Stat501 Hw#5 Solution Hints

9.10 a If the airline is to determine whether or not the flight is unprofitable, they are interested in finding out whether or not m < 60 (since a flight is profitable if m is at least 60). Hence, the alternative hypothesis is

Ha :m < 60 and the null hypothesis is H0 :m = 60 . Formally,

H0 :m = 60 vs. Ha :m < 60

b Since only small values of x (and hence, negative values of z) would tend to disprove H0 in favor of Ha, this is a one-tailed test.

c For this exercise, n=120, x = 58, and s = 11 . One-Sample T

Test of mu = 0.6 vs < 0.6

95% Upper N Mean StDev SE Mean Bound T P 120 0.5800 0.1100 0.0100 0.5966 -1.99 0.024

Since the p-value of 0.024 is small (<0.05) we can reject the null hypothesis and conclude the alternative that this flight is unprofitable.

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Hence, the test statistic is

x-m x - m 58- 60 z =0� 0 = - 1.992 s n s n 11 120

The rejection region with a = .05 is determined by a critical value of z such that P( z< z0 ) = .05 . This

value is z0 = -1.645 and H0 will be rejected if z < -1.645 (compare the right-tailed rejection region in

Exercise 9.6). The observed value of z falls in the rejection region and H0 is rejected. The flight is unprofitable. 2

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9.14 The hypothesis to be tested is

H0 :m= 7 versus H a : m < 7

Insert MINITAB output and make conclusions

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and the test statistic is

x-m x - m 6.7- 7 z =0� 0 = - .992 s n s n 2.7 80

The rejection region with a = .05 is z < -1.645 (similar to Exercise 9.10). The observed value, z = -.992 ,

does not fall in the rejection region and H0 is not rejected. The data do not provide sufficient evidence to indicate that m < 7 .

9.15 a The hypothesis to be tested is

H0 :m= 7.4 versus H a : m > 7.4

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x-m x - m 7.9- 7.4 z =0� 0 = 2.63 s n s n 1.9 100

with p-value= P( z > 2.63) = 1 - .9957 = .0043 . To draw a conclusion from the p-value, use the guidelines for statistical significance in Section 9.3. Since the p-value is less than .01, the test results are highly

significant. We can reject H0 at both the 1% and 5% levels of significance.

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b You could claim that you work significantly fewer hours than those without a college education.

c If you were not a college graduate, you might just report that you work an average of more than 7.4 hours per week..

H0 :m= 98.6 versus H a : m 98.6

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To replace the following.

x-m x - m 98.25- 98.6 z =0� 0 = - 5.47 s n s n .73 130

with p-value= P( z < - 5.47) + P ( z > 5.47)� 2(0) 0 . Alternatively, we could write

p-value= 2 P( z < - 5.47) < 2(.0002) = .0004 With a = .05 , the p-value is less than  and H0 is rejected. There is sufficient evidence to indicate that the average body temperature for healthy humans is different from 98.6.

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b-c Using the critical value approach, we set the null and alternative hypotheses and calculate the test statistic as in part a. The rejection region with a = .05 is |z |> 1.96 . The observed value, z = -5.47 ,

does fall in the rejection region and H0 is rejected. The conclusion is the same is in part a.

d How did the doctor record 1 million temperatures in 1868? The technology available at that time makes this a difficult if not impossible task. It may also have been that the instruments used for this research were not entirely accurate.

H0 :m= 5.97 versus H a : m > 5.97

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Insert MINITAB output and make conclusions 4

To replace the following:

x-m x - m 9.8- 5.97 z =0� 0 = 10.94 s n s n 1.95 31

with p-value= P( z > 10.94) < 1 - .9998 = .0002 (or p-value 0 ). Since the p-value is less than .05, the null hypothesis is rejected . There is sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.97 mm.

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9.18 a-b The hypothesis of interest is one-tailed:

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 > 0

c The test statistic, calculated under the assumption that m1- m 2 = 0 , is

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To replace:

( x- x ) -(m - m ) z = 1 2 1 2 s2 s 2 1+ 2 n1 n 2

2 2 2 2 with s1 and s 2 known, or estimated by s1 and s2 , respectively. For this exercise,

( x- x ) - 0 11.6- 9.7 z � 1 2 = 2.09 s2 s 2 27.9 38.4 1+ 2 + 80 80 n1 n 2

a value which lies slightly more than two standard deviations from the hypothesized difference of zero.

This would be a somewhat unlikely observation, if H0 is true.

d The p-value for this one-tailed test is 5

Since the p-value is not less than a = .01 , the null hypothesis cannot be rejected at the 1% level. There is

insufficient evidence to conclude that m1- m 2 > 0 .

e Using the critical value approach, the rejection region, with a = .01, is z > 2.33 (see Exercise 9.3a).

Since the observed value of z does not fall in the rejection region, H0 is not rejected. There is insufficient

evidence to indicate that m1- m 2 > 0 , or m1> m 2 .

9.21 a The hypothesis of interest is one-tailed:

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 > 0

b The test statistic, calculated under the assumption that m1- m 2 = 0 , is

( x- x ) - 0 6.9- 5.8 z � 1 2 = 2.074 2 2 2 2 s1 s 2 (2.9) ( 1.2) + + n1 n 2 35 35

The rejection region with a = .05 , is z > 1.645 and H0 is rejected. There is evidence to indicate that

m1- m 2 > 0 , or m1> m 2 . That is, there is reason to believe that Vitamin C reduces the mean time to recover.

9.22 a The hypothesis of interest is one-tailed:

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 > 0

MINITAB output ……(use P-value)

The rejection region with a = .01, is z > 2.33 and H0 is rejected. There is evidence to indicate that

m1- m 2 > 0 , or m1> m 2 . The average per-capita beef consumption has decreased in the last ten years. (Alternatively, the p-value for this test is the area to the right of z = 5.33 which is very close to zero and less than a = .01 .)

b For the difference m1- m 2 in the population means this year and ten years ago, the 99% lower

confidence bound uses z.01 = 2.33 and is calculated as

(The lower bound can be found in MINITAB output) 6

Since the difference in the means is positive, you can again conclude that there has been a decrease in average per-capita beef consumption over the last ten years. In addition, it is likely that the average consumption has decreased by more than 5.63 pounds per year.

9.24 The hypothesis of interest is two-tailed:

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 0

(Get a Minitab output and make conclusions)

9.27 a The hypothesis of interest is two-tailed:

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 0

b The 95% confidence interval for m1- m 2 is approximately

(From MINITAB output)

Since the value m1- m 2 = 0 does not fall in the interval in part b, it is not likely that m1= m 2 . There is evidence to indicate that the means are different, confirming the conclusion in part a.

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 0

b Since the p-value = …., we can reject H0 at the 5% level (p-value < .05), but not at the 1% level (p-value > .01). Using the guidelines for significance given in Section 9.3 of the text, we declare the results statistically significant, but not highly significant.

H0 :p= .75 versus H a : p .75

b (Get a Minitab output and make conclusions in stead of using following)

x 58 With x = 58 and n = 100 , so that pˆ = = = .58 , the test statistic is n 100 7

pˆ - p .58- .75 z =0 = = -3.93 p0 q 0 .75( .25) n 100

with p-value= P( z > 3.93) < 2( .0002) = .0004 or p-value 0 . Since this p-value is less than .01, H0 is rejected at the 1% level of significance and the results are declared highly significant. There is evidence that the proportion of red flowered plants is not .75.

9.35 a-b Since the survival rate without screening is p = 2 3 , the survival rate with an effective program may be greater than 2/3. Hence, the hypothesis to be tested is

H0 :p= 2 3 versus H a : p > 2 3

x 164 c With pˆ = = = .82 , the test statistic is n 200

pˆ - p .82- 2 3 z =0 = = 4.6 p0 q 0 (2 3)( 1 3) n 200

The rejection region is one-tailed, with a = .05 or z > 1.645 and H0 is rejected. The screening program seems to increase the survival rate.

d For the one-tailed test,

p-value= P( z > 4.6) < 1 - .9998 = .0002

That is, H0 can be rejected for any value of a .0002 . The results are highly significant.

9.38 a The hypothesis of interest is

H0 :p= .25 versus H a : p .25

x 22 With pˆ = = = .275 , the test statistic is n 80

pˆ - p .275- .25 z =0 = = .52 p0 q 0 .25( .75) n 80

The rejection region is two-tailed a = .05 , or z > 1.96 and H0 is not rejected. There is insufficient evidence to indicate that the claim is incorrect. 8

b The hypothesis of interest is

H0 :p= .24 versus H a : p .24

x 15 With pˆ = = = .1875 , the test statistic is n 80

pˆ - p .1875- .24 z =0 = = -1.10 p0 q 0 .24( .76) n 80

The rejection region is two-tailed a = .05 , or z > 1.96 and H0 is not rejected. There is insufficient evidence to indicate that the claim is incorrect.

c Unless the experimenter had some preconceived idea that the proportion might be greater or less than claimed, there would be no reason to run a one-tailed test.

9.40 The hypothesis of interest is

H0 :p= .35 versus H a : p .35

x 123 with pˆ = = = .41 , the test statistic is n 300

pˆ - p .41- .35 z =0 = = 2.17 p0 q 0 .35( .65) n 300

The rejection region with a =.01 is |z |> 2.58 and the null hypothesis is not rejected. (Alternatively, we could calculate p-value= 2 P( z < - 2.17) = 2(.0150) = .0300 . Since this p-value is greater than .01, the null hypothesis is not rejected.) There is insufficient evidence to indicate that the percentage of adults who say that they always vote is different from the percentage reported in Time.

9.45 a The hypothesis of interest is:

H0 :p 1- p 2 = 0 versus H a : p 1 - p 2 < 0

n1 pˆ 1+ n 2p ˆ 2 18+ 30 Calculate pˆ1 = .36 , pˆ 2 = .60 and pˆ = = = .48 . The test statistic is then n1+ n 2 50 + 50

pˆ- p ˆ .36- .60 z =1 2 = = -2.40 骣1 1 .48( .52)( 1 50+ 1 50) pqˆ ˆ 琪 + 桫n1 n 2 9

The rejection region, with a = .05 , is z < -1.645 and H0 is rejected. There is evidence of a difference in the proportion of survivors for the two groups.

b From Section 8.7, the approximate 95% confidence interval is

pˆ1 q ˆ 1p ˆ 2 q ˆ 2 ( pˆ1- p ˆ 2 ) �1.96 n1 n 2 .36( .64) .60( .40) (.36- .60) � 1.96 50 50

-.24� .19 or < .43 -( p1 < p 2 ) - .05

H0 :p 1- p 2 = 0 versus H a : p 1 - p 2 0

x+ x 123+ 145 123 145 ˆ 1 2 Calculate pˆ1 = = .280 , pˆ 2 = = .259 , and p = = = .268 . The test statistic is then 440 560 n1+ n 2 440 + 560

pˆ- p ˆ .280- .259 z =1 2 = = 0.74 骣1 1 .268( .732)( 1 440+ 1 560) pqˆ ˆ 琪 + 桫n1 n 2

The rejection region, with a = .01 , is z > 2.58 and H0 is not rejected. There is no evidence of a difference in the proportion of frequent moviegoers in the two demographic groups.

b A difference in the proportions might mean that the advertisers would choose different products to advertise before this movie.

9.50 a Since the two treatments were randomly assigned, the randomization procedure can be implemented as each patient becomes available for treatment. Choose a random number between 0 and 9 for each patient. If the patient receives a number between 0 and 4, the assigned drug is aspirin. If the patient receives a number between 5 and 9, the assigned drug is clopidogrel.

b Assume that n1 = 7720 and n2 = 7780 . It is given that pˆ1 = .054 , pˆ 2 = .038 , so that

n pˆ+ np ˆ 7720( .054) + 7780( .038) pˆ =1 1 2 2 = = .046 . n1+ n 2 15,500

The test statistic is then

pˆ- p ˆ .054- .038 z =1 2 = = 4.75 骣1 1 .046( .954)( 1 7720+ 1 7780) pqˆ ˆ 琪 + 桫n1 n 2 10

with p-value= P( z > 4.75) < 2(.0002) = .0004 . Since the p-value is less than .01, the results are statistically significant. There is sufficient evidence to indicate a difference in the proportions for the two treatment groups.

c Clopidogrel would be the preferred treatment, as long as there are no dangerous side effects.

9.51 The hypothesis of interest is

H0 :p 1- p 2 = 0 versus H a : p 1 - p 2 > 0

119 x+ x 93+ 119 93 pˆ =1 2 = = .6625 Calculate pˆ1 = = .769 , pˆ 2 = = .598 , and . The test statistic is then 121 199 n1+ n 2 121 + 199

pˆ- p ˆ .769- .598 z =1 2 = = 3.14 骣1 1 .6625( .3375)( 1 121+ 1 199) pqˆ ˆ 琪 + 桫n1 n 2

with p-value= P( z > 3.14) = 1 - .9992 = .0008 . Since the p-value is less than .01, the results are reported as highly significant at the 1% level of significance. There is evidence to confirm the researcher’s conclusion.

H0 :p= .5 versus H a : p < .5

x 6 with pˆ = = = .222 , and the test statistic is n 27

pˆ - p .222- .5 z =0 = = -2.89 p0 q 0 .5( .5) n 27

Since no value of a is specified in advance, we calculate p-value= P( z < - 2.89) = .5 - .4981 = .0019 .

Since this p-value is less than .10, you can reject H0 at the 1% level (highly significant) and reject the DA’s claim of 50% or greater.

x 455 b If you take x = 455 and n = 503 with pˆ = = = .905 , the 95% confidence interval for p is n 503 approximately

pqˆ ˆ .905( .095) pˆ �1.96 � .905 1.96 .905 .026 n 503 11

or .879

c Even with the conservative value of x in part b, you can see that all the possible values for p are greater than p = .5 . You cannot conclude that the DA’s claim of 50% or greater is wrong – in fact, it appears that the DA is correct!

9.65 a-bThe hypothesis to be tested is

H0 :p 1- p 2 = 0 versus H a : p 1 - p 2 > 0

x+ x 42+ 48 42 48 ˆ 1 2 Calculate pˆ1 = = .636 , pˆ 2 = = .623 , and p = = = .629 . The test statistic is 66 77 n1+ n 2 66 + 77

pˆ- p ˆ .636- .623 z =1 2 = = 0.16 骣1 1 .629( .371)( 1 66+ 1 77) pqˆ ˆ 琪 + 桫n1 n 2

and the p-value is P( z �0.16) - 1 = .5636 .4364 .

c Since the observed p-value, .4364, is greater than a = .05 , H0 cannot be rejected. There is insufficient evidence to support the researcher’s belief.

9.66 Refer to the figure below, which represents the two probability distributions, one assuming that

p1- p 2 = 0 and one assuming that p1- p 2 = .1 . 12

The right curve is the true distribution of the random variable pˆ1- p ˆ 2 and consequently any probabilities that we wish to calculate concerning the random variable must be calculated as areas under the curve to the right. The objective of this exercise is to find a common sample size so that a =P[reject H0 when H 0 is true] = .05 and b = P[accept H0 when H 0 is false] .20 . For a = .05 consider the critical value of pˆ1- p ˆ 2 that separates the rejection and acceptance regions. This value will ˆ ˆ be denoted by ( p1- p 2 )C . Recall that the random variable z=( x - m) s measures the distance from a particular value x to the mean (in units of standard deviation). Since the z-value corresponding to ˆ ˆ ( p1- p 2 )C is z = 1.645 , we have

( pˆ-p ˆ ) - 0 1.645 = 1 2 C p q p q 1 1+ 2 2 n1 n 2 or

p q p q ˆ ˆ 1 1 2 2 ( p1- p 2 )C =1.645 + n1 n 2

Now b =P[accept H0 when p 1 - p 2 = .1] which is the area under the right hand curve to the left of ˆ ˆ b = .20 ( p1- p 2 )C . Since it is required that we must find the z-value corresponding to A= .2 , which is z = -.84 (see Table 3). Then

( pˆ-p ˆ ) -.1 -.84 = 1 2 C p q p q 1 1+ 2 2 n1 n 2

p q p q ˆ ˆ 1 1 2 2 ˆ ˆ where ( p1- p 2 )C =1.645 + . Substituting for ( p1- p 2 )C , n1 n 2

p q p q 1.6451 1+ 2 2 - .1 n n -.84 = 1 2 p q p q 1 1+ 2 2 n1 n 2 .1 2.485 = p q p q 1 1+ 2 2 n1 n 2

The following two assumptions will allow us to calculate the appropriate sample size:

1 n1= n 2 = n . 13

2 The maximum value of p(1- p) will occur when p=1 - p = .5 . Since values of p1 and p2 are unknown, the use of p = .5 will provide a valid sample size, although it may be slightly larger than necessary.

Then, solving for n, we obtain

.1 .1 2.485= � 2.485 骣1 1 .5 2 n .5( .5)琪 + 桫n n

n = 17.57 or n = 308.76 . Hence, a common sample size for the researcher’s test will be n = 309 .

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 > 0

( x- x ) - 0 240- 227 z � 1 2 = 4.33 s2 s 2 980 820 1+ 2 + 200 200 n1 n 2

The rejection region, with a = .05, is one-tailed or z > 1.645 and the null hypothesis is rejected. There is a difference in mean yield for the two types of spray.

b An approximate 95% confidence interval for m1- m 2 is

2 2 s1 s 2 ( x1- x 2 ) �1.96 n1 n 2 980 820 (240- 227) � 1.96 200 200

13� 5.88 or 7.12-(m1 < m 2 ) 18.88

9.72 a Let p1 be the proportion of cells in which RNA developed normally when treated with a .6 micrograms

per millimeter concentration of Actinomysin-D, and p2 be the proportion of normal cells treated with the higher concentration of Actinomysin-D. The hypothesis to be tested is

H0 :p 1- p 2 = 0 versus H a : p 1 - p 2 0

x+ x 55+ 23 55 23 ˆ 1 2 Calculate pˆ1 = = .786 , pˆ 2 = = .329 , and p = = = .557 . The test statistic is 70 70 n1+ n 2 70 + 70 14

pˆ- p ˆ .786- .329 z =1 2 = = 5.44 骣1 1 .557( .443)( 1 70+ 1 70) pqˆ ˆ 琪 + 桫n1 n 2

and the p-value is P( z �5.44) = 2( .0002) .0004 .

b Since the observed p-value < .0004, is must be less than a = .05 , and H0 is rejected. We can conclude that there is a difference in the rate of normal RNA synthesis for cells exposed to the two different concentrations of Actinomysin-D.

H0 :m= 508 versus H a : m 508 .

The test statistic is

x -m 499 - 508 z � = - .92 s n 98 100

and the p-value is

p-value= P( z > .92) + P( z < - .92) = 2(.1788) = .3576

Since the p-value, .3576, is greater than a = .05 , and H0 cannot be rejected and we cannot conclude that the average verbal score for California students in 2005 is different from the national average.

b The hypothesis to be tested is

H0 :m= 520 versus H a : m 520 .

x -m 516 - 520 z � = - .42 s n 96 100

and the p-value is

p-value= P( z > .42) + P( z < - .42) = 2(.3372) = .6744

Since the p-value, .6744, is greater than a = .05 , and H0 cannot be rejected and we cannot conclude that the average math score for California students in 2005 is different from the national average. 15

c Since the same students are used to measured verbal and math scores, there would not be two independent samples, and the two sample z-test would not be appropriate.

H0 :m= 5 versus H a : m > 5

x-m x - m 7.2- 5 z =0� 0 = 2.19 s n s n 6.2 38

The rejection region with a = .01 is z > 2.33 . Since the observed value, z = 2.19 , does not fall in the

rejection region and H0 is not rejected. The data do not provide sufficient evidence to indicate that the mean ppm of PCBs in the population of game birds exceeds the FDA’s recommended limit of 5 ppm.

9.76 Refer to Exercise 9.75, in which the rejection region was given as z > 2.33 where

x - m x - 2.3 z =0 = s n .29 35

Solving for x we obtain the critical value of x necessary for rejection of H0.

x - 5 6.2 >2.33� x + 2.33 = 5 7.34 6.2 38 38

The probability of a Type II error is defined as

b = P(accept H0 when H 0 is false)

Since the acceptance region is x 7.34 from part a, b can be rewritten as

b= P( x�7.34 when H0 is false) � P( x 7.34 when m 5)

Several alternative values of m are given in this exercise.

(Get a Minitab output using power calculation and make conclusions to do the followings)

Insert your output 16

a For m = 6 ,(difference = 6-5=1 in Minitab dialog)

骣 7.34- 6 b=P( x�7.34 when m = 6) P琪 z 桫 6.2 38 = P( z �1.33) .9082

and 1-b = 1 - .9082 = .0918 . b For m = 7 ,

骣 7.34- 7 b=P( x�7.34 when m = 7) P琪 z 桫 6.2 38 = P( z �.34) .6331

and 1-b = 1 - .6331 = .3669 . c For m = 8 ,

1-b = 1 - P( x � 7.34 when m 8) 骣 7.34- 8 =1 - P琪 z 桫 6.2 38 =1 -P( z � = .66) .7454

For m = 9 ,

1-b = 1 - P( x � 7.34 when m 9) 骣 7.34- 9 =1 - P琪 z 桫 6.2 38 =1 -P( z � = 1.65) .9505

For m = 10 ,

1-b = 1 - P( x � 7.34 when m 10) 骣 7.34- 10 =1 - P琪 z 桫 6.2 38 =1 -P( z � = 2.64) .9959

For m = 12 , 17

1-b = 1 - P( x � 7.34 when m 12) 骣 7.34- 12 =1 - P琪 z 桫 6.2 38 =1 - P( z � 4.63) 1

d The power curve is shown on the next page.

0.6 r e w o P 0.4

0.0 6 7 8 9 10 11 12 Mean

You can see that the power becomes greater than or equal to .90 for a value of m a little smaller than m = 9 . To find the exact value, we need to solve for m in the equation:

骣 7.34 - m 1-b = 1 -P( x� 7.34) - 1 � P琪 z .90 桫 6.2 38 骣 7.34 - m or P琪 z � .10 桫 6.2 38

From Table 3, the value of z that cuts off .10 in the lower tail of the z-distribution is z = -1.28 , so that

7.34 - m = -1.28 6.2 / 38 6.2 m =7.34 + 1.28 = 8.63. 38

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 > 0 18

( x- x ) - 0 69.58- 64.43 z � 1 2 = 10.75 s2 s 22.62 2 2.58 2 1+ 2 + n1 n 2 48 77

The rejection region, with a = .01, is one-tailed or z > 2.33 and the null hypothesis is rejected. There is sufficient evidence to indicate that the average height for males is greater than females.

b An approximate 99% one-sided confidence bound for m1- m 2 is approximately

2 2 s1 s 2 ( x1- x 2 ) -2.33 + n1 n 2 2.622 2.58 2 (69.58- 64.43) - 2.33 + 48 77

5.15- 1.12 = 4.03 or(m1 - m 2 ) > 4.03

Males are at least 4.03 inches taller than females on average.

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

2 larger s2 (4.67) = = 1.36 smaller s2(4.00) 2

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 t = .06 with p-value= .95 .

2 c The value of s = 4.38 is labeled “Pooled StDev” in the printout, so that s2 =(4.38) = 19.1844 .

d Since the p-value= .95 is greater than .10, the results are not significant. There is insufficient evidence to indicate a difference in the two population means.

e A 95% confidence interval for (m1- m 2 ) is given as

2 骣1 1 ( x1- x 2) � t .025 s 琪桫n1 n 2 骣1 1 (29- 28.86) � 2.201 19.1844琪桫6 7

.14� 5.363 or< 5.223 -(m1 < m 2 ) 5.503 19

Since the value m1- m 2 = 0 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.

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 0

From the Minitab printout, the following information is available:

2 2 x1=.896 s 1 =( .400) n 1 = 14

2 2 x2=1.147 s 2 =( .679) n 2 = 11

( x- x ) - 0 t =1 2 = -1.16 骣1 1 s2 琪 + 桫n1 n 2

The rejection region is two-tailed, based on n1+ n 2 -2 = 23 degrees of freedom. With a = .05 , from

Table 4, the rejection region is t> t.025 = 2.069 and H0 is not rejected. There is not enough evidence to indicate a difference in the population means.

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:

2 larger s2 (.679) = = 2.88 smaller s2(.400) 2

which is less than three. Therefore, it is reasonable to assume that the two population variances are equal.

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

larger s2 2.78095 = = 16.22 smaller s2 .17143

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 20

H0 :m 1- m 2 = 0 versus H a : m 1 - m 2 0

( x- x ) - 0 3.73- 4.8 t =1 2 = = -2.412 s2 s 2 2.78095 .17143 1+ 2 + 15 15 n1 n 2

2 2 2 骣s1 s 2 琪 + 2 桫n1 n 2 (.185397+ .0114287) df =2 2 = = 15.7 骣s2 骣 s 2 .002455137+ .00000933 琪1 琪 2 桫n1 桫 n 2 + n1-1 n 2 - 1

With df 15 , the p-value for this test is bounded between .02 and .05 so that H0 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.

2 10.40 a-b The table of differences, along with the calculation of d and sd , is presented below.

Week 1 2 3 4 Totals

di –1.77 –15.03 –23.22 127.05 –67.07

d -67.07 d =i = = -16.7675 n 4

( d )2 (-67.07)2 �d 2 i 1499.9047 - i and s = 11.1849 s2 =n =4 = 125.102825 d d n -1 3

The hypothesis of interest is

H0 :m 1- m 2 = 0 or H 0 : md = 0

Ha :m 1- m 2构 0 or H a : md 0

d - m -16.7675 - 0 t =d = = -3.00 11.1849 sd n 4

Since t = -3.00 with df= n -1 = 3 falls between the two tabled values, t.025 and t.05 , 21

for this two tailed test and H0 is not rejected. We cannot conclude that the means are different.

c The 99% confidence interval for m1- m 2 = md is

sd 11.1849 d鞭 t.005 -16.7675 鞭 5.841 - 16.7675 32.666 n 4

or -49.433 <(m1 - m 2 ) < 15.899 .

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

H0 :m 1- m 2 = 0 H a : m 1 - m 2 0

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

Location 1 2 3 4 5 Total

di –.4 –2.7 –1.6 –1.7 –1.5 –7.9

d -7.9 d =i = = -1.58 n 5

2 2 (�d ) ( 7.9) �d 2 - i 15.15 i and s = .8167 s2 =n =5 = .667 d d n -1 4

d - m -1.58 - 0 t =d = = -4.326 .8167 sd n 5

A rejection region with a = .05 and df= n -1 = 4 is t> t.025 = 2.776 , 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 m1- m 2 = md is

sd .8167 d鞭 t.025 -1.58 鞭 2.776 - 1.58 1.014 n 5

or -2.594 <(m1 - m 2 ) < - .566 . 22

c The inequality to be solved is

ta 2 SE B

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

sd t.025 .2 n

With sd = .8167 and n represents the number of pairs of observations, consider the sample size obtained

by replacing t.025 by z.025 = 1.96 .

.8167 1.96 .2 n n侈8.0019 n= 64.03 or n = 65

Since the value of n is greater than 30, the use of za 2 for ta 2 is justified.

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

H0 :m 1- m 2 = 0 or H 0 : md = 0

Ha :m 1- m 2 > 0 or H a : md > 0

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

3 3 2 1 1 3 1 di

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

d - m .857- 0 t =d = = 1.03 2.193 sd n 7

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

p-value> .10

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. 23