pages.uoregon.edupages.uoregon.edu/haowang/teaching/425_FALL2008/SoluManual/C… · 10: Inference from Small Samples 10.1 Refer to Table 4, Appendix I, indexing df along the left

10: Inference from Small Samples 10.1 Refer to Table 4, Appendix I, indexing df along the left or right margin and tα across the top. a with 5 df b .05 2.015t = .025 2.306t = with 8 df c with 18 df c .10 1.330t = .025 1.96t ≈ with 30 df 10.2 The value is the tabled entry for a particular number of degrees of freedom. ( )aP t t a> =

a For a two-tailed test with .01α = , the critical value for the rejection region cuts off 2 .005α = in the two tails of the t distribution shown below, so that .005 3.055t = . The null hypothesis H0 will be rejected if

or (which you can also write as 3.055t > 3.055t < − 3.055t > ).

b For a right-tailed test, the critical value that separates the rejection and nonrejection regions for a right tailed test based on a t-statistic will be a value of t (called tα ) such that ( ) .05P t tα α> = = and 16df = . That is, . The null hypothesis H.05 1.746t = 0 will be rejected if . 1.746t >

c For a two-tailed test with 2 .025α = and 25df = , H0 will be rejected if 2.060t > . d For a left-tailed test with .01α = and 7df = , H0 will be rejected if 2.998t < − . 10.3 a The p-value for a two-tailed test is defined as ( ) ( )-value 2.43 2 2.43p P t P t= > = > so that

( ) 12.43 -value2

P t p> =

Refer to Table 4, Appendix I, with 12df = . The exact probability, ( )2.43P t > is unavailable; however, it is evident that falls between 2.43t = .025 2.179t = and .01 2.681t = . Therefore, the area to the right of

must be between .01 and .025. Since 2.43t =

1.01 -value .0252

p< <

the p-value can be approximated as

.02 -value .05p< <

237

b For a right-tailed test, ( )-value 3.21p P t= > with 16df = . Since the value is larger than , the area to its right must be less than .005 and you can bound the p-value as

3.21t =

.005 2.921t = -value .005p <

c For a two-tailed test, ( ) ( )-value 1.19 2 1.19p P t P t= > = > , so that ( ) 11.19 -value2

P t p> = . From

Table 4 with , is smaller than 25df = 1.19t = .10 1.316t = so that

1 -value .10 and -value .202

p p> >

d For a left-tailed test, ( ) ( )-value 8.77 8.77p P t P t= < − = > with 7df = . Since the value is larger than , the area to its right must be less than .005 and you can bound the p-value as

8.77t =

.005 3.499t = -value .005p < 10.4 a The stem and leaf plot is shown below. Notice the mounded shape of the data, which justifies the

normality assumption. Stem-and-Leaf Display: Scores

Stem-and-leaf of Scores N = 20 Leaf Unit = 1.0 1 5 7 2 6 2 5 6 578 8 7 123 (4) 7 5567 8 8 244 5 8 669 2 9 13

b Using the formulas given in Chapter 2 or your scientific calculator, calculate

1533 76.6520

ixx

n∑

= = =

( ) ( )2 22

2

1533119419

20 100.7658 and 100.7658 10.03821 19

ii

xx

ns sn

∑∑ − −

= = = = =−

c Small sample confidence intervals are quite similar to their large sample counterparts; however, these

intervals must be based on the t-distribution. Thus, the confidence interval for the single population mean described in this exercise will be

2sx tnα±

where 2tα is a value of t (Table 4) based on 1df n= − degrees of freedom that has area 2α to its right. For this exercise, 20, 76.65, 10.0382n x s= = = and with .025t 1 19n − = degrees of freedom is .025 2.093t = . Hence the 95% confidence interval is

210.038276.65 2.093 76.65 4.70

20sx tnα± ⇒ ± ⇒ ±

or 71.95 81.35µ< < . Intervals constructed using this procedure will enclose µ 95% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ .

10.5 a Using the formulas given in Chapter 2, calculate 70.5ix∑ = and . Then 2 499.27ix∑ =

70.5 7.0510

ixx

n∑

= = =

238

( ) ( )2 22

2

70.5499.27

10 .249444 and .49941 9

ii

xx

ns sn

∑∑ − −

= = = =−

b With , the appropriate value of t is 1 9df n= − = .01 2.821t = (from Table 4) and the 99% upper one-sided confidence bound is

.01.2494447.05 2.821 7.05 .446

10sx tn

+ ⇒ + ⇒ +

or 7.496µ < . Intervals constructed using this procedure will enclose µ 99% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ . c The hypothesis to be tested is

0 aH : 7.5 versus H : 7.5µ µ= < and the test statistic is

7.05 7.5 2.849.249444

10

xts n

µ− −= = = −

The rejection region with .01α = and 1 9n − = degrees of freedom is located in the lower tail of the t-distribution and is found from Table 4 as .01 2.821t t< − = − . Since the observed value of the test statistic falls in the rejection region, H0 is rejected and we conclude that µ is less than 7.5.

d Notice that the 99% upper one-sided confidence bound for µ does not include the value 7.5µ = . This would confirm the results of the hypothesis test in part c, in which we concluded that µ is less than 7.5.

10.6 a Using the formulas given in Chapter 2, calculate 12.55ix∑ = and . Then 2 13.3253ix∑ =

12.55 .89614

ixx

n∑

= = =

( ) ( )2 22 12.55

13.325314 .3995

1 13

ii

xx

nsn

∑∑ − −

= =−

=

With , the appropriate value of t is 1 13df n= − = .025 2.16t = (from Table 4) and the 95% confidence interval is

.025.3995.896 2.16 .896 .231

14sx tn

± ⇒ ± ⇒ ±

or .665 1.127µ< < . Intervals constructed using this procedure will enclose µ 95% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ .

b Calculate 4.9 1.2254

ixx

n∑

= = = and

( ) ( )2 22 4.9

6.00584 .0332

1 3

ii

xx

nsn

∑∑ − −

= =−

=

and the 95% confidence interval is

.025.03321.225 3.182 1.225 .053

4sx tn

± ⇒ ± ⇒ ± or 1.172 1.278µ< < .

The interval is narrower than the interval in part a, even though the sample size is smaller, because the data is so much less variable.

c For white tuna in water,

239

10.24 1.288

ixx

n∑

= = = and

( ) ( )2 22 10.24

13.2358 .1351

1 7

ii

xx

nsn

∑∑ − −

= =−

=


.025.13511.28 2.365 1.28 .113

8sx tn

± ⇒ ± ⇒ ± or 1.167 1.393µ< < .

For light tuna in oil,

12.62 1.14711

ixx

n∑

= = = and

( ) ( )2 22 12.62

19.082811 .6785

1 10

ii

xx

nsn

∑∑ − −

= =−

=


.025.67851.147 2.228 1.147 .456

11sx tn

± ⇒ ± ⇒ ± or .691 1.603µ< < .

Cost

White WaterWhite OilLight WaterLight Oil

1.4

1.3

1.2

1.1

1.0

0.9

0.8

The plot of the four treatment means shows substantial differences in variability. The cost of light tuna in water appears to be the lowest, and quite difference from either of the white tuna varieties.

10.7 Similar to previous exercises. The hypothesis to be tested is 0 aH : 5 versus H : 5µ µ= <

Calculate 29.6 4.9336

ixx

n∑

= = =

( ) ( )2 22

2

29.6146.12

6 .01867 and .13661 5

ii

xx

ns sn

∑∑ − −

= = = =−

The test statistic is

4.933 5 1.195.1366

6

xts n

µ− −= = = −

The critical value of t with .05α = and 1 5n − = degrees of freedom is .05 2.015t = and the rejection region is . Since the observed value does not fall in the rejection region, H2.015t < − 0 is not rejected. There is no evidence to indicate that the dissolved oxygen content is less than 5 parts per million.

10.8 Calculate 608 60.810

ixx

n∑

= = =

240

( ) ( )2 22

2

60837538

10 63.5111 and 7.96941 9

ii

xx

ns sn

∑∑ − −

= = = =−

The 95% confidence interval based on 9df = is

.0257.969460.8 2.262 60.8 5.701

10sx tn

± ⇒ ± ⇒ ±

or 55.099 66.501µ< < . 10.9 a Similar to previous exercises. The hypothesis to be tested is 0 aH : 100 versus H : 100µ µ= <

Calculate 1797.095 89.8547520

ixx

n∑

= = =

( ) ( )2 22

2

1797.095165,697.7081

20 222.1150605 and 14.90351 19

ii

xx

ns sn

∑∑ − −

= = = =−


89.85475 100 3.04414.9035

20

xts n

µ− −= = = −

The critical value of t with .01α = and 1 19n − = degrees of freedom is .01 2.539t = and the rejection region is . The null hypothesis is rejected and we conclude that 2.539t < − µ is less than 100 DL.

b The 95% upper one-sided confidence bound, based on 1 19n − = degrees of freedom, is

.0514.9035251189.85475 2.539 98.316

20sx tn

µ+ ⇒ + ⇒ <

This confirms the results of part a in which we concluded that the mean is less than 100 DL. 10.10 a Notice that the stem and leaf plot is fairly mound-shaped, with the peak only slightly to the right of

center and with a small gap. For a small sample of size 18n = , these slight deviations from the normal model are not sufficient to cause us to conclude that there is a problem with the normality assumption.

b Calculate 349 19.38918

ixx

n∑

= = =

( ) ( )2 22

2

3497195

18 25.192810 and 5.019241 17

ii

xx

ns sn

∑∑ − −

= = = =−

c The 95% confidence interval with 17df = is

.0255.0192419.389 2.110 19.389 2.496

18sx tn

± ⇒ ± ⇒ ±

or 16.893 21.885µ< < .

10.11 Calculate 37.82 3.78210

ixx

n∑

= = =

( ) ( )2 22

2

37.82143.3308

10 .03284 and .18121 9

ii

xx

ns sn

∑∑ − −

= = =−

=


241

.025.18123.782 2.262 3.782 .130

10sx tn

± ⇒ ± ⇒ ±

or 3.652 3.912µ< < . 10.12 The problem of selecting a proper sample size to achieve a given bound is now complicated by the fact that

the t value, which is used in calculating the correct sample size, changes as the value of n changes. In Chapter 8, the procedure was to choose n so that the half-width of the ( )100 1 %α− confidence interval was less than some given bound, B. That is,

2 (std error of estimator) Bzα × ≤ Now, the inequality to be solved is 2 (std error of estimator) Btα × ≤

and the 2tα value must be based on 1n − degrees of freedom. Since n is unknown, the procedure is as follows: 1 Calculate n using 2zα instead of 2tα . If the value for n is large (that is, ), this sample size will

be valid. 30n ≥

2 If the value for n is small, we are not justified in using the value 2zα . This value must be replaced by the appropriate t-value with 1n − degrees of freedom. If the inequality holds, the sample size is valid; if not, it is necessary to pick larger values of n until the inequality will hold. This repetitive procedure is usually not necessary, because the 2zα value will usually yield a satisfactory approximation to the required sample size.

In this exercise, we want to estimate µ to within .06 with probability .95. Hence, the following inequality must hold:

.025 .06stn≤

Consider the sample size obtained by replacing with . .025t .025z

.18121.96 .06 5.9192 35.04 or 36n nn

≤ ⇒ ≥ ⇒ ≥ =n

Since this sample size is greater than 30, the sample size 36n = is valid. 10.13 a The hypothesis to be tested is 0 aH : 25 versus H : 25µ µ= < The test statistic is

0 20.3 25 4.31521

xt

s nµ− −

= = = −

The critical value of t with .05α = and 1 20n − = degrees of freedom is .05 1.725t = and the rejection region is . Since the observed value does falls in the rejection region, H1.725t < − 0 is rejected, and we conclude that pre-treatment mean is less than 25. b The 95% confidence interval based on 20df = is

.0257.426.6 2.086 26.6 3.3721

sx tn

± ⇒ ± ⇒ ±

or 23.23 29.97µ< < . c The pre-treatment mean looks considerably smaller than the other two means.

10.14 Calculate 78.6 5.2415

ixx

n∑

= = =

242

( ) ( )2 22

2

78.6412.22

15 .02543 and .159461 14

ii

xx

ns sn

∑∑ − −

= = = =−


.025.159465.24 2.145 5.24 .088

15sx tn

± ⇒ ± ⇒ ±

or 5.152 5.328µ< < . 10.15 a The t test of the hypothesis 0 aH : 1 versus H : 1µ µ= ≠ is not significant, since the -value .113p = associated with the test statistic

1.05222 1 1.64/ .16565 / 27

xts n

µ− −= = =

is greater than .10. There is insufficient evidence to indicate that the mean weight per package is different from one-pound.

b In fact, the 95% confidence limits for the average weight per package are

.025.165651.05222 2.056 1.05222 .06554

27sx tn

± ⇒ ± ⇒ ±

or .98668 1.11776µ< < . These values agree (except in the last decimal place) with those given in the printout. Remember that you used the rounded values of x and s from the printout, causing a small rounding error in the results.

10.16 a Answers will vary. A typical histogram generated by Minitab shows that the data are approximately

mound-shaped.

Serum-Cholesterol

Freq

uenc

y

350300250200150

16

14

12

10

8

6

4

2

0

b Calculate 12348 246.9650

ixx

n∑

= = =

( ) ( )2 22

2

123483,156,896

50 2192.52898 and 46.82441 49

ii

xx

ns sn

∑∑ − −

= = = =−

Table 4 does not give a value of t with area .025 to its right. If we are conservative, and use the value of t with df = 29, the value of t will be .025 2.045t = , and the approximate 95% confidence interval is

.02546.8244246.96 2.045 246.96 13.54

50sx tn

± ⇒ ± ⇒ ±

or 233.42 260.50µ< < .

243

10.17 Refer to Exercise 10.16. If we use the large sample method of Chapter 8, the large sample confidence interval is

.02546.8244246.96 1.96 246.96 12.98

50sx zn

± ⇒ ± ⇒ ±

or 233.98 259.94µ< < . The intervals are fairly similar, which is why we choose to approximate the

sampling distribution of /

xs n

µ− with a z distribution when . 30n >

10.18 The degrees of freedom for s2, the pooled estimator of 2σ are 1 2 2n n+ − . a 1 2 2 16 18 2 22n n+ − = + − = b 1 2 2 10 12 2 20n n+ − = + − = c 1 2 2 15 3 2 16n n+ − = + − =

10.19 a ( ) ( ) ( ) ( )2 2

1 1 2 22

1 2

1 1 9 3.4 3 4.93.775

2 10 4 2n s n s

sn n

− + − += =

+ − + −=

b ( ) ( ) ( ) ( )2 2

1 1 2 22

1 2

1 1 11 18 20 2321.2258

2 12 21 2n s n s

sn n

− + − += =

+ − + −=

10.20 When the actual data are given and s2 must be calculated, the calculation is done by 1 Using your scientific calculator to first obtain s1 and s2 and substituting into the formula for s2, OR 2 Using the computing formula for s2, noting that

( ) ( )22 2 i

i i

xx x x

n∑

∑ − = ∑ −

Hence, for the pooled estimator,

( ) ( )2 21 22 2

1 22 1 2

1 2 2

i ii i

x xx x

n ns

n n

∑ ∑∑ − +∑ −

=+ −

The preliminary calculations are shown below: Sample 1 Sample 2 ∑ = 1 28ix 2 43ix∑ = 2

1 242ix∑ = 22 411ix∑ =

n = 1 4 2 5n = Then

( ) ( )2 2

2

28 43242 411 46 41.24 5 12.45714286

4 5 2 7s

− + − += = =

+ −

b A 90% confidence interval for ( )1 2µ µ− is given as

( )

( )

( )

21 2 .05

1 2

1 2

1 1

1 17 8.6 1.895 12.45714 5

1.6 4.487 or 6.087 2.887

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

− ± − < − <

c As in Chapter 9, the hypothesis to be tested is

244

0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − <

The test statistic, under the assumption that 2 21 2σ σ σ= = is calculated using the pooled value of s2 in the t-

statistic shown below:

( )1 2

2

1 2

0 7 8.6 .6761 11 1 12.45714 5

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

−

The rejection region is one-tailed, based on 1 2 2 7df n n= + − = degrees of freedom. With .05α = from Table 4, the rejection region is . Since the observed value, .05 1.895t t< − = − .676t = − does not fall in the rejection region, H0 is not rejected. We do not have sufficient evidence to indicate that 1 2 0µ µ− < .

10.21 a The hypothesis to be tested is: 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ b The rejection region is two-tailed, based on 1 2 2 16 13 2 27df n n= + − = + − = degrees of freedom. With

.01α = , from Table 4, the rejection region is .005 2.771t t> = .

c The pooled estimator of 2σ is calculated as

( ) ( ) ( ) ( )2 2

1 1 2 22

1 2

1 1 15 4.8 12 5.95.2889

2 16 13 2n s n s

sn n

− + − += =

+ − + −=

and the test statistic is

( )1 2

2

1 2

0 34.6 32.2 2.7951 11 1 5.2889

16 13

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

d The p-value is

( ) ( )-value 2.795 2 2.795p P t P t= > = > , so that ( ) 12.795 -value2

P t p> = .

From Table 4 with , is greater than the largest tabulated value ( ). Therefore, the area to the right of must be less than .005 so that

27df = 2.795t = .005 2.771t =2.795t =

1 -value .005 and -value .012

p p< <

e Comparing the observed to the critical value 2.795t = .005 2.771t = or comparing the p-value ( < .01) to

.01α = , H0 is rejected and we conclude that 1 2µ µ≠ . 10.22 Refer to Exercise 10.21. A 99% confidence interval for ( )1 2µ µ− is given as

( )

( )

( )

21 2 .005

1 2

1 2

1 1

1 134.6 32.2 2.771 5.288916 13

2.4 2.380 or .020 4.780

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± < − <

10.23 a If you check the ratio of the two variances using the rule of thumb given in this section you will find:

( )( )

22

2 2

4.67larger 1.36smaller 4.00

ss

= =

which is less than three. Therefore, it is reasonable to assume that the two population variances are equal. b From the Minitab printout, the test statistic is .06t = with -value .95p = .

245

c The value of 4.38s = is labeled “Pooled StDev” in the printout, so that . ( )22 4.38 19.1844s = =

d Since the is greater than .10, the results are not significant. There is insufficient evidence to indicate a difference in the two population means.

-value .95p =

e A 95% confidence interval for ( )1 2µ µ− is given as

( )

( )

( )

21 2 .025

1 2

1 2

1 1

1 129 28.86 2.201 19.18446 7

.14 5.363 or 5.223 5.503

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

Since the value 1 2 0µ µ− = falls in the confidence interval, it is possible that the two population means are the same. There insufficient evidence to indicate a difference in the two population means.

10.24 a If the antiplaque rinse is effective, the plaque buildup should be less for the group using the antiplaque

rinse. Hence, the hypothesis to be tested is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > The pooled estimator of 2σ is calculated as

( ) ( ) ( ) ( )2 22 2

1 1 2 22

1 2

1 1 6 .32 6 .32.1024

2 7 7 2n s n s

sn n

− + − += =

+ − + −=


( )1 2

2

1 2

0 1.26 .78 2.8061 11 1 .10247 7

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The rejection region is one-tailed, based on 1 2 2 12n n+ − = degrees of freedom. With .05α = , from Table 4, the rejection region is and H.05 1.782t t> = 0 is rejected. There is evidence to indicate that the rinse is effective.

b The p-value is ( )-value 2.806p P t= > From Table 4 with , is between two tabled entries 12df = 2.806t = .005 3.055t = and , we can

conclude that .01 2.681t =

.005 -value .01p< < 10.25 a The hypothesis to be tested is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ From the Minitab printout, the following information is available:

( )( )

221 1

222 2

.896 .400 14

1.147 .679 11

x s n

x s n

= =

= =

1

2

=

=


( )1 2

2

1 2

01.16

1 1

x xt

sn n

− −= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

−

The rejection region is two-tailed, based on 1 2 2 23n n+ − = degrees of freedom. With .05α = , from Table 4, the rejection region is .025 2.069t t> = and H0 is not rejected. There is not enough evidence to indicate a difference in the population means.

246

b It is not necessary to bound the p-value using Table 4, since the exact p-value is given on the printout as P-Value = .260.

c If you check the ratio of the two variances using the rule of thumb given in this section you will find:

( )( )

22

2 2

.679larger 2.88smaller .400

ss

= =

which is less than three. Therefore, it is reasonable to assume that the two population variances are equal. 10.26 a The hypothesis to be tested is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ where 1µ is the average compartment pressure for runners, and 2µ is the average compartment pressure for

cyclists. The pooled estimator of 2σ is calculated as

( ) ( ) ( ) ( )2 22 2

1 1 2 22

1 2

1 1 9 3.92 9 3.9815.6034

2 18n s n s

sn n

− + − += =

+ −=


( )1 2

2

1 2

0 14.5 11.1 1.921 11 1 15.6034

10 10

x xt

sn n

− − −= =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

The rejection region is two-tailed, based on 18df = degrees of freedom. With .05α = , from Table 4, the rejection region is .025 2.101t t> = . We do not reject H0; there is insufficient evidence to indicate a difference in the means. b Calculate

( ) ( ) ( ) ( )2 22 2

1 1 2 22

1 2

1 1 9 3.49 9 4.9518.3413

2 18n s n s

sn n

− + − += =

+ −=

)

A 95% confidence interval for ( 1 2µ µ− is given as

( )

( )

( )

21 2 .025

1 2

1 2

1 1

1 112.2 11.5 2.101 18.341310 10

.7 4.02 or 3.32 4.72

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

c Check the ratio of the two variances using the rule of thumb given in this section:

( )( )

22

2 2

16.9larger 14.29smaller 4.47

ss

= =

which is greater than three. Therefore, it is not reasonable to assume that the two population variances are equal. You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom.

10.27 a Check the ratio of the two variances using the rule of thumb given in this section:

2

2

larger 2.78095 16.22.17143smaller

ss

= =

which is greater than three. Therefore, it is not reasonable to assume that the two population variances are equal. b You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom for testing 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠


247

( )1 2

2 21 2

1 2

0 3.73 4.8 2.4122.78095 .17143

15 15

x xt

s sn n

− − −= = = −

++

with

( )

22 21 2

21 22 22 2

1 2

1 2

1 2

.185397 .011428715.7

.002455137 .00000933

1 1

s sn n

dfs sn n

n n

⎛ ⎞+⎜ ⎟ +⎝ ⎠= =

+⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

− −

=

With , the p-value for this test is bounded between .02 and .05 so that H15df ≈ 0 can be rejected at the 5% level of significance. There is evidence of a difference in the mean number of uncontaminated eggplants for the two disinfectants.

10.28 a Use your scientific calculator or the computing formulas to find:

2

1 1 12

2 2 2

.0125 .000002278 .001509

.0138 .000003733 .001932

x s s

x s s

= = =

= = =

Since the ratio of the variances is less than 3, you can use the pooled t test, calculating

( ) ( ) ( ) ( )2 2

1 1 2 22

1 2

1 1 9 .000002278 9 .000003733.000003006

2 18n s n s

sn n

− + − += = =

+ −


( )1 2

22

1 2

0 .0125 .0138 1.681 11 1

10 10

x xt

ssn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

−

For a two-tailed test with , the p-value can be bounded using Table 4 so that 18df =

1.05 -value .10 or .10 -value .202

p p< < < <

Since the p-value is greater than .10, 0 1 2H : 0µ µ− = is not rejected. There is insufficient evidence to indicate that there is a difference in the mean titanium contents for the two methods.


( )

( )

( )

21 2 .025

1 2

2

1 2

1 1

1 1.0125 .0138 2.10110 10

.0013 .0016 or .0029 .0003

x x t sn n

s

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

− ± − < − <

Since 1 2 0µ µ− = falls in the confidence interval, the conclusion of part a is confirmed. This particular data set is very susceptible to rounding error. You need to carry as much accuracy as possible to obtain accurate results.

248

10.29 a The Minitab stem and leaf plots are shown below. Notice the mounded shapes which justify the assumption of normality.

Stem-and-Leaf Display: Generic, Sunmaid Stem-and-leaf of Generic N = 14 Stem-and-leaf of Sunmaid N = 14 Leaf Unit = 0.10 Leaf Unit = 0.10 1 24 0 1 22 0

4 25 000 1 23 (5) 26 00000 5 24 0000

5 27 00 7 25 00 3 28 000 7 26

7 27 0 6 28 0000 2 29 0 1 30 0

b Use your scientific calculator or the computing formulas to find:

2

1 1 12

2 2 2

26.214 1.565934 1.251

26.143 5.824176 2.413

x s s

x s s

= =

= =

=

=

Since the ratio of the variances is greater than 3, you must use the unpooled t test with Satterthwaite’s approximate df.

22 21 2

1 22 22 2

1 2

1 2

1 2

19

1 1

s sn n

dfs sn n

n n

⎛ ⎞+⎜ ⎟

⎝ ⎠= ≈⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

− −

For testing 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ , the test statistic is

( )1 2

2 21 2

1 2

0 26.214 26.143 .101.565934 5.824176

14 14

x xt

s sn n

− − −= =

++

=

For a two-tailed test with , the p-value can be bounded using Table 4 so that 19df =

1 -value .10 or -value .202

p p> >

Since the p-value is greater than .10, 0 1 2H : 0µ µ− = is not rejected. There is insufficient evidence to indicate that there is a difference in the mean number of raisins per box.

10.30 a The hypothesis of interest is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > and the preliminary calculations are as follows: Sample 1 (Above) Sample 2 (Below) ∑ = 1 25ix 2 24.3ix∑ = 2

1 125.1ix∑ = 22 118.15ix∑ =

n = 1 5 2 5n = Then

249

( ) ( )

( ) ( )

2 21 22 2

1 22 1 2

1 22 2

2

25 24.3125.1 118.15 .1 .0525 5 .019

5 5 2 8

i ii i

x xx x

n ns

n n

∑ ∑∑ − +∑ −

=+ −

− + − += =

+ −=

Also, 125 55

x = = and 224.3 4.86

5x = =


( )1 2

2

1 2

0 5 4.86 1.6061 11 1 .0195 5

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

For a one-tailed test with and 8df = .05α = the rejection region is , and H.05 1.86t t> = 0 is not rejected. There is insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above.


( )

( )

( )

21 2 .025

1 2

1 2

1 1

1 15 4.86 2.306 .0195 5

.14 .201 or .061 .341

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

10.31 a If swimmer 2 is faster, his(her) average time should be less than the average time for swimmer 1.

Therefore, the hypothesis of interest is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > and the preliminary calculations are as follows: Swimmer 1 Swimmer 2 ∑ = 1 596.46ix 2 596.27ix∑ = 2

1 35576.6976ix∑ = 22 35554.1093ix∑ =

1 10n = 2 10n = __________________________________ Then

( ) ( )

( ) ( )

2 21 22 2

1 22 1 2

1 22 2

2

596.46 596.2735576.6976 35554.1093

10 10 .031247225 5 2

i ii i

x xx x

n ns

n n

∑ ∑∑ − + ∑ −

=+ −

− + −= =

+ −

Also, 1596.46 59.646

10x = = and 2

596.27 59.62710

x = =


250

( )1 2

2

1 2

0 59.646 59.627 0.241 11 1 .03124722

10 10

x xt

sn n

− − −= =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

For a one-tailed test with , the p-value can be bounded using Table 4 so that 1 2 2 18df n n= + − =-value .10p > , and H0 is not rejected. There is insufficient evidence to indicate that swimmer 2’s average

time is still faster than the average time for swimmer 1. 10.32 Refer to Exercise 10.31. A 95% lower one-sided confidence bound for ( )1 2µ µ− is given as

( )

( )

( )

21 2 .05

1 2

1 2

1 1

1 159.646 59.627 1.734 .0312472210 10

.019 .137 or .118

x x t sn n

µ µ

⎛ ⎞− − +⎜ ⎟

⎝ ⎠

⎛ ⎞− − +⎜ ⎟⎝ ⎠

− − > −

Since the value 1 2 0µ µ− = is in the interval, it is possible that the two means might be equal. We do not have enough evidence to indicate that there is a difference in the means.

10.33 a The hypothesis of interest is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ and the preliminary calculations are as follows: Favre McNabb_______ ∑ = 1 596.46ix 2 596.27ix∑ = 2

1 35576.6976ix∑ = 22 35554.1093ix∑ =

1349 19.38918

x = = 2316 17.55618

x = =

21 22.26143791s = 2

2 25.19281046s = 1 18n = 2 18n = __________________________________ Then

2 22 1 1 2 2

1 2

( 1) ( 1)2

17(22.26143791) 17(25.19281046) 23.7271241818 18 2

n s n ss

n n− + −

=+ −

+= =

+ −


( )1 2

2

1 2

0 19.389 17.556 1.131 11 1 23.72712

18 18

x xt

sn n

− − −= =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

The degrees of freedom for this test are 1 2 2 34df n n= + − = , we estimate the rejection region using a value of t with df = 29 and the rejection region is | . The null hypothesis is not rejected; there is insufficient evidence to indicate that the average number of completed passes is different for Brett Favre and Donovan McNabb.

| 2.045t >

b Again, we estimate the value of t with df = 29 and the 95% confidence interval for ( )1 2µ µ− is given as

251

( )

( )

( )

21 2 .025

1 2

1 2

1 1

1 119.389 17.556 2.045 23.7271218 18

1.833 3.320 or 1.487 5.153

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

Since the value 1 2 0µ µ− = is in the interval, it is possible that the two means might be equal. We do not have enough evidence to indicate that there is a difference in the means.

10.34 The hypothesis of interest is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ and the preliminary calculations are as follows: Island Thorns Ashley Rails_______ ∑ = 1 90.9ix 2 86.6ix∑ = 2

1 1665.17ix∑ = 22 1510.92ix∑ =

190.9 18.18

5x = = 2

86.6 17.325

x = =

1 1.775387s = 2 1.658915s = 1 5n = 2 5n = __________________________________ Then

2 22 1 1 2 2

1 22 2

( 1) ( 1)2

4(1.775387 ) 4(1.658915 ) 2.9519995 5 2

n s n ss

n n− + −

=+ −

+= =

+ −


( )1 2

2

1 2

0 18.18 17.32 .791 11 1 2.9519995 5

x xt

sn n

− − −= =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

The rejection region with .05α = and 5 5 2 8df = + − = is | and the null hypothesis is not rejected. There is insufficient evidence to indicate a difference in the average percentage of aluminum oxide at the two sites.

| 2.306t >

10.35 a The test statistic is

.3 0 2.372.1610

d

d

dt

s nµ− −

= = =

with degrees of freedom. The p-value is then 1 9n − =

( ) ( )2.372 2 2.372P t P t> = > so that ( ) 12.372 -value2

P t p> =

Since the value falls between two tabled entries for 2.372t = 9df = ( .025 2.262t = and ), you can conclude that

.01 2.821t =

1.01 -value .0252

.02 -value .05

p

p

< <

< <

Since the p-value is less than .05α = , the null hypothesis is rejected and we conclude that there is a difference in the two population means.

252

b A 95% confidence interval for 1 2 dµ µ µ− = is

.025.163 2.262 .3 .28610

dsd t

n± ⇒ ± ⇒ ±

or . ( )1 2.014 .586µ µ< − <

c Using and B = .1, the inequality to be solved is approximately 2 .16ds =

1.96 .1

1.96 .16 7.84 61.47 or 62.1

dsn

n n

≤

≥ = ⇒ ≥ n =

Since this value of n is greater than 30, the sample size, 62n = pairs, will be valid. 10.36 a The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− > >

b The test statistic is

5.7 0 1.51125618

d

d

dt

s nµ− −

= = =

with degrees of freedom. The rejection region is with 1 17n − = .05α = is , and H.05 1.740t t> = 0 is not rejected. We cannot conclude that 0dµ > .

10.37 a It is necessary to use a paired-difference test, since the two samples are not random and independent.

The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− ≠ ≠

The table of differences, along with the calculation of d and 2

ds , is presented below.

id .1 .1 0 .2 –.1 .3id∑ = 2id .01 .01 .00 .04 .01 2 .07id∑ =

.3 .065

idd

n∑

= = = and

( ) ( )2 22

2

.3.07

5 .0131 4

ii

d

dd

nsn

∑∑ − −

= =−

=


.06 0 1.177.013

5

d

d

dt

s nµ− −

= = =

with degrees of freedom. The rejection region with 1 4n − = .05α = is .025 2.776t t> = , and H0 is not rejected. We cannot conclude that the means are different.

b The p-value is ( ) ( ) ( )1.177 2 1.177 2 .10 .20P t P t> = > > =

c A 95% confidence interval for 1 2 dµ µ µ− = is

.025.013.06 2.776 .06 .142

5ds

d tn

± ⇒ ± ⇒ ±

or . ( )1 2.082 .202µ µ− < − <

253

d In order to use the paired-difference test, it is necessary that the n paired observations be randomly selected from normally distributed populations.

10.38 a A paired-difference test is used, since the two samples are not independent (for any given city, Allstate

and 21st Century premiums will be related). b The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− ≠ ≠

where 1µ is the average for Allstate insurance and 2µ is the average cost for 21st Century insurance. The

table of differences, along with the calculation of d and ds , is presented below. City 1 2 3 4 Totals

id −112 −210 −111 39 −394 2id 12544 44100 12321 1521 70486

394 98.54

idd

n∑ −

= = = − and

( ) ( )2 22 394

70,4864 10559 102.75699

1 3

ii

d

dd

nsn

∑∑ − −

= = = =−


98.5 0 1.917102.75699

4

d

d

dt

s nµ− − −

= = = −

with degrees of freedom. The rejection region with 1 3n − = .01α = is .005 5.841t t> = , and H0 is not rejected. There is insufficient evidence to indicate a difference in the average premiums for Allstate and 21st Century.

c ( ) ( )-value 1.917 2 1.917p P t P t= > = > . Since 1.917t = is between .10 .051.638 and 2.353t t= = ,

( )2(.05) -value 2 .10 .10 -value .20p p< < ⇒ < < . d A 99% confidence interval for 1 2 dµ µ µ− = is

.05102.7569998.5 5.841 98.5 300.102

4ds

d tn

± ⇒ − ± ⇒ − ±

or . ( )1 2398.602 201.602µ µ− < − < e The four cities in the study were not necessarily a random sample of cities from throughout the United

States. Therefore, you cannot make valid comparisons between Allstate and 21st Century for the United States in general.

10.39 a The hypothesis of interest is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ for the two independent samples

of runners and cyclists before exercise. Since the ratio of the sample variances is greater than 3, the population variances cannot be assumed to be equal and you must use the unpooled t test with Satterthwaite’s approximate df. Calculate

( )

( ) ( )1 2

2 2 2 21 2

1 2

0 255.63 173.8 1.984115.48 60.69

10 10

x xt

s sn n

− − −= = =

+ +

254

which has a t distribution with

22 21 2

1 22 22 2

1 2

1 2

1 2

13.619 13

1 1

s sn n

dfs sn n

n n

⎛ ⎞+⎜ ⎟

⎝ ⎠= = ≈⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

− −

A two-tailed rejection region is then .025 2.160t t> = and H0 is not rejected. A 95% confidence interval for

1 2µ µ− is given as

( )

( ) ( ) ( )

( )

2 21 2

1 2 .0251 2

2 2

1 2

115.48 60.69255.63 173.8 2.160

10 1081.83 89.11 or 7.28 170.94

s sx x t

n n

µ µ

− ± +

− ± +

± − < − <

b Similar to part a. The hypothesis of interest is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠ for the two independent samples of runners and cyclists after exercise. Since the ratio of the sample variances is greater than 3, you must again use the unpooled t test with Satterthwaite’s approximate df. Calculate

( )

( ) ( )1 2

2 2 2 21 2

1 2

0 284.75 177.1 2.307132.64 64.63

10 10

x xt

s sn n

− − −= = =

+ +


22 21 2

1 22 22 2

1 2

1 2

1 2

13.046 13

1 1

s sn n

dfs sn n

n n

⎛ ⎞+⎜ ⎟

⎝ ⎠= = ≈⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

− −

A two-tailed rejection region is then .025 2.160t t> = and H0 is rejected. A 95% confidence interval for

1 2µ µ− is given as

( )

( ) ( ) ( )

( )

2 21 2

1 2 .0251 2

2 2

1 2

132.64 64.63284.75 177.1 2.160

10 10107.65 100.783 or 6.867 208.433

s sx x t

n n

µ µ

− ± +

− ± +

± < − <

c To test the difference between runners before and after exercise, you use a paired difference test, and the hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− ≠ ≠

It is given that 29.13d = and , so that the test statistic is 21.01ds =

29.13 0 4.3821.01

10

d

d

dt

s nµ− −

= = =

The rejection region with .05α = and 1 9n df .025 2.262t t> = is − = , and H0 is rejected. We can conclude that the means are different.

255

d The difference in mean CPK values for the 10 cyclists before and after exercise uses 3.3d = with . The 95% confidence interval for 6.85ds = 1 2 dµ µ µ− = is

.056.853.3 2.262 3.3 4.90

10ds

d tn

± ⇒ ± ⇒ ±

or . Since the interval contains the value ( )1 21.6 8.2µ µ− < − < 1 2 0µ µ− = , we cannot conclude that there is a significant difference between the means.

10.40 a-b The table of differences, along with the calculation of d and 2

ds , is presented below. Week 1 2 3 4 Totals

di –1.77 –15.03 –23.22 127.05 –67.07

67.07 16.76754

idd

n∑ −

= = = −

( ) ( )2 22

2

67.071499.9047

4 125.1028251 3

ii

d

dd

nsn

∑ −∑ − −

= = =−

and 11.1849ds =

The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− ≠ ≠


16.7675 0 3.0011.1849

4

d

d

dt

s nµ− − −

= = = −

Since with falls between the two tabled values, and , 3.00t = − 1 3df n= − = .025t .05t .05 -value .10p< < for this two tailed test and H0 is not rejected. We cannot conclude that the means are different. c The 99% confidence interval for 1 2 dµ µ µ− = is

.00511.184916.7675 5.841 16.7675 32.666

4ds

d tn

± ⇒ − ± ⇒ − ±

or . ( )1 249.433 15.899µ µ− < − < 10.41 a Each subject was presented with both signs in random order. If his reaction time in general is high,

both responses will be high; if his reaction time in general is low, both responses will be low. The large variability from subject to subject will mask the variability due to the difference in sign types. The paired-difference design will eliminate the subject to subject variability.

b The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− ≠ ≠

The table of differences, along with the calculation of d and 2ds , is presented below.

Driver 1 2 3 4 5 6 7 8 9 10 Totals id 122 141 97 107 37 56 110 146 104 149 1069

1069 106.910

idd

n∑

= = =

( ) ( )2 22

2

1069126,561

10 1364.988891 9

ii

d

dd

nsn

∑∑ − −

= = =−

and 36.9458ds =

256


106.9 0 9.15036.9458

10

d

d

dt

s nµ− −

= = =

Since 9.150t = with is greater than the tabled value , 1 9df n= − = .005t ( )-value 2 .005 .01p < = for this two tailed test and H0 is rejected. We cannot conclude that the means are different. c The 95% confidence interval for 1 2 dµ µ µ− = is

.02536.9458106.9 2.262 106.9 26.428

10ds

d tn

± ⇒ ± ⇒ ±

or . ( )1 280.472 133.328µ µ< − < 10.42 If two measurements are taken on the same person, the measurements are not independent, and a paired

analysis should be used. 10.43 a A paired-difference test is used, since the two samples are not random and independent (at any location,

the ground and air temperatures are related). The hypothesis of interest is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠


Location 1 2 3 4 5 Totals id –.4 –2.7 –1.6 –1.7 –1.5 –7.9

7.9 1.585

idd

n∑ −

= = = −

( ) ( )2 22

2

7.915.15

5 .6671 4

ii

d

dd

nsn

∑ −∑ − −

= =−

= and .8167ds =


1.58 0 4.326.8167

5

d

d

dt

s nµ− − −

= = = −

A rejection region with .05α = and 1 4df n= − = is .025 2.776t t> = , and H0 is rejected at the 5% level of significance. We conclude that the air-based temperature readings are biased.

b The 95% confidence interval for 1 2 dµ µ µ− = is

.025.81671.58 2.776 1.58 1.014

5ds

d tn

± ⇒ − ± ⇒ − ±

or . ( )1 22.594 .566µ µ− < − < − c The inequality to be solved is 2 Bt SEα ≤

We need to estimate the difference in mean temperatures between ground-based and air-based sensors to within .2 degrees centigrade with 95% confidence. Since this is a paired experiment, the inequality becomes

.025 .2dst

n≤

257

With and n represents the number of pairs of observations, consider the sample size obtained by replacing by .

.8167ds =

.025t .025 1.96z =

.81671.96 .2

8.0019 64.03 or 65n

n n

≤

≥ ⇒ = =n

Since the value of n is greater than 30, the use of 2zα for 2tα is justified. 10.44 A paired-difference test is used, since the two samples are not random and independent (within any sample,

the dye 1 and dye 2 measurements are related). The hypothesis of interest is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠


Sample 1 2 3 4 5 6 7 8 9 Totals id 2 1 –1 2 3 –1 0 2 2 10

10 1.119

idd

n∑

= = =

( ) ( )2 22

2

1028

9 2.11111 8

ii

d

dd

nsn

∑∑ − −

= = =−

and 1.452966ds =


1.11 0 2.291.452966

9

d

d

dt

s nµ− −

= = =

A rejection region with .05α = and 1 8df n= − = is .025 2.306t t> = , and H0 is not rejected at the 5% level of significance. We cannot conclude that there is a difference in the mean brightness scores.

10.45 a Use the Minitab printout given in the text below. The hypothesis of interest is 0 aH : 0 H : 0A B A Bµ µ µ µ− = − >


1.4875 0 2.821.49134

8

d

d

dt

s nµ− −

= = =

The p-value shown in the printout is -value .013p = . Since the p-value is less than .05, H0 is rejected at the 5% level of significance. We conclude that assessor A gives higher assessments than assessor B.

b A 95% lower one-sided confidence bound for 1 2 dµ µ µ− = is

.051.491341.4875 1.895 1.4875 .999

8ds

d tn

− ⇒ − ⇒ −

or ( ) . 1 2 .4885µ µ− > c In order to apply the paired-difference test, the 8 properties must be randomly and independently

selected and the assessments must be normally distributed. d Yes. If the individual assessments are normally distributed, then the mean of four assessments will be

normally distributed. Hence, the difference Ax x− will be normally distributed and the t test on the differences is valid as in c.

10.46 The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− ≠ ≠

258

The table of differences is presented below. Use your scientific calculator to find d and ds ,

id 15 15 15 9 16 8 12 8 10 12 9 4 10 4 17 13 4 7 7 10

Calculate 10.25d = , , and the test statistic is 4.051ds =

10.25 0 11.324.051

20

d

d

dt

s nµ− −

= = =

Since 11.32t = with is greater than the largest tabled value , 1 19df n= − = .005t -value .005p <

for this one-tailed test and H0 is rejected. We can conclude that the average recall score is higher when imagery is used. The results are highly significant ( ).01P < .

10.47 A paired-difference analysis must be used. The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− > >


id 3 3 −2 1 −1 3 −1

Calculate .857d = , , and the test statistic is 2.193ds =

.857 0 1.032.193

7

d

d

dt

s nµ− −

= = =

Since 1.03t = with is smaller than the smallest tabled value , 1 6df n= − = .10t -value .10p >

for this one-tailed test and H0 is not rejected. We cannot conclude that the average time outside the office is less when music is piped in.

10.48 It is necessary to test 2 2

0 aH : 15 versus H : 15σ σ= >

This will be done using s2, the sample variance, which is a good estimator for 2σ . Refer to Section 10.6 of the text and notice that the quantity

( ) 2

22

1n sχ

σ−

=

possesses a chi-square distribution in repeated sampling. This distribution is shown below.

259

Notice that the distribution is nonsymmetrical and that the random variable

( ) 2

2

1n sσ−

takes on values commencing at zero (since s2, 2σ and ( )1n − are never negative). The test statistic is

( ) ( )2

220

1 24 21.434.24

15n s

χσ−

= = =

A one-tailed test of an hypothesis is required. Hence, a critical value of 2χ (denoted by 2Cχ ) must be found

such that ( )2 2 .05CP χ χ> =

Indexing 2.05χ with degrees of freedom (see Table 5), the critical value is found to be

(see the above figure). The value of the test statistic does not fall in the rejection region. Hence, H

1 24n − =2.05 36.4151χ =

0 is not rejected . We cannot conclude that the variance exceeds 15. 10.49 For this exercise, and 2 .3214s = 15n = .

A 90% confidence interval for 2σ will be

( ) ( )

( )

2 22

2 22 1 2

1 1n s n s

α α

σχ χ −

− −< <

where 22αχ represents the value of 2χ such that 5% of the area under the curve (shown in the figure above)

lies to its right. Similarly, (21 2αχ − ) will be the 2χ value such that an area .95 lies to its right.

Hence, we have located one-half of α in each tail of the distribution. Indexing 2

.05χ and 2.95χ with

degrees of freedom in Table 5 yields 1 14n − = and 2

.05 23.6848χ = 2.95 6.57063χ =

and the confidence interval is

( ) ( )214 .3214 14 .3214

23.6848 6.57063σ< < or 2.190 .685σ< <

10.50 a Calculate , and 17.7ix∑ = 2 48.95ix∑ = 7n = . Then

260

( ) ( )2 22

2

17.748.95

7 .69904761 6

ii

xx

nsn

∑∑ − −

= = =−

b Indexing 2.025χ and 2

.975χ with 1 6n − = degrees of freedom in Table 5 yields and 2

.025 14.4494χ = 2.975 1.237347χ =


( ) ( )26 .6990476 6 .699047614.4494 1.237347

σ< < or 2.291 3.390σ< <

c It is necessary to test 2 2

0 aH : .8 versus H : .8σ σ= ≠ and the test statistic is

( ) ( )2

220

1 6 .69904765.24

.8n s

χσ−

= = =

The two-tailed rejection region with .05α = and 1 6n − = degrees of freedom is 2 2 2 2

.025 .97514.4494 or 1.237347χ χ χ χ> = < =

and H0 is not rejected. There is insufficient evidence to indicate that 2σ is different from .8. d The p-value is found by approximating ( )2 5.24P χ > and then doubling that value to account for an

equally small value of s2 which might have produced a value of the test statistic in the lower tail of the chi-square distribution. The observed value, , is smaller than in Table 5. Hence, 2 5.24χ = 2

.10 10.6646χ =

( )-value 2 .10 .20p > = 10.51 The hypothesis of interest is 0 aH : .7 versus H : .7σ σ= > or equivalently 2 2

0 aH : .49 versus H : .49σ σ= > Calculate

( ) ( )2 22

2

1036

4 3.66671 3

ii

xx

nsn

∑∑ − −

= = =−


( ) ( )2

220

1 3 3.666722.449

.49n s

χσ−

= = =

The one-tailed rejection region with .05α = and 1 3n − = degrees of freedom is 2 2.05 7.81χ χ> =

and H0 is rejected. There is sufficient evidence to indicate that 2σ is greater than .49. 10.52 Similar to previous exercises. Indexing 2

.05χ and 2.95χ with 1 3n − = degrees of freedom in Table 5 yields

and 2

.05 7.81473χ = 2.95 .351846χ =


( ) ( )2 22

2 2.05 .95

1 1n s n sσ

χ χ− −

< <

261

( ) ( )23 3.6667 3 3.66677.81473 .351846

σ< < or 21.408 31.264σ< <

10.53 a The hypothesis to be tested is 0 aH : 5 H : 5µ µ= ≠

Calculate 19.96 4.994

ixx

n∑

= = =

( ) ( )2 22

2

19.9699.6226

4 .00741 3

ii

xx

nsn

∑∑ − −

= = =−


0 4.99 5 .232.0074

4

xt

s nµ− −

= = = −

The rejection region with .05α = and 1 3n − = degrees of freedom is found from Table 4 as

.025 3.182t t> = . Since the observed value of the test statistic does not fall in the rejection region, H0 is not rejected. There is insufficient evidence to show that the mean differs from 5 mg/cc.

b The manufacturer claims that the range of the potency measurements will equal .2. Since this range is given to equal 6σ , we know that .0333σ ≈ . Then

( )22 20 aH : .0333 .0011 H : .0011σ σ= = >


( ) ( )2

220

1 3 .007420.18

.0011n s

χσ−

= = =

and the one-tailed rejection region with .05α = and 1 3n − = degrees of freedom is 2 2

.05 7.81χ χ> =H0 is rejected; there is sufficient evidence to indicate that the range of the potency will exceed the manufacturer’s claim.

10.54 In order to construct a 95% confidence interval, values of 2

2αχ and (21 2αχ − ) with degrees of

freedom are needed. Using the values for 60 degrees of freedom as approximate values, the confidence interval is approximately

1 63n − =

( ) ( )2 22

2 2.025 .975

1 1n s n sσ

χ χ− −

< <

( ) ( )263 .0063 63 .0063

83.2976 40.4817σ< < or 2.00476 .00980σ< <

10.55 a The force transmitted to a wearer, x, is known to be normally distributed with 800µ = and 40σ = .

Hence,

( ) ( )1000 80001000 5 040

P x P z P z−⎛ ⎞> = > = > ≈⎜ ⎟⎝ ⎠

It is highly improbable that any particular helmet will transmit a force in excess of 1000 pounds. b Since , a large sample test will be used to test 40n = 0 aH : 800 H : 800µ µ= > The test statistic is

0 825 800 3.2622350

40

xt

s nµ− −

= = =

and the rejection region with .05α = is . H1.645z > 0 is rejected and we conclude that 800µ > .

262

10.56 Refer to Exercise 10.55. The hypothesis of interest is 0 aH : 40 H : 40σ σ= > and the test statistic is

( ) ( )2

22 20

1 39 235057.281

40n s

χσ−

= = =

The one-tailed rejection region with .05α = and 1 39n − = degrees of freedom (approximated with 40 degrees of freedom) is , and H2 2

.05 55.7585χ χ> = 0 is rejected. There is sufficient evidence to indicate that σ is greater than 40.

10.57 The hypothesis of interest is 0 aH : 150 H : 150σ σ= < Calculate

( ) ( ) ( )2 22 2 42,812

1 92,305,600 662,232.820

ii

xn s x

n∑

− = ∑ − = − =

and the test statistic is ( ) 2

22 20

1 662,232.8 29.433150

n sχ

σ−

= = = . The one-tailed rejection region with

.01α = and degrees of freedom is , and H1 19n − = 2 2.99 7.63273χ χ< = 0 is not rejected. There is

insufficient evidence to indicate that he is meeting his goal. 10.58 When the assumptions for the F distribution are met, then 2 2

1 2s s possesses an F distribution with and degrees of freedom. Note that df1 1 1df n= − 2 2 1df n= − 1 and df2 are the degrees of freedom associated

with 21s and 2

2s , respectively. The F distribution is non-symmetrical with the degree of skewness dependent on the above-mentioned degrees of freedom. Table 6 presents the critical values of F (depending on the degrees of freedom) such that , respectively. Because right-hand tail areas correspond to an upper-tailed test of an hypothesis, we will always identify the larger sample variance as

( ) for .10,.05,.025,.01 and .005aP F F a a> = =

21s (that is, we will always place the larger sample variance in

the numerator of 2 21 2F s s= ). Hence, an upper-tailed test is implied and the critical values of F will

determine the rejection region. If we wish to test the hypothesis 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ There will be another portion of the rejection region in the lower tail of the distribution. The area to the

right of the critical value will represent only 2α , and the probability of a Type I error is ( )2 2α α= . a In this exercise, the hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ

and the test statistic is 2122

55.7 1.77431.4

sF

s= = = .

The rejection region (two-tailed) will be determined by a critical value of F based on and degrees of freedom with area .025 to its right. That is, from Table 6, . The

observed value of F does not fall in the rejection region, and we cannot conclude that the variances are different.

1 1 1 15df n= − =

2 2 1 19df n= − = 2.62F >

b The student will need to find critical values of F for various levels of α in order to find the approximate p-value. The critical values with 1 15df = and 2 19df = are shown below from Table 6.

α .10 .05 .025 .01 .005 Fα 1.86 2.23 2.62 3.15 3.59

Hence, ( ) ( )-value 2 1.774 2 .10 .20p P F= > > =

263

10.59 Refer to Exercise 10.58. From Table 6, and . The 95% confidence interval for 1 2, 2.62df dfF =

2 1, 2.76df dfF ≈2 21 2σ σ is

( )

2 1

1 2

2 2 21 1 1

,2 2 2,2 2 2

2 21 12 22 2

1

55.7 1 55.7 2.76 or .667 4.89631.4 2.62 31.4

df dfdf df

s sF

Fs sσσ

σ σσ σ

< <

⎛ ⎞ < < < <⎜ ⎟⎝ ⎠

10.60 a The hypothesis of interest is 2 2 2 2

0 1 2 a 1 2H : versus H :σ σ σ= >σ and the test statistic is

2122

18.3 2.3167.9

sF

s= = = .

The rejection region (one-tailed) with .05α = and 1 2 12df df= = degrees of freedom is , and H2.69F > 0 is not rejected. We cannot conclude that the variances are different.

b The critical values of F for various values of α are given below. α .10 .05 .025 .01 .005 Fα 2.15 2.69 3.28 4.16 4.91

Hence, the lies between .05 and .10, and the results are not significant. (-value 2.316p P F= > )

2 10.61 The hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H :σ σ σ= ≠ σ and the test statistic is

2 212 22

71 1.05969

sF

s= = = .

The critical values of F for various values of α are given below using 1 15df = and . 2 14df =α .10 .05 .025 .01 .005 Fα 2.01 2.46 2.95 3.66 4.25

Hence, ( ) ( )-value 2 1.059 2 .10 .20p P F= > > = Since the p-value is so large, H0 is not rejected. There is no evidence to indicate that the variances are

different. 10.62 a The hypothesis of interest is 2 2 2 2

0 1 2 a 1 2H : versus H :σ σ σ= >σ and the test statistic is

2122

92,000 2.48637,000

sF

s= = = .

The rejection region (one-tailed) with .05α = and 1 2 49df df= = degrees of freedom by interpolation in Table 6. The value is roughly halfway between 49,49F 40,40 1.69F = and 60,60 1.53F = ; therefore, we reject H0 if . The observed value of the test statistic falls in the rejection region and we conclude that the “suspect line” possesses a larger variance.

49,49 1.61F F> ≈

b The student must obtain various critical levels of F from Table 6. We “roughly” interpolate as halfway between and .

49,49F

40,40F 60,60Fα .05 .025 .01 .005 Fα 1.61 1.775 1.975 2.13

In any event, ( )-value 2.486 .005p P F= > <

264

10.63 Refer to Exercise 10.62. Noting that , the 90% confidence interval for 1 2 2 1, , 1.61df df df dfF F= ≈ 2 2

1 2σ σ is

( )

2 1

1 2

2 2 21 1 1

,2 2 2,2 2 2

2 21 12 22 2

1

92,000 1 92,000 1.61 or 1.544 4.00337,000 1.61 37,000

df dfdf df

s sF

Fs sσσ

σ σσ σ

< <

⎛ ⎞ < < < <⎜ ⎟⎝ ⎠

10.64 a The assumption of equal variances ( )2 2

1 2σ σ= was made.

b The hypothesis of interest is 2 2 20 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ


2 212 22

.6785 2.88

.3995s

Fs

= = = .

The upper portion of the rejection region with , .05α = , 1 10df = and 2 13df = is and H.025 3.25F F> = 0 is not rejected. There is no reason to believe that the assumption has been violated.

10.65 For each of the three tests, the hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ and the test statistics are

2 212 22

3.98 1.033.92

sF

s= = =

2 212 22

4.95 2.013.49

sF

s= = = and

2 212 22

16.9 14.294.47

sF

s= = =

The critical values of F for various values of α are given below using 1 9df = and . 2 9df =

α .10 .05 .025 .01 .005 Fα 2.44 3.18 4.03 5.35 6.54

Hence, for the first two tests, ( )-value 2 .10 .20p > = while for the last test, ( )-value 2 .005 .01p < = There is no evidence to indicate that the variances are different for the first two tests, but H0 is rejected for

the third variable. The two-sample t-test with a pooled estimate of 2σ cannot be used for the third variable. 10.66 a The hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ and the test statistic is

2122

.273 2.904

.094s

Fs

= = = .

The upper portion of the rejection region with ( )2 .005 .01α = = is (from Table 6) and H.005 6.54F F> = 0 is not rejected. There is insufficient evidence to indicate that the supplier’s shipments differ in variability.

b The 99% confidence interval for 22σ is

( ) ( )2 22 22

22 2.005 .995

1 1n s n sσ

χ χ− −

< <

265

( ) ( )2

2

9 .094 9 .09423.5893 1.734926

σ< < or 22.036 .488σ< <

Intervals constructed in this manner enclose 2

2σ 99% of the time. Hence, we are fairly certain that 22σ is

between .036 and .488. 10.67 A Student’s t test can be employed to test an hypothesis about a single population mean when the sample

has been randomly selected from a normal population. It will work quite satisfactorily for populations which possess mound-shaped frequency distributions resembling the normal distribution.

10.68 As in the case of the single population mean, random samples must be independently drawn from two

populations which possess normal distributions with a common variance, 2σ . Consequently, it is local that information in the two sample variances, 2

1s and 22s , should be pooled in order to give the best estimate of

the common variance, 2σ . In this way, all of the sample information is being utilized to its best advantage. 10.69 Paired observations are used to estimate the difference between two population means in preference to an

estimation based on independent random samples selected from the two populations because of the increased information caused by blocking the observations. We expect blocking to create a large reduction in the standard deviation, if differences do exist among the blocks. Paired observations are not always preferable. The degrees of freedom that are available for estimating

2σ are less for paired than for unpaired observations. If there were no difference between the blocks, the paired experiment would then be less beneficial.

10.70 a The hypothesis to be tested is 0 aH : .05 H : .05µ µ= >


0 .058 .05 2.108.012

10

xt

s nµ− −

= = = .

Critical value approach: The rejection region with .05α = and 1 9df n= − = degrees of freedom is located in the upper tail of the t-distribution and is found from Table 4 as . Since the observed value falls in the rejection region, H

.05 1.833t t> =

0 is rejected and we conclude that µ is greater than .05. p-value approach: The ( )-value 2.108p P t= > . Since the value 2.108t = falls between and , the p-value can be bounded as

.025t .05t

.025 -value .05p< < In any event, the p-value is less than .05 and H0 can be rejected at the 5% level of significance.

10.71 The 90% confidence interval is

.053.411.3 1.746 11.3 1.44017

sx tn

± ⇒ ± ⇒ ±

or 9.860 12.740µ< < . 10.72 Using the formulas given in Chapter 2 or your scientific calculator, calculate

233.29 77.7633

ixx

n∑

= = =

( ) ( )2 22

2

233.2918169.666

3 14.129 and 3.7591 2

ii

xx

ns sn

∑∑ − −

= = = =−

266

The 99% confidence interval is

.0053.75977.763 9.925 77.763 21.540

3sx tn

± ⇒ ± ⇒ ±

or 56.223 99.303µ< < . 10.73 Since it is necessary to determine whether the injected rats drink more water than noninjected rates, the

hypothesis to be tested is 0 aH : 22.0 H : 22.0µ µ= > and the test statistic is

0 31.0 22.0 5.9856.217

xt

s nµ− −

= = = .

Using the critical value approach, the rejection region with .05α = and 1 16n − = degrees of freedom is located in the upper tail of the t-distribution and is found from Table 4 as . Since the observed value of the test statistic falls in the rejection region, H

.05 1.746t t> =

0 is rejected and we conclude that the injected rats do drink more water than the noninjected rats. The 90% confidence interval is

.056.231.0 1.746 31.0 2.62517

sx tn

± ⇒ ± ⇒ ±

or 28.375 33.625µ< < . 10.74 Using the formulas given in Chapter 2 or your scientific calculator, calculate

42.6 4.2610

ixx

n∑

= = =

( ) ( )2 22

2

42.6190.46

10 .998 and .9991 9

ii

xx

ns sn

∑∑ − −

= = =−

=

The 95% confidence interval is then

.025.9994.26 2.262 4.26 .715

10sx tn

± ⇒ ± ⇒ ±

or 3.545 4.975µ< < . 10.75 The hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= <σ and the test statistic is

( )( )

2212 22

28.23.268

15.6s

Fs

= = = .

The critical values of F for various values of α are given below using 1 30df = (since there are no tabled values for ) and . 1 29df = 2 29df =

α .10 .05 .025 .01 .005 Fα 1.62 1.85 2.09 2.41 2.66

Hence, ( )-value 3.268 .005p P F= > < Since the p-value is so small, H0 is rejected. There is evidence to indicate that increased maintenance of the

older system is needed.

267

10.76 The student may use the rounded values for x and s given in the display, or he may wish to calculate x and s and use the more exact calculations for the confidence intervals. The calculations are shown below.

a 1845 184.510

ixx

n∑

= = =

( ) ( )2 22

2

1845344,567

10 462.72221 9

ii

xx

nsn

∑∑ − −

= = =−

21.51 and the 95% confidence interval is 1s =

.02521.5111.845 2.262 184.5 15.4

10sx tn

± ⇒ ± ⇒ ±

or 169.1 199.9µ< < .

b 730 73.010

ixx

n∑

= = =

( ) ( )2 22

2

73053514

10 24.88891 9

ii

xx

nsn

∑∑ − −

= = =−

4.9 and the 95% confidence interval is 89s =

.0254.98973.0 2.262 73.0 3.57

10sx tn

± ⇒ ± ⇒ ±

or 69.43 76.57µ< < .

c 25.42 2.54210

ixx

n∑

= = =

( ) ( )2 22

2

25.4265.8398

10 .135795561 9

ii

xx

nsn

∑∑ − −

= = =−

.3685s = and the 95% confidence interval is

.025.36852.54 2.262 2.54 .26

10sx tn

± ⇒ ± ⇒ ±

or 2.28 2.80µ< < . d No. The relationship between the confidence intervals is not the same as the relationship between the

original measurements. 10.77 From Exercise 10.76, the best estimate for σ is 21.5s = . Then, with B 5= , solve for n in the following

inequality:

.025 5stn≤

which is approximately

( )1.96 21.521.51.96 5 8.4285

71.03 or 72

nn

n n

≤ ⇒ ≥ =

≥ ≥

Since n is greater than 30, the sample size, 72n = , is valid. 10.78 A paired-difference analysis must be used. The hypothesis of interest is

0 1 2 0

a 1 2 a

H : 0 or H : 0H : 0 or H : 0

d

d

µ µ µµ µ µ− = =− < <


id −3 −3 2 −1 −1 1 −3

Calculate 1.1429d = − , , and the test statistic is 2.0354ds =

268

1.1429 0 1.492.0354

7

d

d

dt

s nµ− − −

= = = −

The one-tailed rejection region with 1 6df n= − = is 1.943t < − and H0 is not rejected. There is insufficient evidence to indeicat that the mean reaction time is greater after consuming alcohol.

10.79 Use the computing formulas or your scientific calculator to calculate

322.1 24.77713

ixx

n∑

= = =

( ) ( )2 22

2

322.18114.59

13 11.16191 12

ii

xx

nsn

∑∑ − −

= = =−

and the 95% confidence interval is 3.3409s =

.0253.340924.777 2.179 24.777 2.019

13sx tn

± ⇒ ± ⇒ ±

or 22.578 26.796µ< < . 10.80 a The hypothesis to be tested is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠ The following information is available: 1 27.2x = and 2

1 16.36s = 1 10n = 2 33.5x = and 2

2 18.92s = 2 10n =

Since the ratio of the variances is less than 3, you can use the pooled t test. The pooled estimator of 2σ is calculated as

( ) ( ) ( ) ( )2 2

1 1 2 22

1 2

1 1 9 16.36 9 18.9217.64

2 10 10 2n s n s

sn n

− + − += =

+ − + −=


( )1 2

2

1 2

0 27.2 33.5 3.3541 11 1 17.64

10 10

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

−

Critical value approach: The rejection region is two-tailed, based on 18df = degrees of freedom. With .05α = , from Table 4, the rejection region is .025 2.101t t> = and H0 is rejected. There is sufficient evidence to indicate a difference in the mean absorption rates between the two drugs. b p-value approach: The p-value is ( )2 3.35P t > 4 for a two-tailed test with 18 degrees of freedom. Since exceeds the largest tabled value, 3.354t = .005 2.878t = , we have ( )-value 2 .005 .01p < = Since the p-value is less than .05α = , H0 can be rejected at the 5% level of significance, confirming the results of part a.

c A 95% confidence interval for ( )1 2µ µ− is given as

( )

( )

( )

21 2 .025

1 2

1 2

1 1

1 127.2 33.5 2.101 17.6410 10

6.3 3.946 or 10.246 2.354

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

− ± − < − < −

269

Since the confidence interval does not contain the value 1 2 0µ µ− = , you can conclude that there is a difference in two population means. This confirms the results of Exercise 10.76.

10.81 The inequality to be solved is ( )2 std error of estimator Btα × ≤

In this exercise, it is necessary to estimate the difference in means to within 1 minute with 95% confidence. The inequality is then

2.025

1 2

1 1 1t sn n

⎛ ⎞+ ≤⎜ ⎟

⎝ ⎠

We assume that and from Exercise 10.76. Consider the sample size obtained by replacing by .

1 2n n n= = 2 17.64s =

.025t .025 1.96z =

1 11.96 17.64 1 1.96 35.28

135.53 or 136

nn n

n n

⎛ ⎞+ ≤ ⇒ ≥⎜ ⎟⎝ ⎠

≥ ≥

Since the value of n is greater than 30, the sample size is valid. b Consider the inequality from part a,

22

1 2

1 1 1t sn nα

⎛ ⎞+ ≤⎜ ⎟

⎝ ⎠.

When , this becomes 1 2n n n= =

2 22 2

2 1 2t s n t snα α

⎛ ⎞ ≤ ⇒ ≥⎜ ⎟⎝ ⎠

To reduce the sample size necessary to achieve this inequality, we must reduce the quantity 2

2 2t sα

If we are willing to lower the level of confidence (or equivalently increase the value of α ), the value of 2tα will be smaller. This will decrease the size of n, but at the price of decreased confidence. The only

other quantity in the expression which is not fixed is s2. If this variable can be made smaller, the required sample size will also be reduced. From the definition of s2, it can be seen that s2 includes the variability associated with the difference in drugs A and B as well as the variability in absorption rates among the people in the experiment. The experiment could be run a bit differently by using the same people for both drugs. Each person would receive a dose of drug A and drug B. The difference in absorption rates for the two drugs would be observed within each individual. Such a design would eliminate the variability in absorption rates from person to person. The value of s2 would be reduced and a smaller sample size would be required. The resulting experiment would be analyzed as a paired-difference experiment, as discussed in Section 10.5.

10.82 a To check for equality of variances, we can use the rule of thumb or the formal F-test. First, calculate

2

21

1507022,921,51610 23, 447.3333

9s

−= = and

2

22

1091311,963,75310 6044.0111

9s

−= =

The ratio

2

2larger 23447.3333 3.88

6044.0111smaller sFs

= = =

is greater than 3, which is an indication that the variances are unequal. Using a two-sided F-test with , the p-value for the test is bounded as 1 2 9df df= =

270

1.025 ( -value) .052

.05 -value .10

p

p

< <

< <

Since the p-value is close to .05, we choose to assume that the variances are not equal, and use Satterthwaite’s approximation.

b The hypothesis of interest is 0 1 2 a 1 2H : 300 versus H : 300µ µ µ µ− = − > for the two independent samples of male and female pheasants. Calculate

( )1 2

2 21 2

1 2

0 1507 1091.3 300 2.1323447.3333 6044.0111

10 10

x xt

s sn n

− − − −= = =

++


22 21 2

1 22 22 2

1 2

1 2

1 2

13.35 13

1 1

s sn n

dfs sn n

n n

⎛ ⎞+⎜ ⎟

⎝ ⎠= = ≈⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

− −

A one-tailed rejection region is then and H.05 1.771t t> = 0 is rejected. There is sufficient evidence to indicate that the average weight of male pheasants exceeds that of females by more than 300 grams.

10.83 a The range of the first sample is 47 while the range of the second sample is only 16. There is probably a

difference in the variances. b The hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ

Calculate

( )2

21

838177,294

4 577.66673

s−

= =

( )2

22

1074192,394

6 29.65

s−

= =


2122

577.6667 19.51629.6

sF

s= = = .

The critical values with and 1 3df = 2 5df = are shown below from Table 6. α .10 .05 .025 .01 .005 Fα 3.62 5.41 7.76 12.06 16.53

Hence, ( ) ( )-value 2 19.516 2 .005 .01p P F= > < = 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 Student’s t test requires the assumption of equal variance, it would be inappropriate in this

instance. You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom. 10.84 a Use the computing formulas or your scientific calculator to calculate

.2753 .0275310

ixx

n∑

= = =

( )2

2

.2753.00758155

10 .0000002829

s−

= =

.00053135s = and the 99% confidence interval is

271

.005.00053135.02753 3.25 .02753 .00055

10sx tn

± ⇒ ± ⇒ ±

or .02698 .02808µ< < . b Intervals constructed using this procedure will enclose µ 99% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ .

c The sample must be randomly selected, or at least behave as a random sample from the population of interest. For the chemist performing the analysis, this means that he or she must be certain that there is no unknown factor which is affecting the measurements, thus causing a biased sample rather than a representative sample.

10.85 a The leaf measurements probably come from mound-shaped, or approximately normal populations, since

their length, width, thickness, and so on can be thought of as being a composite sum of many factors which affect their growth (see the Central Limit Theorem). The values of the sample variances are not very different, and we would not question the assumption of equal variances. Finally, since the plants were all given the same experimental treatment, they can be considered random and independent samples within a treatment group.

b The hypothesis to be tested is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠


( ) ( ) ( ) ( )2 22 2

1 1 2 22

1 2

1 1 15 43 14 41.71795.8434

2 29n s n s

sn n

− + − += = =

+ −


( )1 2

2

1 2

0 128 78.7 3.2371 11 1 1795.8434

16 15

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The two-tailed p-value with can be bounded as 29df =

( )-value 2 .005 .01p < = and H0 is rejected. There is a difference in the means. c The hypothesis to be tested is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠

Since the ratio of the variances is slightly greater than 3, you should use the pooled t test with the test statistic

( )

( ) ( )1 2

2 2 2 21 2

1 2

0 46.8 8.1 60.362.21 1.2616 15

x xt

s sn n

− − −= = =

+ +

The p-value is two-tailed, based on

22 21 2

1 22 22 2

1 2

1 2

1 2

25

1 1

s sn n

dfs sn n

n n

⎛ ⎞+⎜ ⎟

⎝ ⎠= ≈⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

− −

degrees of freedom and is bounded as ( )-value 2 .005 .01p < = The results are highly significant and H0 is rejected. There is a difference in the means.

272

10.86 a The hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ The test statistic is

2122

3.1 2.211.4

sF

s= = = .

The two-sided rejection region with 1 24df = and 2 24df = and .05α = is and the null hypothesis is not rejected. There is insufficient evidence to indicate a difference in the precision of the two machines.

.025 2.27F F> =

b The 95% confidence interval for the ratio of the two variances is

( )

2 1

1 2

2 2 21 1 1

,2 2 2,2 2 2

2 21 12 22 2

1

3.1 1 3.1 2.27 or .975 5.031.4 2.27 1.4

df dfdf df

s sF

Fs sσσ

σ σσ σ

< <

⎛ ⎞ < < < <⎜ ⎟⎝ ⎠

Since the possible values for 2 21 2/σ σ includes the value 1, it is possible that there is no difference in the

precision of the two machines. This confirms the results of part a. 10.87 A paired-difference test is used, since the two samples are not random and independent. The hypothesis of

interest is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − > and the table of differences, along with the calculation of d and 2

ds , is presented below. Pair 1 2 3 4 Totals

id –1 5 11 7 22

22 5.54

idd

n∑

= = =

( ) ( )2 22

2

22196

4 251 3

ii

d

dd

nsn

∑∑ − −

= =−

= and 5ds =


5.5 0 2.254

d

d

dt

s nµ− −

= = =

The one-tailed p-value with can be bounded between .05 and .10. Since this value is greater than .10, H

3df =

0 is not rejected. The results are not significant; there is insufficient evidence to indicate that lack of school experience has a depressing effect on IQ scores.

10.88 A paired-difference analysis is used. The differences are shown below. 47, 44, 38, 46, 37, 56, 35, 34 Calculate

337 42.1258

idd

n∑

= = =

( ) ( )2 22

2

33714591

8 56.41071 7

ii

d

dd

nsn

∑∑ − −

= = =−

and 7.5107ds =

and the 95% confidence interval for 1 2 dµ µ µ− = is

.0257.510742.125 2.365 42.125 6.280

8ds

d tn

± ⇒ ± ⇒ ±

or . ( )1 235.845 48.405µ µ< − < 10.89 A paired-difference analysis is used. The hypothesis of interest is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠

273

and the differences are shown below. 156, 447, –3, 42 Calculate

642 160.54

idd

n∑

= = =

( ) ( )2 22 642

2259184 202.3833

1 3

ii

d

dd

nsn

∑∑ − −

= = =−


160.5 0 1.586202.3833

4

d

d

dt

s nµ− −

= = =

The two-tailed p-value with is greater than 3df = ( )2 .10 .20= . Since this value is greater than .05, H0 is not rejected. The results are not significant; there is insufficient evidence to indicate a difference in the average cost of repair for the Honda Civic and the Hyundai Elantra.

10.90 a Because of the mounded shape of the data shown in the stem and leaf plot, you can be assured that the

assumption of normality is valid. b Using the summary statistics given in the printout, the 99% confidence interval is

.005.54625.3953 2.756 5.3953 .2748

30sx tn

± ⇒ ± ⇒ ±

or 5.1205 5.6701µ< < . 10.91 The object is to determine whether or not there is a difference between the mean responses for the two

different stimuli to which the people have been subjected. The samples are independently and randomly selected, and the assumptions necessary for the t test of Section 10.4 are met. The hypothesis to be tested is

0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠ and the preliminary calculations are as follows:

115 1.8758

x = = and 221 2.6258

x = =

( )2

21

1533

8 .696437

s−

= = and

( )2

22

2161

8 .839297

s−

= =


( ) ( )2 2

1 1 2 22

1 2

1 1 4.875 5.875 .76792 14

n s n ss

n n− + − +

= =+ −

=


( )1 2

2

1 2

0 1.875 2.625 1.7121 11 1 .76798 8

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

−

The two-tailed rejection region with .05α = and 14df = is .025 2.145t t> = , and H0 is not rejected. There is insufficient evidence to indicate that there is a difference in means.

10.92 Note that this experiment has been designed differently from the experiment performed in Exercise 10.91.

That is, each of the eight people was subjected to both stimuli (in random order). A paired-difference analysis is used to test the hypothesis

274

0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠ and the table of differences, along with the calculation of d and 2

ds , is presented below. Person 1 2 3 4 5 6 7 8 Totals

id –1 –1 –2 1 –1 –1 0 –1 –6

6 .758

idd

n∑ −

= = = −

( ) ( )2 22

2

610

8 .785711 7

ii

d

dd

nsn

∑ −∑ − −

= = =−

and .88641ds =


.75 0 2.39.88641

8

d

d

dt

s nµ− − −

= = = −

The rejection region with .05α = and 1 7df n= − = is .025 2.365t t> = , and H0 is rejected. There is a significant difference between the two stimuli.

10.93 Refer to Exercise 10.91 and 10.92. For the unpaired design, the 95% confidence interval for ( )1 2µ µ− is

( )

( )

21 2 .025

1 2

1 2

1 1

1 1.75 2.145 .76798 8

.75 .94 or 1.69 0.19

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

− ± − < − <

while for the paired design, the 95% confidence interval is

.025.88641.75 2.365 .75 .74

8ds

d tn

± ⇒ − ± ⇒ − ±

or . Although the width of the confidence interval has decreased slightly, it does not appear that blocking has increased the amount of information by much.

( )1 21.49 .01µ µ− < − < −

10.94 a The analysis is identical to that used in previous exercises. To test

0 1 2 a 1 2H : 0 versus H : 0µ µ µ− = − ≠µ , the test statistic is

( )1 2

2

1 2

09.56

1 1

x xt

sn n

− −= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

with . Since this p-value is very small, H-value .0000p = 0 is rejected. There is evidence of a difference in the means.

b Since there is a difference in the mean strengths for the two kinds of material, the strongest material (A) should be used (all other factors being equal).

10.95 Note that the experiment has been designed so that two cake pans, one containing batter A and one

containing batter B, were placed side by side at each of six different locations in an oven. The two samples are therefore not independent, and a paired-difference analysis must be used. To test

0 2 1 a 2 1H : 0 versus H : 0µ µ µ− = − ≠µ , calculate the differences: –.006, .018, .014, .011, .004, .019

275

Then .06 .016

idd

n∑

= = =

( ) ( )2 22

2

.06.001054

6 .00009081 5

ii

d

dd

nsn

∑∑ − −

= = =−

and .009529ds =


.06 0 2.5706.009529

6

d

d

dt

s nµ− −

= = =

For a two-tailed test with , the p-value is bounded as 5df =

( ) ( )2 .025 -value 2 .05 or .05 -value .10p p< < < < Since the p-value is greater than .10, H0 is not rejected. There is insufficient evidence to indicate a difference between mean densities for batters A and B.

b The 95% confidence interval is

.025.009529.01 2.571 .01 .010

6ds

d tn

± ⇒ ± ⇒ ±

or . ( )1 2.000 .020µ µ< − < 10.96 In order to use the F statistic to test an hypothesis concerning the equality of two population variances, we

must assume that independent random samples have been drawn form two normal populations. 10.97 It is possible to test the null hypothesis 2

0 1 2H : 2σ σ= against any one of three alternative hypotheses: (1) 2 2

a 1 2H :σ σ≠ (2) 2a 1 2H : 2σ σ< (3) 2 2

a 1 2H :σ σ> a The first alternative would be preferred by the manager of the dairy. He does not know anything about

the variability of the two machines and would wish to detect departures from equality of the type 2 21 2σ σ< or 2 2

1 2σ σ> . These alternatives are implied in (1). b The salesman for company A would prefer that the experimenter select the second alternative.

Rejection of the null hypothesis would imply that his machine had smaller variability. Moreover, even if the null hypothesis were not rejected, there would be no evidence to indicate that the variability of the company A machine was greater than the variability of the company B machine.

c The salesman for company B would prefer the third alternative for a similar reason. 10.98 The hypothesis of interest is 2 2 2 2

0 aH : versus H :A B A Bσ σ σ= < σ and the test statistic is

2

2

.065 2.407

.027B

A

sF

s= = = .

The rejection region (one-tailed) will be determined by a critical value of F based on and 1 9df = 2 29df = degrees of freedom, with area .05 to its right. That is, from Table 6, you will reject H0 if . The observed value of F falls in the rejection region, and we conclude that

2.22F >2 2A Bσ σ< .

10.99 a The manufacturer claims that the range of the random variable x (purity of his product) is no more than

2% . In terms of the standard deviation σ , he is claiming that .5σ ≤ since 4 2 .Range 5σ σ≈ = ⇒ = Hence, the hypothesis to be tested is

0 a2 2

0 a

H : .5 H : .5 or equivalently

H : .25 H : .25

σ σ

σ σ

= >

= >

276

Calculate

( )2

2

489.847,982.56

5 .4384

s−

= = and the test statistic is

( ) ( )2

220

1 4 .4387.088

.25n s

χσ−

= = =

The one-tailed p-value with degrees of freedom is 1 4n − = -value .10p > and H0 is not rejected. There is insufficient evidence to contradict the manufacturer’s claim. b Indexing 2

.05χ and 2.95χ with degrees of freedom in Table 5 yields 1 4n − =

and 2

.05 9.48773χ = 2.95 .710721χ =


( ) ( )24 .438 4 .438

9.48773 .710721σ< < or 2.185 2.465σ< <

10.100 To test the hypothesis 0H : 16 versus H : 16aµ µ= < , the test statistic is

0 15.7 16 1.8.5 9

xt

s nµ− −

= = = −

The p-value with 8 degrees of freedom is bounded as .05 -value .10p< < Hence, the null hypothesis is not rejected. There is insufficient evidence to indicate that the mean weight is

less than claimed. 10.101 A paired-difference analysis is used. To test 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > , where 2µ is the mean

reaction time after injection and 1µ is the mean reaction time before injection, calculate the differences

( )2 1x x− : 6, 1, 6, 1

Then 14 3.54

idd

n∑

= = =

( ) ( )2 22

2

1474

4 8.331 3

ii

d

dd

nsn

∑∑ − −

= =−

= and 2.88675ds =


3.5 0 2.4252.88675

4

d

d

dt

s nµ− −

= = =

For a one-tailed test with , the rejection region with 3df = .05α = is , and H.05 2.353t t> = 0 is rejected. We conclude that the drug significantly increases with reaction time.

10.102 Calculate

141.6 5.943

7x = = and 2

112.9 16.1297

x = =

( )2

21

41.6255.96

7 1.456196

s−

= = and

( )2

22

112.91832.11

7 1.865716

s−

= =

277

Since the ratio of the variances is less than 3, you can use the pooled estimator of 2σ calculated as

( ) ( ) ( ) ( )2 2

1 1 2 22

1 2

1 1 6 1.45619 6 1.865711.66095

2 12n s n s

sn n

− + − += =

+ −=

)

A 90% confidence interval for ( 1 2µ µ− is given as

( )

( )

( )

21 2 .05

1 2

1 2

1 1

1 15.943 16.129 1.782 1.660957 7

10.186 1.228 or 11.414 8.958

x x t sn n

µ µ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

− ± − < − < −

10.103 The hypothesis to be tested is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠ and the preliminary calculations are as follows:

1.6 .16

x = = and 2.83 .13836

x = =

( )2

21

.6.0624

6 .000485

s−

= = and

( )2

22

.83.1175

6 .000536675

s−

= =


( ) ( )2 2

1 1 2 22

1 2

1 1 .0024 .00268 .00050832 10

n s n ss

n n− + − +

= = =+ −


( )1 2

2

1 2

0 .0383 2.9451 11 1 .00050836 6

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

−

The p-value for a one-tailed test with 10 degrees of freedom is bounded as .005 -value .010p< < Hence, the null hypothesis H0 is rejected. There is sufficient evidence to indicate that 1 2µ µ< 10.104 a In this exercise, the hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ

and the test statistic is 2122

2.96 1.9221.54

sF

s= = = .

The critical values of F for various levels of α are needed in order to find the approximate p-value. The critical values with and 1 14df = 2 14df = are not found in Table 6. However, they can be approximated as

15,14F and are shown in the following table. α .10 .05 .025 .01 .005 Fα 2.01 2.46 2.95 3.66 4.25

Hence, ( ) ( )-value 2 1.922 2 .10 .20p P F= > > = This is too large to reject H0.

278

b From the table in part a, . The 99% confidence interval for 1 2 2 1, , 4.25df df df dfF F= ≈ 2 2

1 2σ σ is

( )

2 1

1 2

2 2 21 1 1

,2 2 2,2 2 2

2 21 12 22 2

1

11.922 1.922 4.25 or .452 8.16854.25

df dfdf df

s sF

Fs sσσ

σ σσ σ

< <

⎛ ⎞ < < < <⎜ ⎟⎝ ⎠

Intervals constructed using this procedure will enclose the ratio 2 2

1 2σ σ 99% of the time in repeated sampling. Hence, we are fairly confident that the interval, .452 to 8.1685, encloses 2 2

1 2σ σ . 10.105 The underlying populations are ratings and can only take on the finite number of values, 1, 2, …, 9, 10.

Neither population has a normal distribution, but both are discrete. Further, the samples are not independent, since the same person is asked to rank each car design. Hence, two of the assumptions required for the Student’s t test have been violated.

10.106 A paired-difference test is used. To test 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > , where 1µ is the mean

before the safety program and 2µ is the mean after the program, calculate the differences: 7, 6, –1, 5, 6, 1

Then 24 46

idd

n∑

= = =

( )2

2

24148

6 10.45ds−

= = and 3.2249ds =


4 0 3.0383.2249

6

d

d

dt

s nµ− −

= = =

For a one-tailed test with , the rejection region with 5df = .01α = is , and H.01 3.365t t> = 0 is not rejected. We cannot conclude that the drug significantly increases reaction time.

10.107 A paired-difference test is used. To test 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > , calculate the differences: 29, 31, –35, –17, 99, 73, 54 The test statistic is given on the Minitab printout as

1.86d

d

dt

s nµ−

= =

with . Since this value is greater than .10, the results are not significant, and H-value .112p = 0 is not rejected. There is insufficient evidence to indicate a greater mean demand for one of the entrees.

10.108 Indexing 2

.05χ and 2.95χ with degrees of freedom in Table 5 yields 1 19n − =

and 2

.05 30.1435χ = 2.95 10.117χ =


( ) ( )

( ) ( )

2 22

2 2.05 .95

2

1 1

19 39 19 3930.1435 10.117

n s n sσ

χ

σ

− −< <

< <

χ or 224.582 73.243σ< <

279

10.109 The Minitab printout below shows the summary statistics for the two samples:

Descriptive Statistics: Method 1, Method 2 Variable N Mean SE Mean StDev Method 1 5 137.00 4.55 10.17 Method 2 5 147.20 3.29 7.36

Since the ratio of the two sample variances is less than 3, you can use the pooled t test to compare the two methods of measurement, using the remainder of the Minitab printout below:

Two-Sample T-Test and CI: Method 1, Method 2 Difference = mu (Method 1) - mu (Method 2) Estimate for difference: -10.2000 95% CI for difference: (-23.1506, 2.7506) T-Test of difference = 0 (vs not =): T-Value = -1.82 P-Value = 0.107 DF = 8 Both use Pooled StDev = 8.8798

The test statistic is with1.82t = − -value .107p = and the results are not significant. There is insufficient evidence to declare a difference in the two population means.

10.110 .a The hypothesis of interest is 2 2

0 aH : .03 H : .03σ σ= > and the test statistic is

( ) ( )2

220

1 14 .05324.73

.03n s

χσ−

= = =

The one-tailed rejection region with .05α = and 1 14n − = degrees of freedom is , and H

2 2.05 23.6848χ χ> =

0 is rejected. There is sufficient evidence to indicate that 2σ is greater than .03. b Indexing 2

.025χ and 2.975χ with 1 14n − = degrees of freedom in Table 5 yields

and 2

.025 26.1190χ = 2.975 5.62872χ =


( ) ( )214 .053 14 .053

26.1190 5.62872σ< < or 2.0284 .1318σ< <

10.111 a Calculate 68.5 6.8510

ixx

n∑

= = =

( ) ( )2 22

2

68.5478.375

10 1.016667 and 1.00831 9

ii

xx

ns sn

∑∑ − −

= = = =−


.0051.00836.85 3.25 6.85 1.036

10sx tn

± ⇒ ± ⇒ ±

or 5.814 7.886µ< < . b The sample must have been randomly selected from a normal population.

10.112 Calculate 104.9 4.19625

ixx

n∑

= = =

( ) ( )2 22

2

104.9454.81

25 .6104 and .78131 24

ii

xx

ns sn

∑∑ − −

= = = =−

280


.025.78134.196 2.064 4.196 .323

25sx tn

± ⇒ ± ⇒ ±

or 3.873 4.519µ< < . 10.113 The hypothesis to be tested is 0 aH : 280 versus H : 280µ µ= > The test statistic is

358 280 4.575410

xts n

µ− −= = =

The critical value of t with .01α = and 1 9n − = degrees of freedom is .01 2.821t = and the rejection region is . Since the observed value falls in the rejection region, H2.821t > 0 is rejected. There is sufficient evidence to indicate that the average number of calories is greater than advertised.

10.114 a The hypothesis of interest is 2 2 2

0 1 2 a 1 2H : versus H : 2σ σ σ= ≠ σ

and the test statistic is 2 212 22

22 1.2120

sF

s= = = .

The upper portion of the two-tailed rejection region with .05α = is and H19,19 20,19 2.51F F F> ≈ = 0 is not rejected. There is insufficient evidence to indicate that the population variances are different.

b The hypothesis to be tested is 0 1 2 a 1 2H : 0 H : 0µ µ µ µ− = − ≠

Based on the results of part a, you can use the pooled t test. The pooled estimator of 2σ is calculated as

( ) ( )2 2 2 2

1 1 2 22

1 2

1 1 19(22 ) 19(20 ) 4422 38

n s n ss

n n− + − +

= =+ −

=


( )1 2

2

1 2

0 78 67 1.651 11 1 44220 20

x xt

sn n

− − −= =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

The rejection region with .05α = and 20 20 2 38df = + − = (approximated with df = 29) is | and H

| 2.045t >0 is not rejected. There is insufficient evidence to indicate a difference in the two populaton means.

10.115 a A paired difference test is required, since the costs are paired according to the type of drug. To test

0 1 2 a 1 2H : 0 versus H : 0µ µ µ− = − ≠µ , the 9 differences are 111, 201, 45, 14, 175, 105, 288, 94, 79 Calculate

1112 123.55569

idd

n∑

= = =

( )2

2

1112194,614

9 7152.5277788ds−

= = and 84.572618ds =


123.5556 0 4.3884.572618

9

d

d

dt

s nµ− −

= = =

281

For a two-tailed test with , the rejection region with 8df = .01α = is , and H.005| | 3.355t t> = 0 is rejected. There is sufficient evidence to indicate that the average cost of prescription drugs in the United States is different from the average cost in Canada.

b The observed value of the test statistic, t = 4.38 is larger than .005 3.355t = , so that the 1 ( -value) .005 or -value .01.2

p p< < Since .01α = , the p-value would cause us to reject H0, as in part a.

10.116 Use the Student’s t Probabilities applet. Select the proper df using the slider on the right and use either

the one- or two-tailed applet, depending on the probability to be calculated. Type the positive value of t into the box marked “Student t” and press enter. The answers are shown below.

a .1419 b .0054 c .0734 d .2798 10.117 Use the Student’s t Probabilities applet. Select the proper df using the slider on the right and use either

the one- or two-tailed applet, depending on the type of rejection region. Type the positive value of α into the box marked “prob:” and press enter. The answers are shown below.

a b 1.8t > 2.37t > c 2.6t < − 10.118 Use the Interpreting Confidence Intervals applet. a The 95% confidence interval with 10n = or 9df = is calculated as

2.26210sx ±

b Answers will vary. Students should use the formula in part a, substituting the appropriate values for x and s. Remember that the applet rounds to the nearest integer!

10.119 Use the Interpreting Confidence Intervals applet. Answers will vary from student to student. The

widths of the ten intervals will not be the same, since the value of s changes with each new sample. The student should find that approximately 95% of the intervals in the first applet contain µ , while roughly 99% of the intervals in the second applet contain µ .

10.120 Use the Interpreting Confidence Intervals applet. Answers will vary from student to student. The student

should find that approximately 95% of the intervals in the first applet contain µ , while roughly 99% of the intervals in the second applet contain µ .

10.121 Use the Small Sample Test of a Population Mean applet. A screen capture is shown below. The test

statistic

47.1 48 1.4384.720

xts n

µ− −= = = −

has a two-tailed p-value of .1782 and H0 is not rejected. There is insufficient evidence to indicate that µ differs from 48.

282

10.122 Use the Small Sample Test of a Population Mean applet.

a The test statistic for testing 0 aH : 507 versus H : 507µ µ= ≠ is

499 507 .3169815

xts n

µ− −= = = −

has a two-tailed p-value of .7565 and H0 is not rejected. There is insufficient evidence to indicate that µ differs from 507. b The test statistic for testing 0 aH : 519 versus H : 519µ µ= ≠ is

516 519 .1219615

xts n

µ− −= = = −

has a two-tailed p-value of .9054 and H0 is not rejected. There is insufficient evidence to indicate that µ differs from 519.

10.123 Use the Two Sample t Test: Independent Samples applet. The hypothesis to be tested concerns the

differences between mean recovery rates for the two surgical procedures. Let 1µ be the population mean for Procedure I and 2µ be the population mean for Procedure II. The hypothesis to be tested is

0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠

Since the ratio of the variances is less than 3, you can use the pooled t test. Enter the appropriate statistics into the applet and you will find that test statistic is

( )1 2

2

1 2

03.33

1 1

x xt

sn n

− −= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

−

with a two-tailed p-value of .0030. Since the p-value is very small, H0 can be rejected for any value of α greater than .003 and the results are judged highly significant. There is sufficient evidence to indicate a difference in the mean recovery rates for the two procedures.

283

10.124 Use the Two Sample t Test: Independent Samples applet. The hypothesis to be tested is 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − ≠

Since the ratio of the variances is less than 3, you can use the pooled t test. Enter the appropriate statistics into the applet and you will find that test statistic is

( )1 2

2

1 2

04.03

1 1

x xt

sn n

− −= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

−

with a two-tailed p-value of .0004. Since the p-value is very small, H0 can be rejected for any value of α greater than .0004 and the results are judged highly significant. There is sufficient evidence to indicate a difference in the average prices of the two common stocks.

Case Study: How Would You Like a Four-Day Work Week? 1 A paired-difference test is used to test 0 1 2 a 1 2H : 0 versus H : 0µ µ µ µ− = − > , where 1µ is the mean number of personal leave-days taken on the conventional workweek and 2µ is the mean on the four-day workweek. Using Minitab to perform the paired difference test, you obtain the following printout. Paired T-Test and CI: PL-Yr2, PL-Yr1 Paired T for PL-Yr2 - PL-Yr1 N Mean StDev SE Mean PL-Yr2 11 22.3636 7.8648 2.3713 PL-Yr1 11 18.8182 11.2855 3.4027 Difference 11 3.54545 15.59079 4.70080 95% lower bound for mean difference: -4.97456 T-Test of mean difference = 0 (vs > 0): T-Value = 0.75 P-Value = 0.234 The test statistic of with indicates that the results are not significant. There is not enough evidence to indicate that there is a reduction in the average number of personal leave-days.

.75t = -value .234p =

284

2 For the paired difference test of 0 1 2H : 0µ µ− = , where 1µ is the mean number of sick-days taken on the conventional workweek and 2µ is the mean on the four-day workweek, a directional alternative might not be desirable. It could be that working long hours (10 hours per day) might increase rather than decrease the average number of sick-days. A two-tailed test of this hypothesis is provided by Minitab. Paired T-Test and CI: SL-Yr2, SL-Yr1 Paired T for SL-Yr2 - SL-Yr1 N Mean StDev SE Mean SL-Yr2 11 58.4545 25.8587 7.7967 SL-Yr1 11 46.0909 25.8126 7.7828 Difference 11 12.3636 23.0316 6.9443 95% CI for mean difference: (-3.1092, 27.8365) T-Test of mean difference = 0 (vs not = 0): T-Value = 1.78 P-Value = 0.105 The test statistic of with indicates that the results are not significant. There is not enough evidence to indicate that there is a reduction in the average number of sick-days.

1.78t = -value .105p =

3 From the printout in part 2, the 95% confidence interval for the difference in the average number of sick-days in these two years is 1 23.11 27.84µ µ− < − < . Based on this interval, which contains the value 1 2 0µ µ− = , there is no evidence to indicate a difference between the conventional and four-day work week. 4 Based on the analysis of these two variables, there is little or no difference in the two types of work schedules.

285

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pages.uoregon.edupages.uoregon.edu/haowang/teaching/425_FALL2008/SoluManual/C… · 10: Inference from Small Samples 10.1 Refer to Table 4, Appendix I, indexing df along the left