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
#.!
"#
$#.1.##5.91   +±−
( )11.$,"$9.91."".   =±= . Beca&se does not lie inside this interval, we can )e
reasona)l- certain that the tr&e difference "1   µ  µ   − is not and, therefore, that the two
 pop&lation means are not e0&al. 7or a 95A interval, the t val&e increases to a)o&t ".1 or so,
which res&lts in the interval 5$.".   ± . 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&lt
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&lts 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&ltin' 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&lts 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&ltin' 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&ltin' 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.
".$$.4
* .!. The estimated df * "
(&sin' software, and the correspondin' ?val&e is ?(t .! * .1. ence, even at the
1A level, we conc&r with the a&thors6 assessment that this difference is 'reater, on
avera'e, in pitchers than in position pla-ers.
5$
87.  = , 1"1   == σ σ  , d * 1, nn
14".14" ==σ  , so   
14".14 $45.1
  n β  ,
'ivin' =β   .915, .#"$4, ."94, and . for n * "5, 1, "5, and 1, respectivel-. If 
the  si G µ   referred to tr&e avera'e I6s res&ltin' from two different conditions, 1 "1
  =− µ  µ   
wo&ld have little practical si'nificance, -et ver- lar'e sample sizes wo&ld -ield statistical
si'nificance in this sit&ation.
88. : "1   =− µ  µ  H   is tested a'ainst : "1
  ≠− µ  µ a  H   &sin' the twosample t test, re2ectin'
 at level .5 if either 11."15,"5.   =≥  t t   or if 11."−≤t  . 3ith ".11= x ,
$#." 1  = s , !9.9= y , "1.
"  = s , and m * n * #, s p * ".9$, and t * .95, so  is not
re2ected. In the sit&ation descri)ed, the effect of carpetin' wo&ld )e mied &p with an- effects d&e to the different t-pes of hospitals, so no separate assessment co&ld )e made. The
eperiment sho&ld have )een desi'ned so that a separate assessment co&ld )e o)tained (e.'., a
randomized )loc/ desi'n.
89. "1 :   p p H    =  will )e re2ected at level α   in favor of "1:   p p H a   >  if either 
$45.15.   =≥  z  z  . 3ith 1.F "5 "5
1   == p , $$#.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#