, p = 30/1200 and p = . 5. The n
( 5 )° éf 5/> (* = 0 ) = ^ = .61
’ 0 !
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5.83 a The random variable x, the num ber o f tasters w ho pick the correct sam ple, has a
b in om ia l d is tr ib ution w ith n =5 and, if there is no difference in the taste of the thrce
samples, p = P{ taster picks the correct sam ple) = ^
b T he pro babil ity th at exactl y one o f th e five ta s te rs chooses th e la te st batc h as
different from the others is
5.87 The random variable x has a Poisson distribution with p = 2. Use Ta ble 2 in
Appendix I or the Poisson formula to find the following probabilities.
5.91 The random variable x , the num ber of California hom eow ners w ith earthquake
insurance, has a binom ial distr ibution with n = 15 and p = . 1.
a P ( x > 1) = 1- P ( x = 0 ) = 1 - .206 = .794
b P ( x > 4) = 1 - P(x < 3) = 1- .944 = .056
c Calcúlate p = np = 15(. 1) = 1.5 an d a = yfñpq = >/ l5( .l ) (.9) =1.16 19. Then
approximately 95% of the valúes of x should lie in the interval
/ /± 2 c r = > 1 .5 ± 2 ( 1 .1 6 1 9 ) = > - .8 2 to 3 .8 2.
or between 0 and 3.
5 .95 Use the C alcu lat ing Bino m ial P rob ab i l i ties applet . T he correct answers are g iven
below .
a P ( x < 6 ) = 6 .0 ( 1 0) "5 = 0 .0 0 0 0 6 d P ( 2 < x < 6) = .5948
b P ( x = 8 ) = .0 42 e P ( * > 6 ) = 1
c P ( x > 14) = .0207
5.99 Define x to be the num ber of young adu lts who prefer M cD ona ld’s. Then x has a
b in om ia l d is tr ib ution w ith n = 100 and p = .5 . Use the C alcu lat ing Binom ial
P rob ab i l i ties applet.
a P ( 6 l < x < 100) = .0176
a P(jc = 0) = ----------= e~2 = .1353350 !
= . 135335 + .270671 + .270671 = .676676
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b P ( 4 0 < x < 60) = .9648
c If 40 p refer Burger King, then 60 prefer M cDo nalds, and vice versa. The
pro bab il ity is th e sam e as th at calc u la te d in part b. sin ce p = .5.
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6: The Normal Probability Distribution
6.3 The first few exercises are designed to provide pract ice for the student in evaluat ing
areas under the normal curve. The fol low ing notes m ay be of som e assistance.1 Ta ble 3, Ap pendix I tabulates the cum ulat ive area under a standard norma l curve
to the left of a specified valué o f z.
2 Since the total area under the curve is one, the total area lying to the r ight o f a
specified valué of z and the total area to its left m ust add to 1. Th us, in order to
calcúlate a “tail area” , such as the one sho wn in Figure 6.1, the valué o f z = z0 will be
indexe d in T ab le 3, and the area that is obta ined will be subtrac ted from 1. De note
the area obtained by indexing z = z0 in Tab le 3 by / \ (z0) and the desired area by A.
Then, in the above e xam ple, A = 1- A (z 0) .
3 T o find the area under the standard normal curve betw een tw o valúes, z¡ and z2,
calcúlate the difference in their cum ulat ive areas, A = A { z 2 ) - A ( z x).
4 N ote that z, sim ilar to x, is actual ly a random variable which may take on an
infinite num ber of valúes, both posi t ive and negative. N egative valúes of z lie to
the left o f the m ean, z = 0 , and posi t ive valúes l ie to the r ight .
a I t is necessa ry to find the area to the left o f z = 1.6. Th at is, A = A (1.6) = .9452.
b T he area to th e le ft o f z = 1.83 i s A = /ftl .83) = .9664.
c A = A (.90) = .8159
d A = A(4.58) = 1. N otice that the valúes in Tab le 3 app roach 1 as the valué of z
increases. W hen the valué o f z is larger than z = 3.49 (the largest valué in the table) ,
we can assume that the area to its left is approximately 1.
6.7 N ow we are asked to fi nd the z-valu e co rr espond in g to a part ic u la r area.
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a W e need to f ind a zo such that P ( z > z0 ) = .025. T his is eq uiva lent to finding an
indexed area o f 1- .025 = .975 . Search the interior of Ta ble 3 unti l you find the four-
digi t numbe r .9750. Th e correspond ing z-value is 1.96; that is, A ( \ .96) = .975 0.
Therefore, z0 = 1.96 is the de sired z-value (see the f igure below).
b W e need to fi nd a zo s uch th a t P { z < zQ) = .9251 (see below ). U sing Ta ble 3, we
find a valué such that the indexed area is .9251. Th e correspon ding z-value is
* 0 = 1 . 4 4 .
The pth percenti le ,of the standa rd no rmal d istribution is a valué of z wh ich has area
p / 100 to its left. Since all fou r perc entiles in this exerc ise are gre ater than the 50 ,h
percenti le , th e valu é o f z will all lie to the right of z = 0 , as show n for the 90 th
perc enti le in th e fi gure on th e next page.
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a Fro m the figure, the area to the left o f the 90 lh perc entile is .9000. Fro m Ta ble 3,
the appro priate valué o f z is closest to z = 1.28 with area .8997. He nee the 9 0lh
perc enti le is appro x im ate ly z = 1 .28.
b A s in part a, the area to the left of the 95th perce ntile is .9500. From Ta ble 3, the
appropriate valué o f z is found using l inear interpolat ion (see E xercise 6.9b) as
z = 1.64 5. H enee the 95lh percen tile is z = 1 .645.
c Th e area to the left of the 98th percen tile is .9800. From T able 3, the appropriate
valué o f z is closest to z = 2.05 with area .9798. H enee the 98 lh perc entile is
approximately z = 2 .05.
d Th e are a to the left of the 99lh percen tile is .9900. From Ta ble 3, the approp riate
valué of z is close st to z = 2.33 with area .9901. H enee the 9 9,h perc entile is
approximately z = 2 .33.
6.15 Th e 99,h perc entile of the standa rd normal distribution was found in Ex ercise 6.1 Id to
be z = 2 .3 3 . S in ce the re la tionship betw een the genera l norm al random vari able *
x — uand the standard normal z is z = — , the corresponding percenti le for this general
o
normal random variable is found by solving for x = f j + z (J \
10
jc- 35 = 23.3 or x = 58.3
6.19 The random variable jc, the height o f a male hum an, has a normal distr ibution
with fJ = 6 9 and o = 3.5 .
a A height of 6 ’0” represents 6(12) = 72 inches, so that
P ( x > 7 2) = p Í z > — = p { z > -86) = 1 - .805 1 = .1949
b H eights o f 5’8” and 6 ’1” represen t 5(12) -t- 8 = 68 and 6(12) +1 = 73 inches,
respectively. Then
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P (68 < jc < 73) = p \ —— — < z < — j = ^ ( - - 2 9 < z < 1.1 4)
= . 8 7 2 9 - . 3 8 5 9 = . 48 70
c A height o f 6 ’0 ” represents 6(12) = 72 inches, which h as a z-value of
6 8 - 6 9 7 3 -6 9 ^1
3.5
This w ould not be considered an unusu ally large valué, since it is less than two
standard deviations from the mean.
d The probab ility that a man is 6 ’0” o r taller was found in part a to be .194 9, w hich
is not an unusual occurrence. Ho we ver, if you define y to be the number of men in a
random samp le of size « = 36 wh o are 6 ’0” o r taller, then y has a binom ial
distr ibution with mean f i = np = 36(. 1949) = 7.02 and standard deviation
a = y¡ ñp q = y l 3 6 ( . \ 9 4 9 ) ( M 5 \ ) = 2 .3 8 . The valué y = 17 l ies
standard deviations from the mean, and would be considered an unusual occurrence
for the general population of male humans. Perhaps our presidents do not represent a
random sample from this population.
6.23 The random variable jc, total we ight of 8 people, has a m ean o f/r = 1200 and a
variance o 1 = 980 0. I t is necessary to f ind P ( x > 1300) and P ( x > 1500) if the
distr ibution o f jc is approxim ately norm al. Re fer to the next f igurc-
o 2.38
1200 1300 1300
- p 1 3 0 0 - 1 2 0 0 _ 100- 4 - = — . = ------------ = 1.01
o V9800 98.995
P ( x > 1300) = P ( z > 1.01) = 1—A(1.01) = 1—.8438 = .1 5 6 2 .
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6.27
6.31
6.37
Similar ly, the z-value corresponding to x 2 =1 50 0 is
_ *2 - / / _ 1 5 0 0 -1 2 0 0 ^ Zi — — .------- j .V j j .
(7 >/9800
and P ( x > 1500) = P ( z > 3 .03) = 1 - ¿ (3 .03) = 1- .99 8 8 = .0012 .
a It is given that the prime interest rate forecasts, x, are approximately normal with
mean /y = 4.5 and standard d eviation o - 0.1 . I t is necessary to determ ine the
p ro b ahili ty th at x exceed s 4 .7 5 . C alc úla te
x - n 4 . 7 5 - 4 . 5 z = — = --------------- = 2 . 5 . T h e n
o 0.1
P ( x > 4 .75) = P ( z > 2 .5 ) = 1- .9 9 38 = .0062 .
x — u 4 3 7 5 - 4 5b Calcúlate z = ----- -- = -1 -25 . Th en
o 0.1
P ( x < 4 .375) = P ( z < - 1 . 2 5 ) = .1 0 5 6 .
Let w be the num ber of words specif ied in the contract. Then jc, the num ber o f words
in the man uscript, is normally distr ibuted with /y = w + 20 ,000 and cr = 10 ,000 . The
pu b li sh er w ould like to specify w so that
P ( x < 100,000) = .95.
As in Ex ercise 6.30, calcúlate
1 00 ,0 00 0- ( w + 2 0,0 00 ) 8 0 , 0 0 0 - w Z - -
10,000 10,000
8 0 , 0 0 0 - w } —_ T ,= .95 . It is nec essary thatThen P ( x < 100,000 ) = P z <
K ’ 10,000 )
z 0 = ( 8 0 , 0 0 0 - w ) / l 0 , 0 0 0 b e s u ch t ha t
/ , ( z < z 0 ) = -9 5 => A ( z 0 ) = .9500 or z0 = 1.6 45 .
Henee,
8 0 , 0 0 0 - w
10,000= 1.645 or w = 6 3 , 5 5 0 .
a The normal app roxima tion w ill be approp riate if both n p and nq are greater than 5.For this binomial experiment,
np = 2 5(3 ) = 1.5 and nq = 25( .7) = 17.5
and the normal approximation is appropriate.
b For the binomial random variable,
p = n p = 7.5 and o = y jnpq = -v/25 (.3)(.7 ) = 2.2 91 .
c The probahility of interest is the area under the binomial p robah ility histogram
corresponding to the rectangles x = 6,7,8 and 9in the f igure on the next page.
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6.41
6.45
To approximate this area. use the “correction for continuity” and f ind the area under a
normal curve with mean fu = 7.5 and o = 2.291 between x { = 5.5 and x 2 = 9.5 . The
Z-values corresponding to the two valúes of x are
z, = ^ = -. 87 9 . 5 - 7 . 5 ^and z, = ------------- = .87
2 2.2912.291
The approximating probabi l i ty is
P ( 5 .5 < * < 9 .5 ) = P ( - . 8 7 < z < .87) = .80 78 - .19 22 = .6156 .
d From Table 1, Appendix I ,/>(6 < jc < 9 ) = / >( jc < 9 ) —/>( jc < 5 ) = .811 —. 193 = .618
which is not too far from the approximate probability calculated in part c.
Using the binomial tables for n = 20 and p = .3 , you can ver ify that
a P ( .r = 5) = P ( . x < 5 ) - P ( j t < 4 ) = .4 1 6 - .2 3 8 = .178
b P ( x > 7 ) = 1 - P ( a : < 6 ) = 1- .608 = .392
a The approximating probability will be P ( x > 2 0.5) wh ere x has a normal
distribution with ¡a = 50( .32) = 16 and o = >/50( .32 )( .68) = 3.298 . Then
2 0 . 5 - 1 6 ' P ( x > 20.5) = P z > ■
\
b The approximating probability is
3.298
P ( x < 14.5) = P z <1 4 . 5 - 1 6
3.298
= P ( Z > 1.36 ) = 1—.9131 = .0869
= P ( z < - .4 5 ) = .3264
c If few er than 28 students do not prefer cherry, then more than 50 - 28 = 22 do
p re fe r cherr y. T he appro xim atin g p ro babil it y is
P ( x > 22 .5) = p ( z > 22-5 ~ .16.v ’ 1, 3 .298
d As long as your class can be a ssum ed to be a representative sam ple o f all
Americans, the probabilit ies in parts a-c will be accurate.
= P ( z > 1 .97) = 1- .9756 = .0244
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6.49 D efine x to be the num ber of elect ions in wh ich the tal ler cand idate won. If
Am ericans are not biased by height , then the random variable x has a binom ial
distr ibution with n = 31 and p = .5 . Calcúlate
p = np = 31(.5) = 15.5and G = y¡ 3\(.5 )(.5 ) = V7/75 = 2.784
a Using the normal approximation with correction for continuity, we find the area tothe r ight o f x = 16.5 :
P ( x > 16.5) = P\^z > 1625 ~g45 '5 ) = P ( z > -36) = 1 - 6406 = .3594
b S in ce the occurr ence o f 17 ou t o f 31 ta ll er choic es is not unusual, based on th e
results of part a, i t appe ars that Am ericans do not consider height w hen cast ing a vote
for a can didate.
6.53 Re fer to Exercise 6.52, and let x be the num ber of working wom en w ho put in more
than 40 hours per w eek on the job. Then x has a binomial distr ibution with n = 50
and p = .62 .
a The average valué of * is ( i = np = 50(.62) = 31.
b The sta ndard devia tion o f x is G = yjnpq = >/50(.62 )(.38) = 3.432.
c The z-score for x = 25 is z = ——— = — — — = -1 .7 5 w hich is within twoG 3.432
standard deviat ions o f the m ean. This is not considered an u nusual occu rrence.
6 .55 a Th e de sired are A| , as shown in the f igure on the next page, is found by
subtract ing the cum ulat ive areas corresponding to z = 1.56 and z = 0 .3 , respectively.
A, = A ( 1 .56) - A( .3) = .9406 - .6179 = .3227 .
b T he desir ed area is show n on the next page :
A, + A, = A(. 2) - A { - .2 ) = .5793 - .4207 = .1586
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6.59
6.63
desired valué, zo, will be between z, = -67 and z2 = .68 w ith asso ciated p robabilit ies
Px = . 2 5 1 4 a n d P2 = .2483 . Sinc e the de sired tail area, .25 00, is clos er to P[ = .25 14 ,
we approximate zo as z0 = .67 . Th e valú es z = -.6 7 and z = .67 repre sen t the 25 1*1
and 75th percentiles of the standard normal distribution.
It is given that x is normally distr ibuted with fu = 10 and <7 = 3. L et t be the
guaran tee time for the car . I t is necessary that only 5% of the cars fail before time t (see below). That is,
P ( x < í ) = . 05 o r p f z < L - J 5 j = .05
From Table 3, we know that the valué of z that satisf ies the abov e prob ability
statement is z = -1 .645 . Henee,
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6.67
6.71
6.75
-— - = -1 .6 4 5 or t = 5.065 months
W e mu st ha ve
P ( x < X o ) = p [ z < ^ y .
' v - 7 0
l 12
90
= .90
xn —70Consider z0 = — . W ithout interpolat ing, the approxim ate valué for zo is
12
x , - 7 0 _ o
For this exercise ¡u =10 and o = 12. The o bject is to determ ine a part icular valué, x 0,
for the random variable x so that P ( x < x 0 ) = .9 0 (that is, 90% of the students will
f inish the exam ination before the set t im e l imit) . Re fer to the f igure below.
-------------------
I t is given that the ran dom variable * (ounce s of f i ll ) is norm ally distr ibuted with
mean /y and standard deviat ion o = .3 . It is nec ess ary to find a valu é o f /y so that
P { x > 8) = .01 . Th at is, an 8-ounce cup w ill overflow w hen x > 8, and this should
happen only 1% of the t ime. Then
7 - / / ' P ( x > 8) = P ^ z > = .0 1 .
•3
From T able 3 , the valué of z correspond ing to an area ( in the upp er tail o f the
distribution) o f .01 is z0 = 2.3 3. Flence, the valué of /y can be ob tained by solving
for/y in the fol lowing equation:
2 .33 = - ^ ^ or / i = 7.301.3
Def ine x = num ber o f incom ing cal is that are long distance
p = P[inco m ing cal i is long distance] = .3
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n = 200
The desired probab ility is P ( x > 5 0 ) , wh e r e x is a binomial random variable with
H = n p = 200( .3) = 60 and ^ = V2Ó 0(3)C7) = >/42 = 6.481
A correction for continuity is mad e to include the entire area un der the rectangle
cor responding to x = 50 and henee the approximation wil l be
6.79 The random var iable jc,the gestation time for a human baby is normally distr ibutedwith p = 278 and <7 = 12 .
a From Exercise 6.59, the valúes (rounded to two decimal places) z = - .6 7 and
Z = .67 represe nt the 25 lh and VS01 pe rcentiles o f the sta nd ard norm al distribu tion.
Converting these valúes to their equivalents for the general random variable x using
the relationship jc = p + z o , you have:
b If you consider a month to be approximately 30 days, the valué x = 6(30) = 180 is
unusual, since it lies
standard deviations below the mean gestation time.
6.83 In order to implem ent the traditional interpretation of “curving the g rades” , the
p ro port io ns show n in th e ta ble need to be applied to th e norm al curv e, as show n in th e
figure below. _________________________________________________ ______
The lower quartile: jc = - . 6 7 ( 1 2 ) + 2 78 = 26 9 .9 6 a nd
The upper quartile: jc = .67(12) + 2 78 = 286.04
_ x - f j _ 1 8 0 - 2 7 8
o 12= - 8 . 1 6 7
A
0.3-
S 0.2-
0 . 1 -
0.0 -
-3 -2 -1 0 2 3x
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a The C grades constitute the middle 40%, that is , 20% on either side of the mean.
Th e low er bound ary has a rea .3000 to its lef t. From T able 3, we need to Find a valué
o f z such that A(z) = .3000 . The close st valué in the table is .3015 w ith z = - .52 . The
upper boundary is then z = +.52 .
b The cutoff for the lowest D and highest B grades constitute the lower and upper bou ndari es o f th e m id dle 80% , th at is, 40% on e it her sid e o f th e m ean . T he lo w er
boundary has arc a .1 000 to its le ft , so we need to Find a valu é o f z such that A(z) =
.1000. The closest valué in the table is .1003 with z = -1 .28 . The upper boundary is
then z = 1.28.
6.87 Use eithe r the Normal Distribution Probabilit ies or the Normal Probabil i t ies and
z-scores applets.
a P ( - 2 . 0 < z < 2 .0) = .97 72 - .022 8 = .9544
b P ( - 2 3 < z < - 1 . 5 ) = . 0 6 6 8 - . 0 1 0 7 = .0561
6.91 a Use the Norm al D istr ibut ion Probabil i ties applet. En ter 5 as the mean and 2 as
the standa rd deviation, and the ap propriate lower and upp er boun daries for the
pro bab il iti es you need to calc úla te . T he pro babil ity is re ad fr om th e apple t as Prob =
0.9651.
b Use the Norm al Proba bil it ies and z-scores applet. En ter 5 as the mean and 2 as
the standard deviation and x = 7.5 . Ch oose One-tail from the dropdown list and
read the prob ability as Prob = 0.1056.c Use the No rm al P robabil i ties and z-scores applet. En ter 5 as the mean and 2 as
the stand ard deviation and x = 0 . Choose Cumulat ive fromthe dropd ow n list andread the prob ability as Prob = 0.0062.
6.95 a I t is given that the scores on a national achievem ent test were approxim ately
norm ally distr ibuted with a mean of 540 and standard deviation o f 110. I t is
necessary to determine how far. in standard deviations, a score of 680 d eparts from
the mean of 540. Calcúlate
x - / J 6 8 0 - 5 4 0 ,Z = --------- = -------------------= 1 .2 / .
o 110
b To Find the percentage of people who scored h igher than 680 , we f ind the area
und er the standardized no rmal curve greater than 1.27. Using T able 3, this area isequal to
P ( x > 680) = P ( z > 1 .27) = 1- .89 8 0 = .1020
Th us, approx imately 10.2% of the people w ho took the test scored h igher than 680.
(The applet uses three decimal place accuracy and shows z = 1.273 with Prob =
0.1016.)
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7: Sampling Distributions
7.1 You can select a s imple rando m sample o f s ize n = 20u sing T able 10 in Ap pendix I .
First choo se a starting point and consid er the first three digits in each nu m ber. Since
the experimen tal units have already been n um bered from 0 00 to 999, the first 20 can be
used. The three digits OR the (three digits - 500) will identify the prop er experimen tal
unit . For exam ple, if the three digits are 742, you should select the experimen tal unit
num bered 7 4 2 -5 0 0 = 242 . T he probahil ity that any three digit num ber is selected is
2/10 00 = 1/500 . One p ossible selection for the sample size n = 20 is
242 134 173 128 399
056 412 188 255 388
469244
332 439 101399 156 028 238 231
7.5 If a l l o f the town cit izenry is l ikely to pass this com er, a samp le obtained by selecting
every tenth person is probably a fairly random sample.
7.9 Use a random ization schem e similar to that used in Exercise 7 .1. N um ber each of the
50 rats from 01 to 50. To ch oose the 25 rats who will receive the dose of MX , select
25 two -digit rando m nu m bers from Table 10. Each tw o-digit num ber OR the (two
digits - 50) will identify the prop er experimental unit .
7.13 a Th e firs t question is more unbiased. b N oti ce th at th e percenta ge favorin g th e new sp ace pro gram drops dram ati call y
wh en the phrase “spending b il lions o f dollars” is added to the question.
7.19 Re gardless o f the shape of the population from wh ich we are sampling, the sam pling
distribution o f the sample m ean will have a mean / / equ al to the mean o f the
popu la ti on fr om w hic h we are sam pling, and a st andard devia tion equal to o ¡ \ f ñ .
a /y = 10; ( j / y fñ =3/>/36 =.5
b ju = 5; ( j / y f ñ = 2 / -n/ToO =.2
c n = 120; cr/Vñ = l/>/8 =.3536
7.22-23 For a popula t ion wi th o = 1 , the standard error of the mean is
<7 / y fñ = l / y f ñ
The va lúes of <7 / y fñ for various valúes of n are tabulated below and plotted on the next
page. N oti ce th at th e sta ndard e rro r decreases as the sample size increases. _ n ______________________ 1 __________ 2 __________ 4 _________ 9 _________ 16 __________ 25 ___________ 100
SE(x ) = a /4 ñ L(X) _7Q7 _5(X) J33 _25Q _20Q 100
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7.25
7.29
7.33
a Ag e o f equipm ent, technician error, technician fat igue, equipm ent fai lure,
difference in Chemical puri ty, contam ination from outside sources, a nd so on.
b T he vari abil it y in th e avera ge m easure m ent is m easured by th e sta ndard err or,
o / y f ñ . In order to decrease this variabil ity you should increase the sam ple size n.
a The po pulat ion f rom w hich we are random ly sampl ing n = 35 measurem ents i s not
necessari ly normally distr ibuted. Ho wev er, the sam pling distr ibution of x does have
an approximate normal distr ibution, with mean ¡a and standard deviat ion <r/ y fñ . T he p robab il it y o f in te re st is
P ( \ x - j u \ < l ) = P ( - 1 < ( * - / / ) < l ) .
Since z = X , has a stand ard norm al distribution, we need only find o / y f ñ to( j /y jn
approx ima te the above probabili ty. Th ough o is unknown, i t can be approximated by
s = 12 and o / y f ñ = 12 /^35 = 2 .028 . T hen
P ( \ x - f i \ < 1) = P ( - 1/2.028 < z < 1/2.028)
= P ( - A 9 < z < .49) = .6879 - .31 21 = .3758
b N o. T here are m any possib le valú es fo r* , th e actu al percen t tax savin gs, as giv en
by th e probabil it y d is tr ib u tion fo r * .
a Since the-eriginal populat ion is norm ally distr ibuted, the sam ple m ean * is also
norm ally distr ibuted (for any sample size) with mean /y and standard d eviat ion
(7/y fñ ~ 0 .8 /V l30 = .07016
The z-value corresponding to * = 98.25 is
9 8 . 2 5 - 9 8 .6 _
• Cj/yfñ 0.8A/I3Ó
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7.37
7.41
7.45
an d
P ( x < 98.25) = P ( z < - 4 . 9 9 ) = O
b Sin ce th e pro babil ity is ex tr em ely sm all , th e avera ge te m pera ture o f 98.2 5 degre es
is very unlikely.
a p = .3; S E ( p ) = = .0458V ' V n V 100
b p = .l ; S£ ( p ) = ̂ = J | | = .015
c p = . * 0310
The valúes S E = p q / n for n = 100 and v arious valúes o f p are tabulated and graphed
belo w . N oti ce th at S E is máximum for p = .5 and becom es very small for p nea r zero
and one. p .01 .10 .30 .50 .70 .90 .99
SE (P ) .0099 .03 .0458 .05 .0458 .03 .0099
a The random variable p , the sample propor t ion o f brown M &M s in a package of
n = 55 , has a binomial distr ibution with n = 55 and p = . 13. Since n p = 7.15 and
nq = 47.85 are both greater than 5, this binomial distr ibution can be approx imated by a
I 13( .87 )normal distr ibution with mean p = . 13 and S E = . : 1 = .04535.
F V 55
b / - ( p < . 2 ) = / > ^ < | ^ | j = P ( c < 1 .5 4 ) = .9 382
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c P (p > . 3 5 ) = ̂ > ^ ^ j = /> (j > 4 .8 5 ) = 1 -1 = 0
d From the Empirical Rule (and the general properties of the normal distr ibution) ,approximately 95% o f the m easurem ents will l ie w ithin 2 (or 1.96) standard d eviations
of the mean:
p ± 2 S E => . 1 3 ± 2 ( . 0 4 5 3 5 )
.13 ±.0 9 or .04 to .22
7.49 aThe upp er and lowe r control l imits are
ÜCL = x + 3-^= = 155.9 + 3 ^ 2 = 155.9 + 5.77 = 161.67
L C L = x - 3 - ^ = = 1 5 5 . 9 - 3 ^ = 1 5 5 .9 -5 .7 7 = 150.13 y/5
b The control chart is constructed by plotting two horizontal l ines, one the upper
con trol limit and one the low er control l imit (see Figure 7.15 in the text) . V alúes of
x are plotted, and should remain within the control limits . I f not, the proces s should be
checked.
7.55 Calcúla te p = — — + -26 - 1 9 7 The upper and lower control l imitsk 30
for the p chart are then
UCL = p + 3 J P (1 P) = . 197 + 3 J - 1- - =. 197 + . 119 = .316
L C L = p - 3 l-P(1- = . 1 9 7 - 3.1' 19 7 (8 0 ^ = . 19 7 - . 1 19 = .0 78V n V 100
7.60 a C \ = = 6 samp les are p ossible.
b-c T h e 6 sam ples a long with the sample means for each are shown below.
Sample _________________ Observat ions ________________ x_
1 6 , 1 3.5
2 6 ,3 4.5
3 6 , 2 4.0
4 1, 3 2 . 0
5 1 , 2 1.5
6 3 ,2 2.5
d Since each of the 6 distinct valúes of x are equally likely (due to random
sampling) , the sampling distr ibution of x is given as
p ( T ) = - fo r x = 1 .5 ,2 ,2 .5 ,3 .5 ,4 ,4 .56
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7.65
7.71
7.75
The s am pling dis t r ibut ion is shown be low.
1.5 2.0 2.5 3.0 3.5 4.0 4.5
*
e The popu lation mean is p = (6 + 1+ 3 + 2 ) /4 = 3 . Not ice that none o f the samples
of size n = 2 produce a valué o f x exactly equal to the population mean.
a To divide a grou p o f 20 people into two groups of 10, use Ta ble 10 in A ppen dix I.
Assign an iden tif ication n um ber from 01 to 20 to each person. Th en select ten two digit
num bers from the random num ber table to identify the ten people in the f irst group. (If
the number is greater than 20, subtract múltiples of 20 from the random number untilyou obtain a number between 01 and 20.)
b A lt hough it is not possib le to select an actu al ra n do m sam ple from th is hypo th eti ca l
popula ti on , th e re searcher m ust ob ta in a sa m ple th at behaves l ike a random sample . A
large datab ase o f som e sort should be u sed to ensure a fair ly represen tative sample.
c The researcher has actually selected a convenience sample-, however, it will
p ro bably behave like a sim ple ran do m sam ple , s in ce a p e rso n ’s en th us ia sm fo r a paid
jo b should not aff ect his re sponse to th is psychologic al experi m ent.
a Since each cluster (a city block) is censused, this is an exam ple o f cluster sampling.
b T his is a 1 - in -10 syste m ati c sa m ple .
c The wards are the strata, and the sample is a stratif ied sam ple.
d This is a l- in-10 sys tematic sample .
e This is a simple rando m sample fromthe popu lation o f all tax returns f iled in the
city of San Bernardino , California.
a The average propor t ion o f defect ives is
_ .04 + .0 2 H— + .03 ^ p = --------------------------- = .032
25
and the control l imits are
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U CL = p + 3 j — — — = -032 +n ■/ 100
and L C L = p - 3 P(1 P) = . 0 3 2 - 3n
If subsequ ent sam ples do not stay w ithin the limits, UCL = .0848 and L C L = 0 , the
p ro cess should be checked .
b From par t a, we must have p > .0848 .
c An erroneou s conc lusión will have occurred if in fact p < .0848 and the sam ple has
p ro d u ced p = .15 by chance. On e can obtain an upper bound on the proba bility of this
p arti cu lar ty pe o f e rr o r by calc u la ti ng P ( p > .1 5 when p = . 0848) .
7.79 A nsw ers will vary from student to student. Paying cash for opinions will notnecessa rily produ ce a random sample o f opinion s of all Pepsi and Coke drinkers.
7.83 a Th e average propor t ion of inoperable com ponents is
_ _ 6 + 7 h h5 _ 75 _ ]0
P ~ 50(15) _ 75 0 ~~
and the control l imits are
and
If subseq uen t samples do not stay w ithin the limits , UCL = .2273 and L C L = 0 , the
pro cess should be checked.
7.87 a Th e theoretical mean and standard dev iation of the sampling distr ibution o f x when
ti = 4 are
p = 3.5 and a / y [ ñ = 1.708 / \[ Á = .854
b-c An swers will vary from student to student. The distr ibution sho uld be relatively
uniform with mean and standard deviation cióse to those given in part a.
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8: Large-Sample Estimation
The margin o f error in estimation prov ides a practical up per boun d to the difference betw een a part ic u la r estím ate and th e param ete r w hic h it estim ate s. In th is ehapte r,
the margin of error is 1.96x (standard error of the estimator) .
Th e margin of error is 1.96 S E = 1.96-^L , where o can be estimated by the sample yjn
standard deviation s for large valúes o f n.
a 1 . 9 6 , — = . 5 5 4 b 1 .9 6, — = . 1 7 5 c 1 .9 6 ./—̂ — = .055 V 50 y 50 0 V 50 00
For the estím ate of p given as p = x /n , the margin o f error is 1.96 S E = 1.96. — .V n
Use the estimated valué given in the exercise for p.
a 1.96 = .0588 b 1 . 9 6 = .0898 c 1. 9 6 = .098\ 100 V 100 V 100
d 1.96 = .0898 e 1 .9 6 J— ^ — =.0 58 8V 100 V 100
f Th e largest margin o f error occurs w hen p = .5 .
The point est ímate o f p is x = 39.8° and the m argin of error with 5 = 17.2 and n = 5 0
is
1.96 S E = 1 . 9 6 = 1 .96-4= = 1,9 6 - 4 ¿ = 4 .768 yfn yjn V50
X a Th e point estímate for p is given as p = — = . lS and the m argin o f error is
n
approximately
1.96 J — = 1 .9 6 ,f 7 8 ^ 2 2 ^ = .026n V 1000
b Th e p oll’s margin o f error does not agree w ith the results of part a, because the
samp ling error was reported using the máxim um margin o f error using p = .5 :
1.96 E = 1.96 = .0 3 1 o r ± 3 .1 %\ n \ 1000
A p oint estímate for the mean length o f t ime is x = 19.3, w ith m argin of error
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1.96 S E = 1 .9 6 -^ r « 1 . 9 6 - ^ = l . 9 6 - 5 ¿ = 1.86 yjn yj n >/30
.23
.27
.31
.35
The 90% confidence interval for/y given as
3c ± 1 .6 4 5 y¡n
w here <7 can be est im ated by the sam ple standard dev iat ion 5 for large v alúes of n.
a .84 + 1 .6 45 ./— = . 8 4 ± . 0 4 3 o r .79 7 < u < .883V 125 ^
J
3 44 — = 2 1 .9 ± .4 3 1 or 21.4 69 < /i < 22.331
50
c Intervals constructed in this m anne r wil l en do se the true valué o f p 90% of the
t im e in repeated sam pling. Henee, w e are fair ly confiden t that these pa rt icular
in tervals wi l l end ose p .
The width of a 95% confidenc e interval for p is g iven as 1 .9 6 -y r . Henee , yjn
a Wh en n = 100, the width is 2L 96j y = 2 ( i -9 6 ) = 3 -9 2 -
b W hen n = 20 0, the width is 2[ 1.96—¡Í£ = = 2(1.38 6) = 2.772 .I > / 2 0 0 j
c Wh en n = 40 0, the width is 2'■967 m ) - - 2 { -9 S ) - - L 9 6 -
With n = 40, x = 3 .7and s = .5 and a = .01 , a 99% conf idence in terval for ¡u is
approximated by
x ± 2 .5 8 -4 = = 3 .7 ± 2 .5 8 —¿ = = 3 .7 ± .2 0 4 o r 3 .4 9 6 < lí < 3.904
•Jn V 40In repeated sam pling, 99% of al l intervals construc ted in this m ann er wil l en do se / i .
Henee, we are fairly certain that this part icular interval con tains f j . (In orde r for this
to be true, the sample m ust be random ly selected.)
a The point est imate o f p is p = — = = . 136, and the approxim ate 95%" n 500
confiden ce interval for p is
p ± 1 .96 = . 136 ± 1 .96 | - 36 ( :864) = . 136 ± .030500or . 106 < /? < . ! 66 .
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b In ord er to in cre ase th e accura cy o f th e confi dence in te rv al, you m ust decre ase its
width. You can accom plish this by (1) increasing the sample size n, or (2) decreasing
Za/2 by de creasing the co nfidence coeff icient.
8.39
8.43
a W hen estimating the difference /y, - / y 2, the (1 - a ) 100% co nfide nce ijiterval is
(3c. - x ,) ± za/2 — + — . Estimating o \ and o \ w ith s] and s2 , the approximate1 " , " 2
95% confidence interval is
l¡ 38 4 14( 1 2 .7 -7 .4 )± 1 .9 6 J— + — = 5 .3 ± .6 9 0 o r 4 .61 < / / , - f i 2 < 5.99 .
b S in ce th e valu é // , - / y , = 0 is not in th e confi dence in te rv al, it is not li kely th at
/y, = /y , . You should conc lude that there is a difference in the two po pulation m eans.
a Th e param eter to be estimated is /y , the mean score for the posttest for all BA CC
classes. The 95% con fidence interval is approximately
I ± 1 . 9 6 - ^ = 1 8 .5 ± 1 .9 6 -? = £ L = 1 8 .5 ± .8 2 4 o r 1 7 . 6 7 6 < /y < 1 9.3 24V365
b T h e p a ra m ete r to be esti m ate d is / y , th e m ean score for th e postt est fo r all
traditional classes. Th e 95% co nfidence interval is approximately
5 6 963c ± 1 . 9 6 4 = = 1 6 . 5 ± 1 . 9 6 - t = = 1 6 . 5 ± .7 9 0 o r 1 5 .7 10 < /y < 17 .2 90
yf n V298
c No w we are interested in the difference between posttest means, /y, - / y , , for
BA CC v ersus traditional classes. Th e 95% con fidence interval for /y, ~/y 2 is
approximately
( ,8 .5 -1 6 .5 , ± l . 9 6 j M * + « « lv 7 v 365 298
2.0± 1.14 2 or .858 < (/y , - / y 2)< 3.142
d Since the confidence interval in part c has two positive endp oints, i t does not
contain the valué /y, - / y 2 = 0 . H enee, it is not likely that the mean s are equal. It
appears that there is a real difference in the mean scores.
8.47 Refer to Exercise 8.18.
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a The 95% conf idence interval for / /, - / / 2is approximately
(170 —160) ± 1.96.
1 0 ± 6.667 or 3 .333 < ( / /, - / / , )< 16 .667
b The 99% confidence in te rv al fo r is appro xim ate ly
- 1 5 ± 7 . 0 4 0 o r - 2 2 .0 4 0 < ( // , - / ¿ 2 ) < - 7 . 9 6 0
c N either of the intervals contain the valué ( /v ,- // 2) = 0 . I f ( / / , —//2) = 0 is
contained in the confidence interval, then it is not unlikely that H\ could equal / /2 ,
implying no difference in the average room rates for the two hotels . Th is wo uld be of
interest to the experimenter.
d Since neither confide nce interval con tains the valué //, = 0 , it is not likely
that the me ans are equal. Yo u should con clud e that there is a differenc e in theaverage room rates for the Marriott and Wyn dham and also for the R adisson and the
Wyndham chains .
x 337 x 37 48.51 a Calcúlate A = — = 1— = .42 and p 2 = — = - — = .5 8 . T h e a p p ro x im a t e 90 %
n, 800 ' n 2 640
confidence interval is
b The tw o bin om ia l sam ple s m ust be ra ndom and in dependent and th e sa m ple sizes
must be large enoug h that the distr ibutions of p xand p 2 are approximately normal.
Assum ing that fí le sam ples are random, these conditions are met in this exercise.
- .1 6 ± . 0 4 3 o r - .2 0 3 117
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8.55 a W ith p. = — — = .45 and p , = -~ 2— = .51. The approxim ate 99% conf idence1 1001 1001
interval is
( A - p ; ) i 2 . 5 8 j M + M
.45 .55 ) .51 .49( .4 5 -.5 1 1 2 . 5 8 . — i ------^ + — i----
v y V 1001 1001
- . 0 6 1 . 0 5 8 o r - .1 1 8 < ( p , - p 2) < - .002
b Sin ce th e in te rv al in part a co n ta ins only negati ve valú es o f p , - p 2 , it is likely
that p| - p 2 < 0 P| < p 2 . Th is wo uld indicate that the proportion of adults who
claim to be fans is higher in November than in March.
r 120 x , 548.59 Ca lcúlate p, = — = ----- = .7 and P7 = — = ------ = .54 . The approximate 90%
1 n, 180 2 n 2 100
confidence interval is
( p , - p 2) ± l - 6 4 5 j ^ +
( .7 - .54) ± 1.645 1v ' V 180 100
.1 6 1 .0 9 9 o r . 0 6 1 < ( p , - p 2) < .2 5 9
Intervals constructed in this man ner w ill en do se the true valué of p , - p 2 95% of the
time in repeated samp ling. Hen ee, we are fairly certain that this particular interval
e n d o s e s p, - p 2 .
8.63 Follow the instructions in the My P ersonal Traine r section. The answ ers are show n in
the table below.
T ype of D ata O ne or Tw o
S a m p l e s
M a r g i n ofe r r o r
p o r a B o u n d ,B
Solve thisinequal i ty
Sam ple s i ze
Binomial One p ~ . 5 .051.96 ^ ^ < . 0 5
V n
n > 3 8 5
Quantitative O ne1 . 9 6 - i -
yjn
( 7 * 1 0 21.9641
« > 9 7
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8.67 For the difference p x - /y , in the popu lation m eans for two quant i ta t ive p opulat ions ,
the 95% u pper confidence bound uses z ^ = 1.645 and is calculated as
( * , - J 2) + 1.545 + ̂ = ( 1 2 - 1 0 ) + 1 . 6 4 5 J | - + ̂ -\ n, n2 V50 50
2 + 2 .0 0 o r ( / / , - / / , ) < 4
8.71 In this exercise, the param eter o f interest is p¡ - p 2 , nx= n 2 = n , and B = .05 . Since
we have no pr ior knowledge about p xand p 2 . we assum e the largest possible
variation, which occurs if p x = p 2 = .5 . Then
za/2 x(s td er ror of p x- p 2) < B
, & ^ r £ 05 ^ o s
\ nx n2 V n n
y f ñ > —— — => n > 1 08 5.7 8 o r « , = « , = 1086.05 1 2
8.75 The standard deviation is estimated as R / 4 = 104/4 = 26 , and .
2.58.1— + — < 5 => 2.58,1— + — < 5n, V n n
^ > 2 .5 8V Í3 52 ^ ^ > 3 59 9 g 0f ^ = ^ = 3 60
8.76 a Fo r the difference /y, - f i2 in the population means this year and ten years ago, the
99% lower conf idence bound uses z 01 = 2.33 and is ca lcula ted as
K s,2 _ 2 5 2 2 8 :
( x - x , )- 2 .3 3 í— + — =(73 —63) —2.33A/------+ ■V 1 ^ nx n2 ’ \ 4 00 4 00
1 0 - 4 . 3 7 o r ( jjx - / / , ) > 5.63
b S in ce th e dif ference in th e m eans is posit ive, you can co n c lu d e th at th ere has been
a decrease in the average per-capita beef consumption over the last ten years.
8.83 a The point es tímate of p is x = 29.1 and the margin o f error in estimation with
5 = 3.9 an d n = 64 is
1.96(7, = 1 .96-?= * 1.96 -7= = 1.963.9
y fñ y fñ y¡64, b T he appro xim ate 90% confi dence in te rv al is
= .9555
x ± 1 .6 4 5 -4 = = 2 9 . 1 ± 1 . 6 4 5 - ^ L = 29 .1 ± . 8 0 2 o r 2 8 .2 9 8 < p < 29.902 yf n V64
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Intervals constructed in this man ner end os e the true valué of /y 90% o f the t ime in
repeate d sam pling. Th erefore , we are fairly certain that this particular interval
e n d o se s / y .
c The approximate 90% lower conf idence bound is
1 - 1 . 2 8 - ^ = 2 9 . 1 - 1 . 2 8 ^ = 28.48 o r /y > 28.48-Jn V64 _
d W ith B = .5, o ~ 3 . 9 , a nd 1 - a = .95 , we mu st solve for n in the following
inequality:
cr 3.9\ . 9 6 —j= < B => \. 9 b —j= < . 5
■Jn yjn
4 ñ > 15.288 => n > 233.723 or n > 234
8.87 Assum ing máximum var ia t ion wi th p = .5 , solve
1.645J - ^ < .0 2 5
4~n >. i W M =3 2 , n > 1082 .41 o r ai > 1083.025
8.91 a Define sam ple #1 as the samp le o f 482 w omen and samp le #2 as the sample of
356 men. The n p, = .5 and p : = .75.
b T he approxim ate 95% confid ence in te rv al is
( P \ - M ± 1.96 +n, n 2
( .5 - .7 5 ) ± 1 .9 6 M +v ’ \ 482 356
- .2 5 ± . 0 6 3 o r - .3 1 3 < ( p , - p 2) < - .1 87
c Since the valué p, - p 2 = 0 is not in the confidence interval , i t is unlikely that
p l = p 2 . You should not conclude that there is a difference in the proportion of
wo me n and men on Wall Street who have children. In fact, since all the probable
va lúes of p , - p 2 are negative, the proportion of men o f Wall Street who have
children appears to be larger than the proportion o f women.
8.95 Assume that o = 2.5 and the desired bou nd is .5. Then
1 . 9 6 < B 1 .9 6 ^ p < .5 => n > 9 6.0 4 or n > 9 7 y¡n y¡n
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8.99
8.103
8.107
8.111
8.115
a If you use p = .8 as a conservative est íma te for p , the margin o f error is
approximately
± 1 . 9 6 . ® ^ = +.029V 750
b To reduce the m argin o f error in part a to ± . 0 1 , sol ve for « in the equ ation
1 . 9 6 . p í ^ = , 0 1 => V ^ = L 9 6 ( '4 > = 7 8.4 => « = 6 14 6 .5 6 o r « = 6147V « .01
It is assumed that p = .2 and that the desired boun d is .01. Hen ee,
1 .9 6 J — < .01 => yj n > í = 42 .72.01
« > 1 8 2 4 . 7 6 o r « > 1 8 2 5
a The approximate 95% conf idence in terval for p is
í 529I ± 1 . 9 6 - = = 2 .9 62 ± 1 .9 6- = = = 2 .962 ± . 125
yjn V69
or 2 .837 < p < 3.087 .
b In orde r to cut the interval in half, the sam ple size m ust increas e by 4. If this is
done, the new half-width of the confidenc e interval is
l . 9 6 - í = = - ( l . 9 6 - í \ . y[4n 2 ^ y fñ
Henee, in th is case, the new samp le s ize is 4( 69 ) = 276 .
The approximate 98% confidence interval for p is
T ± 2 .3 3 -4 = = 2 .7 05 ± 2 . 3 3 ^ 2 = 2 .7 05 ± .0 11V« V36
or 2 .694 < p < 2 . 7 1 6 .
For this exercise, B = .08 for the binomial est ima tor p , where S E ( p ) = A — . If V «
p = .3 , we have
1.96 A ^ ~ < B => 1 .9 6,p í í ! < .08«
r 1 .9 6 Í2 ( \8 )V« > ----------------- = > « > 9 . 8 o r « > 9 6 . 0 4
.08or « > 97.
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8.117 Use the Interpret ing Co nf ídence Intervals applet. A nsw ers will vary, but the
w idths of all the intervals should be the same. M ost of the simu lations will show
betw een 8 and 10 in terv als th at w ork corr ectly .
8.121 Use the Ex ploring Conf ídence Intervals applet.a-b M ove the slider on the r ight side o f the applet to change the samp le size.
Increasing the sam ple size results in a sma ller standard error and in a narrower
interval.c By increasing the sam ple size n, you obtain more information and can obtain this
more precise es tímate o f (J without sacrif icing confídence.
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9: Large-Sample Tests of Hypotheses
a Th e cri t ical valué that separates the reject ion and nonreject ion regions for a r ight-
tai led test based o n a z-statist ic w ill be a valué of z (cal led za ) such that
/ >( z > z a ) = tf = .01 . T hat is, z 0I = 2.33 (see the figure below). The nuil hypothesis
H0 will be rejected if z > 2 .33 .
b F or a tw o-ta il ed te s t w ith a = .05 , the critical va lué for the rejection región cuts
off oc¡2 = .025 in the tw o tai ls o f the z distributio n in F igu re 9 .2, so th at z 025 = 1.96 .
The nuil hypothesis H0 will be rejected if z > 1.96 or z < -1 .96 (w hich you can also
write as Id > 1.9 6 ).
c Sim ilar to part a, with the reject ion región in the low er tail o f the z distribution.
The nuil hypothesis H0 wil l be rejected i f z < -2 .3 3 .d Sim ilar to part b, w ith a / 2 = .005 . Th e nuil hyp othes is H0 will be rejec ted if
z > 2.58 or z < -2 .5 8 (w hich you can also wri te as |z| > 2.58 ).
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In this exercise, the parameter of interest is fJ , the pop ulation mean. The objective of
the experimen t is to show that the mean ex ceeds 2.3.
a W e want to prov e the alternative hyp othe sis that /v is, in fact, grea ter then 2.3.
Henee, the alternative hypo thesis is
H a : / y > 2 . 3
and the nuil hypothesis is
H 0 : / / = 2.3 .
b T he best estim ato r fo r f i is the samp le averag e x , and the test statistic is
I z ñ ." a / J ¡
w hich represents the distance (mea sured in units of standard dev iations) from x to the
hyp othesize d m ean /y . H enee, if this valué is large in abso lute valué, one of two
conclusions may be drawn. Either a very unlikely even t has occurred, or the
hyp othesize d mean is incorrect. Re fer to part a. If a - .05, the critical valué o f z that
separates the rejection and non-rejection regions w ill be a valué (denoted by z0 ) such
that
P ( z > zn ) = a = .05
That is, z0 = 1.645 (see below). H enee, H0 will be rejected if z > 1-645 .
a - 05
Reject H0
c The standard error o f the mean is found using the sample standard deviation 5 to
approxim ate the pop ulation standa rd deviation c r :
nr, cr 5 .29 _
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d To cond uct the test, calcúlate the valué of the test statistic using the inform ation
con tained in the sam ple. No te that the valué of the true standa rd deviation , o , is
approximated using the sample standard deviat ion 5 .
a ¡ 4 ñ s / y f ñ .049
The observed valué of the test statistic, z = 2.04 , falls in the rejection región and ti
nuil hypothesis is rejected. Th ere is sufficient evidenc e to indíca te that p > 2 .3 .
9.11 a In order to make sure that the average w eight was one po und, you wo uld test
H 0 : / / = 1 v ers us H a : p * 1
b -c T he te st st atist ic is
z = x-Mo a *~Po _ 1 . 01 -1 = 33
a / y f ñ s / y f ñ .1 8/ >/35
with p-value = ^ ( |z | > -33) = 2(.3 70 7) = .7414 . Since the /?-value is greater than .(
the nuil hypothesis should not be rejected. The m anage r should repo rt that there is
insufficient evidence to indícate that the mean is different from 1.
9.15 a The hypothesis to be tested is
H0 :/y = 110 versus H a : /y < 110
and the test statistic is
x - H o = 1 0 7 - 1 1 0 _
O \ [ ñ s / y f ñ 13/Vi 00
with p-value = P ( z < -2 .31 ) = .0104. To draw a conclusión f rom the p - valué, use t
gu idelines for statistical significance in Sec tion 9.3. Since the p-v alue is betw een .(
and .05, the test results are sign ifícant at the 5% level, but not at the 1% level.
b I f a = .05 , H0 can be rejected and you can conclude that the average score
improve m ent is less than claimed. This w ould be the most ben eficial way for thecompetitor to State these conclusions.
c If you worked for the Princeton Review, i t would be more beneficial to concludi
that there was insuf f icient evidence at the 1% level to conclud e that the ave rage scoi
improve m ent is less than claimed.
9.19 The hypothesis of interest is one-tailed:
H 0 :/A = 0 v ersu s H a : / / , - / / 2 < 0
The test statistic, calculated under the assumption that // , —Jl^ = 0 , is
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with the unknown cr2and a \ estimated by s f and s \ , respect ively. The s tudent can
use one o f two m ethods for decis ión m aking.
p-value approach: Calcúlate p-value = P ( z < -1 .3 3 ) = .0918 . S ince this p-value is
grea ter than .05, the nuil hy po thesis is not rejected. Th ere is insufficient evid enc e to
indicate that the mean for population 1 is smaller than the m ean for population 2.
Critical valué approach; The rejection región with a = .05 . is z < -1.6 45 . S ince
the observed valué o f z does no t fall in the rejection reg ión, H 0 is not rejected. Th ere
is insuff icient evidence to indicate that the mean for popu lation 1 is smaller than themean fo r population 2.
a The hypothesis o f interest is two -tailed:
H 0 : p , - p 2 = 0 versus H a :p , - p 2 * 0
The test s tatistic, calculated un der the assumption that / / , - p , = 0 , is
w ith p-valu e = P ( |z | > 2.26 ) = 2 ( .0 1 19) = .0238 . Since the p-va lue is less than .05,
the nuil hypothesis is rejected. There is evidence to indicate a difference in the mean
lead levels for the two sections o f the city.
b From Section 8.6, the 95% confidenc e interval for Pj - p , is approximately
c Since the valué p, - p 2 = 5 or p 1- p , = -5 is not in the conf ídence interval in par t
b, it is not likely that the difference will be more than 5 ppm, and henee the statistical
signif ícance o f the difference is not of practical im portance to the engineers.
a Th e hy pothesis o f interest is two-tailed:
H 0 : p , - p 2 = 0 versus H a : p , - p 2 ^ 0
-1 .9 ± 1 .6 5 or - 3 .5 5 < ( p , - p , ) < - . 2 5
and the test statistic is
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9.29
9.33
B - ^ ) - 0 ,9 4 - 2 .8 = _ 3 1 g
1-22 2.82
n, + n2 V 36 + 26
with p-va lue = P ( |z | > 3 .18) = 2( .0007 ) = .001 4. Since the p -\a lue is less than .05,the nuil hy poth esis is rejected. Th ere is evide nce to indícate a differen ce in the mean
concentrations for these two types o f s ites,
b T he 95% confid ence in te rv al fo r/y , - / y , is approxim ate ly
(*i “ * 2 ) ± 1-96 + ~V "i n2
( . 9 4 - 2 . 8 ) ± 1 . 9 6 J — + — v 1 V 36 26
-1 .8 6 + 1.15 or —3.01 < (//, —// 2) < —.71Since the valué p ^ - p 2 = 0 d oes not fal l in the interval in part b, it is not likely that
/y, = p 2 . Th ere is evidence to indícate that the mean s are different, con firm ing the
conclusión in part a.
a Th e hyp othesis of interest is two-tailed:
h o : ^ i - ^ 2 = 0 v e r s u s H a -Mi-Mi 5 6 0
and the test statistic is
_ (x , - x2) - 0 _ 9 8 .1 1 -9 8 .3 9 _ „ 22 j s [ + s [ I. 72 | ,742
n, n2 V65 65
with p-value = P(\z \ > 2 .22 ) = 2 ( 1 - .9 86 8) = .0264 . S ince thep -va lue i s be tween .01
and .05, the nuil hypo thesis is rejected, and the results are significan t. Th ere is
evidence to indicate a difference in the mean tem peratures for men v ersus wom en.
b Since the p-value = .0264, we can reject H0 at the 5% level (p-value < .05), but not
at the 1% level (p-value > .01). Using the guidelines for significance g iven in Section
9.3 o f the text, we declare the results statistically s ignif icant , but not highly
sig nif ic ant.
a The two sets of hypothesis both involve a different binom ial param eter p:
H 0 : p = .6 versus H a : p .6 (part c)
H 0 : p = .5 ve rsu s H a : p < .5 (p art b)
x 35b For the second test in part a, x = 35 and n = 75 , so that p = — = — = .4667 , the
n 75
test statistic is
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z = P Z P g = A 6 V - ¿ = _ 5S
M o -5( '5)
n V 75Since no valué of a is specif ied in advance, we calcúlate
p -v alu e = P ( z < - .5 8 ) = .281 0. S ince this p-value is greater than .10, the nuil
hyp othesis is not rejected. The re is insuff icient eviden ce to contradict the claim.
x 49c Fo r the first test in part a, x = 49 and n = 75 , so tha t p = —= — = .6533 , the te st
n 75
statistic is
_ _ P - P o _ .6 5 3 3 - . 6 _ n i
M o /.6 (.4 )
75
with p-v alue = P ( |z | > .94) = 2( . 1736) = .3472 . Since the p-va lue is greater than .10,
the nuil hyp othesis is not rejected. Th ere is insuff icient evidenc e to con tradict theclaim.
The hypo thesis o f interest is
H 0 : p = .45 vers us H a : p * .45
x 32W ith p = —= — = .4 , the tes t s ta tis tic is
ai 80 z = P S P ^ = - 4 0 - .4 5 - 90
Po lo .4 5 (.55)
" V 80
The rejection región is two-tailed a = .01 , or |z¡ > 2.58 and H 0 is not rejected. Th ere
is insuff icient evidence to dispute the newspaper’s claim.
The hypo thesis of interest is
H 0 : p = .40 versu s H a : p * .40
x 114with p = — = ------ = .38 , the test s tatistic is
ai 300
P - P o _ -3 8 - . 4 0 _ n]
M o 1.40 (.60 )
300
The re ject ion región w kh a =.05 is | z |> 1.96 and the nuil hypothesis is not rejected.
(Alternatively, we cou ld calcúlate p-va lue = 2 P ( z < - .71 ) = 2( .2389) = .477 8. S ince
this p-va lue is greater than .05, the nuil hypo thesis is not rejected.) Th ere is
insuff icient evide nce to indicate that the proportion o f hou seholds w ith at least onedog is different from that reported by the H um ane Society.
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9.45
then
9.47
9.51
a Th e hypothesis of interest is:
H 0 : P\ - p 2 = 0ve rsu s H a : p, - p 2 < 0
Calcúlate p , = .36 , p , = .60 and p = + n2 P i _ 18 + 3 0 = 4 3 . T he test statistic is
n, + n2 50 + 50
T - P ' - P > - -3 6 - - 6 0
l - f l + j J l V-4 8 (-5 2 ) ( 1/ 5 ° -+ 1/ 5 0 )
V U i n2 J
Th e reject ion región, w ith a = .05 , is z < -1 .645 and H 0 is rejected . Th ere is
evidenc e o f a difference in the proport ion o f survivors for the two groups.
b F rom Section 8.7 , the appro xim ate 95% confíd ence in te rval is
( Á - P : ) ± 1 .9 6 M + MV ni n2
( .3 6 - .6 0 ) ± 1 . 9 + Mv 7 V 50 5 0
- .2 4 ± .1 9 o r - .4 3 < ( p , - p , ) < - . 0 5
The hypothesis o f interest is
H 0 : p, - p, = 0 versus H a : p, - p 2 ^ 0
Ca lcúlate p, = — = .214 , p 1 = — = .25 , and p = + * 2 = + ^ = .227 .56 - 32 n, + n 2 56 + 32
Th e test statistic is then
T=h z h - -2 1 4 ~ -2 5 - ™ yj.2 21 ( .773) ( l / 56 + 1 /32)í 1 O— + —
n. n ,v i 2 y
Th e rejection reg ión, w ith or = .05 , is |z| > 1.96 and H 0 is no t rejected . Th ere is
insufficient evidenc e to indicate a difference in the p roport ion o f red M & M s for the
p la in and peanu t vari eti es. T hese re sults m atc h th e conclu sio ns o f E xercis e 8.53.
Th e hy pothesis o f interest is
H 0 : p, - p 2 = 0 versus H a : p, - p , > 0
93 119 x + x 93 + 119C a lc úla te p, = — = . 7 6 9 , p 0 = ------ = .5 9 8 , and p = — - = ---------------= .6625 .
121 199 n ,+ n 2 121 + 199
The test statistic is then
^/.6625 (.3 37 5) (1/121 + 1/199)
z = , -7 6 9 - 5 9 8 = 3 . 1 4I
1 1w — + — « 2 /
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9.55
9.59
9.63
9.67
with p-value = P { z > 3.14) = 1- .9 99 2 = .0008 . Since the p-valu e is less than .01, the
results are repo rted as highly sign ificant at the 1% level o f significan ce. The re isevidence to confirm the researcher 's conclusión.
The power of the test is 1 - / ? = ^( re jec t H0 when H0 is fa ls e) . As p gets farther
from p 0 , the pow er of the test increases.
a-b Since it is nec essary to próve that the averag e pH level is less than 7.5, the
hypo thesis to be tested is one-tailed:
H n : p = 7.5 versus H a : p < 7.5
d Th e test statistic is
1 Z _ÍL x x - E = - .2
a / 4 7 t s / J ¡ .2/V3Ó
and the rejection región w ith a = .05 is z < -1.6 45 . The observed valué ,
z = -5 .4 7 7 , falls in the rejection región and H0 is rejected. W e conclude that the
average pH level is less than 7.5.
Let p, be the proportion of defectives produced by m achine A and p, be the
p ro port io n o f defe ctiv es p ro du ced by m achin e B. T he hypoth esis to be te ste d is
H(>: P\ ~ Pi - 0 ve rsus Ha : p , - p 2 * 0
Calcúla te p , = = .08 , p , = —— = .04, and p = — - .06 . The20 0 200 K ni + n 2 200 + 200
test statistic is
then
, •Q8~ 04 = 1.684 N 0 6 ( .9 4 ) (V 200 + 1 /2 0 0 )
\pq1 1
— H-----
n. n 2
T h e rejection región, with a = .05, is |z| > 1.96 and H0 is not rejected. Th ere is
insuff icient evidenc e to indícate that the mach ines are performing differently in terms
o f the percentage o f defect ives being produced.
The hypo thesis to be tested is
H n : p , - p : = 0 v ers us H a : p , - p : > 0
and the test statistic is
( I , - T 2 ) - 0 _ 1 Q -8 _ /l
n, n , V 40 40
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The reject ion región, w ith a = .05, is one -tailed or z > 1 645 and the nuil hypothesis
is rejected. There is sufficient evidence to indícate a difference in the two means.
Henee, we conclude that diet I has a greater mean weight loss than diet II.
9.71 a The hypothesis to be tested is
H 0 : / y , - / / 2 = 0 versus H a > 0
and the test statistic is
The rejection región, with a = .05, is one -tailed or z > 1.645 and the nuil hypo thesis
is rejected. T here is a difference in mean yield for the two typ es o f spray.b An approximate 95% confidence interval for/y, - is
The rejection región w ith a = .01 is z > 2.33 . Since the observ ed valué, z = 2.19 ,
doe s not fall in the rejection región and H 0 is not rejected. Th e data do not provide
sufficient evidence to indicate that the mean ppm o f PCB s in the po pulat ion o f game
b irds exceeds the F D A ’s re com m ended lim it o f 5 ppm .
9.79 The hypothesis to be tested is
13 + 5.88 or 7.12 < (// , - f r ) < 18.88
9.75 The hypothesis to be tested isH 0 : // = 5 versus H a : // > 5
and the test statistic is
_ _ x Mo x Mo _ 7.2 5 _ . . .
a / J ñ s / ^ J ñ 6 .2 /V38
H 0 : P\ - Pi ~ 0 vers us H a : p x - p 2 * 0
Calcúlate p , = — = .40 , p , = -1 6124 5
. _ x, + x 2 _ 6124(.4) + 5512(.37)
^ n i + n 2 11636
= .40 , p 2 = — = .37 , and5512
1) + 551 2(.37 ) _ ■phe test statistic is then11636
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z =• P \ - P l . 4 0 - . 3 7
>/ .386 ( .61 4)( l /61 24 + l /5512 )= 3.32
The rejection región for a = .01 is | z \> 2.58 and the nuil hyp othesis is rejected.
Th ere is sufficient evidenc e to indícate that the percen tage o f studen ts who are fluent
in English differs for these two districts.
9.83 a The param eter of interest is / / , the average daily wage o f wo rkers in a given
industry. A samp le o f n = 40 w orkers has been drawn from a part icular company
within this indu stry and x , the sam ple average, has been calculated . Th e objective is
to determine w hether this com pany p ays wages different from the total industry. That
is, assume that this sample o f forty wo rkers has been drawn from a hypo thetical popula tion o f w ork ers . D oes th is popu la ti on have as an average wag e / / = 54 , o r is
/ / different from 54 ? Thu s, the hy pothesis to be tested is
H0 : ¡x = 54 versus H : / / * 54 .
b-c T he te st sta ti st ic is
x - M 5 1 . 5 0 - 5 4 z ~
s / y /ñ 11 .88 /740= -1 .331
and the Larg e-S am ple T es t of a P op ula t io n M ean applet g ives p-va lue = . 1832 .
(Using Table 3 w ill produce a p - \ a lue o f . 1836.)
d Since a = .01 is sma ller than the p-valu e, . 1832, H0 can not b e rejected and w e
canno t conclude that the com pany is paying w ages different from the industry
average.
e Since n is greater than 30, the Central Limit Theorem will guarantee the norm ality
o f x regardless of w hether the original population w as normal o r not.
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10: Inference from Small Samples
10.1
10.5
10.9
Refer to Table 4, Appendix I , indexing d f a long the left o r r ight m argin and ta across
the top.
a tM = 2.015 with 5 d f b t 02S = 2.306 with 8 d f
c / 10 = 1.33 0 w ith 18 d f c t 02i » 1 .96 w ith 30 d f
a U sing the form ulas given in Ch apter 2, calcúlate X * (. = 70.5 and X * 2 = 49 9.27 .
Then
y . Z i . T O í . ™ ,
n 10
4 9 9 . 2 7 - ^/ = --------------2— = ------------------1 ^— = . 24 9 44 4 a n d 5 = .4 9 94
n - \ 9
b W ith d f = n -1 = 9 , the appropr ia te valué o f t is / 01 = 2.821 (from Ta ble 4 ) and
the 99% upper one-s ided conf idence bound is
c 19 4 9 4 4 4 x + t 0i- f = => 7 .05 + 2 .8 2 1J: => 7.05 + .446
01 V io
or /y < 7 .4 96 . Intervals constructed using this procedure wil l end os e /y 99% o f the
time in repeated sam pling. H enee, we are fairly certain that this particular interval
e n d o s e s / y .
c The hypothe sis to be tested is
H 0 : /y = 7.5 versu s H a : /y < 7.5
and the test statistic is
7 0 5 - 7 .5 ,
s / J ñ 1.249444
10
Th e re ject ion región with a = .01 and « - 1 = 9 degrees o f f reedom is located in the
lower ta il of the /-dist ribut ion and is found f rom Table 4 as / < - / 01 = -2.8 2 1 . S ince
the observ ed valué o f the test s tatistic falls in the rejection región, H 0 is rejected an d
we conclude that /y is less than 7.5.
d N otice that the 99% up per one-sided confidence bou nd for /y does not include the
valué /y = 7.5 . Th is would con firm the results o f the hy pothe sis test in pa rt c, in
w hich w e conclud ed that /y is less than 7.5.
a Sim ilar to previous exercises. The hypothesis to be tested is
H 0 :/ y = 1 00 v ers us Ha : / y < 1 0 0
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Calcúlate x = — Ln
Ijc,. _ 1797.095
n ~ 20= 89.85475
165,697.7081-(1797.095 )2
= 222.115 0605 and 5 = 14.9035n -1
The test statistic is
19
V2Ó
The cri t ical valué of t with or = .01 and n - \ = 19 degrees o f freedom is
r01 = 2 .539 and the reje ctio n reg ión is t < -2.53 9 . The nuil hypothesis is rejected and
we conclude that /J is less than 100 DL.
b T he 95% upper one-s id ed confi dence bound, based on n -1 = 19 degrees of
freedom, is
x + í 0 => 8 9.8 54 75 + 2 .5 3 9 14' 9Q^ 12 5 11 => / / < 9 8.3 16 yj n V 20
This confirms the results o f part a in w hich we con cluded that the mean is less than
100 DL.
10.13 a Th e hypo thesis to be tested is
H 0 : n = 25 vers us H a : ¡u < 25
The test statistic is
The cri t ical valué o f t with a = .05 and n - 1 = 20 degrees of f reedom is
t 05 =1 .725 and the rejection región is t < -1 .7 2 5 . Since the observed valué does falls
in the rejection región, H0 is rejected, and we conclude that pre-treatment mean is lessthan 25.
b The 95% confid ence in te rv al based on d f = 20 is
o r 2 3 . 2 3 < 29.97 .
c The pre-treatment mean looks considerably sm aller than the other two means.
10.17 Refer to Exercise 10.16. If we use the large sam ple method o f Ch apter 8, the large
sample con fidence interval is
=> 2 6.6 ± 2 . 0 8 6 - = => 26 .6 ± 3 . 3 77.4
V 2 l
=> 246.96±12.98
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10.19
10.25
10.27
or 233.98 < p < 259.94 . The intervals are fair ly similar, w hich is why we choo se to
x — u.approxim ate the samp l ing dis tr ibut ion o f p- wi th a z distr ibution wh en n > 3 0 .
s / y j n
a _ (/,, + ( „ , - 1 ) k _ 9 ( 3 .4 ) + 3 ( 4 .9 ) 2 7 ? ;
n ] + n 2 — 2 10 + 4 - 2
b + ( n 2 - \ ) s ; l l (1 8 ) + 2 0 ( 2 3 ) ^
nt + n 2 —2 12 + 21 — 2
a Th e hypo thesis to be tested is
H 0 : // , ~ n 2 = 0 v ers us H a : p , - p 2 * 0
From the M in i ta b printout , the fol lowing inform ation is available:
^ = .896 s f = (.4 0 0 )2 n, = 1 4
x 2 = 1 . 1 4 7 s \ = ( 6 79)2 rc2 = 11
and the test statistic is
(T, - x 2 ) - 0t = U = - 1 .1 6
\2
= 2.1
«2
T h e rejection región is two-tailed, based on n, + n 2 - 2 = 23 degrees o f f reedom. W i th
a = .05 , from Table 4, the rejection región is \t\ > tm5 = 2.06 9 and Hq is not rejected.
Th ere is not enough e vidence to indícate a difference in the popu lat ion m eans.
b It is not necessary to bound the p -va lu e usin g T able 4, sin ce th e exact p -v a lu e is
given on the p rintout as P-Value = .260.
c If you check the ratio of the two variances using the rule of thumb g iven in thissection you will find:
l a r g e r r _ ( .6 7 9)2
sm aller s2 ( .400 )2
w hich is less than three. The refore, i t is reasonable to assum e that the two popu lat ion
variances a re equal.
a Ch eck the ratio of the two variances using the rule of thum b given in this section:
J a r g e r r _ _ 2/78095 _ lf i2 2
smaller 5 .17143
w hich is greater than three. Therefore, i t is not reasona ble to assum e that the two
popu la ti on varia nces are equal.
b Y ou shou ld use the unpoole d t test with Satterthwaite’s approximation to the
degree s of freedom for testingH 0 : p , ~ n 2 = 0 versu s H a : //, - ¡i2 * 0
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10.35
10.37
For a on e-tai led test with d f = «, + « , - 2 = 18 , t he p -va lue can be bounded us ing
Ta ble 4 so that p-value > .10 , and H 0 is not rejected. Th ere is insu fficient evidence to
indícate that sw imm er 2’s average t ime is sti ll faster than the averag e t ime for
swimmer 1.
a Th e test statistic is
d - n d . 3 - 0t =
J y f ñ= 2.372
wi th « -1 = 9 degrees of f reedom. The p-value is then
1 P(\t\ > 2 .372) = 2 P ( t > 2 .372) so that P ( t > 2 . 3 7 2 ) = - p - va lu e
Since the valué t = 2.372 fal ls betwee n two tabled entr ies for d f = 9 ( f 025 = 2.26 2 an d
f01 = 2.8 21 ), you can co nclu de that
.01 < —p-v alue < .0252
.02 < p-v alue < .05
Since the p-va lue is less than a = .05 , the nuil hypo thesis is rejected and we conclud e
that there is a difference in the two po pulat ion m eans.
b A 95% con fid ence in te rv al fo r p 1 - p , = p^ is
d ± t a a -^= => 3 ± 2.262. — => ,3±.286
025 V IOo r ‘.0 1 4 < ( p , - p 2 ) < .5 8 6 .
c Using sf¡ = .1 6 and B = . 1, the inequali ty to be solved is approxim ately
1 . 9 6 - ^ < . l■Jn
r 1 .9 6V 46 f n > -------------- = 7.84
.1« > 6 1 . 4 7 o r « = 6 2
Since this valué of n is gre ater than 30, the samp le size, « = 62 pairs, w ill be v alid.
a I t is necessary to use a paired-difference test , s ince the two sam ples are not
rando m and indepen dent . The hyp othesis o f interest is
H 0 : P] - P 2 = 0 or H0 : p d = 0
H a : p , - p 2 * 0 or H a : p ^ 0
The table of differences, along w ith the calculation o f d and s] , is presented below.
d¡ .1 .1 0 .2 -.1 I d, = .3
d f .01 .01 .00 .04 .01 Z d f = .07
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10.41
d = ^ - = - = . 06n 5
The test statistic is
and
. 0 7 -(■3):
JJ _ = .013
d - j U j . 0 6 - 0 ,t = — = =1 .177
s j y j n I m 3
with n - 1 = 4 degrees o f freedom . Th e rejection región w ith a = .05 is
H > ¿025 = 2 .77 6, and H0 is not rejected. W e cannot conclude that the m eans are
different.
b T he p -va lu e is
Pf l íl > 1 .177) = 2 />( f >1 .177 ) > 2( .10 ) = .20
c A 95% con fidenc e interval for // , —// 2 = / id is
¿ ± '0 2 5 - fv n
.06 + 2.776..013
.06 ± .14 2
o r - . 0 8 2 < ( / / , - / / , ) < .2 02 .
d In orde r to use the paired-difference test, it is necessary that the n paired
obse rvations be random ly selected from norm ally distr ibuted populations.
a Each sub ject wa s presented w ith both signs in random order. If his react ion timein general is high, bo th respon ses will be high; if his reaction time in gene ral is low,
both responses w ill be lo w . T he la rge vari abil ity fr om subje ct to subje ct will mask
the variabil ity due to the difference in sign types. The pa ired-difference design will
eliminate the subject to subject variability.
b The hypo th esis o f in te rest is
h o : ^ i - / /2 = ° or H (): / rd = 0
H a : / r , - / r 2 * 0 o r H a : / / r f* 0
The table of differences, along with the calculat ion o f d and s] , is presented below.
Driver 1 2 3 4 5 6 7 8 9 10 Totalsd, 122 141 97 107 37 56 110 146 104 149 1069
j =I 4 =1069 =106.9n 10
Z d ? - .( £ 4 ) 2
126,561 -(1069)2
10“ 1n - \
and the test statistic is
= 1364. and = 36.9458
J - ^ = 1 0 6 .9 -0
sd /4~n 36.9458
Vio
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10.45
10.49
Since t = 9.150 w ith d f = n -1 = 9 is greater than the tabled v alué t m ,
p -valu e < 2 (.0 0 5 ) = .01
for this two tailed test and Ho is rejected. W e canno t conclude that the m eans are
different.c The 95% conf idence interval for / i, - / / , = fi d is ~~~
d ± t m, - ^ = => 106.9 ± 2 .2 6 2 --6-^i 5-8 => 106.9 ±2 6.4 28' yfn Vio
o r 8 0 . 4 7 2 < ( f i x ~ n 2 ) < 133 .328 .
a Use the M in i ta b pr intout given in the text below . Th e hypo thesis of interest is
H 0 ' V a ~ M b = 0 H a : ^ - / y B > 0
and the test statistic isd - n d 1 . 4 8 7 5 - 0
r - ^ 7 X - T 4 9 Í 3 T - 2 -82
V8
Th e p-valu e sho wn in the printout is p-valu e = .013 . Since the p-valu e is less than
.05, H0 is rejected at the 5% level o f signif icance. W e conclude that assessor A gives
higher assessments than assessor B.
b A 95% lo w er one-s id ed co nfi dence bound for p , - / / 2 = n d is
- s 1 4 9 1 3 4
d - 105 —j= => 1 .487 5-1 .895 ' => 1 .48 75 - .999V» V8
or ( // , - / / 2 ) > .4885 .
c In order to apply the paired-difference test, the 8 prop erties must be random ly and
independently selected and the assessments must be normally distr ibuted.
d Yes. I f the individual assessments are normally distr ibuted, then the mean of four
assessments will be norm ally distr ibuted. Henee, the difference x A —x will be
normally distr ibuted and the t test on the differen ces is valid as in c.
Fo r this exe rcise, s 2 = .3214 and n = 15 . A 90% co nfidence interval for c r will be
( n - l ) s 2 2 ( m —1)< o~ <
Xa/ 2 X(i-a/2)
where Xa /i represents the valué o f x~ such that 5% o f the area un der the curve
(shown in the figure on the next pag e) lies to its right. Sim ilarly, x]\-a¡i) he the
X~ valué such that an area .95 lies to its right.
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10.53
Henee, we have located o ne-ha l f of a in each tail o f the distribution . Index ing
X 2m and x l s with n -1 = 14 degrees o f f reedom in Table 5 yields
X ls = 23.6848 and x l s = 6.57063
and the confíden ce interval is
14 ( .3214) , 14Í .3214)
— -------- - < a < — ---------- o r . 1 9 0 < <T 2 < . 6 8 523.6848 6.57063
a The hypo thesis to be tested is
H 0 :/¿ = 5 H a : / y * 5
T x 19 96C alcú late J = ± ^ = — - = 4.99
n 4
9 9 . 6 2 2 6 - .( 1 9 % >2
s~ = ---------------2— = -------------------- = = .0074
n - 1 3and the test statistic is
4 . 9 9 - 5t = . ^ , = - .23 2
s /y fn .0074
4
The reject ion región with a = .05 an d n -1 = 3 degrees o f f reedom is found f rom
Table 4 as~~|f| > tms = 3.182 . Since the observed valué o f the test statistic does not
fall in the rejection región , Ho is not rejected. Th ere is insufficient evide nce to show
that the me an differs from 5 mg/cc.
b The m an ufactu rer c la im s th a t th e ra nge o f the po te ney m easurem ents will equal .2.
Since this range is given to equal 6cr . we know that o ~ .0333 . Th en
H 0 :<72 = ( .03 3 3)2 =.0011 H 3 :c72 >.0011
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The test statistic is
, ( n - l ) í 2 3 Í.0 07 4 ) y = 4 J - = - ± -------- = 20.18
a - .0011
and the one-tailed rejection región with a = .05 and n - 1= 3 degrees of f reedom is
r > * 2 5 = 7.81
H0 is rejected; there is sufficient evidence to indicate that the range of the potency
will exceed the m anu facturer’s claim.
10.57 The hypothesis of interest is H 0 : a = 150 H a : a < 150
Calcúlate
(n - 1)s 2 = I * 2 - = 9 2 , 3 0 5 , 6 0 0 - = 662,232.8
and the test statistic is y 2 = —— íp—- = ^> 2,232 .8 = 29.43 3. The o ne-tailed rejection<t(; 150*
región with or = .01 and n - 1= 19 degrees o f f reedom i s # 2 < Z .99 = 7.63 273 , and H 0
is not rejected . Th ere is insu fficient evide nce to indicate that he is mee ting his goal.
10.61 The hypothesis of interest is H 0 : o f = o f versus Ha :a2 * of
and the test statistic is
s 2 712 F = A r = — 7 = 1.059 .
s¡ 69Th e crit ical valúes of F for various valúes o f a are given below using df¡ = 15 and
d f 2 = 14 .
a .1 0 .05 .025 .01 .005
Fa 2.01 2.46 2.95 3.66 4.25
Henee,
p -valu e = 2 P ( F > 1.059) > 2 (. 10) = .20
Since the p -\a lu e is so large, H0 is not rejected. Th ere is no evidenc e to indicate that
the variances are different.
10.65 For ea ch o f the three tests, the hyp othesis of interest is
H 0 : of = of ve rsu s H a : of * of
and the test statistics are
F 1.03 F = 4 = 4 ^ = 2.01 and F = ¿ = ^ = 14.29S-s-2 3 .922 ' s i 3 .4 92 ' 5 ,2 4 .4 7 2
The cri tical valúes of F for various valúes of a are given on the following page,
using d f = 9 and d f2 = 9 .
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10.69
10.73
10.76
a .10 .05 .025 .01 .0052.44 3.18 4.03 5.35 6.54
Henee, for the first two tests,
p-va lu e > 2 (. 10) = .20
wh ile for the last test,
p-va lu e < 2 (.0 05) = .01
Th ere is no e vidence to indícate that the variances are d ifferent for the f irst two tests,
but H0 is reje cte d fo r th e th ird vari able . The tw o-s am ple r- te st w ith a poo le d estím ateo f o 1 cannot be used for the third variable.
Paired observations are used to est ímate the difference between two populat ion means
in preference to an est imation based on independent random samples selected from
the two populat ions beca use o f the increased information caused by blocking the
observations. W e expect blocking to create a large reduction in the standard
deviat ion, i f differences do ex ist amon g the blocks.
Paired observations are not alway s preferable. The de grees of freedom that are
available for est im ating o 2are less for paired than for unpaired observations. If there
were no difference between the blocks, the paired experiment would then be less benefi cia l.
Since i t is necessary to determine w hether the injected rats drink m ore w ater than
noninjected rates, the hyp othesis to be tested is
H 0 : p = 22.0 H a : p > 22.0
and the test statistic is
f = 3 1 . 0 - 2 2 . 0 = 5 9 8 5
s/ 'y jn 6 2
Using the cri tical valué approach , the rejection región with a = .05 and
n - 1 = 16 degrees of freedom is located in the uppe r tai l of the r-distribution and is
found from Table 4 as t > r05 = 1.746 . Sinc e the ob ser ve d valu é of the test statistic
falls in the rejection región, H0 is rejected and we conclude that the injected rats do
drink more w ater than the noninjected rats. The 90% confidence interval is
I ± 7 ;, ~ => 3 1 . 0 ± 1 . 7 4 6 4 ¿ 3 1 .0 ± 2 .6 2 5 yf n V Í7
or 28.375 < p < 33.625 .
The student may use the rounded valúes for x and s given in the display, or he may
wish to calcúlate x and s and use the more ex act calculat ions for the confidence
intervals. The ca lculat ions are shown on the next page.
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10.79
_ I jc¿ 1845a x = L = --------= 184.5
n 10
3 4 4 , 5 6 7 - í ^s 2 = ---------- *— = ---------------------------------= 462.7222
n - 1 9
5 = 21.511 and the 95% co nfidence interval is
5 _ ____________ 21.511 x ± t m - = => 1 .845 ± 2 .262 — ^= — => 184 .5 ±15 .4
yj n -s/l 0
or 169.1 < /u < 199.9 .
v- 2 (X - v.) 2 ( 7 3 0 ) ‘Z x ,2 - V ^ - 5 3 5 1 4 - > '
b I = ^ Í L = — = 73.0 52 = — -5— = —— = 24.8889n 10 n - 1 9
s = 4.989 and the 95 % confidence interval is
j 4 989 x ± t m. - 7= => 7 3 . 0 ± 2 . 2 6 2 - = => 73 .0 ±3 .5 7
J ü VTÓ
or 69.43 < /J < 76.57 .
_ I x 25.42c x = i- = --------- = 2.542
n 10
S x j _ ( £ ü ) _ 6 5 . 8 3 9 8 _ ( 2 5 ' 4 2 )
^ ------= .13579556n - 1 9
5 = .3685 and the 95% confidence interval is
x ± tms => 2 . 5 4± 2 . 2 6 2 = > 2 .5 4 ± .2 6Vn VlO
or 2.28 < f i < 2.80 .
d No. Th e relationship between the confidence intervals is not the same as the
relationship between the original m easurements .
U se the com puting formulas or you r scientific calculator to calcúlate
_ I x , . 322.1v = -----= ----------- = 24.777
n 13
„ 2 ( I ± ) 2 (322.1)2Ijc ,2 - V 8 114.59 — } —
52 = --------------- '-1— = — — = 1 1.1 61 9n - 1 12
5 = 3.3409 and the 95% confidence interval is
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c t . "unoI ± t m5 —= 24 .777 ± 2 . 1 7 9 ^ = ^
yjn v i 324 .777 ±2 .019
o r 2 2 .5 7 8 < m < 26 .796 .
a The range o f the f irst samp le is 47 while the range o f the second samp le is only
16. Th ere is proba bly a differen ce in the variances.
b The hypoth esis o f in te re st is
H (l : ay = o \ ver sus Ha : a ,2 * a \
Calcúlate s i =
1 7 7 , 2 9 4 -(838)2
= 577.6667
1 9 2 , 3 9 4 - M
*,2= 6 = 29.6
and the test statistic isF = ii_ = 5TL666T_ _
s i 29.6
Th e cri tical valúes with d f x = 3 an d d f2 = 5 are shown be low from T able 6 .
a .10 .05 .025 .01 .005
Fa 3.62 5.41 7.76 12.06 16.53
Henee,
p -v a lu e = 2 P ( F > 19.516) < 2(.0 05) = .01
Since the p-value is smaller than .01, H0 is rejected at the 1% level of significance.There is a difference in variability.
c Since the Stu den t’s t test requires the assump tion o f equal variance, i t would be
inapp ropriate in this instance. You should use the unpooled t test with Satterthwaite’sapproximation to the degrees of freedom.
A paired-difference test is used, since the two sam ples are not random and
indepen dent . The hypo thesis of interest is
H 0 : A ~ = 0 H a : / i l - l i 2 > 0
and the table o f differences, along w ith the calculat ion o f d and s] , is presented
belo w .
Pair 1 2 3 4 Total s
4 -1 5 11 7 22
d- = M = ̂ = 5.5
n 4 sd = 5
and the test statistic is
Z d > -( 1 4 )2
1 9 6 -(22)2
n - 1
= 25 and
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10.91
10.97
d - n (i 5 . 5 - 0 I = y-£= = = 2.2
s J y J n A
The one-tai led p-value with d f = 3 can be bounde d betw een .05 and .10. Since thi
valué is greate r than .10. H0 is not rejected. Th e results are not significan t; there is
insufficient evidence to indicate that lack of school experience has a d epressing efl
on IQ scores.
Th e object is to determine w hether or not there is a difference betwe en the m ean
responses for the two different st imuli to which the people have been sub jected. T
samples are independently and randomly selected, and the assumptions necessary 1
the t test of Sec tion 10.4 are met. Th e hypo thesis to be tested is
H o ; A i - / / 2 = 0 H a : //, - f ¿ 2 * 0
and the preliminary calculations are as follows:
15 21T, = — = 1.875 and x 2 = — = 2.625
(15)2 (21)23 3 - - — — 61 - - —
= .6964 3 and s22 = = .8392 9
Since the ratio o f the varian ces is less than 3, you can use the p ooled t test. Th e
poole d estim ato r o f <r: is calc u la te d as
2 (rh - \ ) s ; + ( n 2 - \ ) s ¡ 4.875 + 5.875= .7679
n , + n 2 - 2 14
and the test statistic is
. .. . 1 . 8 7 5 - 2 .6 2 5 , m
1.7679 ÍI+H
Ui n 2J V
o o
o c
Th e two -tailed reject ion región with a = .05 and d f = 14 is |r| > t m¡ = 2.1 45 , and 1
is not rejected. Th ere is insufficient evide nce to indicate that there is a difference i
means.
It is possible to test the nuil hypo thesis H 0 : tJ,‘ = cr2a gain st any one o f three
alternative hypotheses:
(1) H a io f * o \ (2) H a : <t2 < o \ (3) H a : o \ > o \
a The first al ternat ive wou ld be preferred by the m anage r of the dairy. He doe s n
know anything about the variabil ity o f the two m achines and would wish to detectdepa rtures from equ ali ty o f the type o \ < o \ o r o \ > o \ . These alternatives are
implied in (1).
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b T he sale sm an fo r com pany A w ould p re fe r th at the experim en te r sele ct th e second
alternative. Re jection of the nuil hypothesis w ould im ply that his m achin e had
sm aller variability. M oreo ver, even if the nuil hypothesis were not rejected, there
wo uld be no evidence to indicate that the variabili ty o f the compa ny A m achine wasgreater than the variabili ty o f the compa ny B m achine.
c Th e salesm an for com pany B would prefer the third al ternat ive for a similarreason.
101 A pa ired -diff ere nc e an alys is is used. To test H (): //, —// 2 = 0 versus H a : /v, > 0 ,
w here is the mea n reaction time after injection and // , is the m ean reac tion time
before in je ction, calc úla te the d if ferences ( x 2 - x , ) :
6, 1,6, 1
( W 7 4 _ M l
Then d = ^ = — = 3.5 = — * — = ------------ 4— = g .33 andn 4 d n - 1 3
sd = 2.88 675 and the test statistic is
d - n d 3 . 5 - 0t = — = -------------- = 2 425
s J J i 2.88675 '
Fo r a one-tai led test with d f = 3 , the rejection región with a = .05 is t > tM = 2.35 3 ,
and H 0 is rejected . W e conc lude that the drug significantly increases w ith reaction
time.
.105 Th e un derlying populat ions are rat ings and can only take on the f inite num ber of
valúes, 1, 2, ..., 9, 10. N either population has a norm al distribution, but both are
discrete . Further, the sam ples are not indepe nden t, since the same person is asked to
rank each car design. Henee, two of the assumptions required for the S tuden t’s t test
have been violated.
.109 The M in i tá b p rintout below sho ws the sum ma ry stat ist ics for the two samples:
Descriptive Statistics: Method 1, Method 2Variable N Mean SE Mean StDev
Method 1 5 137.00 4.55 10.17Method 2 5 147.20 3.29 7.36
Since the rat io of the tw o sam ple variances is less than 3, you can use the pooled t test
to com pare the tw ojne thod s o f measurement , using the rem ainder of the M in i tab
pri n to ut on the next pa ge:
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Two-Sample T-Test and Cl: Method 1, Method 2Difference = mu (Method 1) - mu (Method 2)Estímate for difference: -10.200095% CI for difference: (-23.1506, 2.7506)
T-Test of difference = 0 (vs not =): T-Value = -1.82 P-Value = 0.107 DFBoth use Pooled StDev = 8.8798 —
The test statistic is t = -1 .8 2 w ith p-value = . 107 and the results are not significant
Th ere is insufficient evidence to declare a difference in the two popu lat ion m eans.
10.113 The hyp othesis to be tested is
The cri t ical valué of t w ith a = .01 an d n - 1 = 9 degrees of freedom is í0l = 2.821
the reject ion región is t > 2.821. Since the observed va lué falls in the reject ion
región, H0 is rejected. The re is sufficient evidenc e to indicate that the ave rage
num ber o f calories is greater than advert ised.
10.119 Use the Interpreting Confidence Intervals applet . An sw ers will vary from studt
to student. Th e w idths of the ten intervals will not be the sam e, since the valué o f
chan ges w ith eac h new sam ple. Th e student should f ind that approxim ately 95% i
the intervals in the first app let contain // , wh ile rou ghly 99 % o f the intervals in tf
second applet contain / / .
10.123 Use the Tw o Sam ple t Test: Independent Sam ples applet . The hyp othesis to b(
tested concerns the differences between mean recovery rates for the two surgical
p ro cedures. L et // , be th e popu la tion m ean fo r Pro cedure I and f i 2 be th e popula ti
m ean for Procedure II . The hyp othesis to be tested is
Since the rat io of the variances is less than 3, you can use the po oled t test. En ter
appropriate statistics into the applet and you will find that test statistic is
with a two-tai led p-value of .0030. Since the p-va lue is very sm all, H0 can be reje
for any valué of a g reater than .003 and the results are judg ed highly significant.
The re is sufficient evidence to indicate a difference in the me an recov ery rates fo r
two procedures.
H 0 : n = 280 versus H a : a > 280
The test statistic is
Vio
H0 : A, - A2 = 0 vers us H a : A, - A2 ^ 0
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11: The Analysis of Variance
1 1 . 1
11.4
11.7
In comp aring 6 popu lat ions, there are k -1 degrees o f f r eedo m fo r tr ea tmen t s and
n = 6 (10) = 60 . The AN OV A table is shown be low.
Source d f Treatments 5
E rror 54
Total 59
Sim ilar to Exercise 11.1. W ith n = 4 (6 ) = 24 and k = 4 , the sources o f variat ion
associated d f are show n below.Source d f Treatm ents 3
Erro r 20
Total 23
The fol low ing prel iminary calculat ions are necessary:
7¡ = 1 4 7; = 1 9 7¡ = 5 G = 3 8
( 3 8 ) :a C M =
14
= 103.142857
Tota l SS = S x l - CM = 32 + 22 + • • • + 22 + 12 - CM = 130 -10 3.1 42 85 7 = 26 .8í
S S T = Z - — C M = - + — + - — C M = 1 1 7 .6 5 -1 0 3 .1 4 2 8 5 7 = 14.5071n, 5 5 4
, SS T 14.5071a nd M S T = -------= ------------- = 7.2536
k - 1 2
c By sub trac tion, SS E = To ta l S S -S S T = 26 .8571 -14 .5071 = 12 .3500and the
degrees of f reedom, by subt ract ion, are 1 3 -2 = 11. Then
M S E = SS E = 1 2 3 5 0 011 11
d The information obtained in parts a-c is Consolidated in an A N O V A table.
Source d f SS M S
Treatm ents 2 14.5071 7.2536
Error 11 12.3500 1.1227
Total 13 26.8571
e The hypothesis to be tested is
H 0 : //, = = fiy versus H a : at least one pair of m eans are differen t
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1.10
.1 3
f The reject ion región for the test statistic F = ---------= —;— — = 6.46 is based on anMSE 1.1227
F-distr ibution w ith 2 and 11 degrees of freedom. The cri t ical valúes of F for
boundin g th e p -valu e fo r th is one-t a iled te st are show n belo w .
a .10 .05 .025 .01 .005
2.86 3.98 5.26 7.21 8.91
Since the observed valué F = 6.46 is be tw een F 01a nd F 025 ,
.01 < p -va lue < .025
and H0 is rejected at the 5% level o f signif icance. There is a difference amo ng the
means.
a The following prel iminary calculat ions are necessary:
7; = 3 8 0 T2 = 199 F, = 261 G = 840
( I jÜ 2 (8 4 0 )2C M = — = --------- = 64,145.4545
n 11
T o ta l SS = I x¡ - C M = 6 5 ,2 8 6 - C M = 1 14 0.5 45 5
SST = Z —— C M = ^ 5 _ + — + — — C M = 641.87883n, 5 3 3
Calcúla te MS = S S / d f and consolídate the information in an ANOVA table.
Source df SS M S
Treatm ents 2 641.8788 320.939E rror 8 498 .6667 62.333
Total 10 1140.5455
b T he hypoth esis to be te ste d is
H 0 : //, = fi2 = p , versus H a : at least one pair of me ans are different
and the F test to detect a difference in m ean student response is
F = M 5 I = 5 , 5 .M SE
The reject ion región w ith a = .05 and 2 and 8 d f i s F > 4.46 and H0 is rejected.
The re is a signif icant difference in m ean respon se due to the three different methods.
a W e would be reasona bly confiden t that the data sat isf ied the norm ality assum ption
because each m easure m ent re presen ts th e avera ge o f 10 continuous m easure m ents .
Th e Central Limit Th eorem assures us that this mean w ill be app roximately norm ally
distributed.
b W e have a com ple te ly random iz ed desig n w ith fo ur tr eatm ents , each con ta in in g 6
m easuremen ts. The analysis o f variance table is given in the M in i ta b printout . The F
test is
f = M S T = ^ 5 8 0 = 5 7 38
MSE .115
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11.17
w ith p-va lue = .000 (in the colum n m arked “P” ). Since the p-va lue is very small i
than .01), H0 is rejected . Th ere is a signifícant difference in the me an le af length
among the four locations with P < .01 or even P < .001.
c The hypothesis to be tested is H0 : p, = p 4 versus H a : p, * p 4 and the teststatis tic is ~~~
x , - x á 6 . 0 1 6 7 - 3 . 6 5t =
M SE1 1
n, n.
= 12.091 1
6 6
The p-value with d f = 20 is 2 P ( t > 12.09) is bounded (using Tab le 4) as
p-valu e < 2 ( .0 0 5 ) = .01
and the nuil hypo thesis is reject. W e con clude that there is a differe nce b etw een t
means.d The 99% conf idence interval for p , - p 4is
1 1 — + — ( x ¡ - x 4 ) ± r m ^ M S E
(6 .0 16 7 - 3 .6 5) ± 2 . 8 4 5 ^ . 1 1 5 ^ + ^ )
2 .367 ± .557 o r 1 .810 < p , - p 4 < 2 .924
e W hen conduc ting the t tests, remem ber that the stated con fidence co efficients
based on ra ndom sam pling. If you lo oked at the data and on ly com pared th e la rg eand smallest sample means, the randomness assumption would be disturbed.
a Th e design is a com pletely random ized design (four independ ent samp les).
b T he fo llow in g pre lim in ary calc ula tions are necess ary :
7; =1211 T 2 = 1074 r 3 =11 58 T4 =1 243 G = 4686
CM =( I * , ) 2 (4 6 8 6 )2
20= 1,097,929.8
T o ta l S S = - C M = 1 ,1 0 1,8 6 2 - C M = 3 93 2.2
SST = Z— — CM =n; 5
12112 10742 11582 12432+ - - C M = 3 27 2.2
5 5 5
Calcú la te MS = S S / d f and consolídate the information in an ANOVA table.
Source d f SS MS
Treatments 3 3272.2 1090.7333
Error 16 660 41.25
Total 19 3932.2
The hypothesis to be tested is
H0 :p , = p 2 = p 3 = p 4 versus Ha : a t least one pai r of m eans are d i f fer
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and the F test to detect a difference in average price s is
M S T o a F = ---------= 26.44 .M SE
The rejection región w ith a = .05 and 3 and 16 d f is approximately F > 3.24 and Ho isrejected. There is enough ev idence to indicate a difference in the average p rices for
the four States.
11.21 a c o = q0, ( 4 , 1 2 ) - ^ = 4 . 2 0 - ^ = 1.878sV5 V 5
b c o= q m ( 6 , 1 2 ) - ^ = 6 . 1 0 - ^ = 2.1567sV 8 V 8
11.25 The d esign is completely random ized w ith 3 treatm ents and 5 replications pertreatment. The M in i ta b printout below shows the an alysis of variance for this
experiment.
One-way ANOVA: mg/dl versus LabSource DF SS MS F PLab 2 42.6 21.3 0.60 0.562Error 12 422 . 5 35 .2Total 14 465.0
S = 5. 933 R-Sq = 9.15% R-Sq(adj)
Individual 95% CIs For Mean Based on
Pooled StDevLevel N Mean StDev . . +------........+----------+--------- +-----
1 5 108.86 7.47 (--------- -)2 5 105.04 6 . 01 (-------3 5 105.60 3 .70 (------ ---)
100.0 104.0 108.0 112.0Pooled StDev =5.93
Tukey 95% Simultaneous Confidence IntervalsAll Pairwise Comparisons among Levels of LabIndividual confidence level = 97.94%
Lab = 1 subtracted from:Lab Lower Center Upper
2 -13.824 -3.820 6.1843 -13.264 -3.260 6.744
14.0 -7.0 0.0 7.0
Lab = 2 subtracted from:Lab Lower Center Upper +--------- +----------+----------+-----
3 -9.444 0.560 10.564 (...... -..... -*...............)
-14.0 -7.0 0.0 7.0
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11.29
11.33
a The analysis o f var iance F test for H 0 : //, = //2 = //? is F = .60 with
p -va lu e = .562 . T he re sults are not sig n if ic an t and H 0 is not reje cte d . T here is
insufficient evidence to indícate a difference in the treatme nt means.
b Sin ce th e tr eatm ent m eans are not sig n if ic an tly dif ferent, th ere is no need to us eT uk ey ’s test to search for the pairw ise difference s. NotiCe that all three intervals
generated by M in i tab contain zero, indicating that the pairs cannot be judg eddifferent.
Re fer to Ex ercise 11.28. The given sum s of squares are inserted and m issing entries
found by subtraction. The me an squares are found as M S = S S / d f .
Source d f SS M S F
Treatments 2 11.4 5.70 4.01
Blocks 5 17.1 3.42 2.41Error 10 14.2 1.42
Total 17 42.7
Use M in i tab to obtain an ANOVA printout , or use the fol lowing calculat ions:
( I * y ) ( 1 1 3 ) “CM = 4 ^ = 1064.08333
n 12
To ta l S S = X x 2 - C M = 6 2 + 1 0 2 +••■ + 14 2 - C M = 1 2 1 3 - C M = 1 4 8 .9 1 6 6 7
T 2 2 2 2 + 3 4 2 + 2 7 2 + 3 0 2S ST = I - C - C M = - ~U - C M = 25.583333 3
SSB = I í - - C M = 3 3~ + 2 5 2 + 5 5 ~ - C M = 1 2 0.6 66 67 an d4 4
SSE = To ta l SS - SST - SSB = 2 .6667
Calcúlate MS = S S / d f and consol ídate the informat ion in an A N OV A table.
Source d f SS M S F
T reatm ents 3 25.5833 8.5278 19.19
Blocks 2 120.6667 60.3333 135.75Error 6 2.6667 0.4444
Total 11 148.9167
a To test the difference amon g treatm ent means, the test stat istic is
P _ M S T _ ^ 5 2 8 _ ]9 19
MSE .4444
and the reject ion región w ith a = .05 and 3 and 6 d f is F > 4 .76 . There i s a
significant difference am ong the treatm ent m eans.
b T o te s t th e d if fere nce am ong b lo ck m eans, th e te st sta ti sti c is
F = M S B = 6 0 3 3 3 3
MSE .4444
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1.37
and Ihe rejection región with a = .05 and 2 and 6 d f \s F > 5.14 . Th ere is a
significant difference am ong the block m eans.
c With k = 4 , d f = 6 , n, = 3 ,
ÍA VÑÍSE 1.4444a)=<im (4*6) r = 7.03 J — r— = 2.71
yjn, V 3
The ranked means are shown below.
7.33 9.00 10.00 11.33
-*l - 4̂ X2
d Th e 95% confidence interval is
( x A - x B) ± t M SE
( 7 .3 3 3 - 1 1 .3 3 3 ) ± 2 .4 4 7 ^ .4 4 4 4
- 4 1 1 . 3 3 2 o r - 5 .3 3 2 < / i A - f i B < - 2 .6 6 8
e Since there is a significant difference amo ng the block m eans, blocking has been
effective. Th e variation due to block differences can be isolated using the rando m ized
blo ck desig n.
Sim ilar to previous exercises. The M in i tab printout for this random ized blockexperiment is shown below.
Two-way ANOVA: Measurements versus Blocks, ChemicalsSource DF SS MS F PBlocks 2 7.1717 3.58583 40 .21 0.000Chemicals 3 5.2000 1.73333 19.44 0.002Error 6 0.5350 0.08917Total 11 12 . 9067S = 0.2986 R-■Sq = 95. 85% R-:Sq (adj) = 92.40%
Individual 95% CIs For Mean Based on PooledStDev
Blocks123
Mean10.875
12.70012.225
10.50 11.20 11.90 12.60
Individual 95% CIs For Mean Based onPooled StDev
Chemicals1234
Mean11.400012.3333
11 .'ZhOO12.8000
1 1 . 2 0 11. 90 12 .60 13 .30
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Bo th the treatm ent and b lock m eans are signif icantly different. Since the four
Chemicals represent the treatments in this experiment, T uk ey 's test can be used to
determ ine w here the differences lie:
, . > / M S É 1 .0 89 17 -
m= q<* = y ~ r ~ =
The ranked means are show n below. 11.20 11.40 12.33 12.80
X, Xj x^
The Chemical falls into two signif icantly different groups - A and C versus B and D.
11.41 A random ized block design has been used with “estim ators” as treatm ents and
“construction jo b ” as the block factor. The a nalysis of variance table is found in the
M in i ta b pr intout below.Two-way ANOVA: Cost versus Estimator, JobSource DF SS MS F PEstimator 2 10. 8617 5.4308 7.20 0.025Job 3 37. 6073 12.5358 16.61 0.003Error 6 4 .5283 0.7547Total 11 52 .9973S = 0.8687 R-Sq = 91.46% R-Sq(adj)
Individual 95% CIsPooled StDev
= 84.34%
For Mean
Estimator Mean ------- +----------+ _
A 32 ..6125 (-------- *---------)
B 34 .. 8875 (--------
C 34..1875 (----- --
★-----
3 2 . 4 3 3 . 6 3 4 . 8 3 6 . 0
Both treatments and blocks are signif icant. The treatment m eans can be further
compared using Tukey’s test with
a> = q m ( 3 , = 434 = 1 -885
The ranked means are shown below.
32.6125 34.1875 34.8875
Estimators A and B show a signif icant difference in average costs .
11 .45 a -b There a re 4 x 5 = 20 t rea tments and 4 x 5 x 3 = 60 total observat ions .
c In a factorial experimen t, variation due to the interaction A x B is isolated from
SSE. The sources of variation and associated degrees of freedom are given on the
next page.
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Source df
A 3
B 4A x B 1 2
Error 40
Total 59
49 a Based on the fact that the mean respon se for the two levels o f factor B behav es
very differently dependíng on the level of factor A under investigation, there is a
strong interaction present between factors A and B.
b T he te st st ati st ic for in te racti on is
F = M S(A B ) /M SE = 37 .85 wi th p -va lue = .000 f rom the M in itab printout. Th ere is
eviden ce of a significant interaction. Th at is, the effect of factor A depen ds upon thelvel of factor B at which A is measured.
c In light of this type of interaction, the main effect means (averaged ov er the levels
o f the other factor) differ only slightly. Henee, a test o f the main -effect terms
produces a non-s ig nif ic ant re su lt .
d No. A significant interaction indicates that the effect of one factor dep end s upon
the level of the other. Each factor-level combination should be investigatedindividually.
e A nsw ers will vary.
53 a The design is a 2 x 4 factorial experiment with r = 5 replications. T here are two
factors, Gender and School, one at two levels and one at four levels.
b T he analy sis o f vari ance ta ble can be fo und usin g a Com pute r prin to ut o r the
following calculations:
Schools
Gender 1 _____ 2 ______ 3 ______ 4 Tota l
M ale 2919 3257 3330 2461 11967
Fem ale 3082 3629 3344 2410 12465
Total 6001 6886 6674 4871 24432
? 4 4 3 2 2C M = ----------- = 14923065.6
40Tota l SS = 15281392 - CM = 358326.4
20
6 00 12 + 68 862 + 66 742 + 48 712- C M = 24 67 25 .8iC =
10
- S S G - S S ( S c ) - C M = 10574.9
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Source d f SS M S F
G 1 6200.1 6200 .100 2.09
Se 3 246725.8 82241.933 27.75G x S c 3 10574.9 3524.967 1.19
E rro r 32 94825.6 2963.300
Total 39 358326.4
c Th e test statistic is F = M S (G S c) /M S E = 1 .19 and the reject ion región is
F > 2 .92 (with a = .05 ). Al ternately , you can bound the p-value > .1 0 . H en ee ,?
is not rejected. Th ere is insuffícient evidence to indícate interact ion betw een gende
and schools.
d You can see in the interact ion plot that there is a small differenc e betw een the
averag e scores for male and fem ale students at schoo ls 1 and 2, but no difference te
e Th e test statistic for testing gen der is F = 2 .09 wi th F os = 4 .1 7 (or p-value > .11
The test statistic for schools is F = 27.75 with F 05 = 2.92 (or p-value < .005 ) . The
is a significant effect due to schools.
Using T ukey ’s method of paired compar isons wi th a = .01, calcúlate
„ = í „ , ( 4 , 3 2 ) ^ 4 , o J f L 82.63 yjn, V 10
The ranked m eans are shown below.
487.1 600.1 667 .4 688.6
*4 *3 x2
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11.59 The objective is to determine w hether or not mean reaction time differs for the f ive
stimuli. The fou r peo ple used in the experim ent act as block s, in an attem pt to isolate
the variation from person to person. A random ized block design is used, and the
analysis of variance table is given in the printout.
a The F statistic to detect a difference due to stimuli is
f = ̂ I = 2 7 .7 8
M SE
with p-value = .00 0. Th ere is a signif icant difference in the effect of the f ive stimuli.
b T he tr eatm ent m eans can be fu rt her com pared usin g T u k ey ’s te st w ith
, c i M > /M S É „ C1 ¡.00708 i n nú) = q os(5,12 ) — — = 4 .5 l J — — = . '9 0
V"r V 4
The ranked means are shown below.E A B D C
.525 .7 .8 1.025 1.05
c The F test for blocks produ ces F = 6.59 with p-value = .007 . The block
differences are significant; blocking has been effective.
11.63 Th is is similar to previous exercises. Th e com plete AN OV A table is shown below.
Sourced f
SS MS F
A 1 1.14 1.14 6.51
B 2 2.58 1.29 7.37
A x B 2 0.49 0.245 1.40
Error 24 4 .20 0.175
Total 29 8.41
a The test s tatistic is F = MS ( A B )/M S E = 1.40 and the rejection región is
F > 3.4 0. There is insuff icient evidence to indicate an interaction.
b U sin g T able 6 w ith d f = 2 and d f 2 = 24 , the following critical valúes are
obtained.
a .10 .05 .025 .01 .005
2.54 3.40 4.32 5.61 6.66
The observed valué of F is less than F 1(), so that p-v alu e > . 10 .
c Th e test statistic for testing factor A is F = 6.51 with F ()5 = 4.2 6. The re is
evidence that factor A affeets the response.
d Th e test statistic for factor B is F = 7.37 with Fm = 3.4 0. Factor B also affeets
the response.
11.67 a The design is a random ized block design, with week s representing blocks and
stores as treatments.
b T he M in i tab Computer printout is shown on the next page.
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11.71
Two-way ANO VA: Total ve rsu s Week, StoreSource DF SS MS F PWeek 3 571.71 190.570 8 .27 0.003Store 4 684.64 171.159 7 .43 0 .003Error 12 276.38 23.032
Total 19 1532.73S = 4.799 R-Sq = 81.97% R-Sq(adj) = 71.45%
c The F test for treatments is F = 7.43 with p-value = .003 . Th e p-value is small
enoug h to allow rejection o f H0. There is a signif icant difference in the average
w eekly totals for the f ive supermarkets.
d Wi t h k= 5, df = \2, n = 4 ,
t e , ^ V M S E , (2 3.0 32a> = qtí5 (5,12) - j = = 4 .5 i y — - — = 10.82
The ranked means are shown below.
1 5 4 3 2
240.23 249.19 252.18 254.87 256.99
a The des ign is a comp lete ly random ized des ign with three samples , each having a
dif ferent num ber o f measurements .
b Use the computing formulas in Section 11.5 or the M in i ta b pr intout below.
One-way ANOVA: Iron versus SiteSource DF SS MS F P
Site 2 132.277 66.139 126.85 0.000Error 21 10.950 0.521Total 23 143.227S = 0.7221 R-Sq = 92.36% R-Sq(adj) = 91.63%
Th e F test for treatm ents has a test s tatistic F = 126.85 w ith p-va lue = .000. Th e nuil
hyp othesis is rejected and we conclude that there is a signif icant difference in the
average p ercentage o f iron ox ide at the three sites.
c Th e diagno stic plots are show n on the next page. There appears to be no violation
o f the norm ality assum ptions; the variances may be unequal, jud gin g by the differ ing
bar w id th s above and belo w th e cen te r line.
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Normal Probability Plotof the Residual!( respo nse is I ron)
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s ** = Z x ? ~ ^ X¡^ = 21,066.82n
2
b-c
= 88-800033 = ft0()42
S_„ 21066.82
and a = y - b x = 0 .41722 -0 .004215 16(94 .53 33) = 0 .187
The least squares line is y = a + b x = 0 .187 + 0 .0042*. The graph o f the least squares
l ine and the nine data points are show n below.
d Wh en x = 100, the valué for y can be pred icted using the least squares l ine as
y = 0.187 + 0.0042(100) = 0.44
Calcúlate
an d
SSR = ( ^ = m 0 3 3 3 3 ) l = 3743064
S„ 21,066.82
( 5 l 2
S S E = T o ta l SS - S S R = S - ± - 2 - L = .3747976 - .374306417 = .00049118$xx
The A NO VA table with 1 d f for regression and n - 2 d f for error is show n on the next
page. R em em ber th at th e m ean square s are calc ula te d as M S = S S / d f .
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Source di SS M S
R egression 1 .3743064 .374306
Error 7 .0004912 .0000010
Total 8 .3747976
12.13 a The scatterplot gene rated by M in i ta b i s shown below. The assumption of
b C alc ú la te
I - * = 1 1 9 2 I * = 7 2 5 = 59 ,3 2 9
£ x 2 = 9 6 ,9 9 0 £ y 2 = 3 6,4 61 n = 15
Then
5 = Y x y -Í5^1í5iil = 1710.6667
,2
S a = y x f = 2265.7333
n2
1419.3333
b = - 2 - = --------= .75502S„ 2265 .7333
and a = y - b x = 4 8 . 3 3 3 3 - ( 0 . 7 5 5 0 2 ) ( 7 9 . 4 6 6 7 ) = - 1 1 . 6 6 5
(using full accurac y). Th e least squares line is
y = a + fox = -11 .66 5 + 0.755*.
c W hen x = 85, the valué for y can be predicted using the least squares line as
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y = a + bx = - \ \ .665 + .755( 85) = 52 .51 .
12.17
12.21
a The hypothesis to be tested is
H0 : (1 = 0 versus H a : /? * 0
and the test statistic is a Student's t, calculated as
b - / 3 0 _ 1 .2 - 0
Vmse J s 2 V 0-533/10
The cr it ical valué o f t is based on n - 2 = 3 degrees o f freedom and the rejection
región for a = 0.0 5 is |?| > t m = 3.182 . Since the observed valué o f t falls in the
rejection región, we reject H0 and conclude that /? * 0 . Th at is , x is useful in the pred ic ti on o f y.
b From the AN O VA table in Exercise 12.6, calcúlate
f = M S R = J ± 4 _ = 2 7 0 ()
MSE 0.5333
which is the square of the t statistic from p art a: r = (5 . 2 0 ) 2 = 2 7 . 0 .
c The cr itical valué of t from part a was t m5 = 3.182 , while the critical valué of F
f rom part b with d f = 1and d f 2 = 3 is F 05 = 10.13 . N otice th at the relation ship
betw een th e tw o cri ti cal valú es is
F = 10.13 = (3 .182 )2 = t 2
a Th e depen dent v ariable ( to be predicted) is >’= cost and the indepen dent va riable
is x = distance.
b Preliminary calculations:
: = 2 1 , 5 3 0 = 5 0 5 2 ^ x j , = 7 , 5 6 9 ,9 9 9
Y , x f = 3 7 , 7 6 3 , 3 1 4 = 1 ,6 9 5 ,9 3 4 n = 18
Then
x iy¡ ~ ̂ ) ( ^ ^ } = ̂ 5 2 7 ^245.667
\ 2
=12,011 ,041 .78n
S /, = - 5 1 = 0 ^ 2 7 1 5 3
a = y - b x = 280.6667 -0.1271 53(119 6.1111) = 128.57699
and the least squares line is y = a + bx = 128.57699 + 0.127153*.
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c Th e plot is show n below . Th e line app ears to fit well through the 18 data poii
12.25
Fitted Line Ploty = 128.6 +0.1272 x
s 72.3755
R-Sq 69.9%
R-Sq(adJ) 68.0%
d Ca lcúlate Total SS = S yy= ' £ y ? - ̂ i L = 1 ,695 ,934 - = 278 ,006 .n 18
Then
(5 „ ) (1 ,527 ,245 .667)"S S E = 5 = 2 7 8 , 0 0 6 - — ------ ------------- ¿- = 83,811.41055
” S 12,011,041.78
SS F 83 811 41055and MS E = -------- = — ------ 1---------- = 5238.213. The hypothesis to be tested is
n - 2 16
H 0 : [3 = 0 versus H a : /? * 0
and the test statistic is
O 197151 — 0
= 6.09t _ b - P Q 0 . 1 2 7 1 5 3 - 0
^ M S E / S „ V5238-213/12,011,041.78
The cri tical valué of t is based on « - 2 = 16 degrees of f reedom and the rejectionregión for a = 0.05 is |/| > t m = 2.1 20 , and H0 is rejected. Th ere is evide nce at tf
5% level to indícate that x and y are linearly related. T hat is, the regress ion m odel
y = a + f i x + e is useful in predicting co st y.
a Th e scat terplot generated by M in i tab is show n on the next page. Th e
assum ption o f l ineari ty is reasonable.
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Scatter plot of Final Exam vs Pos ttest
10 0 •• •
• • •90 •
•• • •80 •
E •
5
i ”•
60• •
50•
4 0 -
7 0 7 5 8 0 8 5 9 0 9 5 10 0
Posttest
b U sin g th e M in i ta b printout, the equation of the regression line isy = - 2 6 .8 2 + 1 .2 6 1 7 * .
c T he hypothesis to be tested is
H n : p = 0 versus H a : /? * 0
and the test statistic is a Student’s t, calculated as
t = - r b ~ P Q = = 1.49 Vmse f s „
with p-value = .000 . Since the p-va lue is less than a = .01, we reject H0 and
con clude that /? * 0 . Th at is, final exam scores and po sttest score s are linearly
related.
d Th e 99% confidence interval for the slope /? is
b ± t a / 2 ylM S E / S a => 1.2617 ± 2.878(0.1253) => 1.2617 + 0.4849
or 0.7768 < /? < 1.7466 .
2.29 Use a plot of residuals versus f its. The plot should appear as a random sc at ter of poin ts , fr ee o f any pattern s.
2.33 a If you look careful ly, there appea rs to be a sl ight curve to the f ive points.
b T he fi t o f th e regressio n line, m easure d as r 2 = 0.959 indicates that 95.9% of the
overal l variat ion ca n be e xplained b y the straight l ine m odel .
c W hen we look at the residuals there is a strong curvil inear pat tern that has not
been expla in ed by the str aig ht line m odel. T he re la ti onsh ip be tw een tim e in m onth s
and num ber of beok s appears to be curvil inear .
2.39 a Although very sl ight , the student m ight notice a sl ight curva ture to the data
poin ts . b T he fit o f th e li near m odel is very good, assum in g th at th is is indeed the correct
mo del for this data set.
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12.42
c Th e normal probability plot follows the correct pattern for the assum ption of
norm ality. Ho wev er, the residuals show the pattern of a quadratic curve, indicatin¿
that a quadratic rathcr than a l inear model may have been the co rrect model for this
data.
a a Use a Computer program or the hand calculations show n below.
£ > , = 1 1 6 ¿ * = 1 4 8 0 ¿ > ,* = 3 6 ,1 3 3
= 2 8 18 £ y 2 = 4 6 7 ,6 0 0 n = 5
Then
n
■ S „ = 2 > ,2 - — ^ - = 126.8n2
S = V v2 - ^ —^ 2 - = 29 ,52 0n
1797b = - 2 - = _ _ = 14.17192
S„ 126.8
a = y - b x = 2 9 6 - 1 4 .1 7 1 9 2 ( 2 3 . 2 ) = -3 2 . 7 8 9
and the least squares line is
y = -32 .78 9 + 14.172*.
b T he pro port io n o f th e to ta l vari ation expla ined by re gre ssio n isS~ 17972
r 2 = — 3 - = — ---------- = 0.8627S S m (126.8X29,520)
XX yv
c The diagnostic plots , gene rated by M in i ta b are shown below. The plots do m
show any strong violation of assumptions.
Res iduals Versu s the Fitted Valúes
(response i s Total Yante)
50- •
40
30-
20-
s io - •
-10- •
-20
-30 •
150 200 250 300 350Htted Valué
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12.43 Re fer to Exe rcise 12.42 and calcúlate
1797 '(SJ SSE = S . - ^ L L . = 2 9 5 2 0 -
an d M S E =
1268SSE 4053.05205
= 4053.05205
= 1351.01735.ti —2 3
a The point est imator for E (y) when x = 21 is
y = -3 2 .78 9 +14 .172(21) = 264 .823
and the 95% confidence interval is
$ ± f 0 2 5 j M S E
2 A
264.823 ± 3.182. / l 351.017351 (2 1- 23 .2 )
5 + 126.8
264 .823± 57 .079
or 207.744 < E ( y ) < 32 1.902 .
b T h e po in t estim ato r fo r y when x = 21 is still y = -32 .78 9 + 14.172(21) = 264.823
and the 95% predict ion interval is
M SE, 1 ( t p - x )1 + - + — ---------
n S „
2 A
264 .823 ±3 .182^135 1 .01735
264.823 ±1 30.143
or 134.680 < > '<39 4 .96 6 .
1 (21 - 2 3 .2 )2 ̂1+ - + - -
5 126.8
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12.47
12.51
c Th is w ould not be advisable, since you are t rying to est ímate outside the range
of exper imen taron .
a Re fer to f igure below. The sam ple correlat ion coefficient wil l be posi t ive.
Scatterplot of y vs x
4.0- •
3.5-
>. 3.0- •
2 .5 -
2 .0- • •
- 2 - 1 0 1 2X
b C alc úla te
Then r =
S „ = ^ - ^ - 1 0 - y = 10
= « - í f = 4
= 0 .9487 and r 2 = (0.9487)2 = 0.9000 . A pproxim atelyV io
90% of the total sum o f squares of deviat ions wa s reduced by u sing the least squares
equa tion instead o f y as a predictor of y.
W hen the pre-test score x is high, the post-test score y should also be high. The re
should be a posi t ive correlation.
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Calcúlate
. 70 , 0 0 6 - ^ Z p = 468.42857
s *x = 6 5 , 9 9 3 - - ^ - = 517.42857^ n 7
Syy = Z X2 - ^ y‘ ̂ = 7 4 , 5 8 5 = 733.42857n 7
T , S v 468.42857 n _ Then r = - = . . .. = 0 .760 .
^517 .42857(733 .42857)
The test of hypothesis is
H 0 : p = 0 versus Ha : p > 0
and the test statistic is
, _ r V ^ 2 _ 0 .7 6 0 ,/5 „í r i — «Z.O I ^
VI - r 2 y l \ - ( 0 . 7 6 0 ) 2
Th e reject ion región for a = 0.05 is t > tM = 2 .01 5 and H0 is rejected. Th ere is
sufficient evidence to indícate posi t ive correlation.
12.55 a Since nei ther of the two var iables , amo unt of sodium or num ber of calor ies , is
control led, thejnethods of correlat ion rather than l inear regression analysis should beused.
b U se a Com pute r program , y ou r scie nti fi c calc u la to r o r th e co m puti ng form ula s
given in the text to calcúlate the correlation coefficient r. The M in i ta b printout for
this data set is show n on the next page.
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Correlations: Sodium, CaloriesPearson correlation of Sodium and Calories = 0.981P-Value = 0.003
The re is evidence o f a highly signifícant correlat ion, since the p-valu e is so smalThe correlat ion is posi tive.
12.61 An sw ers will vary. The M in i ta b output for this l inear regression p roblem is sho belo w .
Regression Analysis: y versus xThe regression equation isy = 21.9 + 15.0 x
Predictor Coef SE Coef T PConstant 21.867 3.502 6.24 0.000x 14.9667 0.9530 15.70 0.000
S = 3.69098 R-Sq = 96.1% R-Sq(adj) = 95.7%
Analysis of Variance
Source DF SS MS F PRegression 1 3360.0 3360.0 246.64 0.000Residual Error 10 136.2 13.6Total 11 3496.2
Correlations: x, yPearson correlation of x and y = 0.980P-Value = 0.000
a The correlat ion coefficient is r = 0.980.
b The coeffic ie nt o f dete rm in ation is r 2 = 0.961 (or 96.1% ).
c T he least squa res line is y = 2 1.867 + 1 4:9667 x .
d W e wish to es t ím ate the mean percentage of k il l for an appl ication of 4 poi
o f nematicide per acre. Since the percent kil l y is actually a binomial percentage
var iance o f y wil l change depending on the valué of p , the propor t ion o f nematoi
ki lled for a part icular applicat ion rate. Th e residual plot versus the f í t ted valúes
shows this phen om eno n as a “footbal l-shaped ” pat tern. Th e normal probahil i ty
also shows som e deviat ion from no rmali ty in the tai ls of the plot. A transformat
may be neede d to assure that the regression a ssum ptions are sat isfied.
12.65 a Use a Computer program, your scientific calculator or the com puting form i
given in the text to calcúlate the correlat ion co efficient r.
= | , 233. 9 8 7 - 5028i( f 5 6 ) = 3 7 ,3 2 3
= Y x ? - = 2,21'2, i 7 8 - = 1<0 5 , 4 4 6^ n 12
Syy = Z y f ~ ~ ~ ~ = 723’882 ~ = 44 ’154
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T hen r = ^ = _____ ^ l 3 23 = 0 .5 4 70 .^ V 1'°5 ,446(44,154)
The test of hypothesis isH0 : p = 0 versus H a : p > 0
and the test statistic is
r y f ñ — 2 0 .5470VÍ0 „l = —■ = .. — = z.Uoo
VI - r y j \ - ( 0 M 1 0 ) 2
with p-value = P (t > 2.066) bounded as
0 .05 < p -va lue <0 .10
If the exp erim enter is willing to tolérate a p-va lue this large, then Ho can be rejected.
Otherwise, you would declare the results not significant; there is insufficient evidenceto indicate that bending stiffness and twisting stiffness are positively correlated.
b r~ = (0.5470 )2 = 0.2992 so that 29.9% o f the total variation in y can be
explained by the independent variable x.
12.69 a Th e plot is shown below. N otice that the relationship is fairly weak.
Scat terp lo t o f Per-S i te Av ermge v s W eek s i n R e í ca se
7000 ----------------------------------------------------------------------------------------
6000 *
5000
i« 4000<9Í 3000
* 2000 -
1000
o6 8 1 0 12
Weeks h Retease
Regression Analysis. ,The regression equation isv = 3091 - 98 x
Predictor Coef SE Coef T PConstant 3091.4 701.0 4.41 0.000x -97.7 107.2 -0.91 0.374S = 1657.23 R-Sq = 4.4% R-Sq(adj) = 0.0%
Analysis of VarianceSource DF SS MS F PRegression 1 2282896 2282896 0.83 0.374Residual Error 18 49435210 2746401Total 19 51718106
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12.73
Unusual ObservationsObs X y Fit SE Fit Residual St Resid
1 1.0 6551 2994 613 3557 2. 3 IR
10 16.0 2538 1527 1180 1011 0 .87 X
13 5.06219 2603 375 3616 2 . 24R
b From the printout , r 2 = 0.044 . Only about 4% o fth e overal l variation in y is
explained by using the l inea r model .
c From the printout , the regression equa tion is y = 3091.4 - 97 .7* and the
regression is not significant (t = -0 .91 wi th p -va lue = 0 .374) .
d Since the regression is not significant , i t is not appropriate to use the regressio
line for estimation or prediction.
a Th e calculat ions shown below are done using the com puting formulas. An
appropriate Com puter program wil l provide identical results to within roun ding e rre
X * , = 1 5 ° 2 > , = 91 2 > f = 986
x,2 = 2 7 5 0 £ y? = 1 1 2 0 .0 4 n = 10
Then
1 5 0 ( 91 ) = _ 3 79
* n 10
5 = £ x2 - = 2750 - — = 500“ ' n 10
S = y y ? J 1 * ) = 1 1 2 0 .0 4 - — = 291.94» n 10
- 3 7 9b = — = — — = - .7 58
S „ 5 0 0
a = y - b x = 9 .1 - ( - .758) (15) = 20 .47
and the least squares line is y = a + bx = 2 0 . 4 7 - .7 5 8 x .
b Since Total SS = S = 291.94 andyy
2
SSR = = ( 3?9) - = 287 .282S „ 5 0 0
\2
(SJ T h en S S E = T o ta l S S - S S R = 5 = 4.658
The AN OV A table wi th 1 d f for regression and n - 2 d f for error is show n below.
R em em ber that the mean squares are calculated as M S = SS¡ d f .
Source d f SS M S
Regression 1 287.282 287.282E rror 8 4.658 .58225
Total 9 291.940
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c T o test H 0 : /? = O, H a : /? * O , the test statistic is
The reject ion región for a = 0 .05 is \t\ > t M5 = 2.306 and we reject H0. Th ere is
sufficien t eviden ce to indícate that x and >' are linea rly related.
d Th e 95% confid enc e interval for the slope (3 is
b ± ía/ 27 M S E ¡ s 2 => - .75 8 ± 2 .896^ .58225 /500 => - .758 ± .099
o r - . 8 5 7 < < - .6 5 9 .
e W hen x = 14, the est ímate o f exp ected freshness E(y) is
y = 20 .47 - .75 8(1 4) = 9 .858 and the 95% con f idence in terval is
The total variat ion has been reduced by 98.4%% by using the l inear model .
12.76-77 Use the H ow a Line W ork s applet . The l ine y = 0.5x + 3 has a slope of 0.5 and a y-
intercept of 3, while the line y = -0 .5 x + 3 has a s lope of -0 .5 and a y- in tercept of 3.
Th e second l ine slopes dow nw ard at the sam e rate as the f irst line slopes upw ard.
They both cross the y axis at the sam e point.
12.80 a Use a Computer, your scientif ic calculator or the com puting form ulas to f ind the
correlat ion between x and y. The M in i ta b correlat ion printout below shows r = 0.231
with p-va lue = 0.549 w hich is not significant at the 5% level of significance. You
cann ot conclude that there is a significant posi tive correlat ion be tween m edian rate
and score for “budg et” hotels.
i , o ,
9 .8 5 8 ± .5 6 2
or 9 .296 < E { y ) < 10.420.
f Calcúlate
_ _ S S R _ _ 287 28 2 _ Q gg4
Total SS 291.94
Correlations: Median Rate, ScorePearson correlation of Median Rate and Score = 0.231
P-Value = 0.549
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b-c Use the Correlat ion an d the Scatterplot applet . The re is a rando m ser
o f points, with no outl iers. The studen t’s plot should look similar to the M in i ta b
shown below.
Scatte rplot of Sc ore vs Median Rate~
80- •
75- •
v 70-
•
••
65-
••
•60-
•
40 42 44 46 48 50Median Rate
52 54 56
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13: Múltiple Regression Analysis
13.1 a W h e n = 2 , £ (> • ) = 3 + * , - 2 ( 2 ) = - 1 .
W he n jc2 = 1,E ( y ) = 3 + * , - 2 ( 1 ) = x, + 1 .
W h e n = 0 , E ( y ) = 3 + x l - 2 ( 0 ) = x l + 3 .
These three straight lines are graphed below.
x1
b N otice th at th e lines are para ll el (th ey have th e sam e slope).
13.5 a The model is quadratic.
b Sin ce R~ = . 8 1 5 , the sum o f squares of deviations is reduced by 81.5% using the
quadratic model rather than y to predict y.
c The hypothesis to be tested is
H 0 : /?, = p : = 0 H a: at least one /?( diffe rs from zero
and the test statistic is
F = ̂ = 37.37M SE
which has an F distribution with df¡ = k = 2 and d f 2 = n - k - \ = 2 0 - 2 - 1 = 1 7 . T he
p -valu e g iv en in th e prin to u t is P = .0 00 an d H 0 is reje cte d. T here is evid ence th at th e
model contributes information for the prediction o f y.
13.9 a Rate of increase is measure d by the slope o f a line tangent to the curve; this line is
given by an equ ation obtained as d y / d x , the derivative of y w ith respect to x. In part ic ula r,
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13.15
13.19
^ = ̂ ( A , + / ? , * + A * 2 ) = f l + 2 / ¡ ,*dx dx
which has slope 2/3-,. If /?, is negative , then the rate o f increase is dec reasing .Henee, the hypothesis of interest is
H n : /?2 = 0 , H a: £ < 0
b The in div id ual r- te st is / = -8 .1 1 as in E xerc is e 13 .8b. H ow ever, th e te st is one-
tailed, which m eans that the p-v alue is half of the am ount given in the printout. That
is, p-v alue = -^(.000 ) = .000 . Hen ee, H0 is again rejected. Th ere is evidence to
indícate a decreasing rate of increase.
a The M in itab printou t fitting the m odel to the data is show n on the next page. Theleast squares line is
y = -8 .1 77 + 0.292a:, + 4.434a:2
Regression Analysis: y versus x1, x2The regression equation isy = - 8.18 + 0.292 xl + 4.43 x2
Predictor Coef SE Coef T PConstant -8.177 4.206 -1.94 0.093xl 0.2921 0.1357 2.15 0.068x2 4.4343 0.8002 5.54 0.001
S = 3.30335 R-Sq = 82.3% R-Sq(adj) = 77.2%
Analysis of VarianceSource DF SS MS F PRegression 2 355.22 177.61 16.28 0.002Residual Error 7 76.38 10.91Total 9 431.60
Source DF Seq SSxl 1 20.16x2 1 335.05
b T he F test for the overall u tility o f the m odel is F = 16.28 with P = .002 . The
results are highly significant; the model contributes significant information for the
pre dic tion o f y.c T o test the effect of adve rtising expen diture, the hypo thesis o f interest is
' H 0 : /?, = 0, H a: & * 0
and the test statistic is t = 5.54 with p-value = .001. Since a = .01, H0 is rejected.
W e conclude that advertis ing exp enditure contributes s ignificant information for the
pre dic tion o f y, g iv en th at capit al in vestm ent is alr eady in th e mode l.
d From the M in i tab printout, R -S q = 82 . 3%. which means that 82.3% of the
total variation can be explained by the quadratic m odel. Th e m odel is very effective.
a The variable x 2must be the quantitative variable, since it appears as a quadratic
term in the model. Q ualitative variables appear only with exp onen t 1, although they
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ma y ap pea r as the coefficient of another quanti tat ive variable w ith expo nen t 2 or
greater.
b W hen x, = 0 , y = 12.6 + 3.9x2 w hile w hen x, = 1,
y = 12.6 + .5 4 ( l ) - 1 .2 x 2 +3 .9x 22 _
= 1 3 . 1 4 - 1 . 2 x2 + 3 . 9 x22
c The graph below shows the two parábolas.
x2
13.23 Th e basic response equation for a specifíc type of bonding comp oun d wo uld be
E ( y ) = A , + P \ X\ + P 2X\
Since the quali tat ive variable “bonding com pou nd” is at tw o levels, one dum m y
variable is needed to incorpórate this variable into the m odel . D efine the dum m y
variable x2 as follow s:
x2 = 1 i f bonding compound 2
= 0 otherwise
The expanded model is now writ ten as
£ (> 0 = A ) + $ x i + M 2 + P i* 2 + P*x \x i + M 2*2
13.25 a From the printout , the predict ion equation is y = 8 . 58 5 + 3 .8 2 0 8 x - 0 .2 1 6 6 3 x 2 .
b R 2 is labeled “R -sq” or R 2 = .94 4. He nee 94.4% o f the total variation is
acco unted for by using x an d x2 in the m odel.
c The hypo thesis o f interest is
H 0 : /?, = p 2 = 0 H a: at least one differs from zero
and the test statistic is F = 33.44 with p-value = .003 . H enee, H 0 is rejec ted, an d we
conc lude that the m odel contributes significant information for the pred ict ion o f y.
d The hypothesis of in terest isH 0 : A = 0 H a:/? 2 * 0
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13.29
13.31
and the test statistic is t = -4.9 3 w ith p-value = .008 . Henee, H0 is rejected, and we
conclude that the quadratic model provides a better fit to the data than a simple linear
model.
e The pat tern of the diagno st ic plots does not indícate any obvious violation of theregression assum ptions.
a The model is
>’ = fio +M*i+ PiX 2+ M 2+ + A*í*2 + £ and the M initab printout is show n below.
Regression Analysis: y versus x1, x2, xlsq, x1x2, x1sqx2The regression equation isy = 4.5 + 6 .39 xl - 50.9 x2 + 0.132 xlsq ■
Predictor Coef SE Coef T PConstant 4 .51 42 .24 0.11 0.916xl 6.394 5 . 777 1.11 0 .275x2 -50.85 56 .21 -0.90 0.371xlsq 0.1318 0.1687 0.78 0.439xlx2 17.064 7 .101 2.40 0.021xlsqx2 -0.5025 0.1992 -2.52 0.016S = 71.6891 R-Sq = 76.8% R-Sq(adj) = '
Analysis of VarianceSource DF SS MS FRegression 5 664164 132833 25.85Residual Error 39 200434 5139Total 44 864598
P0.000
b The fi tted pre dic tion m odel uses th e coeffic ie nts giv en in th e colu m n m ark ed
“C o e f’ in the printout:
y = 4.51 -t- 6.394jc, - 5 0 .8 5 x , + 17.0 64x,;t2 + .131 8.x,2 - . 5 0 2 5 x f x 2
The F test for the m od el’s utility is F = 25.85 with P = .000 and R~ = .768 . T he
model fits quite well.
c If the dolphin is fema le, x2 = 0 and the predict ion equation becom es
y = 4.51 + 6.394.V, + .1 31 8jc2
d If the dolphin is male, x 2 = 1and the prediction equ ation become s
y = -4 6.3 4 + 23.458.x, - .3707a:2
e The hypo thesis of interest isH „ : / ? 4 = 0 H a : / ? 4 * 0
and the test statistic is t = .78 with p-value = .43 9. H0 is not rejected an d we con clude
that the quadratic term is not important in predicting mercury concentration for
female dolphins.
a- b Th e dató-is plotted on the next page. It app ears to be a curv ilinear relationship,
which cou ld be desc ribed using the quadratic m odel y = /?0 + ¡3xx + f i2 x 2+ £ .
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X
— -----------------
c The M in i ta b pr intout is shown below.
Regression Analysis: y versus x, x_sqThe regression equation isy = 4114749 - 4113 x + 1.03 x_sq
Predictor Coef SE Coef T PConstant 4114749 343582 11.98 0.001x -4113.4 343.2 -11.99 0.001x_sq 1.02804 0.08568 12.00 0.001S = 0.523521 R-Sq = 99.7% R-Sq(adj) = 99.5%
Analysis of VarianceSource DF SSRegression 2 297.16Residual Error 3 0.82
Total 5 297.98
d The hypo thesis of interest is
H 0 : f l = A = 0
and the test statistic is F = 542.11 with p-value = .00 0. H0 is rejected an d we
conclud e that the model p rovides valuable information for the prediction o f y.
e R 2 = .99 7. Henee, 99.7% o f the total variation is acco unted for by using x and x 2
in the model.
f Th e residual plots are shown below. Th ere is no reason to dou bt the validity of
the regression assumptions. ____
Res iduals V ersus the F itted Valúes(raparse • y)0M -------------------------------------------
| a oo
•0.25
0 5 10 15 20ntted Valué
MS F P148.58 542.11 0.000
0.27
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13.35 a R 2 = .999 . Henee, 99.9% of the total variation is acco unted for by using x and x 2in the m odel.
b T he hypoth esis o f in te re st is
H 0 : / ? , = A = 0
and the test statistic is F = 1676.61 with p-valu e = .00 0. H0 is rejected and we
conclude that the model provides valuable information for the prediction of y.
c The hypo thesis o f interest is
H 0 : f l = 0
and the test statistic is t = -2.6 5 with p-value = .045 . H0 is rejected and we conclu de
that the linear regression c oeff icient is s ignif icant when x 1 is in the m odel.
d The hypo thesis of interest is
H (, : A = 0
and the test statistic is t = 15.14 with p-valu e = .00 0. H0 is rejected and we conclude
that the quadratic regression coe ff icient is s ignif icant w hen x is in the model.
e W hen the quad ratic term is rem ove d from the model, the valué of R2 decrease s by
9 9 .9 -9 3 .0 = 6.9% . This is the addi tional contribut ion o f the quadrat ic term w hen i t
is included in the model.
f The clear pattem o f a curve in the residual plot indicates that the quadratic term
should be included in the model.
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14: Analysis of Categorical Data
14.3 Fo r a test of specified cel l probabil i ties, the deg rees o f freedom are k - 1 . U se Ta ble
5, Appendix I:
a d f = 6 ; x l s = 12.59; reje ct H0 if X 2 > 12.59
b d f = 9 ; X m = 21.666; reject H0 if X 2 > 21.666
cd f = 13; x%a = 29.814; reject H0 if X 2 > 29.8194
d d f = 2; Xas = 5-99; reje ct H 0 if X 2 > 5.99
14.7 One thousand cars were each classif ied according to the lañe which they occupied
(one throu gh four). If no lañe is preferred over another, the prob ability that a car will be driven in la ñe i, i = 1,2,3,4 is V a. The nuil hypo thesis is then
H n : Px = P 2 = Pi = P< = j
and the test statistic is
with £, = npi = 1000(1/4) = 250 for i = 1 ,2 ,3 ,4 . A table of observed and expected
cell counts follows:Lañe 1 2 3 4
o¡ 294 276 238 192
E, 250 250 250 250
Then
v 2 _ ( 2 9 4 - 2 5 0 ) 2 i ( 2 7 6 - 2 5 0 ) 2 _ ( 2 3 8 - 2 5 0 ) 2 _ ( 1 9 2 - 2 5 0 ) 2
250 250 250 250
_ 6 H 0 _ 2 4 4 g
250
The reject ion región with k - 1 = 3 d f is X 2 > x l s = 7.81. Since the observed valuéo f X 2fa lls in the rejection región, we reject H0. Th ere is a differen ce in prefe renc e
for the four lañes.
14.11 Sim ilar to previous exercises. The h ypothesis to be tested is
H 0 : P\ = Pi = " ‘ = Pn ~
ve rsus H a : at least one p¡ is different from the others
with
E. = nPi = 4 0 0 ( 1 /1 2 ) = 3 3 .3 3 3 .The test statistic is
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, ( 3 8 - 3 3 . 3 3 ) ( 3 5 - 3 3 . 3 3 )X 2 = - ---------------- ¿ - + - + ̂ -------------- — = 13.58
33.33 33.33
Th e up per tai led reject ion región is with a = .05 and k - 1 = 11 d f is
X 2 > Xas - 19.675 . The nuil hypothesis is not rejected and we ca nno t con clude that
the proport ion o f cases varies from m onth to month.
14.15 It is necessary to determ ine whether adm ission rates differ from the previously
reported rates. A table of observed and ex pected cel l counts fol lows:
U nco ndition al T rial Refused T otals
O, 329 43 128 500 E¡ 300 25 175 500
The nuil hypothesis to be tested is
H 0 : P, = -60; p 2 = .05; = .35
ag ains t the alternative that at least one of these prob abilities is incorrect. T he test
statistic is
x 2 = ( 3 2 9 - 3 0 0 ) - + ( « z ^ l + ( 1 2 L - 17 S y
300 25 175
Th e num ber of degrees of f reedom is k - 1= 2 and the r ej ect ion r eg ión2 2
X > X as = ^ -99. The nuil hypothesis is rejected, and we conclude that there has been a departu re fr om previo us adm is sio n ra te s. N otice th a t the percen ta ge o f
unco ndit ional adm issions has r isen sl ightly, the num ber of con dit ional adm issions has
increased, and the percentage refused adm ission has dec reased at the expen se o f the
fist two categories.
14.17 Re fer to Section 14.4 o f the text. For a 3 x 5 contingency table with r = 3 and c = 5 ,
th er e are ( r - l ) ( c - l ) = ( 2 ) ( 4 ) = 8 d eg re es o f fr ee do m .
14.21 a Th e hypothesis o f independ ence between at tachm ent pat tern and child care t ime is
tested using the chi-square stat ist ic. The c ontingency table, including colum n and
row totals and the est imated ex pected cel l counts, fol lows. ________
C h i l d C a r e
A t t a c h m e n t Low M odérate H igh T o t a l
Secure 24
(24.09)
35
(30.97)
5
(8.95)
64
A nxious 11
(10.91)
10
(14.03)
8
(4.05)
29
T o ta l 111 51 297 459
The test statistic isv 2 _ ( 2 4 - 2 4 .0 9 ) 2 | ( 3 5 -3 0 . 9 7 )2 <
24.09 30.97
( 8 - 4 . 0 5 )+ - ------------ — = 7.2 67
4.05
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and the rejection región is X 2 > X m = $- 99 w ' t h 2 d f . is rejected . Th ere is
evidence of a dependence between at tachm ent pat tern and c hi ld care t ime,
b The valu é X 2 = 7.2 67 is betw een x l ¡ and Xm s so l^at < p-v alue < .05 .
resu lts are significant.
14.25 a The hypothesis of independence between salary and numb er of workdays at he
is tested using the chi-square stat ist ic. The co ntingency table, including colum n a
row totals and the est im ated exp ected cel l counts, generated by M in i ta b follows.
Chi-Square Test: Less than one, At least one, not all, All at homeExpected counts are printed below observed countsChi-Square contributions are printed below expected counts
Atleast
1
Lessthanone38
36.270.083
one,notall16
21.081.224
Allat
home14
10.651.051
Total68
2 5449.070.496
2628.520 .223
1214 .410.404
92
3 3535.200 . 001
2220 .460.116
910 . 340 . 174
66
4 33
39.471. 060
29
22 . 941.601
12
11.590.014
74
Total 160 93 47 300
Chi-Sq = 6.447, DF = 6, P-Value = 0.375
Th e test statistic is
v 2 _ ( 3 8 - 3 6 .2 7 ) 2 | ( 1 6 - 2 1 .0 8 ) 2 ( | ( 1 2 - 1 1 .5 9) 2 _ ^
36.27 21.08 11.59
and the reject ion región with a = .05 and d f = 3(2) = 6 is X 2 > Xa s = 12.59 and
nuil hyp othes is is not rejected . Th ere is insufficien t evid enc e to indícate that salai
dependent on the num ber of w orkdays spent at home.
b T he observed valu é o f th e te st sta ti stic , X 2 = 6.447 , is less than Xa o = 10.64¿
that the p-value is more than .10. This w ould confirm the non-reject ion of the nul
hypothesis from part a.
14.29 Be cause a set num ber of A m ericans in each sub-populat ion were each fíxed at 20'
we have a contingen cy table with fixed rows. The table, with estima ted expected
counts appea ring in parentheses, is show n on the next page.
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Yes No T otal
W hite-A m erican 40
(62)
160
(138)
200
A frican-A merican 56
(62)
144
(138)
200
H ispanic-A merican 68
(62)
132
(138)
200
A sian-A m erican 84
(62)
116
(138)
200
Total 248 552 800
The test statistic is
v 2 _ ( 4 0 - 6 2 ) 2 t ( 5 6 - 6 2 ) 2 | , ( 1 1 6 - 1 3 8 ) 2 21
62 62 138
and the rejection región w ith 3 d f '\s X 2 >1 1.34 49 . H0 is rejected and we conclud e
that the incidence o f parental support is depend ent on the sub-po pulation o f
Amer icans .
14.33 Th e number of observations per column were se lected pr ior to the exper imen t . The
test proce dure is identical to that used for an r x c con tingenc y table. The
contingency table, including column and row totals and the estimated expected cell
counts, follows.
Type
FamilyMembers
Apar tment D úplex SingleResidence
Total
1 8 20 1 29
(9.67) (9.67) (9 .67)
2 16 8 9 33
(11) (11) (11)3 10 10 14 34
(11.33) (11.33) (11.33)
4 or more 6 2 16 24
(8 ) (8) (8)
Total 40 40 40 120The test statistic is
. (8 —9.67 )2 (2 0 - 9 .6 7 )2 ( 1 6 - 8 )2X - ------------ ’ - + ±----------------- + + ± — = 36.499
9.67 9.67 8
u sin g C om p ute r a cc ura cy . W i th ( r - l ) ( c - l ) = 6 d fa n d a = .01 , the rejection región
is X 2 > 16.8449 . The nuil hypo thesis is rejected. There is suff icient eviden ce to
indicate that fam ily size is depende nt on type of family residence. I t appears that as
the family size increases, it is more likely that people will live in single residences.
14.35 If the hou sekee per actually has no preference, he or she has an equ al chan ce of
p ic k ing any o f th e five floor poli shes. H enee, the nui l hypoth esis to be te ste d is
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14.39
H, P\ = Pi = Py = P\ = Ps = ^
The valúes of O, are the actual counts observed in the experim ent, and
E. = / i p , = 1 0 0(1 /5 ) = 2 0 . -
Polish A B C D E
o¡ 27 17 15 22 19
E¡ 20 20 20 20 20
Then X =( 2 7 - 2 0 ) ' (1 7 -2 0 )" ( 1 9 - 2 0 ) 2
= 4.4020 20 20
The p-value with d f = k -1 = 4 is greater than .10 and Ho is not rejected. W e cannot
con clude that there is a difference in the preferenc e for the five floor polishes. Even
if this hyp othesis h ad been rejected, the conclusión wo uld be that at least one o f thevalué of the p, was signif icantly different from 1/6. H ow ever, this does not imply that
p¡ is necessarily greater than 1/6. Hen ee, we cou ld not con clude that polish A is
superior.
If the objective of the exp erimen t is to show that polish A is superior , a better
p ro cedure would be to te st an hypoth es is as fo llow s:
H 0 '■P\ = 1/6 H a : p, >1 /6
From a sample of n = 100 hou sewives, x = 27 are found to prefer polish A. A s- test
can be performed on the single binomial parameter p¡.
a To test for hom ogen eity of the five binom ial pop ulations, we use chi-square
s ta tist ic and the 5x 2c on t ing en cy table shown below T he nuil hypothes is is that voter
choice and chu rch attenda nce are independent, w ith p be the proportion o f voters who
intend to vote for G.W. Bush in the 20 04 election for a particular church attendance
group. The contingency table generated by M in i ta b is shown below
Chi-Square Test: G.W. Bush, DemocratExpected counts are printed below observed countsChi-Square contributions are printed below expected counts
Total142
155
178
248
G.W. Bush Democrat
i 89 5373 .64 68.36
3 .205 3 .453
2 87 6880.38 74 . 620.546 0.588
3 93 8592.30 85.700.005 0.006
4 114 134128.60 119.401.658 1.786
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5 22 36 5830.08 27.922.169 2.336
Total 405 376 781
Chi-Sq = 15.752, DF = 4, P-Value = 0.003
The observed valué of the test statistic is X 2 = 15.752 with p-v alue = .003 and the
nuil hy poth esis is rejected at the 5% level o f signifícance. Th ere is sufficient
evide nce to indicate that the proportion o f adults wh o intend to vote for G .W . Bush in
the 2004 election is depend ent on church attendance.
14.43 The flower fall into one of four classifications, with theoretical ratio 9:3:3:1.
Co nverting these ratios to probabil it ies ,
Pl = 9 /1 6 = .5625 p2
=3 /16 = . 1875
p , = 3 /1 6 = .1 875 p 4 = 1 /1 6 = .0625
We will test the nuil hypothesis that the probabilities are as above against the
alternative that they differ. Th e table of obse rved and expected cell cou nts follows:
AB Ab aB aa
o¡ 95 30 28 7
E, 90 30 30 10
T he tes t statistic is
v 2 _ ( 9 5 - 9 0 ) 2 ^ 3 0 - 3 0 ) * t ( 2 8 - 3 0 ) 2 , ( 7 - 1 0 ) 2 _ ^ ^ ]
90 30 30 10
The number o f degrees of f reedom is k - 1= 3 and the re jec t ion región w i th a = .01 is
X 2 > x \ \ = 1 1-3449 . Since the observed v alué of X 2 does no t fall in the rejection
región, we do not reject H0. W e do not have enough information to contradict the
theoretical model for the classification of flower color and shape.
14.48 In order to perform a chi-square “goodness of fit” test on the given data, it is
necessa ry that the va lúes O, and E¡ are know n for each o f the fíve cells. T he O, (the
num ber of measurements falling in the i-th cell) are given. H ow ever, £. = np( must
be calc ula te d . R em em ber th at p , is th e pro babil ity th at a m easurem ent fa ll s in th e j'-th
cell. Th e hypothesis to be tested is
H0 : the experim ent is binom ial versus Ha : the ex perim ent is not binom ial
Let x be the n um ber o f successes and p be the probabil i ty o f success on a s ingle tr ial.
Then, assuming the nuil hypothesis to be true,
^ = p ( x = 0 ) = Coy (1 - P ) 4 p l = p ( x = l ) = c t 4 p ' ( l - p )3
p 2 = /> ( * = 2 ) = c 24p 2( l - p ) 2 p 3 = p ( x = 3 ) = C * p 3( l - p ) ‘
p 4 = p ( x = 4 ) = C ; p 4( l - p ) °
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H enee, once an est ím ate for p is obtained, the expec ted cel l frequen cies can be
calculated using the above proba bil i ties. No te that eac h o f the 100 experim ents
consists of four t r iáis and henee the comp lete expe rime nt involves a total o f 400 triáis.
14.51
The best est imator of p is p = x / n (as in Ch apter 9). Then,
„ _ x _ num ber of successes _ 0(1 l ) + l ( l7 ) + 2( 42 ) + 3(12)-t -'4‘(9) 1
n num ber of t riá is 400 2
The ex perimen t (consisting of four tr iáis) was repeated 100 t imes. Th ere are a total of
40 0 tr iáis in which the result “no succes ses in four tr iáis” was obse rved 11 t imes, the
result “one succes s in four t r iáis” was observed 17 t ime s, and so on. Then
Po = c * ( 1 / 2 ) 0 ( l / 2 ) 4 = 1/16 p , = c ; (1 /2 )' (1 /2 )3 =4 /16
p 2 = < T Í ( l / 2 ) ' ( l /2 ) ’ = 6 /1 6 p , = C, ( l /2 )? (1 /2) ' =4 /16
P ,= C 44( l /2 )4 (1 /2 ) ° =1 /16
The observed and expected cel l frequencies are shown in the fol lowing table.
X 0 1 2 3 4
o , 11 17 42 21 9
E, 6.25 25.00 37.50 25.00 6.25
and the statistic is
X =( 1 1 - 6 .2 5 )2 ( 1 7 - 2 5 .0 0 ) 2 ( 9 - 6 .2 5 )
= 8.566.25 25.0 0 6.25
In orde r to bound the p-value or set up a rejection región, i t is necessary to determ inethe appropriate degrees of freedom associated w ith the test stat ist ic. Tw o degrees of
freedom are lost because:
1 The cel l probab il it ies are restr icted by the fact that Z p, = 1.
2 The binomial param eter p i s unknown and m ust be es timated before calculat ing
the expected cel l counts. The num ber of degrees o f freedom is equa l to
k - \ - \ = k - 2 = 3 . Wi th d f = 3 , the p-value for X 2 = 8.56 is betwe en .025 and
.05 and the nuil hypothesis can be rejected at the 5% level of significance. We
conclude that the exp erimen t in quest ion do es not fulfi ll the requireme nts for a
b in om ia l experi m ent.
The nuil hypothesis to be tested is
u 1H 0 : p , = p 2 = p 3 = -
and the test statistic is
x 2 = I
with E j = np¡ = 200(1/3) = 66.67 for / = 1 ,2 ,3 . A table of observed and expected cell
counts follows:
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E ntrance 1 2 3
o¡ 83 61 56
E¡ 66.67 66.67 66.67Then
v 2 _ ( 8 4 - 6 6 . 6 7 ) 2 | ( 6 1 - 6 6 . 6 7 ) 2 | ( 5 6 -6 6 .Ó 7 ) 2 _ 6 ^
66.67 66.67 66.67
Wi th d f = k - 1= 2 , the p-v alue is between .025 and .05 ánd we can reject H 0 at the
5% level of significance . Th ere is a difference in preferen ce for the three doors. A
95% confidence interval for p \ is given as
± Z o 2 5 \ r ^ - => — ± l - 9 6 j 4 1 5^'5 85 ^ => .4 15 ±.0 68n 025V n 200 V 200
or .347 < p, < .483.
14.55 Since the cards for each of the three holidays will be either “hum orous” or “not
hum orou s”, the table actually consis ts of two rows and three colum ns, and is shown
with estimated expected and observed cell counts in the M in itab printout below.
Chi-Square Test: Fathers, Mothers, ValentinesExpected counts are printed below observed countsChi-Square contributions are printed below expected counts
Fathers Mothers Valentines Total
i 100 125 120 345
115.00 115.00 115.001. 957 0.870 0 .2172 400 375 380 1155
385.00 385.00 385.000 .584 0 .260 0 . 065
Total 500 500 500 1500Chi-Sq = 3.953, DF = 2, P-Value = 0. 139
The test statistic for the equality of the three population proportions is
X 2 = 3.953 with p-value = . 139 and H0 is not rejected. There is insufficient evidence
to indícate a difference in the proportion o f hum orous cards for the three holidays.
14.59 a The 2 x 3 cont ingency table i s ana lyzed as in previous exerc ises . The Min i tab
p rin to ut be lo w sh ow s th e observed and estim ate d expecte d cell counts , th e te ststatistic and its associated p-value .
Chi-Square Test: 3 or fewer, 4 or 5, 6 or moreExpected counts are printed below observed countsChi-Square contributions are printed below expected counts
3 orfewer 4 or 5 6 or more Total
1 49 43 34 12637.89 42 .63 45.47
T . 254 0.003 2 .8952 31 47 62 140
42.11 47.37 50.532 . 929 0.003 2 .605
Total 80 90 96 266
Chi-Sq = 11.690, DF = 2, P-Value = 0.003
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14.65
14.69
The results are highly significant ( p-value = .003 ) and we conclude that there is a
difference in the suscep tibili ty to colds dep ending on the num ber of relationships you
have.
b The proportion o f people with colds is calculated cond itionally for each of the
T h ree o r few er F ou r o r fiv e Six or more
Coid i » = .6180
« = . 4 890
— = .3596
No coid — = .3980
— = .5290
— = .6596
Total 1.00 100 1.00
As the researche r suspects, the susceptibility to a co id seem s to decrease as the
num ber o f relationships increases!
U se the first Goodness-of-Fit applet. Enter the observed valúes into the three cells in
the first row. and the expected cell counts will automatically appear in the second
row. The valué of X : = 18.5 with p-value = .0001 provide suff icient evidence to
reject H0 and conclude that custom ers have a preference for one o f the three brands
(in this case, Brand II).
Oreen R e d B iue To«*l
« 1 1 5 120 65 300
..............
Uoo o
■ i
100.0 100 0 30 0
Obse ivéd Fr«Qu«nel*s
ChiSq(2) = 18.5, p-value = 1 OE-4
The data is analyzed as a 2 x 3 contingency table with estimated expected cell counts
shown in parentheses. Use the Chi-Square Tes t o f Independence applet. Your
Opinión
Party 1 2 3 T otal
R epublican 114
(120.86)
53
(48.10)
17
(15.03)
184
Dem ocrat 87
(80.14)
27
(31.89)
8
(9.97)
122
Total 201 80 25 306
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The test statistic is
y 2 _ ( 1 1 4 - 1 2 0 . 8 6 ) 2 t ( 5 3 - 4 8 . 1 0 ) 2 [ ( ( B - 9 . 9 7 ) 2 = 2
120.86 48.1 0 9.97
With d f = 2 , the p-valu e is greater than .10 ( the applet reports p-valu e = .237 8 ) and
H0 is not rejected . Th ere is no eviden ce that party affiliation has any effect onopinión.
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15: Nonparametric Statistics
15.1 a If distr ibution 1 is shifted to the r ight of distr ibution 2, the rank sum for sam ple 1
(T¡) will tend to be large. T he test statistic w ill be T ¡ , the rank sum for sample 1 if
the obse rvations had been ranked from large to small . Th e nuil hyp othesis wil l be
rejected if 7j’ is unusually small.
b F rom T ab le 7 a with n, = 6 , n 2 = 8 and a = .05 , H0 will be rejected if 7'* <31.
c From Table 7c wi th n, = 6 , n2 = 8 and a = .0 1, H0 will be reje cte d if 7”, < 2 1 .
15.5 If Ha is t rue and populat ion 1 l ies to the r ight o f popu lat ion 2, then 7, w ill be large and
7* will be sm all. H enee, the test statistic w ill be 7* and the large sa m ple
approximation can be used. Calcúlate
7* = n, (rtj + n2 + 1) - 7 , = 1 2( 27 ) —193 = 131
_ n , ( t t i + n 2 + l ) _ 1 2 (2 6 + 1) ^
Mt 2 2
= nin2( n ] + n 2+ 1) = 12 (14 )(27 ) = 3?g
r 12 12T he tes t statistic is
7 , - A r _ 13 1 -1 6 2 _ Z — / v.jyGr V378
The reject ion región with a = .05 is z < -1 .6 4 5 and Hq is not rejected. Th ere is
insufficient evidence to indicate a difference in the two po pulat ion distr ibutions.
15.9 Sim ilar to previous exercises. The data, with correspo nding ranks, are shown in the
fol lowing table. ___________ ______________
D eaf (1) H ea r i n g (2 )
2 . 7 5 ( 1 5 ) 0 .8 9 ( 1 )
2 .1 4(1 1) 1.43 (7)
3 .2 3 ( 1 8 ) 1.06 (4)
2 .0 7(1 0) 1.01 (3)
2 . 4 9 ( 1 4 ) 0.94 (2)
2 .1 8(1 2) 1.79 (8)
3 .1 6(1 7 ) 1 .1 2 (5 .5 )
2 .9 3(1 6) 2.01 (9)
2 .2 0(1 3 ) 1 .1 2 (5 .5 )
7, =126
Calcúlate
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T{ = 126
T¡ = « , ( « , + n 2 + l ) - r , = 9 (1 9 ) —126 = 4 5
The test statistic is
T = m in( 7J ,7’1‘ ) = 45
With n, = n 2 = 9 , the tw o-tailed rejection re gión with ar = .05 is found in Tab le 7b to
be 7j’ < 62 . T he observed valu é, T = 45, falls in the rejection región and H0 is
rejected . W e conclude that thé d ea f chi ldren do di f fer from the hear ing children in
eye-mov emen t rate .
15.13 a If a paired difference expe riment has been used and the sign test is one-tai led
(H a : p > .5) , then the experimenter would l ike to show that one populat ion of
m easurem ents lies above the other populat ion. An exac t pract ical statem ent of theal ternat ive hyp othesis would depen d on the expe rimental si tuat ion.
b I t is necessary th at a ( the probab ili ty of reject ing the nuil hyp othesis w hen i t is
true) take valúes less than a = .15 . Assum ing the nuil hypothesis to be true, the two
popu la tions are id entical and consequen tly ,
p = P { A exceeds B for a given pair of observations) is 1/2. The binom ial probabil i ty
wa s discussed in Cha pter 5. In particular , i t was noted that the distr ibution o f the
random variable * is symmetrical about the mean np when p = 1/2 . For exam ple,
with n = 25, P ( x = 0 ) = P ( x = 2 5 ) . Simi larly , P { x = 1) = P ( x = 24) and so on.
He nee, the lower tailed probabil i ties tabulated in Tab le 1, Ap pendix I wil l be identicalto their upper tailed equivalent probab ili t ies. The valúes of a avai lable for this upper
tai led test and the correspon ding reject ion regions are show n below.
Reject ion Región a
* > 2 0 .002
* > 1 9 .007
* > 1 8 .022
* > 1 7 .054
* > 1 6 .115
15.17 a If assessors A and B are equal in their property assessm ents, then p , the
p robab il it y th at A ’s assessm en t exceeds B ’s assessm en t fo r a g iv en property , should
equal 1/2. If one of the assessors tends to be m ore conservative than the other, then
either p > 1/2 or p < 1/2 . Henee, we can test the equivalence o f the two assessors by
testing the hypothesis
H n : p = l / 2 versus H a : p í ¿ l / 2
using the test stat ist ic *, the nu m ber of t im es that assessor A exce eds a ssessor B for a
parti cu la r property assessm ent. To find a tw o-t ail ed re je cti on re g ió n w ith a cióse to.05, use Table 1 wi th n = 8 and p = .5 . Fo r the rejec tion región {* = 0, * = 8) the
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valué of a is .004 + .004 = .008 , while for the rejec tion región { x = 0,1,7,8} the
valué of a is .03 5+ .035 = .070 w hich is closer to .05. H enee, using the reject ion
región {jc < 1 or * > 7 } , the nuil hypoth esis is not rejected, since x = number o f
p roperties fo r w hic h A exceeds B = 6 . T he p -v alu e fo r th is tw o-t ailed te s t is
p -va lu e = 2 P ( x > 6 ) = 2 (1 - .8 5 5 ) = .290
Since the p-valu e is greater than .10, the results are no t significant; H0 is not rejected
(as with the critical valué approach).
b T he t statistic use d in Exe rcise 10.45 allows the exp erim en ter to rejec t Ho, wh ile
the sign test fails to rejec t H0. Th is is beca use the sign test usedless inform ation and
makes few er assumpt ions than does the t test. If al l norm ali ty assum ptions are met,
the t test is the m ore pow erful test and can reject when the sign test cannot .
15.21 a Ho'. po pula tion distributions 1 and 2 are identical
H a: the distribu tions d iffer in location
b S in ce T ab le 8, A ppendix I giv es cri ti cal valú es fo r re je ction in th e lo w er ta il o f the
distribution, we use the smaller of T +and T~ as the test statistic.
c From Table 8 wi th n = 3 0 , a = .05 and a two-tai led test, the reject ion región is
T < 137 .
d Since T + = 24 9, we can calcúlate
n ( n + 1) . 3 0( 3 1)T~ = — - - T * = — - 2 4 9 = 2 1 6 .
2 2The test statistic is the sm aller of T + and T~ or T = 216 and H0 is not rejected. The re
is no evidenc e o f a difference b etween the two distr ibutions.
15.25 a Th e hypo thesis to be tested is
H 0: po pu lation dis tribu tions 1 and 2 are iden tical
H a: the d istributions diffe r in location
and the test statistic is T, the rank sum o f the posit ive (or negative) differences. The
ranks are obtained by ordering the differences according to their absolute valué.Def ine d¡ to be the d i fference between a pai r in populations 1 and 2(i.e., x u - x 2¡).
The differences, along with their ranks (according to absolute magnitude), are shown
in the following table. _______
di .1 .7 .3 -.1 .5 .2 .5
R a n k | d¡ \ 1.5 7 4 1.5 5.5 3 5.5
The rank sum for posi t ive differences is T + = 26.5 and the rank sum for negative
differences is T~ =1.5 wi th n = 7 . Con sider the sm aller rank sum an d determine the
appropriate low er portion of the two-tai led reject ion región. Indexing n = l and
a = .05 in Table 8, the rejection región is T < 2 and Hq is rejected. Th ere is adifference in the two p opu lat ion locations.
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b The results do not agree with those obtained in Exercise 15.16. W e are able
to reject H0 with the more p ow erful W ilcoxon test.
15.29
15.31
a The paired data are given in the exercise. The d ifferences, along w ith their ranks
el i 2 -1 1 3 1 -1 3 -2 3 1 0
Rank
K l
3.5 7.5 3.5 3.5 10 3.5 3.5 10 7.5 10 2.5 -
Let p = P ( A exceeds B for a given intersec tion) and x = num ber of intersect ions a t
w hich A exceeds B. The hyp othesis to be tested is
H 0 : p = 1/2 versu s H a : p * 1/2
using the sign test with x as the test statistic.
Critical valué approach: Various two tailed rejection regions are tr ied in order to
f ind a región w ith a ~ .05 . T he se are show n in the follow ing table.Rejection Región a
jc < 1;jc > 10 .012
x < 2 , x > 9 .066
* < 3 ;jc > 8 .226
W e ch oo se to reject H o i f x < 2 o r ; t > 9 w ith a = .066 . Since x = 8 , H qis not
rejected. There is insufficient evidence to indícate a difference betw een the two
methods.
p-value approach: For the observed valué x = 8 , calcúlate the tw o-tailed p-value:
p -v alu e = 2 P ( x > 8 ) = 2(1 - .88 7 ) = .226Since the p-value is greater than .10, H0 is not rejected.
b To use the W ilcoxon signed rank test, we use the ranks o f the absolute differences
show n in the table above. The n T + = 5 1 . 5 a n d T~ =14.5 with n = 11 . I ndexing
« = 11 and a = .05 in Table 8, the low er portion of the tw o-tailed rejection región is
T < 11 and H0 is not re jecte d, a s in pa rt a.
a Since the experimen t has been design ed as a paired exp eriment, there are three
tests available for testing the differences in the distributions with and without imagery
- (1) the paired difference t test; (2) the sign test or (3) the Wilcoxon signed rank test.
In order to use the paired difference t test, the scores must be ap proxim ately normal;since the num ber of words recalled has a binomial d istr ibution with n = 25 and
unknown recall probahility, this distr ibution may not be approximately normal,
b Using the sign test, the hypothesis to be tested is
H 0 : p = 1/2 versus H a : p > 1/2
For the observed valué x = 0 we calcúlate the tw o-tailed p-value:
- p -v alu e = 2 P ( x < 0 ) = 2 (.000 ) = .000
Th e results are highly signif icant; Ho is rejected and we co nclud e there is a difference
in the recall scores with and without imagery.
Using the W ilcoxon s igned-rank test , the differences w ill all be positive ( x = 0 for
the sign test), so that and
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15.35
15.39
. n ( n + 1) 20(21)T + = — -------- = — -— - = 210 and T ~ = 2 1 0 - 2 1 0 = 0
2 2
Indexing n = 20 and a = .01 in Table 8, the lower po rtion o f the two -tailed rejection
región is T < 37 and H0 is rejected.
The d ata w ith corresponding ranks in parentheses are shown b elow.
A ge
1 0 - 1 9 20 - 39 4 0 - 5 9 6 0 - 6 9
29 (21) 24 (8) 3 7 (3 9 ) 28 (18)
33 (29.5) 2 7 (1 5 ) 2 5 (1 0 .5 ) 29 (21)
2 6 (1 2 .5 ) 33 (29.5) 22 (5.5) 34 (34)
27 (15) 31 (24) 33 (29.5) 36 (37.5)
39 (40) 21 (3) 2 8 ( 1 8 ) 2 1 ( 3 )
35 (36) 2 8 (1 8 ) 2 6 (1 2 .5 ) 2 0 ( 1 )33 (29.5) 24 (8) 30 (23) 2 5 (1 0 .5 )
29 (21) 34 (34) 34 (34) 2 4 (8 )
36 (37.5) 2 1 ( 3 ) 2 7 (1 5 ) 33 (29.5)
22 (5.5) 32 (25.5) 33 (29.5) 32 (25.5)
7j = 247.5 T2 = 168 T 3 = 2 1 6 .5 7; = 188
n, = 1 0 n 2 = 10 n 3 = 1 0 n4 = 10
Th e test statistic, based on the ran k sums, is
12 H = 3 ( „ + l)
n ( n + 1) /i,
12
4 0 ( 4 1 )
(2 47 .5 )2 (1 68 )2 (2 16 .5 )2 (1 88 )2
10 10 10 10- 3 ( 4 1 ) = 2.6 3
Th e rejection región w ith a = .01 and k -1 = 3 d f is based on the chi-square
dis tribution, o r H > = 11.35 . The nuil hyp othesis is not rejected. Th ere is no
evidence of a difference in location.
b S in ce th e obse rved valu é H = 2.63 is less than %2l0 = 6.25 , the p-v alue is greater
than .10.
c-d From Ex ercise 11.60, F = .87 w ith 3 and 36 df . Again, the p-v alue is greaterthan .10 and the results are the same.
Th e ranks o f the data are shown on the next page.
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Treatment
Block 1 2 3 4
1 4 1 2 3
2 4 1.5 1.5 3
3 4 1 3 2
4 4 1 2 3
5 4 1 2.5 2.5
6 1 2 3
7 4 3 2
8 4 1 2 3
7; =32 T 2 = 8.5 7 ; = 1 8 7 ; =21 .5
a Th e test statistic is
Fr = ,12 , y T - - ? , b { k + \)
r bk(lc + 1 ) ^ ' V ’
= 8( ^ 5 )[ (32)2 +(8*5)2+182 +(21-5)2] - 3(8)(5) = 2119
and the rejection región is Fr > ^ 205 = 7.81. Hen ee, Ho is rejected and we conclude
that there is a difference am ong the four treatments .
b The ob served valué, Fr = 2 1 . 1 9 , e x c e e d s x 2m , p-valu e < .005 .
c-e The analysis of variance is perform ed as in Ch apter 11. The A NO VA table is
shown below.
Source d f SS M S F
T reatm ents 3 198.34375 66 .114583 75.43B locks 7 220.46875 31.495536
E rro r 21 18.40625 0.876488
Total 31 437 .40625
The analysis of variance F test for treatm ents is F = 75.43 and the appro xim ate p-
value with 3 and 21 d f is p-v alue < .005 . Th e result is identical to the param etric
result.
15.43 Ta ble 9, A ppen dix I gives critical valúes r0 such that P ( r s >r0 ) = a . Hen ee, for an
upper-tailed test, the critical valué for rejection can be read directly from the table.a rs > . 4 2 5 b a; >.601
15.47 a The two variables (rating and dis tance) are ranked from low to high, and the
results are shown in the following table.
Voter X v Voter X y
1 7.5 3 7 6 4
2 4 7 8 11 2
3 3 12 9 1 10
4 12 1 10 5 9
5 10 8 11 9 5.5
6 7.5 11 12 2 5.5
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Calcúlate I j c ¡y, = 4 4 2 .5
n = \2
I x f = 6 4 9 .5 X y f = 649.5
I x , = 7 8 X y, = 78
Then
S n, = 42 2.12
7 8 :S a = 6 4 9 .5 ---------= 142.5
12
S yy = 6 4 9 .5 -12
and
b T he h yp oth esis o f in tere st is H 0: no co rr e la ti on vers us H a: n egati ve corr e la ti on.
Co nsul t ing Table 9 for a = .05 , the critical valué of rs, deno ted by r0 is - .49 7. Since
the valué of the test statistic is less than the
critical valué, the nuil hyp othesis is rejected. Th ere is eviden ce o f a significant
negative correlation between rating and distance.
15.51 Re fer to Ex ercise 15.50. To test for positive correlation with a = .05 , index .05 in
Table 9 and the rejection región is rs > .600 . W e reject the nuil hypothe sis of no
association and conclude that a positive correlation exists between the teacher’s ranks
and the ran ks o f the IQs.
15.55 a Def ine p = P (res po ns e for stimulus 1 exceed s that for stimulus 2 ) and x = n u m b e r
of t im es the respon se for stimulus 1 exceed s that for stimulus 2. Th e hyp othesis to be
tested is
using the sign test with x as the test s tatistic. N otice that for this exercise n = 9 , and
the ob served valué o f the test s tatistic is x = 2 . Various two tailed rejection regions
are tr ied in order to f ind a región w ith a ~ .05 . Th ese are show n in the following
table.
W e choose to reject Hq if x < 1or x > 8 w ith a = .04 0 . S ince x = 2 , H0 is not
rejected. T here is insuff icient eviden ce to indícate a difference between the two
stimuli.
b T h e ex perim en t has been d esig ned in a pair ed m anner, and th e pair ed dif fe re nce
test is used. The differences are show n below.
d, - . 9 - 1 .1 1.5 - 2 . 6 - 1 . 8 - 2 . 9 - 2 . 5 2 .5 - 1 . 4
The hypothesis to be tested isH 0 - / / 2 = 0 H a : / / , - / / 2 * 0
Calcúlate
H 0 : p = 1/2 ve rsu s H a : p 1/2
Reject ion Región a_
x = 0 , x = 9 .004
jc < 1;jc > 8 .040
x < 2 - , x > 7 .180
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15.59
- = I < = - 9 2 = _ 1
n 'd = ^ L = — = - \ m i
n 9
, > „ ' 2
5: = Z d ? -
n - 1
and the test statistic is
—— = 3 7 -1 4 -9 .4 0 4 = 3 467
The re jection región with a = .05 and 8 d f is |/| > 2.3 06 and H 0 is not rejec ted.
The data, with corresponding ranks, are shown in the following table.
A (1) B (2)
6.1 (1) 9.1 (16)
9 . 2 ( 1 7 ) 8.2 (8)
8 .7 (1 2 ) 8 .6 (1 1 )
8 .9 (1 3.5 ) 6 .9 (2)
7.6 (5) 7.5 (4)
7.1 (3) 7.9 (7)
9 . 5 ( 1 8 ) 8.3 (9.5)8.3 (9.5) 7.8 (6)
9.0 (1.5) 8.9(13.5)
7¡ =94
The d ifference in the brightness levels using the two p rocesse s can be tested using the
nonparametr ic W ilcoxon rank sum test , or the parametr ic tw o-samp le t test.
1 To test the nuil hypo thesis that the two pop ulation distr ibution s are identical,
calcúlate
7¡ =1 + 1 7 h------h 1.5 = 94
7 T = * , ( n , + « 2 + l ) - 7 ¡ = 9 ( 18 + l ) - 9 4 = 77
The test statistic is
T = m in(7¡ ,r ,* ) = 77
With n x = n 2 = 9, the two-tailed rejection región with a = .05 is found in Ta ble 7 b to
be Tx < 62 . The obse rved valué, T = 77 , doe s not fall in the rejection región and H 0
is not rejected. We canno t conclud e that the distribution s of br ightne ss measu reme nts
is different for the two processes.
2 To test the nuil hypo thesis that the two pop ulation m eans are identical, calcúlate
_ I * , ; 74.4= 8.2667
_ 1 * 2 , 73.2 jc, = ------ - ---------
n, 9= 8.1333
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15.61
15.65
2 _ (w, -1 )J ,2 + (n 2 - l ) s 22 _
n\ + n 2 - 2 and the test statistic is
*1 ~*2
6 2 5 . 0 6 -( 7 4 . 4 ) 2
+ 5 9 9 . 2 2 -(73 .2)2
t =8 . 2 7 - 8 . 1 3
1 1 — +n, n2
- i - J . 8 6 75
16
= .304
- — = .8675
The rejection región with a = .05 and 16 deg rees o f freedo m is |f| > 1.746 and H0 is
not rejected. Th ere is insuff icient eviden ce to indicate a difference in the average
bri gh tn ess m easure m ents fo r th e tw o pro cesses.N oti ce th at th e nonpara m etr ic and
para m etr ic te sts re ach th e sam e conclu sio ns.
Since this is a paired exp erimen t, you can cho ose either the sign test, the Wilcoxon
signed rank test, or the param etr ic paired t test. Since the tenderizers have been
score d on a scale of 1 to 10, the param etric test is not applicab le. Sta rt by using the
easiest of the two nonp aram etr ic tests - the sign test.
Def ine p = ^( te nd er iz er A exceeds B for a given cut) and x = num ber of times that A
excee ds B. The hy pothes is to be tested is
H 0 : p = l / 2 v e rs us H a : p * 1/2
using the sign test w ith x as the test statistic. N otice that for this exe rcise n = 8 (there
are tw o ties), and the observe d valué o f the test s tatistic is x = 2 .
p -v a lu e a p p ro a c h : For th e observ ed valu é x = 2, calcúlate
p -v alu e = 2 P ( x < 2 ) = 2 ( .145 ) = .290
Since the p-value is greater than .10, H0 is not rejected. The re is insuff icient eviden ce
to indicate a difference betw een the two tenderizers.
If you use the W ilcoxon signed rank test, you will f ind T* = 7 and T~ = 29 which w ill
not allow rejection o f H0 at the 5% level of signif icance. The results are the same.
The hypothesis to be tested isH0: pop ulation distr ibutions 1 and 2 are identical
Ha: the distributions differ in location
and the test statistic is T, the rank sum o f the positive (or negative) differences. The
ranks are o btained by o rdering the differences according to their absolute valué.
Def ine d¡ to be the difference b etween a p air in populations 1 and 2 ( i .e. , x u - x 2i).
The differences, along with their ranks (according to absolute magnitude) , are shown
di -31 -3 1 - 6 -1 1 - 9 - 7 7
R a n k | d¡ | 14.5 14.5 4.5 12.5 10.5 7 7
di -1 1 7 - 9 - 2 - 8 -1 - 6 - 3
R a n k | d¡ \ 12.5 7 10.5 2 9 1 4.5 3
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15.69
15.73
15.77
The rank sum for positive differences is T * = 14 and the rank sum for negative
differences is T~ = 106 with n = 15 . Co nsider the smaller rank sum and determ ine
the appropriate lower portion of the two-tailed rejection región. Indexing n = 15 and
a = .05 in Table 8, the rejection región is T < 25 and H 0 is rejected. W e concludethat there is a difference betw een m ath and ar t scores.
a-b Since the exper iment is a complete ly random ized des ign, the K ruskal W all is H
test is used. The com bined ranks are show n below.
P la n t R a n k s
A 9 12 5 1 7 34
B 11 15 4 19 14 63
C 3 13 2 9 6 33
D 20 17 9 16 18 80
The test statistic, based on the rank sums, is
H =12
/ t ( n + l )
12
20(21)
Z ^ ~ 3 ( « + l)
(34)2 , (63 )2 , (33 )2 _ (80)- 3 ( 2 1 ) = 9.0 8
Wi t h d f = k -1 = 3 , the observed valué H = 9.08 is between %m5 and X 05so l^at
.025 < p - \alu e < .05 . The n uil hyp othesis is rejected and we con clude that there is a
difference am ong the fou r plants.
c From Exercise 11.66, F = 5.2 0 , and H0 is rejected. The results are the same.
Th e data are already in rank form. The “substantial experien ce” sample is design ated
as sample 1, and n, = 5 ,n2 = 7 .C alcúla te
7, =19
T¡ = « , ( / ! , + n 2+ \ ) - T { = 5 (13 ) —19 = 46
The test statistic is
T = m in(7 ^,7 'l*) = 19
With n, = n 2 = 12, the one-tailed rejection región w ith a = .05 is found in Table 7a to
be Tx < 21 . The ob served valué, 7 = 19, falls in the rejection región and H 0 is
rejected. There is suff icient eviden ce to indicate that the review boa rd considers
expe rience a pr ime factor in the selection of the best cand idates.
The data with corresponding ranks in parentheses are shown on the next page.
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Training Per iods (hours)
.5 1.0 1.5 -2:0
8 (9.5) 9( 1 1 . 5 ) 4 ( 1 . 5 ) 4( 1 . 5 )
14( 14) 7 ( 7 ) 6 ( 5 ) 7 ( 7 )
9 ( 1 1 . 5 ) 5 (3.5) 7 ( 7 ) 5 (3.5)
1 2( 1 3) 8 (9.5)
7 , =48 T 2= 2 2 T, =23 7, =12
« , = 4 n2 = 3 * , = 4 n4 = 3
The test statistic, based on the rank sums, is
12 H =
n ( n + \
12
14(15)
' (48 ) : _ (22 ) : 1 (2 3) : , (12)2
- 3 ( 1 5 ) = 7 .4 3 33
The re jection región with a = .01 and k - 1= 3 d f is based on the chi-square
distribution, or H > %20l = 11.34 . The nuil hyp othesis is not rejected and we
conclude that there is insufficient evidence to indícate a difference in the distribution
of times for the four groups.
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