brase_9e_ism_ch09

486 Part IV: Complete Solutions, Chapter 9

Chapter 9: Hypothesis Testing

Section 9.1Note: For all graphs provided, the P value is indicated by the shaded portion in the tails.

1. See text for definitions. Essays may include(a) A working hypothesis about the population parameter in question is called the null hypothesis. The

value specified in the null hypothesis is often a historical value, a claim, or a production specification.(b) Any hypothesis that differs from the null hypothesis is called an alternate hypothesis.(c) If we reject the null hypothesis when it is in fact true, we have an error that is called a type I error. On

the other hand, if we fail to reject the null hypothesis when it is in fact false, we have made an error that is called a type II error.

(d) The probability with which we are willing to risk a type I error is called the level of significance of a test. The probability of making a type II error is denoted by

2. The alternate hypothesis is used to determine which type of test is used. 3. No, if we fail to reject the null hypothesis, we have not proven it to be true beyond all doubt. The evidence

is not sufficient to merit rejecting

4. No, if we reject the null hypothesis, we have not proven it to be false beyond all doubt. The test was conducted with a level of significance that is the probability with which we are willing to risk a type I error (rejecting when it is in fact true).

5. (a) The claim is = 60 kg, so you would use = 60 kg.(b) We want to know if the average weight is less than 60 kg, so you would use < 60 kg.(c) We want to know if the average weight is greater than 60 kg, so you would use > 60 kg.(d) We want to know if the average weight is different from 60 kg, so you would use 60 kg.(e) Since part (b) is a left-tailed test, the critical region is on the left. Since part (c) is a right-tailed test, the

critical region is on the right. Since part (d) is a two-tailed test, the critical region is on both sides of the mean.

6. (a) The claim is = 8.3 min, so you would use = 8.3 min. If you believe that the average is less than 8.3 min, then you would use < 8.3 min. This is a left-tailed test.

(b) The claim is = 8.3 min, so you would use = 8.3 min. If you believe the average is different from 8.3 min, then you would use 8.3 min. This is a two-tailed test.

(c) The claim is = 4.5 min, so you would use = 4.5 min. If you believe the average is more than 4.5 min, then you would use > 4.5 min. This is a right-tailed test.

(d) The claim is = 4.5 min, so you would use = 4.5 min. If you believe the average is different from 4.5 min, then you would use 4.5 min. This is a two-tailed test.

7. (a) The claim is = 16.4 ft, so = 16.4 ft.(b) You want to know if the average is getting larger, so > 16.4 ft.(c) You want to know if the average is getting smaller, so < 16.4 ft.(d) You want to know if the average is different from 16.4 ft, so 16.4 ft.(e) Since part (b) is a right-tailed test, the area corresponding to the P value is on the right. Since part (c) is

a left-tailed test, the area corresponding to the P value is on the left. Since part (d) is a two-tailed test, the area corresponding to the P value is on both sides of the mean.

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8. (a) The claim is = 8.7 s, so = 8.7 s.(b) You want to know if the average is longer, so > 8.7 s.(c) You want to know if the average is reduced, so < 8.7 s.(d) Since part (b) is a right-tailed test, the P value area is on the right. Since part (c) is a left-tailed test, the

P value area is on the left.

9. (a)

Since > is in use a right-tailed test.(b) Use the standard normal distribution. We assume x has a normal distribution with known standard

deviation Note that is given in the null hypothesis.

(c) P value = P(z > 0.90) = 0.1841

(d) Since a P value of 0.1841 > 0.01 for , we fail to reject The data are not statistically significant.(e) There is insufficient evidence at the 0.01 level to reject the claim that average yield for bank stocks

equals average yield for all stocks.

10. (a)

Since > is in use a right-tailed test.(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard


(c) P value = P(z > 1.99) = 0.0233


Part IV: Complete Solutions, Chapter 9 488

(d) Since a P value of 0.0233 0.05, reject Yes, the data are statistically significant.(e) The sample evidence is sufficient at the 0.05 level to justify rejecting It seems Gentle Ben’s

glucose is higher than average.

11. (a)

Since < is in use a left-tailed test.(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard


(c) P value = P(z < 2.80) = 0.0026

(d) Since a P value of 0.0026 0.01, we reject Yes, the data are statistically significant.



(e) The sample evidence is sufficient at the 0.01 level to justify rejecting It seems the humming birds in the Grand Canyon region have a lower average weight.

12. (a)

Since < is in use a left-tailed test.(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard


(c) P value = P(z < 1.58) = 0.0571

(d) Since a P value of 0.0571 > 0.05 for , we fail to reject The data are not statistically significant.(e) There is insufficient evidence at the 0.05 level to reject It seems the average P/E for large banks is

not less than that of the S&P 500 Index.

13. (a)

Since is in use a two-tailed test.(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard


(c) P value = 2P(z > 1.20) = 2(0.1151) = 0.2302



(d) Since a P value of 0.2302 > 0.01 for , we fail to reject The data are not statistically significant.(e) There is insufficient evidence at the 0.01 level of significance to reject It seems the average hail

damage to wheat crops in Weld County matches the national average.

14. (a)

Since is in use a two-tailed test.(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard


(c) P value = 2P(z > 2.62) = 2(0.0044) = 0.0088

(d) Since a P value of 0.0088 0.01, we reject The data are statistically significant.



(e) At the 1% level of significance, the sample average is sufficiently different from = 28 that we reject It seems Roger’s average red cell volume is different from the average for healthy adults.

Section 9.21. The P value for a two-tailed test of μ is twice the P value for a one-tailed test based on the same sample

data and null hypothesis.

2. If σ is known, use the standard normal distribution. If σ is not known, use the Student’s t distribution with n – 1 degrees of freedom.

3. Use n – 1 degrees of freedom.

4. Not necessarily. If the P value = 0.04, you would reject at the α = 0.05 level but not at the α = 0.01 level. On the other hand, if the P value is 0.003, you would reject at both the α = 0.05 and α = 0.01 levels.

5. Yes. If the P value is less than α = 0.01, then it also will be less than α = 0.05. In both cases, reject the null hypothesis.

6. No. The P value for the two-tailed test is twice the P value for the one-tailed test. If the P value < 0.01, then 2 × P value might be greater than or less than 0.01.

7. (a)

(b) Use the standard normal distribution. The sample size is large, n 30, and = 3.5.

(c) P value = P(z > 1.54) = 0.0618

(d) Since a P value of 0.0618 > 0.01, we fail to reject The data are not statistically significant.(e) At the 1% level of significance, there is insufficient evidence to say average storm level is increasing.



8. (a)

(b) Use the standard normal distribution. The sample size is large, n 30, and σ = 1.2 hours.

(c) P value = P(z < 2.86) = 0.0021

(d) Since a P value of 0.0021 0.01, we reject The data are statistically significant.(e) At the 1% level of significance, evidence shows average assembly time is less than 38 hours.

9. (a)

(b) Use a standard normal distribution. The sample size is large, n 30, and σ = 18.5 e-mails.

(c) P value = 2P(z < 1.99) = 2(0.0233) = 0.0466



(d) Since a P value of 0.0466 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, there is sufficient evidence to say the average number of e-mails is

different than 41.7 under the new priority system.

10. (a)

(b) Use the Student’s t distribution with d.f. = n 1 = 31 1 = 30. The sample size is large, n 30, and is unknown.

(c) If d.f. = 30, 2.051 falls between the entries 2.042 and 2.457, use two-tailed areas to find that 0.020 < P value < 0.050. Using a TI-84, P value ≈ 0.0491.

(d) Since the entire P-value interval 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to say that the drug has changed the mean pH

level.



11. (a)


(c) For d.f. = 45, 2.481 falls between entries 2.412 and 2.690. Use one-tailed areas to find that 0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0084.

(d) Since the entire P-value interval 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, sample data indicate that the average age of Minnesota region coyotes

is greater than 1.75 years.

12. (a)


(c) For d.f. = 50, 1.116 falls between entries 0.679 and 1.164.Use one-tail areas to find that 0.125 0.05, we fail to reject The data are not statistically significant.(e) At the 5% level of significance, the sample data do not indicate that the average fish length is less than

19 inches.

13. (a)


(c) For d.f. = 35, 1.731 falls between entries 1.690 and 2.030. Use two-tailed areas to find that 0.05 0.05, we fail to reject The data are not statistically significant.(e) At the 5% level of significance, the sample evidence does not support rejecting the claim that the

average P/E for socially responsible funds is different from that of the S&P stock index.



14. (a)


(c) In Table 6, use the closest d.f. smaller than 36 or d.f. = 35. Since 2.561 falls between entries 2.438 and 2.724, use one-tailed areas to find 0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0074.

(d) Since the entire P-value interval 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, sample data support the claim that the average arsenic content is less

than 8.0 ppb.

15. (i) Use a calculator to verify. Rounded values are used in part (ii).

(ii) (a)

(b) Use the Student’s t distribution with d.f. = n 1 = 6 1 = 5. We assume that x has a distribution that is approximately normal and that is unknown.




(d) Since the entire P-value interval 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, sample evidence supports the claim that the average RBC count

for this patient is less than 4.8.


(ii) (a)



(d) Since P-value interval > 0.01, we fail to reject No, the data are not statistically significant.



(e) At the 1% level of significance, the sample data do not support the claim that the average HC for this patient is higher than 14.


(ii) (a)



(d) Since P-value interval > 0.01, we fail to reject The data are not statistically significant.(e) At the 1% level of significance, sample evidence does not support a claim that the average

thickness of slab avalanches in Vail is different from that of those in Canada.


(ii) (a)





(d) Since the P-value interval > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, evidence is not strong enough to conclude that the population mean

life span is less than 77 years.


(ii) (a)





(d) Since P-value interval > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, we cannot conclude that the catch is different from 8.8 fish per day.


(ii) (a)


(c) For d.f. = 9, 2.714 falls between entries 2.262 and 2.821. Use two-tailed areas to find 0.020 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, there is not enough evidence to conclude that the population mean of tree ring dates is different from 1,300.



21. (a) The P value of a one-tailed test is smaller. For a two-tailed test, the P value is double because it includes the area in both tails.

(b) Yes; the P value of a one-tailed test is smaller, so it might be smaller than , whereas the P value of a two-tailed test is larger than .

(c) Yes; if the two-tailed P value is less than , the one-tail area is also less than .(d) Yes, the conclusions can be different. The conclusion based on the two-tailed test is more conservative

in the sense that the sample data must be more extreme (differ more from ) in order to reject

22. Essay or class discussion.

23. (a)

For = 0.01, c = 1 0.01 = 0.99, and

The hypothesized mean = 20 is not in the interval. Therefore, we reject(b) Because n = 36 is large, the sampling distribution of is approximately normal by the central limit

theorem, and we know σ.

From Table 5, P value = 2P(z < 3.00) = 2(0.0013) = 0.0026. Since 0.0026 0.01, we reject The results are the same.

24. (a)

For = 0.01, c = 1 0.01 = 0.99, and

The hypothesized mean = 21 falls into the confidence interval. Therefore, we do not reject(b) For = 0.01, the two-tailed test’s critical values are Because n = 36 is large, the sampling

distribution of is approximately normal by the central limit theorem, and we know σ.

From Table 5, P value = 2P(z < 1.50) = 2(0.0668) = 0.1336. Since 0.1336 > 0.01, we do not rejectThe results are the same.

25. For a right-tailed test and = 0.01, the critical value is critical region is values to the right of 2.33. Since the sample statistic z = 1.54 is not in the critical region, fail to reject At the 1% level, there is insufficient evidence to say that the average storm level is increasing. Conclusion is the same as with the P-value method.



26. For a left-tailed test and = 0.01, the critical value critical region is values to the left of 2.33. Since the sample statistic z = 2.86 is in the critical region, reject At the 1% level, evidence shows that average assembly time is less than 38 hours. The conclusion is that same as with the P-value method.

27. For a two-tailed test and = 0.05, the critical value critical regions are values to the left of 1.96 and values to the right of 1.96. Since the sample test statistic z = 1.99 is in the critical region, reject

At the 5% level, there is sufficient evidence to say the average number of e-mails is different with the new priority system. The conclusion is the same as with the P-value method.

28. For a two-tailed test and = 0.05, critical values are with d.f. = 30; critical regions are values to the right of 2.042 and those to the left of 2.042. Since the sample test statistic t = 0.051 is in the critical region, reject At the 5% level, the evidence is sufficient to say that the drug has changed the mean pH level. The conclusion is the the same as with the P-value method.

29. For a right-tailed test and = 0.01, critical value is with d.f. = 45. Critical region is values to the right of 2.412. Since the sample test statistic t = 2.481 is in the critical region, reject At the 1% level, sample data indicate that the average age of Minnesota region coyotes is higher than 1.75 years. The conclusion is the same as with the P-value method.

30. For a left-tailed test and = 0.05, critical value is with d.f. = 50. Critical region is values to the left of 1.676. Since the sample test statistic t = 1.116 is not in the critical region, fail to reject At the 5% level of significance, the sample data do not indicate that the average fish length is less than 19 inches. The conclusion is the same as with the P-value method.

Section 9.31. The value of p comes from H0. Note that q = 1 – p.

2.

3. Yes. The corresponding P value for a one-tailed test is half that of a two-tailed test. Thus the P value for the one-tailed test is also less than 0.01.

4. Answers vary. First, we don’t know if the study is based on population data or sample data. If the study is based on population data, we do not need to conduct a hypothesis test. However, assuming that the study was based on sample data, testing H0: p = 0.15 versus H1: p > 0.15 seems appropriate. Without a specific source for the study, we do not know how reliable the information in the sample is. Also, we are not given information about the sample size. Level of significance could be one of the common values, α = 0.01 or α = 0.05. Finally, if the conclusions are based on sample data, we cannot conclude that the conclusions are absolutely true.

5. (i) (a)



(b) Use the standard normal distribution. The distribution is approximately normal when n is sufficiently large, which it is here, because np = 215(0.301) 647 and nq = 215(0.699) 150.3 are both > 5.

(c) P value = P(z < 2.78) = 0.0027

(d) Since a P value of 0.0027 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the sample data indicate that the population proportion of numbers

with a leading 1 in the revenue file is less than 0.301 predicted by Benford’s law.(ii) Yes. The revenue data file seems to include more numbers with higher first nonzero digits than

Benford’s law predicts.(iii) We have not proved to be false. However, because our sample data lead us to reject and

conclude that there are too few numbers with a leading digit 1, more investigation is merited.

6. (i) (a)

(b) Use the standard normal distribution. The distribution is approximately normal when n is sufficiently large, which it is, because np = 228(0.301) 68.6 and nq = 228(0.699) 159.4 are both > 5.

(c) P value = P(z > 3.39) = 0.0003



(d) Since a P value of 0.0003 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the sample data indicate that the proportion of numbers in the

revenue file with a leading digit 1 exceed the 0.301 predicted by Benford’s law.(ii) Yes. There seem to be too many entries with a leading digit 1.(iii) We have not proved to be false. However, because our data led us to reject and conclude that

there are “too many” numbers with a leading digit 1, more investigation is merited.

7. (a)

(b) Use the standard normal distribution. The distribution is approximately normal when n is sufficiently large, which it is here, because np = 32(0.7) = 22.4 and nq = 32(0.3) = 9.6, and both are greater than 5.

(c) P value = 2P(z > 0.62) = 2(0.2676) = 0.5352

(d) Since a P value of 0.5352 > 0.01, we fail to reject No, the data are not statistically significant.



(e) At the 1% level of significance, we cannot say that the population proportion of arrests of males aged 15 to 34 in Rock Springs is different than 70%.

8. (a)


(c) P value = P(z < 1.54) = 0.0618

(d) Since a P value of 0.0618 > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, there is insufficient evidence to say that the proportion of women

athletes who graduate is less than 67%.

9. (a)


(c) P value = P(z < 2.65) = 0.0004



(d) Since a P value of 0.0004 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the data show that the population proportion of driver fatalities related

to alcohol is less than 77% in Kit Carson County.

10. (a)


(c) P value = 2P(z < 0.45) = 2(0.3264) = 0.6528

(d) Since a P value of 0.6528 > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the data do not indicate that the proportion of college students favoring

the color blue is different from 0.24.



11. (a)

(b) Use the standard normal distribution. The distribution is approximately normal when n is sufficiently large, which it is here, because np = 34(0.50) = 17 and nq = 17, and both are greater than 5.

(c) P value = P(z < 2.40) = 0.0082

(d) Since a P value of 0.0082 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the data indicate that the population proportion of female wolves is

now less than 50% in the region.

12. (a)


(c) P value = 2P(z > 0.44) = 2(0.3300) = 0.6600



(d) Since a P value of 0.6600 > 0.05, we fail to reject No, the data are not statistically different.(e) At the 5% level of significance, there is insufficient evidence to indicate that the population proportion

of guests who catch pike of 20 pounds is different from 75%.

13. (a)


(c) P value = 2P(z < 2.78) = 2(0.0027) = 0.0054

(d) Since a P value of 0.0054 0.01, we reject Yes, the data are statistically significant.



(e) At the 1% level of significance, the sample data indicate that the population proportion of the five-syllable sequence is different from the text of Plato’s Republic.

14. (a)


(c) P value = P(z > 3.35) = 0.0004

(d) Since a P value of 0.0004 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the sample data indicate that the population proportion of the five-

syllable sequence is higher than that found in Plato’s Symposium.

15. (a)


(c) P value = P(z > 1.09) = 0.1379



(d) Since a P of 0.1379 > 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, there is insufficient evidence to support the claim that the population

proportion of customers loyal to Chevrolet is more than 47%.

16. (a)

(b) Use the standard normal distribution. The distribution is approximately normal when n is sufficiently large, which it is here, because np = 115(0.8) = 92 and nq = 115(0.2) = 23, and both are greater than 5.

(c) P value = P(z < 0.93) = 0.1762

(d) Since a P value of 0.1762 > 0.05, we fail to reject No, the data are not statistically significant.



(e) At the 5% level of significance, there is insufficient evidence that the population proportion of prices ending with digits 9 or 5 is less than 80%.

17. (a)


(c) P value = P(z > 2.71) = 0.0034

(d) Since a P value of 0.0034 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the data indicate that the population proportion of students with

hypertension during final exams week is higher than 9.2%.

18. (a)


(c) P value = P(z < 1.93) = 0.0268



(d) Since a P value of 0.0268 > 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, the data are insufficient to conclude that the population proportion of

patients having headaches is less than 0.12.

19. (a)


(c) P value = 2P(z 1.18) = 2(0.1190) = 0.2380



(d) Since a P value of 0.2380 > 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, the evidence is insufficient to indicate that the population proportion

of extroverts among college student government leaders is different from 82%.

20. For a two-tailed test and = 0.01, critical values are The critical regions are values greater than 2.58 and values less than 2.58. The sample test statistic z = 0.62 is not in the critical region, so we do not reject Result is consistent with the P-value conclusion.

21. For a left-tailed test and = 0.01, critical value is The critical region consists of values less than 2.33. The sample test statistic z = 2.65 is in the critical region, so we reject Result is consistent with the P-value conclusion.

22. For a two-tailed test and = 0.01, critical value is The critical region consists of values greater than 2.33. The sample test statistic z = 1.09 is not in the critical region, so we fail to reject Result is consistent with the P-value conclusion.

Section 9.41. Paired data are dependent.

2. Take the difference of corresponding paired data values. The sample test statistics is which is the mean of the differences.

3. H0: . We test that the mean difference is 0.

4. Here, n is the number of data pairs.

5. Here, d.f. = n – 1.

6. (a) For a right-tailed test that the before score is higher, use d = B – A.(b) For a left-tailed test that the before score is higher, use d = A – B.

7. (a)

Since is in a two-tailed test is used.(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,

symmetric distribution.

Pair 1 2 3 4 5 6 7 8d = B A 3 2 5 4 10 15 6 7

(c) d.f. = n 1 = 8 1 = 7From Table 6 in Appendix II, 0.818 falls between entries 0.711 and 1.254. Using two-tailed areas, find that 0.250 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to claim a difference in population mean

percentage increases for corporate revenue and CEO salary.

8. (a)



Pair 1 2 3 4 5 6 7d = B A 0.1 0.4 0.4 1.0 0.6 0.6 0.5

(c) d.f. = n 1 = 7 1 = 6From Table 6 in Appendix II, 2.08 falls between entries 1.943 and 2.447. Using two-tailed areas, find that 0.050 0.01, we do not reject No, the data are not statistically significant.(e) At the 1% level of significance, the evidence is insufficient to claim that there is a difference in

population mean hours per fish between boat fishing and shore fishing.

9. (a)

Since > is in a right-tailed test is used.(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,


Pair 1 2 3 4 5d = Jan April 35 9 26 24 17

(c) d.f. = n 1 = 5 1 = 4From Table 6 in Appendix II, 1.243 falls between entries 0.741 and 1.344. Using one-tailed areas, find that 0.125 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, the evidence is insufficient to claim average peak wind gusts are

higher in January.

10. (a)



Pair 1 2 3 4 5 6 7 8 9 10d = B A 1.2 1.2 2.9 1.7 13.4 2.8 5.3 7.9 6.7 4.3

(c) d.f. = n 1 = 10 1 = 9From Table 6 in Appendix II, 3.452 falls between entries 3.250 and 4.781. Using two-tailed areas, find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0036.



(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to claim that the January mean population of

deer has dropped. Note that this test does not determine the cause of the drop. Development around the highway, disease, etc. may have affected the deer population.

11. (a)



Pair 1 2 3 4 5 6 7 8d = winter summer 19 4 17 7 9 0 9 10

(c) d.f. = n 1 = 8 1 = 7From Table 6 in Appendix II, 1.76 falls to the between entries 1.617 and 1.895. Using one-tailed areas, find that 0.05 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to indicate that the population average

percentage of male wolves in winter is higher.

12. (a)



Pair 1 2 3 4 5 6 7 8 9d = A B 2.9 1.1 2.1 2.1 1.4 1.8 3.3 5.1 1.6

10 11 12 13 14 15 161.3 4.0 5.6 7.6 0.9 4.1 6.5

(c) d.f. = n 1 = 16 1 = 15From Table 6 in Appendix II, 1.175 falls between entries 0.691 and 1.197. Use two-tailed areas to find that 0.250 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, the evidence is insufficient to claim that the population average birth

and death rates are different in this region.

13. (a)



Pair 1 2 3 4 5 6 7 8d = houses hogans 5 2 22 23 4 19 33 32

(c) d.f. = n 1 = 8 1 = 7From Table 6 in Appendix II, 0.789 falls between entries 0.711 and 1.254. Use one-tailed areas to find that 0.125 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to show that the mean

number of inhabited houses is greater than that of hogans.

14. (a)



Pair 1 2 3 4 5 6 7 8d = flaked nonflaked 4 1 9 2 3 6 21 13




(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to indicate that the population average

number of flaked stone tools is higher.


(ii) (a)

Since > is in a right-tailed test is used.(b) Use the Student’s t distribution. The sample size is greater than 30.

(c) d.f. = n 1 = 36 1 = 35 From Table 6 in Appendix II, 1.223 falls between entries 1.170 and 1.306. Use one-tailed

areas to find that 0.100 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to claim that the population mean cost

of living index for housing is higher than that for groceries.


(ii) (a)

Since < is in a left-tailed test is used.(b) Use the Student’s t distribution. The sample size is greater than 30.



(c) d.f. = 45 From Table 6 in Appendix II, 2.447 falls between entries 2.412 and 2.690. Use one-tailed

areas to find that 0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0092.

(d) Since the P-value interval 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to claim that the population mean cost of

living index for utilities is less than that for transportation in this region.

17. (a)



Pair 1 2 3 4 5 6 7 8 9d = B A 7 2 9 0 6 1 3 3 3




(d) Since the P-value interval is > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to claim that the population score on the

last round is higher than that on the first.

18. (a)



Pair 1 2 3 4 5 6d = 1 5 0.6 0.5 0.1 0.2 0.7 0.9




(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to claim that the population mean time for

rats receiving larger rewards to run the maze is less.

19. (a)



Pair 1 2 3 4 5 6 7 8d = time 1 time 5 1.4 1.7 0.8 1.5 0.5 0.1 1.7 1.3

(c) d.f. = n 1 = 8 1 = 7From Table 6 in Appendix II, 2.080 falls between entries 1.895 and 2.365. Use one-tailed areas to find 0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0380.

(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to claim that the population mean time for

rats receiving larger rewards to climb the ladder is less.

20. (a) ; t0.95 = 2.365 with d.f. = 7; E 2.365; 95% confidence interval for d from –4.25 to 8.75. We are 95% confident that the difference in population mean percentage increase between company revenue and CEO salaries is between –4.25% and 8.75%.

(b) = 0.05; c = 1 – = 0.95; H0: d = 0; H1: d ≠ 0; Since d = 0 from the null hypothesis is in the 95% confidence interval, do not reject H0 at the 5% level of significance. The data do not indicate a difference in population mean percentage increases between company revenue and CEO salaries. This result is consistent with the conclusion reached using the P-value method of testing for = 0.05 in Problem 7.



21. For a two-tailed test with = 0.05 and d.f. = 7, critical values are The sample test statistic t = 0.818 is between 2.365 and 2.365, so we do not reject This conclusion is the same as that reached by the P-value method.

22. For a right-tailed test with = 0.01 and d.f. = 4, the critical value is The sample test statistic t = 1.243 is to the left of so we do not reject This conclusion is the same as that reached by the P-value method.

Section 9.5

1. The test statistic is .

2. Use the Student’s t distribution.

3. H0: μ1 = μ2 or H0: μ1 – μ2 = 0.

4. The test statistic is where .

5. The best estimate is .

6. (a) Satterthwaite’s gives the larger degrees of freedom. It is more conservative to use a smaller degrees of freedom because there will be more area in the tails of the Student’s t distribution, resulting in larger P values.

(b) The P value will be larger. Using a larger P value might mean that the test is not significant, whereas a smaller P value might result in significance.

7. H1: μ1 > μ2 or H1: μ1 – μ2 > 0.

8. H1: μ1 < μ2 or H1: μ1 – μ2 < 0.

9. (a)

(b) Use the standard normal distribution. We assume that both population distributions are approximately normal and that and are known.

(c) P value = P(z > 2.57) = 0.0051



(d) Since P value = 0.0051 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the evidence is sufficient to indicate that the population mean REM

sleep time for children is more than for adults.

10. (a)


(c) P value = 2P(z > 0.96) = 2(0.1685) = 0.3370

(d) Since P value = 0.3370 > 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, the evidence is insufficient to claim that there is a difference in mean

population pollution index for Englewood and Denver.



11. (a)


(c) P value = 2P(z > 2.16) = 2(0.0154) = 0.0308

(d) Since P value = 0.0308 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to show that there is a difference between

mean response regarding preference for camping or fishing.

12. (a)


(c) P value = P(z < 2.91) = 0.0018



(d) Since P value = 0.0018 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to show that the population mean percentage

of young adults who attend college is higher.

13. (i) Use a calculator to verify. Use rounded results to compute t.

(ii) (a)

(b) Use the Student’s t distribution. We assume that both population distributions are approximately normal and that and are unknown.

(c) Since In Table 6 in Appendix II, 0.965 falls between entries 0.703 and 1.230. Use one-tailed

areas to find that 0.125 0.01, we fail to reject No, the data are not statistically significant.



(e) At the 1% level of significance, the evidence is insufficient to indicate that violent crime in the Rocky Mountain region is higher than in New England.


(ii) (a)


(c) Since .In Table 6 in Appendix II, 2.053 falls between entries 1.771 and 2.160. Use one-tailed areas to

find that 0.025 < P value < 0.050. Using a TI-84, d.f. ≈ 27.42, and P value ≈ 0.0249.

(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to indicate that the population mean rate of

hay fever is lower for the age group over 50.

15. (a)

(b) Use the Student’s t distribution. Both sample sizes are large, and and are unknown.

(c) Since d.f. = n 1 = 30 1 = 29.From Table 6 in Appendix II, 0.751 falls between entries 0.683 and 1.174. Use two-tailed areas to find 0.250 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to indicate there is a difference between the

control and experimental groups in the mean score on the vocabulary portion of the test.

16. (a)

(b) Use the Student’s t distribution. Both sample sizes are large, and and are unknown.

(c) Since d.f. = n 1 = 30 1 = 29.From Table 6 in Appendix II, 1.524 falls between entries 1.479 and 1.699. Use one-tailed areas to find 0.050 0.01, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, the evidence is insufficient to claim that the population mean score for

the experimental group was higher than for the control group.




(ii) (a)

(b) Use the Student’s t distribution. We assume that the population distributions for both are approximately normal and mound-shaped and that and are unknown.

(c) Since From Table 6 in Appendix II, 0.869 falls between entries 0.692 and 1.200. Use two-tailed

areas to find 0.250 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, the evidence is insufficient to indicate that there is a difference in

the mean number of cases of fox rabies between the two regions.


(ii) (a)

(b) Use the Student’s t distribution. We assume the population distributions for both are approximately normal and mound-shaped and that and are unknown.

(c) Since From Table 6 in Appendix II, 2.008 falls between entries 1.771 and 2.160. Use one-tailed




(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to indicate that population mean soil

water content is higher in Field A.


(ii) (a)

(b) Use the Student’s t distribution. We assume the population distributions for both are approximately normal and mound-shaped and that and are unknown.

(c) Since From Table 6 in Appendix II, 1.041 falls between entries 0.718 and 1.273. Use two-tailed




(d) Since the P-value interval is > 0.05, we fail to reject No, the data are not statistically significant.

(e) At the 5% level of significance, the evidence is insufficient to indicate that the mean time lost owing to hot tempers is different from that lost owing to technical workers’ attitudes.

20. (a) t0.95 = 2.365 with d.f. = 7; E 2.365; 95% confidence interval for d from –4.25 to 8.75. We are 95% confident that the difference in population mean percentage increase between company revenue and CEO salaries is between –4.25% and 8.75%.

(b) = 0.05; c = 1 – = 0.95; H0: d = 0; H1: d ≠ 0; since d = 0 from the null hypothesis is in the 95% confidence interval, do not reject H0 at the 5% level of significance. The data do not indicate a difference in population mean percentage increases between company revenue and CEO salaries. This result is consistent with the conclusion reached using the P-value method of testing for = 0.05 in Problem 7.

21. (a)

(Note: Some software will truncate this to 19.)(b) In Problem 13,

The convention of using the smaller of and leads to a d.f. that is always less than or equal to that computed by Satterthwaite’s formula.

22. (a)

From Table 6 in Appendix II, 0.865 falls between entries 0.683 and 1.174. Using two-tailed areas, find 0.250 0.05, we fail to reject

(b) Using unpooled sample standard deviation, t 0.869, d.f. = 14, and fail to reject Conclusions are the same, however; the exact P value using the pooled standard deviation method and larger d.f. will be slightly smaller than that using unpooled methods.

23. (a)

(b) Use the standard normal distribution. The number of trials is sufficiently large because and are each larger than 5.



(c) P value = 2P(z < 1.13) = 2(0.1292) = 0.2584

(d) Since the P value = 0.2584 > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 1% level of significance, there is insufficient evidence to conclude that the population

proportion of women favoring more tax dollars for the arts is different from the proportion of men.

24. (a)




(c) P value = P(z < 0.61) = 0.2709


proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.

25. (a)


(c) P value = 2P(z > 0.79) = 2(0.2148) = 0.4296




proportion of high-school dropouts on Oahu is different from that of Sweetwater County.

26. (a)


(c) P value = P(z < 1.98) = 0.0239



(d) Since the P value = 0.0239 < 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, there is sufficient evidence to conclude that the population proportion

of voter turnout in Colorado is greater than that in California.

27. (a)


(c) P value = P(z < -1.43) 0.0764


proportion of adults believing in extraterrestrials who attended college is higher than the proportion who did not attend college.

28. (a)




(c) P value = P(z > 2.13) = 0.0166


of conservative voters who prefer art with fully clothed people is higher.

29. (a)


(c) P value = P(z < 2.04) = 0.0207




of trusting people in Chicago is higher in the older group.

30. for = 0.01 and a right-tailed test, the critical value sample test statistic z = 2.57 is in the critical region, reject This result is consistent with the P-value method.

31. for d.f. = 9, = 0.01 in the one-tail area row, the critical value Sample test statistic t = 0.965 is not in the critical region, fail to reject This result is consistent with the P-value method.

32. for = 0.05 and a two-tailed test, the critical values are sample test statistic z = 1.13 is not in the critical region, fail to reject This result is consistent with the P-value method.

Chapter 9 Review1. Look at the original x distribution. If it is normal or n ≥ 30, and σ is known, use the standard normal

distribution. If the x distribution is mound-shaped or n ≥ 30, and σ is unknown, use the Student’s t distribution. The degrees of freedom are determined by the application.

2. A significant test means we reject H0. Statistically significant results are not necessarily practically significant results.

3. A larger sample size increases the value of z or t.

4. A larger value of z or t will have a smaller P value.

5. (a)

Since is in use a two-tailed test.



(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard deviation Note that is given in the null hypothesis.

(Note: 600 miles = 0.6 thousand miles)(c) P value = 2P(z < 3.00) = 2(0.0013) = 0.0026

(d) Since a P value of 0.0026 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to say that the miles driven per vehicle in

Chicago is different from the national average.

6. (a)


(c) P value = P(z > 2.48) = 0.0066



(d) Since a P value of 0.0066 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence indicates that more than 35% of the students have jobs.

7. (a)


(c) For d.f. = 8, 4.390 falls between entries 3.355 and 5.041. Using one-tailed areas, 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0012.

(d) Since the entire P-value interval 0.01, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to say that the Toylot claim of 0.8 A is too

low.



8. (a)


(c) Since d.f. = n 1 = 12 1 = 11.In Table 6 in Appendix II, 2.986 falls between entries 2.718 and 3.106. Use one-tailed areas to find that 0.005 < P value < 0.010. Using a TI-84, d.f. ≈ 21.95, and P value ≈ 0.0034.

(d) Since the P-value interval is 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the evidence shows that the yellow paint has less visibility after

1 year.

9. (a)

(b) Use the standard normal distribution. The distribution is approximately normal when n is sufficiently large, which it is here, because np = 90(0.6) = 54 and nq = 90(0.4) = 36, and both are greater than5.

(c) P value = P(z < 3.01) = 0.0013



(d) Since a P value of 0.0013 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the evidence is sufficient to show the mortality rate has dropped.

10. (a)


(c) Since In Table 6 in Appendix II, 2.735 falls between entries 2.678 and 3.496. Use two-tailed areas to find that 0.001 < P value < 0.010. Using a TI-84, d.f. ≈ 94.53, and P value ≈ 0.0074.

(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence shows that there is a significant difference in average off-

schedule times.



11. (a)

Since > is in use a right-tailed test.(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard


(c) P value = P(z > 3.34) = 0.0004

(d) Since a P value of 0.0004 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, the evidence is sufficient to say that the population average number of

matches is larger than 40.

12. (a)


(c) P value = P(z < 0.91) = 0.1814



(d) Since the P value = 0.1814 > 0.05, we fail to reject No, the data are not statistically significant.(e) At the 5% level of significance, there is insufficient evidence to say that the population proportion of

suburban residents subscribing to Sporting News is higher.

13. (a)


(c) Since In Table 6 in Appendix II, 1.808 falls between entries 1.676 and 2.009. Use two-tailed areas to find that 0.050 0.05, we fail to reject No, the data are not statistically significant.



(e) At the 5% level of significance, the evidence is insufficient to show that there is a difference in population mean length of the two types of projectile points.

14. (a)


(c) P value = P(z < 1.94) = 0.0262

(d) Since a P value of 0.0262 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, the evidence is sufficient to show that the population percentage of

bachelor or higher degree holders in the private sector is less than in the federal civilian sector.

15. (a)


(c) For d.f. = 7, 1.697 falls between entries 1.617 and 1.895. Using two-tailed areas, 0.100 0.05, we fail to reject No, the data are not statistically significant.

(e) At the 5% level of significance, the evidence is insufficient to show that the population mean amount of coffee per cup is different from 7 ounces.

16. (a)

(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped, symmetric distribution.

Pair A B C D E Fd = exp cont 14 14 7 7 12 5

(c) d.f. = n 1 = 6 1 = 5From Table 6 in Appendix II, 6.07 falls between entries 4.032 and 6.869. Using one-tailed areas, find 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0009.



(d) Since the P-value interval is 0.01, we reject Yes, the data are statistically significant.(e) At the 1% level of significance, there is sufficient evidence to indicate that the program of the

experimental group promoted creative problem solving.

17. (a)

(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped, symmetric distribution.

Pair 1 2 3 4 5d = before after 6.4 7.2 8.6 3.8 1.3




(d) Since the P-value interval is 0.05, we reject Yes, the data are statistically significant.(e) At the 5% level of significance, there is insufficient evidence to claim that the injection system lasts

less than an average of 48 months.

18. (i) Use a calculator to verify. The rounded values are used in part (ii).

(ii) (a)


(c) For d.f. = 9, 1.396 falls between entries 1.383 and 1.574. Use one-tailed areas to find that0.075 0.05, we fail to reject No, the data are not statistically significant.

(e) At the 5% level of significance, there is insufficient evidence to claim that the injection system lasts less than an average of 48 months.

Cumulative Review Chapters 7–91. (a) Answers vary. A population is the entire group of interest. Random sampling is accomplished through

the assigning of numbers to members of the population and then by using a random-number table or a computer package to draw the simple random sample.

(b) A sample statistic is a numerical value calculated from the sample. A sampling distribution is the probability distribution associated with a sample statistics such as the sample mean.

(c) See Chapter 7.(d) Answers vary.

2. (a) Since the sample size is large, the central limit theorem describes the distribution of (b)



(c) The probability that the 50 workers’ average white blood cell count is that low or lower, if they were healthy adults, is extremely low. It would be reasonable to gather additional facts.

3. We are given p = 0.45 and n = 32. Both np > 5 and nq > 5.(a)

(b)

(c) Yes. Here, np = 32 × 0.45 = 14.4 and nq = 32 × 0.55 = 17.6, both greater than 5.The expected value of .

The standard deviation of .

4. (a) H0: μ = 2.0H1: μ > 2.0α = 0.01Since σ is known and x is normal, we will use the standard normal distribution.

The P value will be P(z > 2.53) = 0.0057.

Since the P value = 0.0057 < 0.01, we reject H0. Yes, the data are statistically significant.



At the 1% level of significance, the evidence is sufficient to say that the population mean discharge level of lead is higher.

(b)

(c) so use n = 48 water samples.

5. (a) Use a calculator to confirm that and s ≈ 1.36.(b) H0: μ = 10%

H1: μ > 10%α = 0.05Since x is approximately normal and σ is unknown, we will use a Student’s t distribution.d.f. = 12 – 1 = 11

Using a TI-84, the P value will be P(t > 1.248) = 0.119.Using Table 6 in Appendix II, 0.100 0.05, we fail to reject H0. The data are not statistically significant.At the 5% level of significance, the evidence does not indicate that the patient is asymptotic.

(c) For a 99% interval, tc = 3.106.



6. (a) Check conditions: Since both estimates are reasonably greater than 5, the conditions are satisfied.H0: p = 0.10H1: p ≠ 0.10α = 0.05Since the conditions are satisfied, follows a normal distribution.

The P value will be 2 × P(z > 1.29) = 0.197.

Since the P value > 0.05, we fail to reject H0. No, the data are not statistically significant.At the 5% level of significance, the data do not indicate any difference from the national average for the population proportion of crime victims.

(b) For a 95% confidence interval, zc = 1.96.

(c) so use n = 193 students.



7. H0: μd = 0H1: μd ≠ 0α = 0.05

Pair 1 2 3 4 5 6 7d = A – B –0.001 –0.028 –0.002 0.016 –0.01 0.001 –0.003

Here, , sd = 0.013, and we will use the Student’s t distribution with d.f. = 7 – 1 = 6.

Using a TI-84, the P value will be 2 × P(t < –0.771) = 0.470.Using Table 6 in Appendix II, 0.250 0.05, we fail to reject H0. No, the data are not statistically significant.At the 5% level of significance, the evidence does not show a difference in phosphorous reduction between the two methods.

8. (a) H0: μ1 = μ2

H1: μ1 ≠ μ2

α = 0.05We will use the Student’s t distribution with d.f. = 16 – 1 = 15.

Using the TI-84, the P value will be 2 × P(t > 1.952) = 0.070.Using Table 6 in Appendix II, 0.050 0.05, we fail to reject H0. No, the data are not statistically significant.At the 5% level of significance, the evidence does not show any difference in the population mean proportion of on-time arrivals in summer versus winter.

(b)

(c) We assume that x1 and x2 are approximately normal or at least mound-shaped and symmetric.

9. (a) H0: p1 = p2

H1: p1 > p2

α = 0.05We check that and all are greater than 5. It thus seems reasonable to use the normal distribution.

The P value will be P(z > 0.58) = 0.2810.



Since P value > 0.05, we fail to reject H0. No, the data are not statistically significant.At the 5% level of significance, the evidence does not indicate that the population proportion of single men is greater than the population proportion of single women.

(b)

10. (a) Essay. Refer to Chapters 7, 8, and 9.(b) Answers will vary.

11. Answers will vary. Salient points include that the hypotheses of interest areH0: μ1 = μ2

H1: μ1 < μ2, with μ1 representing the average time to recovery under the new methodAt the 1% level of significance, the data are statistically significant, indicating that the new method leads to shorter recovery times, on average.

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