CHAPTER 9
Section 9.1
1.
a. ( ) ( ) ( ) 4.5.41.4 −=−=−=−
Y E X E Y X E
, irrespective of sample sizes.
b. ( ) ( ) ( ) ( ) ( ) !"4.
, and the s.d. of
"$91.!"4. ==−Y X .
c. % normal c&rve with mean and s.d. as 'iven in a and
b ()eca&se m * n * 1, the C+T
implies that )oth X and Y have
approimatel- normal distri)&tions, so
Y X − does
also. The shape is not necessaril- that of a normal c&rve when
m * n * 1, )eca&se the C+T cannot )e invo/ed. So if the two
lifetime pop&lation distri)&tions are not normal,
the distri)&tion of Y X − will
t-picall- )e 0&ite complicated.
2. The test statistic val&e is
n
s
m
s
9$.1−≤ z . 3e comp&te #5.4
.4
"1
45
19
45
. Since 4.#5
1.9$, re2ect and concl&de that the two )rands differ with
respect to tr&e avera'e tread lives.
3. The test statistic val&e is
( )
."≥ z . 3e comp&te
, which is
not "., so we don6t re2ect and concl&de that the tr&e
avera'e life for radials does not
"
4.
( ) ( ) ( ) .#49"1.49$.1"19$.1
s y x
( )."949,$!.1"5= . In the contet of this pro)lem sit&ation, the
interval is
moderatel- wide (a conse0&ence of the standard deviations )ein'
lar'e, so the
information a)o&t 1 µ and " µ is
not as precise as mi'ht )e desira)le.
b. 7rom 8ercise , the &pper )o&nd is
( ) 95.$5"95.$5"5!9.9$$45.15! =+=+ .
5.
a. a sa-s that the avera'e calorie o&tp&t for
s&fferers is more than 1 calcm"min )elow that
for nons&fferers. ( ) ( )
<".9 = "., re2ect .
6.
( ) ."5.
b. ( ) ( ) #5.5. 59.
n * #.
d. Since n * " is not a lar'e sample, it wo&ld no lon'er )e
appropriate to &se the lar'e
sample z test of Section 9.1. % small sample t
proced&re sho&ld )e &sed (Section 9.",
and the appropriate concl&sion wo&ld follow. >ote,
however, that the test statistic of .5 wo&ld not chan'e, and
th&s it sho&ldn6t come as a s&rprise that we wo&ld
still re2ect at
the .1 si'nificance level.
7.
1 ?arameter of interest: =− "1 µ µ
the tr&e difference of means for males and
females on the Boredom ?roneness @atin'. +et = 1
µ men6s avera'e and =" µ
women6s avera'e.
−− = z
! @e2ect . The data indicates the avera'e Boredom ?roneness @atin'
is hi'her for
males than for females.
8.
a.
1 ?arameter of interest: =− "1 µ µ
the tr&e difference of mean tensile stren'th of the
1$4 'rade and the 1!# 'rade wire rod. +et = 1 µ
1$4 'rade avera'e and =" µ
1!# 'rade avera'e.
4
−−− = z
! 7or a lowertailed test, the pval&e * ( ) 5!."# ≈−Φ ,
which is less than an- α , so re2ect . There is ver-
compellin' evidence that the mean tensile stren'th of the
1!# 'rade eceeds that of the 1$4 'rade )- more than 1.
b. The re0&ested information can )e provided )- a 95A
confidence interval for "1 µ µ −
:
( ) ( ) ( ) ( )5##.15,41".1$"1.9$.11$9$.1
9.
a. point estimate ".$!.19.19 =−=− y x . It
appears that there co&ld )e a
difference. b.
( ) 14.1
44.5
, and the
pval&e * "?(z 1.14 * "( .1"!1 * ."54". The p val&e is
lar'er than an- reasona)le
α, so we do not re2ect . There is no si'nificant difference.
c. >o. 3ith a normal distri)&tion, we wo&ld epect
most of the data to )e within " standard
deviations of the mean, and the distri)&tion sho&ld )e
s-mmetric. " sd6s a)ove the mean is 9#.1, )&t the
distri)&tion stops at zero on the left. The distri)&tion is
positivel-
s/ewed.
d. 3e will calc&late a 95A confidence interval for D, the
tr&e avera'e len'th of sta-s for
patients 'iven the treatment. ( )#."1,.19.99.19 $
1.9 9$.19.19 =±=±
Chapter 9: Inferences Based on Two Samples
a. The h-potheses are : 5"1 =− µ µ
and a: 5"1 >− µ µ . %t level
.1, sho&ld )e
re2ected if #.≥ z . Since ( )
−− = z , cannot )e
re2ected in favor of a at this level, so the &se of the
hi'h p&rit- steel cannot )e 2&stified.
b. 1"1 =−− o µ µ , so ( )
"#91.5. ""!".
1 #. =−Φ=
( ) ( ) ( ) ""
"
1" SE SE z y x +±−
α . Esin' ,5.=α 9$.1"
=α z , so
( ) ( ) ( ) ( )41.",99."..9$.1#.5.5 "" =+±− . 3e are 95A confident
that the
tr&e avera'e )lood lead level for male wor/ers is )etween .99
and ".41 hi'her than the
correspondin' avera'e for female wor/ers.
12. The C.I. is ( ) ( ) 4$."!!.#914.5#."!!.#5#." "
"
"
s y x
( )1.$,".11 −−= . 3ith 99A confidence we ma- sa- that the
tr&e difference )etween
the avera'e !da- and "#da- stren'ths is )etween 11." and $.1
>mm".
13. 5."1 ==σ σ , d * .4, 5.,1. ==
β α , and the test is onetailed, so
( ) ( ) #.49
1$.
=n , so &se n * 5.
14. The appropriate h-potheses are : =θ vs. a:
<θ , where "1" µ µ θ
−= . ( <θ is
e0&ivalent to "1 " µ µ < , so
normal is more than twice schizophrenic The estimator of
θ is
Y X −= "Fθ , with ( ) ( ) ( )
nm
Y Var X Var Var "
"
"
(θ FVar , and θ σ F is o)tained )- replacin'
each "
iσ with "
θ σ
( ) 9!.5.$$9.""F −=−=θ and ( ) ( )
15.
Chapter 9: Inferences Based on Two Samples
a. %s either m or n increases, σ decreases, so
σ
µ µ o−− "1 increases (the
n&merator is
positive, so
−−
decreases.
b. %s β decreases, β z
increases, and since β z is the
n&merator of n , n increases also.
16.
nn
s
n
s
=
+
− =
. 7or n * 1, z * 1.41 and pval&e * ( )[ ] 15#$.41.11"
=Φ− .
7or n * 4, z * ".# and pval&e * .4$. 7rom a practical point of
view, the closeness of
x and y s&''ests that there is
essentiall- no difference )etween tr&e avera'e
fract&re
to&'hness for t-pe I and t-pe I steels. The ver- small
difference in sample avera'es has )een ma'nified )- the lar'e
sample sizes < statistical rather than practical si'nificance.
The p
val&e )- itself wo&ld not have conve-ed this messa'e.
Section 9.2
d.
+ =ν
( ) ( ) ( )
$#.$
== +
− =t leads to a pval&e of " ?(t $.1! = "(.5
*.1, which is less than most reasona)le sGα , so we
re2ect and concl&de that there is a
difference in the densities of the two )ric/ t-pes.
19. 7or the 'iven h-potheses, the test statistic ".1 !.
$.1.1"9!.115
;!$4."9,1. −=−≤ t t since <1."
".!$4, we don6t re2ect .
20. 3e want a 95A confidence interval for "1
µ µ − . "$"."9,"5. =t , so the
interval is
( ) ( )#.$,4."!."$"."$.1 −−=±− . Beca&se the interval is
so wide, it does
not appear that precise information is availa)le.
21. +et =1 µ the tr&e avera'e 'ap detection
threshold for normal s&)2ects, and =" µ the
correspondin' val&e for CTS s&)2ects. The relevant
h-potheses are : "1 =− µ µ vs.
a:
"1 <− µ µ , and the test statistic
4$." "9.
#".
"9
Chapter 9: Inferences Based on Two Samples
$"."15,1. −=−≤ t t . Since <".4$ is not
$"."−≤ , we fail to re2ect . 3e have
ins&fficient evidence to claim that the tr&e avera'e 'ap
detection threshold for CTS s&)2ects
eceeds that for normal s&)2ects.
Chapter 9: Inferences Based on Two Samples
22. +et = 1 µ the tr&e avera'e stren'th for
wire)r&shin' preparation and let =" µ the
avera'e
stren'th for handchisel preparation. Since we are concerned
a)o&t an- possi)le difference
)etween the two means, a twosided test is appropriate. 3e
test : "1 =− µ µ H
vs.
( ) ( ) ( )
we re2ect if 145."14,"5. =≥ t t . The
test statistic is
( ) 159.
"44".1
− =t , which is 145."−≤ , so we re2ect and
concl&de that there does appear to )e a difference )etween the
two pop&lation avera'e
stren'ths.
23.
a.
Esin' Hinita) to 'enerate normal pro)a)ilit- plots, we see that
)oth plots ill&strate s&fficient linearit-. Therefore, it
is pla&si)le that )oth samples have )een selected from
normal pop&lation distri)&tions.
1
&:
Normal Pro%a%ility Plot 'or &ig( )uality *a%ri+
A!erage: ". #$,#0
P:
b.
0.# ".# .#
-omarati!e /o Plot 'or &ig( )uality and Poor )uality
*a%ri+
)uality
Poor
)uality
&ig(
etensi%ility 12
The comparative )oplot does not s&''est a difference )etween
avera'e etensi)ilit- for
the two t-pes of fa)rics.
c. 3e test : "1 =− µ µ H
vs. : "1 ≠− µ µ a H .
3ith de'rees of freedom
( ) 5.1
1!9$.
4"$5. "
==ν , which we ro&nd down to 1, and &sin'
si'nificance level .
5 (not specified in the pro)lem, we re2ect if ""#."1,"5.
=≥ t t . The test
statistic is ( )
#. 4"$5.
#. −=
− =t , which is not ""#."≥ in a)sol&te val&e,
so we
cannot re2ect . There is ins&fficient evidence to claim that
the tr&e avera'e
etensi)ilit- differs for the two t-pes of fa)rics.
24.
a. 95A &pper confidence )o&nd: x
t .5,$51SE * 1.4 1.$!1(".5 * 1$.# seconds
b. +et μ1 and μ" represent the tr&e avera'e
time spent )- )lac/)irds at the eperimental and
nat&ral locations, respectivel-. 3e wish to test
: μ1 < μ" * v.
a: μ1 < μ" . The
relevant test statistic is ""
49
!$.1
$4
5."
"""
+
+ J 11".9. @o&ndin' to t * 1.4 and df * 1",
the ta)&lated P val&e is
ver- ro&'hl- .#". ence, at the 5A si'nificance level, we fail
to re2ect the n&ll
h-pothesis. The tr&e avera'e time spent )- )lac/)irds at the
eperimental location is not
statisticall- si'nificantl- hi'her than at the nat&ral
location.
"
Chapter 9: Inferences Based on Two Samples
c. 95A CI for silvere-es6 avera'e time < )lac/)irds6 avera'e
time at the nat&ral location:
(#.4 < 9.! K (". "" $.5!$.1 + * (1!.9$ sec, 9.44
sec. The t val&e ". is
)ased on estimated df * 55.
25. 3e calc&late the de'rees of freedom
( ) ( ) ( )
, or a)o&t 54
(normall- we wo&ld ro&nd down to 5, )&t this n&m)er
is ver- close to 54 < of co&rse for this lar'e n&m)er of
df, &sin' either 5 or 54 won6t ma/e m&ch difference in the
critical t val&e
so the desired confidence interval is ( ) 1
#.!
"#
$#.1.##5.91 +±−
( )11.$,"$9.91."". =±= . Beca&se does not lie inside
this interval, we can )e
reasona)l- certain that the tr&e difference "1
µ µ − is not and, therefore, that the
two
pop&lation means are not e0&al. 7or a 95A interval,
the t val&e increases to a)o&t ".1 or so,
which res<s in the interval 5$.". ± . Since this
interval does contain , we can no
lon'er concl&de that the means are different if we &se a
95A confidence interval.
26. +et =1 µ the tr&e avera'e potential drop for
allo- connections and let =" µ the tr&e
avera'e potential drop for 8C connections. Since we are interested
in whether the potential
drop is hi'her for allo- connections, an &pper tailed test is
appropriate. 3e test
: "1 =− µ µ H vs. :
"1
>− µ µ a H . Esin' the
S%S o&tp&t provided, the test statistic,
when ass&min' &ne0&al variances, is t * .$$", the
correspondin' df is !.5, and the pval&e
for o&r &pper tailed test wo&ld )e L (twotailed
pval&e * ( ) 4.#. " 1 = . M&r p
val&e of .4 is less than the si'nificance level of .1, so we
re2ect . 3e have s&fficient
evidence to claim that the tr&e avera'e potential drop for
allo- connections is hi'her than that for 8C connections.
( ) ( ) ( )
( ) ( )4$!4.5$."9.1#$."4."1.4 #
±=+±−
( )5.1,.$$!.1"9.1# =±= . Beca&se is not contained in
this interval, there is
Chapter 9: Inferences Based on Two Samples
e0&al. Calc&latin' a confidence interval for 1"
µ µ − wo&ld chan'e onl- the order
of
s&)traction of the sample means, )&t the standard error
calc&lation wo&ld 'ive the same res<
as )efore. Therefore, the 95A interval estimate of 1"
µ µ − wo&ld )e ( 1.5, $.,
2&st the
ne'atives of the endpoints of the ori'inal interval. Since is not
in this interval, we reach
eactl- the same concl&sion as )efore; the pop&lation means
are not e0&al.
( )
( ) #."
( ) ( ) ( )
, and the pval&e from ta)le %.# is
appro .45, which is = .1 so we re2ect and concl&de that
the tr&e avera'e lean an'le for
older females is more than 1 de'rees smaller than that of
-o&n'er females.
29. +et =1 µ the tr&e avera'e compression stren'th
for straw)err- drin/ and let =" µ the
tr&e
avera'e compression stren'th for cola. % lower tailed test is
appropriate. 3e test
: "1 =− µ µ H vs. :
"1 <− µ µ a H . The test
statistic is 1." 154."9
14 −=
=ν , so &se df*"5. The pval&e
".1."( =−<≈ t P . This pval&e
indicates stron' s&pport for the alternative
h-pothesis. The data does s&''est that the etra car)onation of
cola res<s in a hi'her avera'e compression stren'th.
30.
( ) ( ) ( )
+ =ν
, which we wo&ld ro&nd down to !, ecept that
there is no df * ! row in Ta)le %.5. Esin' $ de'rees of freedom (a
more conservative
choice, !19."$,5. =t , and the 99A C.I. is
4
( ) ( )#.$,9#.115!$."4.9!19."#.4"4. "$
−−=±−=+±− . 3e
are 99A confident that the tr&e avera'e load for car)on )eams
eceeds that for fi)er'lass
)eams )- )etween $.# and 11.9# />.
b. The &pper limit of the interval in part a does not 'ive
a 99A &pper confidence )o&nd.
The 99A &pper )o&nd wo&ld )e ( ) 9.!94!.44."4.9
−=+− , meanin' that the
tr&e avera'e load for car)on )eams eceeds that for fi)er'lass
)eams )- at least !.9 />.
5
31.
a.
The most nota)le feat&re of these )oplots is the lar'er
amo&nt of variation present in the
midran'e data compared to the hi'hran'e data. Mtherwise, )oth loo/
reasona)l- s-mmetric with no o&tliers present.
b. Esin' df * ", a 95A confidence interval for
rangehighrangemid −− −
µ µ is
( ) ( )54.9,#4.!$9.##5.$9."45.4!.4# 11
#.$
−=±=+±− .
Since pla&si)le val&es for rangehighrangemid
−− − µ µ are )oth positive and
ne'ative (i.e.,
the interval spans zero we wo&ld concl&de that there is not
s&fficient evidence to s&''est that the avera'e val&e
for midran'e and the avera'e val&e for hi'hran'e differ.
32. +et =1 µ the tr&e avera'e proportional stress
limit for red oa/ and let =" µ the tr&e
avera'e proportional stress limit for No&'las fir. 3e test 1:
"1 =− µ µ H vs.
1: "1 >− µ µ a H . The
test statistic is
( ) #1#.1
=ν , the pval&e * ?(t 1.# * .
4#. 3e wo&ld re2ect at si'nificance levels 'reater than
.4$ (e.'., the standard 5A si'nificance level. %t O * .5, there is
s&fficient evidence to claim that tr&e avera'e
$
Chapter 9: Inferences Based on Two Samples
33. +et μ1 and μ" represent the tr&e mean
)od- mass decrease for the ve'an diet and the control
diet, respectivel-. 3e wish to test the h-potheses
: μ1 < μ" P 1 v.
a: μ1 < μ" 1. The
relevant test statistic is
* 1., with estimated df * $ &sin' the
form&la. @o&ndin' to t * 1., Ta)le %.# 'ives a
onesided ?val&e of .9# (a comp&ter will
'ive the more acc&rate ?val&e of .94. Since o&r
?val&e O * .5, we fail to re2ect at the
5A level. 3e do not have statisticall- si'nificant evidence that
the tr&e avera'e wei'ht loss
for the ve'an diet eceeds that for the control diet )- more than 1
/'.
34.
a. 7ollowin' the &s&al format for most confidence
intervals: statistic ± (critical value)
(standard error) a pooled variance confidence interval for the
difference )etween two
means is ( ) nm pnm st y x 11
"," +⋅±− −+α .
b. The sample means and standard deviations of the two samples are
9.1= x ,
""5.11 = s , ".1"= y , 1.1" = s . The
pooled variance estimate is ="
p s
nm
m
"$.1= , so 1""!.1= p s . 3ith df * mn1 * $ for this
interval, 44!."$,"5. =t
and the desired interval is ( ) ( ) ( ) 4 1
4 11""!.144!."".1"9.1 +±−
( )$4.,"4.94.1!.1 −=±= . This interval contains , so it does
not s&pport the
concl&sion that the two pop&lation means are
different.
c. Esin' the twosample t interval disc&ssed earlier, we &se
the CI as follows: 7irst, we need
to calc&late the de'rees of freedom.
( ) ( ) ( )
interval is
1.1
4
""5.1 ""
!
Chapter 9: Inferences Based on Two Samples
35. There are two chan'es that m&st )e made to the
proced&re we c&rrentl- &se. 7irst, the
( ) ( )
where s p is
defined as in 8ercise 4 a)ove. Second, the de'rees of freedom * m n
< ". %ss&min' e0&al variances in the sit&ation from
8ercise , we calc&late s p as follows:
( ) ( ) 544."5." 1$
! "" =
+
= p s . The val&e of the test statistic
is, then,
( ) ( ) ".""4."
1
1
. The de'rees of freedom * 1$, and the p
val&e is ? ( t = "." * ."1. Since ."1 .1, we fail to re2ect
.
Section 9.3
36. "5.!=d , #$"#.11= ! s
1 ?arameter of Interest: = ! µ tr&e avera'e
difference of )rea/in' load for fa)ric in
&na)raded or a)raded condition.
$ !.1 ##$"#.11
"5.! =
− =t
! 7ail to re2ect . The data does not indicate a si'nificant mean
difference in )rea/in'
load for the two fa)ric load conditions.
#
37.
a. This eercise calls for paired anal-sis. 7irst, comp&te the
difference )etween indoor and
o&tdoor concentrations of heavalent chromi&m for each of
the ho&ses. These
differences are s&mmarized as follows: n * , 4"9.−=d ,
#$#.=d s , where d *
(indoor val&e < o&tdoor val&e. Then !."","5.
=t , and a 95A confidence interval
for the pop&lation mean difference )etween indoor and
o&tdoor concentration is
( ) ( )"#$#.,5$11.1!15.4"9.
±− . 3e can
)e hi'hl- confident, at the 95A confidence level, that the
tr&e avera'e concentration of
heavalent chromi&m o&tdoors eceeds the tr&e avera'e
concentration indoors )-
)etween ."#$# and .5$11 nano'ramsm.
b. % 95A prediction interval for the difference in concentration
for the 4th ho&se is
( ( ) ( ( )!5#,.""4.11#$#.!."4"9.1 11
","5. −=+±−=+± nd st d .
This prediction interval means that the indoor concentration ma-
eceed the o&tdoor concentration )- as m&ch as .!5#
nano'ramsm and that the o&tdoor concentration ma-
eceed the indoor concentration )- a m&ch as 1.""4 nano'ramsm,
for the 4th ho&se.
Clearl-, this is a wide prediction interval, lar'el- )eca&se of
the amo&nt of variation in
the differences.
38.
a. The median of the Q>ormalR data is 4$.# and the &pper and
lower 0&artiles are 45.55 and 49.55, which -ields an I@ of
49.55 < 45.55 * 4.. The median of the Qi'hR data
is 9.1 and the &pper and lower 0&artiles are ##.55 and
9.95, which -ields an I@ of
9.95 < ##.55 * ".4. The most si'nificant feat&re of these
)oplots is the fact that their
locations (medians are far apart.
Normal:&ig(:
9
Chapter 9: Inferences Based on Two Samples
b. This data is paired )eca&se the two meas&rements are
ta/en for each of 15 test conditions.
Therefore, we have to wor/ with the differences of the two samples.
% normal
pro)a)ilit- plot of the 15 differences shows that the data
follows (approimatel- a
strai'ht line, indicatin' that it is reasona)le to ass&me that
the differences follow a normal
distri)&tion. Ta/in' differences in the order Q>ormalR <
Qi'hR, we find ".4"−=d ,
and 4.4=d s . 3ith 145."14,"5. =t , a 95A
confidence interval for the difference
)etween the pop&lation means is
( ) ( )#.9,$.4444."".4" 15
±− . Beca&se
is not contained in this interval, we can concl&de that the
difference )etween the
pop&lation means is not ; i.e., we concl&de that the
two pop&lation means are not e0&al.
39.
a. % normal pro)a)ilit- plot shows that the data co&ld easil-
follow a normal distri)&tion.
b. 3e test : =d H µ vs.
: ≠d a H µ , with test
statistic
!."!4." 14""#
. The twotailed pval&e is " ?( t ".! *
".9 * .1#. Since .1# = .5, we can re2ect . There is stron'
evidence to s&pport
the claim that the tr&e avera'e difference )etween inta/e
val&es meas&red )- the two
methods is not . There is a difference )etween them.
40. 7rom the data, n * 1, d * 15.!, sd *
1.#45.
a. +et μd * tr&e mean difference in TBBHC,
postweanin' min&s lactation. 3e wish to test
the h-potheses : μd P "5 v. a: μd
"5. The test statistic is 1#45.1
"5!.15 − =t *
".4$; at 9df, the correspondin' ?val&e is aro&nd .1#. ence,
at the 5A si'nificance level, we re2ect and concl&de that
tr&e avera'e TBBHC d&rin' postweanin' does
eceed the avera'e d&rin' lactation )- more than "5 'rams.
b. % 95A &pper confidence )o&nd for μd *
d t .5,9 sd " n *
15.! 1.#(1.#45
1 * 1$5.#9 'rams.
c. >o. If we pretend the two samples are independent, the
new standard error is is ro&'hl-
"5, far 'reater than 1.#45 1 . In t&rn, the res<in'
t statistic is 2&st t * .45, with
estimated df * 1! and ?val&e * ."9 (all &sin' a
comp&ter.
41. 3e test 5: =d H µ
vs. 5: >d a H µ . 3ith
$.!=d , and 1!#.4=d s ,
9.1#!.1 9.1
− =t . 3ith de'rees of freedom n < 1 * #, the
correspondin' pval&e is ?( t 1.9 * .4!. 3e wo&ld re2ect
at an- alpha level 'reater
than .4!. So, at the t-pical si'nificance level of .5, we wo&ld
re2ect , and concl&de that
the data indicates that the hi'her level of ill&mination -ields
a decrease of more than 5
seconds in tr&e avera'e tas/ completion time.
4
42.
1 ?arameter of interest: d µ denotes the
tr&e avera'e difference of spatial a)ilit- in
)rothers eposed to N8S and )rothers not eposed to N8S.
+et
.epep osed unosed d
µ µ µ −=
" : = ! H µ
$ ( )
%.#
! @e2ect . The data s&pports the idea that epos&re to N8S
red&ces spatial a)ilit-.
43.
a. %ltho&'h there is a Q2&mpR in the middle of the
>ormal ?ro)a)ilit- plot, the data follow a
reasona)l- strai'ht path, so there is no stron' reason for
do&)tin' the normalit- of the
pop&lation of differences.
b. % 95A lower confidence )o&nd for the pop&lation mean
difference is:
( ) 14.4954.1$.# 15
.
3e are 95A confident that the tr&e mean difference )etween a'e
at onset of C&shin'6s
disease s-mptoms and a'e at dia'nosis is 'reater than 49.14.
c. % 95A &pper confidence )o&nd for the correspondin'
pop&lation mean difference is
#.$ 1.54 * 49.14.
44. 3e need to chec/ the differences to see if the ass&mption
of normalit- is pla&si)le. % normal pro)a)ilit- plot
validates o&r &se of the t distri)&tion. % 95A
&pper confidence )o&nd for DN
is ( ) 91."""$."$5 1$
+=
+
* "#5#.54.
3e are 95A confident that the tr&e mean difference )etween
mod&l&s of elasticit- after 1
min&te and after 4 wee/s is at most "#5#.54.
45. 7rom the data, n * 1", d *
<.!, sd * ".#1.
a. +et μd * the tr&e mean difference in
stren'th )etween c&rin' &nder moist conditions and
la)orator- dr-in' conditions. % 95A CI for μd is
d K t ."5,11 sd " n *
<.! K "."1(".#1
1 * (<".5" H?a, 1.5 H?a. In partic&lar, this interval
estimate incl&des the val&e
zero, s&''estin' that tr&e mean stren'th is not
si'nificantl- different &nder these two
conditions.
b. Since n * 1", we m&st chec/ that the differences are
pla&si)l- from a normal pop&lation.
The normal pro)a)ilit- plot )elow stron'l- s&)stantiates that
condition.
41
Diferences
t
5.02.50.0-2.5-5.0-7.5
99
95
90
80
70
60
50
40
30
20
10
5
1
Normal Probability Plot o Diferences Normal
46. 3ith ( ) ( )5,$, 11 = y x , ( ) ( )14,15, ""
= y x , ( ) ( ),1, = y x , and (
) ( )","1, 44 = y x ,
1=d and =d s (the dI6s are 1, 1, 1,
and 1, while s 1 * s" * #.9$, so s p * #.9$ and
t * .1$.
Section 9.4
47. will )e re2ected if ."1. −=−≤
z z . 3ith 15.F 1 = p , and .F
" = p ,
( )( ) ( ) 1#.4
− = z . Beca&se ."1#.4 −≤− , is
re2ected; the proportion of those who repeat after ind&cement
appears lower than those who repeat after no ind&cement.
48.
$ F 1 == p , and
( ) ( )[ ] ( ) ( )[ ] =
+−
49.
1 ?arameter of interest: p1 < p" * tr&e difference
in proportions of those respondin' to
two different s&rve- covers. +et p1 * ?lain, p" *
?ict&re.
" : "1 =− p p H
4 ( )
$ ( )( )( )
− = z ; pval&e * .4"4!
! 7ail to @e2ect . The data does not indicate that plain cover
s&rve-s have a lower
response rate.
# p
m
# p
9$.1 "$$ 14
51.
a. +et p1 and p" denote the tr&e incidence
rates of I pro)lems for the olestra and control
'ro&ps, respectivel-. 3e wish to test
: p1 < μ" * v.
a: p1 < p" U . The pooled
proportion is 5$5"9
15#(.5$1!$(.5"9 F
15#.1!$.
− * .!#. The twosided ?val&e is
"?(V W .!# * .4 O * .5, hence we fail to re2ect the n&ll
h-pothesis. The data do not
s&''est a statisticall- si'nificant difference )etween the
incidence rates of I pro)lems )etween the two
'ro&ps.
b. ( )
=n , so a
common sample size of m $ n $ 1"11 wo&ld )e re0&ired.
52. +et p1 * tr&e proportion of irradiated )&l)s that
are mar/eta)le; p" * tr&e proportion of
&ntreated )&l)s that are mar/eta)le; The h-potheses are :
"1 =− p p H vs.
: "1 >− p p H . The test statistic
is ( )
nm # p
4
$$1. 1#
− = z .
The pval&e * ( ) ".41 ≈Φ− , so re2ect at an-
reasona)le level. @adiation appears to
)e )eneficial.
53.
a. % 95A lar'e sample confidence interval form&la for (
)θ ln is
( ) ny
yn
mx
− +
− ± "
Fln α θ . Ta/in' the antilo's of the &pper and lower
)o&nds
'ives the confidence interval for θ itself.
b. #1#.1F
( ) ( ) ( )( ) 1"1.
#45,1 =+ , so the CI for ( )θ ln is
( ) ( )#$,.$.1"1.9$.159#. =± . Then ta/in' the antilo's of
the two )o&nds
'ives the CI for θ to )e ( )1.",4.1 . 3e are 95A
confident that people who do not
ta/e the aspirin treatment are )etween 1.4 and ".1 times more
li/el- to s&ffer a heart
attac/ than those who do. This s&''ests aspirin therap- ma- )e
effective in red&cin' the
ris/ of a heart attac/. 54.
a. The QafterR s&ccess pro)a)ilit- is p1 p while the
Q)eforeR pro)a)ilit- is p1 p", so p1
p p1 p" )ecomes p p"; th&s we wish
to test " : p p H = vers&s
": p p H a > .
b. The estimator of (p1 p < (p1 p" is ( ) ( )
n
+−+ .
p p
"
"
"
d. The comp&ted val&e of V is $#." 15"
15" =
, so ( ) !.$#."1 =Φ−= P .
%t level .1, can )e re2ected )&t at level .1
wo&ld not )e re2ected.
44
55. 55. 4
!15 F 1 =
( ) ( ) ( )!,.5."1.14.1$.9$.1$9.55. −=±−=±− .
% !!19.""5."5.
9$.1" =
a. 7rom Ta)le %.9, col&mn 5, row #, $9.#,5,1.
= & .
b. 7rom col&mn #, row 5, #".45,#,1. = &
.
c. "!. 1
& & .
h. Since 1!!. $4.5
58.
a. Since the 'iven f val&e of 4.!5 falls )etween .1,5,5.
= & and $4.51,5,1.
= &
, we can sa- that the &ppertailed pval&e is )etween .1 and
.5.
b. Since the 'iven f of ". is less than 5"."1,5,1.
= & , the pval&e .1.
c. The two tailed pval&e * ( ) ".1(."$4.5"
==≥ & P .
45
Chapter 9: Inferences Based on Two Samples
d. 7or a lower tailed test, we m&st first &se form&la
9.9 to find the critical val&es:
. 1
()&t o)vio&sl- closer to .5.
e. There is no col&mn for n&merator d.f. of 5 in Ta)le %.9,
however loo/in' at )oth df *
and df * 4 col&mns, we see that for denominator df * ", o&r
f val&e is )etween 7.1 and 7.1. So we can sa- .1= pval&e =
.1.
59. 3e test ""
""
( )
"
== ' . 3ith n&merator d.f. * m < 1 *
1 < 1 * 9, and denominator d.f. * n <
1 * 5 < 1 * 4, we re2ect if .$4,9,5. =≥
& ' or
"!5. $.
& ' . Since .#4 is in neither
re2ection re'ion, we do
not re2ect and concl&de that there is no si'nificant
difference )etween the two standard
deviations.
60. 3ith = 1σ tr&e standard deviation for
notf&sed specimens and ="σ tr&e standard
deviation for f&sed specimens, we test "1 :
σ σ = H vs. "1:
σ σ >a H . The calc&lated
test
statistic is ( )
== ' . 3ith n&merator d.f. * m < 1 *
1 < 1 * 9, and
denominator d.f. * n < 1 * # < 1 * !, !,9,1.!"."#14.1
& ' =<= . 3e can sa-
that the p
val&e .1, which is o)vio&sl- .1, so we cannot re2ect .
There is not s&fficient evidence that the standard deviation of
the stren'th distri)&tion for f&sed specimens is
smaller
than that of notf&sed specimens.
61. +et ="
1σ variance in wei'ht 'ain for lowdose treatment, and
="
"σ variance in wei'ht
"
"
"
"
is "
( )
"
≥== ' , so re2ect at level .5. The
data does s&''est that there is
more varia)ilit- in the lowdose wei'ht 'ains.
62. 7or the h-potheses "1 : σ σ
= H vers&s "1: σ σ
≠a H , we find a test statistic of ' *
1."".
%t df * (4!,44 ≈ (4,4, 1."" = 1.51 indicates the ?val&e is
'reater than "(.1 * .".
4$
Chapter 9: Inferences Based on Two Samples
ence, is not re2ected. The data does not s&''est a
si'nificant difference in the two
pop&lation variances.
63. α σ
σ α α −=
parentheses is clearl- e0&ivalent to "
1
1,1,"
1 s and "
1
σ , and
"
we need "#.9,,5. = & and 1#. "#.9
1 ,,95.
!4, the C. I. for "
1
1
"
& s . 3e are
confident that the ratio of the standard deviation of triacetate
porosit- distri)&tion to that of
the cotton porosit- distri)&tion is at most #.1.
Supplementary Exercises
( ) ( ) "".
( )
"
=
+
=ν , which we ro&nd down to 15. The pval&e for a
two
tailed test is approimatel- "?(T ."" * "( . * .$. This small of a
pval&e 'ives
stron' s&pport for the alternative h-pothesis. The data
indicates a si'nificant difference. N&e
to the small sample sizes (1 each, we are ass&min' here that
compression stren'ths for )oth fied and floatin' test platens are
normall- distri)&ted. %nd, as alwa-s, we are ass&min'
the
data were randoml- sampled from their respective
pop&lations.
4#
66.
a.
%ltho&'h the median of the fertilizer plot is hi'her than that
of the control plots, the
fertilizer plot data appears ne'ativel- s/ewed, while the opposite
is tr&e for the control
plot data.
b. % test of : "1 =− µ µ H
vs. : "1 ≠− µ µ a H
-ields a t val&e of .", and a two
tailed pval&e of .#5. (d.f. * 1. 3e wo&ld fail to re2ect ;
the data does not indicate a
si'nificant difference in the means.
c. 3ith 95A confidence we can sa- that the tr&e avera'e
difference )etween the tree densit-
of the fertilizer plots and that of the control plots is somewhere
)etween <144 and 1".
Since this interval contains , is a pla&si)le val&e for the
difference, which f&rther
s&pports the concl&sion )ased on the pval&e.
67. +et p1 * tr&e proportion of ret&rned
0&estionnaires that incl&ded no incentive; p" *
tr&e
proportion of ret&rned 0&estionnaires that
incl&ded an incentive. The h-potheses are
: "1 =− p p H vs. : "1
<− p p H . The test statistic is ( )
nm # p
$$ F "
== p . %t this point we notice that since "1 FF
p p > , the
n&merator of the z statistic will )e , and since we have a
lower tailed test, the pval&e will
)e .5. 3e fail to re2ect . This data does not s&''est
that incl&din' an incentive increases
the li/elihood of a response.
49
-omarati!e /olot o' Tree Density /et7een
*ertilier Plots and -ontrol Plots
Chapter 9: Inferences Based on Two Samples
68. S&mmar- 0&antities are m * "4, $$.1= x , s1 *
.!4, n * 11, 11.11= y , s" * .$.
3e &se the pooled t interval )ased on "4 11 < " * d.f.; 95A
confidence re0&ires
.","5. =t . 3ith $#.1" = p s and
!.= p s , the confidence interval is
( ) ( ) ( )"#.5,1#.!."55."!.."55." 11 1
"4 1 −=±=+± . 3e are confident that
the difference )etween tr&e avera'e dr- densities for the two
samplin' methods is )etween
.1# and 5."#. Beca&se the interval contains , we cannot sa-
that there is a si'nificant
difference )etween them.
69. The center of an- confidence interval for "1
µ µ − is alwa-s "1 x x − ,
so
.$9 "
9.1$91.4! "1 =
+− =− x x . 7&rthermore, half of the width of this
interval is
( ) $.1#"
.4!9.1$91 =
−− . 80&atin' this val&e to the epression on the ri'ht of
the
95A confidence interval form&la, we find 5.55" 9$.1
$.1#"
s . 7or a 9A
interval, the associated z val&e is 1.$45, so the 9A confidence
interval is then
( ) ( ) $.9#.$95.55"$45.1.$9 ±=± ( )9.151!,."99−=
.
70.
a. % 95A lower confidence )o&nd for the tr&e avera'e
stren'th of 2oints with a side coatin'
is ( ) !#.5945.".$ 1
s t x . That is,
with a confidence level of 95A, the avera'e stren'th of 2oints with
a side coatin' is at
least 59.!# (>ote: this )o&nd is valid onl- if the
distri)&tion of 2oint stren'th is normal.
b. % 95A lower prediction )o&nd for the stren'th of a sin'le
2oint with a side coatin' is
( ) 1 11
9,"5. 19$.5#.1".$1 +−=+− n
st x
!!.514$.11".$ =−= . That is, with a confidence level of 95A,
the stren'th of a
sin'le 2oint with a side coatin' wo&ld )e at least 51.!!.
c. 7or a confidence level of 95A, a twosided tolerance interval for
capt&rin' at least 95A
of the stren'th val&es of 2oints with side coatin' is ± x
(tolerance critical val&es. The
tolerance critical val&e is o)tained from Ta)le %.$ with 95A
confidence, / * 95A, and n
* 1. Th&s, the interval is
( )( ) ( )!.#,9.414."".$9$.5!9.".$ =±=± . That is, we can
)e
hi'hl- confident that at least 95A of all 2oints with side coatin's
have stren'th val&es
)etween 4.9 and #.!.
d. % 95A confidence interval for the difference )etween the
tr&e avera'e stren'ths for the
two t-pes of 2oints is ( ) ( ) ( )
1
9$.5
1
approimate de'rees of freedom is
( )
( ) ( ) ( )."5,11.1$1.!!".1!5!.11."!".1! =±=± . 3ith
95A
confidence, we can sa- that the tr&e avera'e stren'th for
2oints witho&t side coatin'
eceeds that of 2oints with side coatin' )- )etween 1.11 and "5.
l)in.in.
71. m * n * 4, .9!5= x , s1 * "45.1, ."!95= y ,
s" * "9.!. The lar'e sample 99A
confidence interval for "1 µ µ −
is ( ) 4
!."9
4
""
+±−
( ) ( )14,1"15$.11# ≈± . The val&e is not contained in
this interval so we can
state that, with ver- hi'h confidence, the val&e of "1
µ µ − is not , which is e0&ivalent
to
concl&din' that the pop&lation means are not
e0&al.
72. This eercise calls for a paired anal-sis. 7irst comp&te the
difference )etween the amo&nt of
cone penetration for comm&tator and pinion )earin's for each of
the 1! motors. These 1!
differences are s&mmarized as follows: n * 1!, 1#.4−=d ,
#5.5=d s , where d *
(comm&tator val&e < pinion val&e. Then 1"."1$,"5.
=t , and the 95A confidence
interval for the pop&lation mean difference )etween penetration
for the comm&tator armat&re
)earin' and penetration for the pinion )earin' is:
( ) ( )"5.14,$1.""4.1#1#.4 1!
#5.5 1"."1#.4 −=±−=
±− . 3e wo&ld have
to sa- that the pop&lation mean difference has not )een
precisel- estimated. The )o&nd on the
error of estimation is 0&ite lar'e. %lso, the confidence
interval spans zero. Beca&se of this,
we have ins&fficient evidence to claim that the pop&lation
mean penetration differs for the two
t-pes of )earin's.
Chapter 9: Inferences Based on Two Samples
73. Since we can ass&me that the distri)&tions from which
the samples were ta/en are normal, we
&se the twosample t test. +et 1 µ denote the
tr&e mean heada)ilit- ratin' for al&min&m
/illed steel specimens and " µ denote the tr&e mean
heada)ilit- ratin' for silicon /illed steel.
Then the h-potheses are : "1
=− µ µ H vs. : "1
≠− µ µ a H . The test statistic
is
"5." #$#.
( )
###.
#$#. ""
"
=
+
=ν , so we &se 5!. The twotailed pval&e
( ) "#.14." =≈ , which is less than the specified
si'nificance level, so we wo&ld re2ect
. The data s&pports the article6s a&thors6 claim.
74. +et 1 µ denote the tr&e avera'e tear
len'th for Brand % and let " µ denote the tr&e
avera'e
tear len'th for Brand B. The relevant h-potheses are : "1
=− µ µ H vs.
: "1 >− µ µ a H .
%ss&min' )oth pop&lations have normal distri)&tions,
the twosample t
test is appropriate. m * 1$, .!4= x , s1 * 14.#, n * 14,
.$1= y , s" * 1".5, so the
approimate d.f. is
14 5.1"
1$ #.14
− =t . 7rom Ta)le %.!, the pval&e * ?( t ".$
* .!. %t a si'nificance level of .5, is re2ected and we
concl&de that the avera'e tear
len'th for Brand % is lar'er than that of Brand B.
75.
a. The relevant h-potheses are : "1
=− µ µ H vs. : "1
≠− µ µ a H . %ss&min'
)oth
pop&lations have normal distri)&tions, the twosample
t test is appropriate. m * 11,
1.9#= x , s1 * 14.", n * 15, ".1"9= y , s" *
9.1. The test statistic is
#4." "5".1"
1.1
freedom
=ν , so we &se 1#. 7rom Ta)le %.#,
the twotailed pval&e ( ) 1".$." =≈ . >o,
o)vio&sl-, the res<s are different.
5"
Chapter 9: Inferences Based on Two Samples
b. 7or the h-potheses "5: "1
−=− µ µ H vs. "5: "1
−<− µ µ a H , the test
statistic
chan'es to ( )
55$. "5".1"
"51.1 −=
−−− =t . 3ith de'rees of freedom 1#, the pval&e
( ) "!#.$. =−<≈ t P . Since the
pval&e is 'reater than an- sensi)le choice of α ,
we
fail to re2ect . There is ins&fficient evidence that the
tr&e avera'e stren'th for males
eceeds that for females )- more than "5>.
76.
a. The relevant h-potheses are : "1 =− ∗∗
µ µ H (which is
e0&ivalent to sa-in'
"1 =− µ µ vers&s : "1
≠− ∗∗ µ µ a H
(which is the same as sa-in' "1
≠− µ µ .
="
.4. !4".4
y x t . 7rom Ta)le %.!, the pval&e
associated with t * . is "?( t . * "(.4 * .#. %t si'nificance level
.5, is
∗ "
µ , which is
e0&ivalent to sa-in' that there is a difference )etween
1 µ and " µ .
∗
σ are the
parameters of the lo'normal distri)&tion (i.e., the mean
and standard deviation of ln(.
So when ∗∗ = "1 σ σ , then
∗∗ = "1 µ µ wo&ld impl- that "1
µ µ = . owever, when
∗∗ ≠ "1 σ σ , then even if ∗∗ = "1
µ µ , the two means 1 µ
and " µ ('iven )- the form&la
a)ove wo&ld not )e e0&al.
77. This is paired data, so the paired t test is emplo-ed. The
relevant h-potheses are
: =d H µ vs. :
< d a H µ , where
d µ denotes the difference )etween the
pop&lation
avera'e control stren'th min&s the pop&lation avera'e
heated stren'th. The o)served
differences (control < heated are: .$, .1, .", , and .5. The
sample mean and standard
deviation of the differences are "4.−=d and
5.=d s . The test statistic is
#.1!$.1 "4.
5 5.
−≈−= −
=t . 7rom Ta)le %.#, with d.f. * 5 < 1 * 4, the lower
tailed p
val&e associated with t * 1.# is ?( t = 1.# * ?( t 1.# * .!. %t
si'nificance level .5,
sho&ld not )e re2ected. Therefore, this data does not show that
the heated avera'e stren'th
eceeds the avera'e stren'th for the control pop&lation.
78. +et 1 µ denote the tr&e avera'e ratio for
-o&n' men and " µ denote the tr&e avera'e ratio
for
elderl- men. %ss&min' )oth pop&lations from which these
samples were ta/en are normall-
5
Chapter 9: Inferences Based on Two Samples
distri)&ted, the relevant h-potheses are : "1
=− µ µ H vs. : "1
>− µ µ a H . The val&e
of the test statistic is
( )
. The d.f. * " and the pval&e is ?(t
!.5 J . Since the pval&e is 5.=<α , we re2ect . 3e
have s&fficient evidence to claim
that the tr&e avera'e ratio for -o&n' men eceeds that for
elderl- men.
79. The normal pro)a)ilit- plot )elow indicates the data for 'ood
visi)ilit- does not follow a
normal distri)&tion, th&s a ttest is not appropriate for
this small a sample size. (The plot for
poor visi)ilit- isn6t as )ad.
Good
3210-1
99
95
90
80
70
60
50
40
30
20
10
5
1
80.
a. % 95A CI for μ!,dr- * "5.! K t ."5,5(4.9! $
* "5.! K ".5!1(14."!$ * ("#9.,
$".4. 3e are 95A confident that the tr&e avera'e )rea/in' force
in a dr- medi&m at
!Y is )etween "#9. > and $".4 >.
$
9!.4 ""
+ * (<.#1,!.9. 3e are 95A confident that the tr&e
avera'e
)rea/in' force in a dr- medi&m at !Y is )etween .#1 >
less and !.9 > more than the
tr&e avera'e )rea/in' force in a wet medi&m at !Y.
*
$
* ".5#. The estimated df * 9 a'ain, and the
approimate ?val&e is .15. ence, we re2ect and
concl&de that tr&e avera'e force in
a dr- medi&m at !Y is indeed more than 1 > 'reater than the
avera'e at ""Y.
54
Chapter 9: Inferences Based on Two Samples
81. 3e wish to test : "1 µ µ =
vers&s a: "1 µ µ ≠
Enpooled:
( ) ( ) ( )
− =t
leads to a pval&e of a)o&t "?(t15 1.#
*"(.4$
* .9".
?ooled:
The de'rees of freedom are "4"1"14" =−+=−+= nmν
and the pooled variance is
( ) ( ) 9!.15".1 "4
11 !9.
#9.1 4$5.
. The pval&e * "?( t"4 1.9 * "( .5 * .!.
3ith the pooled method, there are more de'rees of freedom, and the
pval&e is smaller than with the &npooled method. That is,
if we are willin' to ass&me e0&al variances (which
mi'ht
or mi'ht not )e valid here, the pooled test is more capa)le of
detectin' a si'nificant
difference )etween the sample means.
82. Beca&se of the nat&re of the data, we will &se a
paired t test. 3e o)tain the differences )-
s&)tractin' inta/e val&e from ependit&re val&e. 3e
are testin' the h-potheses : Dd * vs
a: Dd U . Test statistic ##. !5!.1
! 19!.1
==t with df * n < 1 * $ leads to a pval&e
of
"?(t.## J .#. Esin' either si'nificance level .5 or .1, we
wo&ld re2ect the n&ll
h-pothesis and concl&de that there is a difference )etween
avera'e inta/e and ependit&re. owever, at si'nificance level
.1, we wo&ld not re2ect.
83.
a. 3ith n denotin' the second sample size, the first is m * n. 3e
then wish
( ) nn
4
b. 3e wish to find the n which minimizes ( ) nn
z 4
n which minimizes nn
55
e0&atin' to -ields ( ) 449 "" =−− −− nn , whence ( )
""
449 nn −= , or
,$4"5 " =−+ nn . This -ields n * 1$, m * 4 < n *
"4.
84. +et p1 * tr&e s&rvival rate at ( 11 ;
p" * tr&e s&rvival rate at ( ; The
h-potheses are
: "1 =− p p H vs. : "1 ≠−
p p H a . The test statistic is
( ) nm
# p
11
− = z . The pval&e *
( ) #$.4(."$."" ==−Φ , so re2ect at most reasona)le
levels (.1, .5, .1.
The two s&rvival rates appear to differ.
85. 3e want to test the h-pothesis : μ1 P
1.5 μ" v. a: μ1 1.5 μ" < or,
&sin' the hint, : P v.
a: . M&r point estimate of is "1 5.1F
X X −=θ , whose estimated
standard error
e0&als
"
" ""
5.1(F( nn
V σ σ
θ += . ?l&' in
the val&es provided to 'et a test statistic t *
#9!5."
15.14(5.1$."" −− J .#. %
conservative df estimate here is Z * 5 < 1 * 49, and
t .5,49 J 1.$!$. Since .# = 1.$!$, we fail
to re2ect at the 5A si'nificance level. The data does not
s&''est that the avera'e tip after an introd&ction is more
than 5A 'reater than the avera'e tip witho&t
introd&ction.
86.
a. 7or the paired data on pitchers, n * 1!, d * 4.$$,
and sd * .955. t ."5,1$ * ".1", and
the
res<in' 95A CI is ("., $.1. 3e are 95A confident that the
tr&e mean difference
)etween dominant and nondominant arm translation for pitchers
is )etween ". and
$.1.
b. 7or the paired data on position pla-ers, n * 19, d *
.", and sd * 1.$. t ."5,1# * ".11,
and the res<in' 95A CI is (<.54, 1.1. 3e are 95A confident
that the tr&e mean
difference )etween dominant and nondominant arm translation for
position pla-ers is )etween ". and $.1.
c. +et μ1 and μ" represent the tr&e mean
differences in sidetoside %? translation for
pitchers and position pla-ers, respectivel-. 3e wish to test
the h-potheses : μ1 < μ" * v.
a: μ1 < μ" . The data for this anal-sis are
precisel- the differences &tilized in parts a
and b. ence, the test statistic is t *
19
$.1
1!
955.
".$$.4
* .!. The estimated df * "
(&sin' software, and the correspondin' ?val&e is ?(t .! *
.1. ence, even at the
1A level, we conc&r with the a&thors6 assessment that this
difference is 'reater, on
avera'e, in pitchers than in position pla-ers.
5$
87. = , 1"1 == σ σ , d * 1, nn
14".14" ==σ , so
14".14 $45.1
n β ,
'ivin' =β .915, .#"$4, ."94, and . for n * "5, 1, "5,
and 1, respectivel-. If
the si G µ referred to tr&e avera'e I6s
res<in' from two different conditions, 1 "1
=− µ µ
wo&ld have little practical si'nificance, -et ver- lar'e sample
sizes wo&ld -ield statistical
si'nificance in this sit&ation.
88. : "1 =− µ µ H is
tested a'ainst : "1
≠− µ µ a H &sin'
the twosample t test, re2ectin'
at level .5 if either 11."15,"5. =≥
t t or if 11."−≤t . 3ith ".11= x
,
$#." 1 = s , !9.9= y , "1.
" = s , and m * n * #, s p * ".9$, and t *
.95, so is not
re2ected. In the sit&ation descri)ed, the effect of carpetin'
wo&ld )e mied &p with an- effects d&e to the different
t-pes of hospitals, so no separate assessment co&ld )e made.
The
eperiment sho&ld have )een desi'ned so that a separate
assessment co&ld )e o)tained (e.'., a
randomized )loc/ desi'n.
89. "1 : p p H = will )e
re2ected at level α in favor of "1:
p p H a > if either
$45.15. =≥ z z . 3ith 1.F "5 "5
1 == p , $$#.F "5 1$!
" == p , and #4.F = p ,
".4 !9.
". == z , so is re2ected . It appears that a
response is more li/el- for a white
name than for a )lac/ name.
90. The comp&ted val&e of V is 4.1 4$4
4$4 −=
= z . % lower tailed test wo&ld )e
appropriate, so the pval&e ( ) 5.91.4.1 >=−Φ= , so we
wo&ld not 2&d'e the dr&'
to )e effective.
91.
a. +et 1 µ and " µ denote the tr&e
avera'e wei'hts for operations 1 and ", respectivel-.
The relevant h-potheses are : "1
=− µ µ H vs. : "1
≠− µ µ a
test statistic is
=ν , so &se df * 5!. ."5!,"5. ≈t
, so we can re2ect at level .5. The data indicates that there
is a si'nificant difference
)etween the tr&e mean wei'hts of the pac/a'es for the two
operations.
b. 14: 1 = µ H will )e tested
a'ainst 14: 1 > µ a H &sin'
a onesample t test with
test statistic
14− = . 3ith de'rees of freedom * "9, we re2ect
if
$99.1"9,5. => t t . The test statistic
val&e is 1.1 ."
"4."14"4.14"
=t .
Beca&se 1.1 = 1.$99, is not re2ected. Tr&e avera'e
wei'ht does not appear to eceed
14.
variance )- the pooled estimate nm
Y n X m
+ +
=λ F . 3ith the o)vio&s point estimates X =1
Fλ ,
Y = " Fλ , we have a lar'esample test statistic of
m
Y
n
X
λ . 3ith
$1$.1= x and 55!."= y , z * 5. and pval&e * (
)( ) $..5" <−Φ , so we
wo&ld certainl- re2ect "1 : λ λ
= H in favor of "1: λ λ
≠a H .
93. % lar'esample confidence interval for [ 1 <
[ " is nm
z "1 ""1
n
y
m
x z y x +±− "( α . 3ith
$".1= x and 5$."= y , the 95A confidence
interval
for [ 1 < [ " is .94 K 1.9$(.1!! * .94 K .5
* (1."9,.59.
5#