Chapter 8: Hypothesis Testing

Chapter 8: Hypothesis Testing Section 8.1 Note: 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 .β 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 0.H 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 significanceα that is the probability with which we are willing to risk a type I error (rejecting 0H when it is in fact true). 5. The probability of rejecting the null hypothesis when it is true is called the level of significance. The symbol is α , also the probability of a type I error. 6. This is the power of the test. 7. Fail to reject 0H . 8. Reject 0H . 9. ( )value 2 0.0092 0.0184P − = =

10. ( )1value 0.0134 0.00672P − = =

11. (a) 0 : 40H µ = (b) 0 : 40H µ ≠ (c) 0 : 40H µ > (d) 0 : 40H µ < 12. (a) 0 : 30H µ = (b) 0 : 30H µ ≠ (c) 0 : 30H µ > (d) 0 : 30H µ < 13. (a) Yes, because the x distribution is normal.

(b) 8 7 1.124

20

xzn

µσ− −

= = ≈

(c) ( ) ( )value 2 2 1.12 0.2627xP P z z P z− = > = > ≈

(d) Since value 0.2627 0.05,P α− = > = do not reject the null hypothesis. 14. (a) Yes, because the x distribution is normal.

(b) 4.5 6.3 2.403

16

xzn

µσ− −

= = ≈ −

(c) ( ) ( )value 2.40 0.0082xP P z z P z− = < = < − ≈

(d) Since value 0.0082 0.01,P α− = < = reject the null hypothesis. 15. (a) The claim is µ = 60 kg, so you would use 0:H µ = 60 kg. (b) We want to know if the average weight is less than 60 kg, so you would use 1:H µ < 60 kg. (c) We want to know if the average weight is greater than 60 kg, so you would use 1:H µ > 60 kg. (d) We want to know if the average weight is different from 60 kg, so you would use 1:H µ ≠ 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. 16. (a) The claim is µ = 8.3 min, so you would use 0:H µ = 8.3 min. If you believe that the average is less than 8.3 min, then you would use 1:H µ < 8.3 min. This is a left-tailed test. (b) The claim is µ = 8.3 min, so you would use 0:H µ = 8.3 min. If you believe the average is different from 8.3 min, then you would use 1:H µ ≠ 8.3 min. This is a two-tailed test. (c) The claim is µ = 4.5 min, so you would use 0:H µ = 4.5 min. If you believe the average is more than 4.5 min, then you would use 1:H µ > 4.5 min. This is a right-tailed test. (d) The claim is µ = 4.5 min, so you would use 0:H µ = 4.5 min. If you believe the average is different from 4.5 min, then you would use 1:H µ ≠ 4.5 min. This is a two-tailed test. 17. (a) The claim is µ = 16.4 ft, so 0:H µ = 16.4 ft. (b) You want to know if the average is getting larger, so 1:H µ > 16.4 ft. (c) You want to know if the average is getting smaller, so 1:H µ < 16.4 ft. (d) You want to know if the average is different from 16.4 ft, so 1:H µ ≠ 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. 18. (a) The claim is µ = 8.7 s, so 0:H µ = 8.7 s. (b) You want to know if the average is longer, so 1:H µ > 8.7 s. (c) You want to know if the average is reduced, so 1:H µ < 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. 19. (a)

01

0.01: 4.7%: 4.7%

HH

==>

αµµ

Since > is in 1,H use a right-tailed test.

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

2.410

5.38 4.7 0.90n

xz σµ− −

= = ≈

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

(d) Since a P value of 0.1841 > 0.01 for α, we fail to reject 0.H 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. 20. (a)

01

0.05: 85 mg/100 mL: 85 mg/100 mL

HH

αµµ

==>

Since > is in 1,H use a right-tailed test. (b) Use the standard normal distribution. We assume that x has a normal distribution with known standard deviation .σ Note that µ is given in the null hypothesis.

12.58

93.8 85 1.99n

xz σµ− −

= = ≈

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

(d) Since a P value of 0.0233 ≤ 0.05, reject 0.H Yes, the data are statistically significant. (e) The sample evidence is sufficient at the 0.05 level to justify rejecting 0.H It seems Gentle Ben’s glucose is higher than average.

21. (a) 01

0.01: 4.55 g: 4.55 g

HH

αµµ

==<

Since < is in 1,H use a left-tailed test. (b) Use the standard normal distribution. We assume that x has a normal distribution with known standard deviation .σ Note that µ is given in the null hypothesis.

0.706

3.75 4.55 2.80n

xz σµ− −

= = ≈ −

(c) P value = P(z < −2.80) = 0.0026

(d) Since a P value of 0.0026 ≤ 0.01, we reject 0.H Yes, the data are statistically significant. (e) The sample evidence is sufficient at the 0.01 level to justify rejecting 0.H It seems the humming birds in the Grand Canyon region have a lower average weight. 22. (a)

01

0.05: 19: 19

HH

αµµ

==<

Since < is in 1,H use a left-tailed test. (b) Use the standard normal distribution. We assume that x has a normal distribution with known standard deviation .σ Note that µ is given in the null hypothesis.

4.514

17.1 19 1.58n

xz σµ− −

= = ≈ −

(c) P value = P(z < −1.58) = 0.0571

(d) Since a P value of 0.0571 > 0.05 for α, we fail to reject 0.H The data are not statistically significant. (e) There is insufficient evidence at the 0.05 level to reject 0.H It seems the average P/E for large banks is not less than that of the S&P 500 Index. 23. (a)

01

0.01: 11%: 11%

HH

αµµ

==≠

Since ≠ is in 1,H use a two-tailed test. (b) Use the standard normal distribution. We assume that x has a normal distribution with known standard deviation .σ Note that µ is given in the null hypothesis.

5.016

12.5 11 1.20n

xz σµ− −

= = =

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

(d) Since a P value of 0.2302 > 0.01 for α, we fail to reject 0.H The data are not statistically significant. (e) There is insufficient evidence at the 0.01 level of significance to reject 0.H It seems the average hail damage to wheat crops in Weld County matches the national average. 24. (a)

01

0.01: 28 mL/kg: 28 mL/kg

HH

αµµ

==≠


4.757

32.7 28 2.62n

xz σµ− −

= = ≈

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

(d) Since a P value of 0.0088 ≤ 0.01, we reject 0.H The data are statistically significant. (e) At the 1% level of significance, the sample average is sufficiently different from µ = 28 that we reject 0.H It seems Roger’s average red cell volume is different from the average for healthy adults. Section 8.2 1. 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) 0.010 value 0.020P< − < ; technology gives value 0.015P − ≈ . (b) 0.005 value 0.010P< − < ; technology gives value 0.0075P − ≈ . 8. (a) 0.050 value 0.100P< − < ; technology gives value 0.08594P − ≈ . (b) 0.025 value 0.050P< − < ; technology gives value 0.04297P − ≈ . 9. (a) Yes, because the x distribution is mound-shaped and symmetric and σ is unknown; 1 25 1 24df n= − = − = (b) 0 1: 9.5; : 9.5;H Hµ µ= ≠

(c) 10 9.5 1.252

25

xt sn

µ− −= = =

(d) Using Table 6, 0.200 value 0.250P< − < .

(e) Since value 0.05,P α− > = do not reject the null hypothesis. (f) The sample evidence is insufficient at the 0.05 level to reject the null hypothesis. 10. (a) Yes, because the sample size is large enough and σ is unknown; 1 49 1 48df n= − = − =

(b) 0 1: 9.2; : 9.2;H Hµ µ= <

(c) 8.5 9.2 3.2671.5

49

xt sn

µ− −= = = −

(d) Using Table 6, 0.0005 value 0.005P< − < . (e) Since value 0.01,P α− < = reject the null hypothesis. (f) The sample evidence is sufficient at the 0.01 level to reject the null hypothesis. 11. (a)

01

0.01: 16.4 ft: 16.4 ft

HH

αµµ

==>

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

3.536

17.3 16.4 1.54n

xz σµ− −

= = ≈

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

(d) Since a P value of 0.0618 > 0.01, we fail to reject 0.H The data are not statistically significant. (e) At the 1% level of significance, there is insufficient evidence to say average storm level is increasing. 12. (a)

01

0.05: 7.4 pH: 7.4 pH

HH

αµµ

==≠

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

1.931

8.1 7.4 2.051sn

xt µ− −= = ≈

(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 0.H 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. 13. (a)

01

0.01: 1.75 years: 1.75 years

HH

αµµ

==>


0.8246

2.05 1.75 2.481sn

xt µ− −= = ≈

(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 0.H 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. 14. (a)

01

0.05: 19 in.: 19 in.

HH

αµµ

==<


3.251

18.5 19 1.116sn

xt µ− −= = ≈ −

(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 0.H 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. 15. (a)

01

0.05: 19.4: 19.4

HH

αµµ

==≠


5.236

17.9 19.4 1.731sn

xt µ− −= = ≈ −

(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 0.H 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. 16. (a)

01

0.01: 8.0 ppb: 8.0 ppb

HH

αµµ

==<


1.937

7.2 8.0 2.561sn

xt µ− −= = ≈ −

(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 0.H 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. 17. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.05: 4.8: 4.8

HH

αµµ

==<

(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 σ is unknown.

0.286

4.40 4.8 3.499sn

xt µ− −= = ≈ −


3.5002.0001.0000.000-1.000-2.000-3.499

Student's t Distribution with d.f. = 5

(d) Since the entire P-value interval ≤ 0.05, we reject 0.H 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.

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

01

0.01: 14: 14

HH

αµµ

==>


2.5110

15.1 14 1.386sn

xt µ− −= = ≈


(d) Since P-value interval > 0.01, we fail to reject 0.H 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. 19. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.01: 67: 67

HH

αµµ

==≠


10.616

61.8 67 1.962sn

xt µ− −= = ≈ −


(d) Since P-value interval > 0.01, we fail to reject 0.H 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. 20. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.05: 77 years: 77 years

HH

αµµ

==<


20.6520

71.4 77 1.213sn

xt µ− −= = ≈ −


(d) Since the P-value interval > 0.05, we fail to reject 0.H 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. 21. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.05: 8.8: 8.8

HH

αµµ

==≠


4.0314

7.36 8.8 1.337sn

xt µ− −= = ≈ −


(d) Since P-value interval > 0.05, we fail to reject 0.H 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. 22. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.01: 1,300: 1,300

HH

αµµ

==≠


37.2910

1268 1300 2.714sn

xt − −= = ≈ −

µ

(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 0.H 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. 23. (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 α, whereas the P value of a two-tailed test is larger than α. (c) Yes; if the two-tailed P value is less than α, the one-tail area is also less than α. (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 0H ) in order to reject 0.H 24. Essay or class discussion.

25. (a) 0

1

: 20: 20

HH

µµ=

≠

For α = 0.01, c = 1 − 0.01 = 0.99, 4,σ = and 2.58.cz =

42.58 1.7236cE z

nσ

≈ = =

22 1.72 22 1.72

20.28 23.72

x E x Eµµµ

− < < +− < < +

< <

The hypothesized mean µ = 20 is not in the interval. Therefore, we reject 0.H (b) Because n = 36 is large, the sampling distribution of x is approximately normal by the central limit theorem, and we know σ.

22 20 3.00/ 4/ 36

xznµ

σ− −

= = =

From Table 5, P value = 2P(z < −3.00) = 2(0.0013) = 0.0026. Since 0.0026 ≤ 0.01, we reject 0.H The results are the same. 26. (a) 0

1

: 21: 21

HH

µµ=

≠

For α = 0.01, c = 1 − 0.01 = 0.99, 4,σ = and 2.58.cz =

42.58 1.7236cE z

nσ

≈ = =

22 1.72 22 1.72

20.28 23.72

x E x Eµµµ

− < < +− < < +

< <

The hypothesized mean µ = 21 falls into the confidence interval. Therefore, we do not reject 0.H (b) For α = 0.01, the two-tailed test’s critical values are 0 2.58.z = ± Because n = 36 is large, the sampling distribution of x is approximately normal by the central limit theorem, and we know σ.

22 21 1.50/ 4/ 36

xznµ

σ− −

= = =

From Table 5, P value = 2P(z < −1.50) = 2(0.0668) = 0.1336. Since 0.1336 > 0.01, we do not reject 0.H The results are the same. 27. For a right-tailed test and α = 0.01, the critical value is 0 2.33;z = 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 0.H 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. 28. For a two-tailed test and α = 0.05, critical values are 0 2.042t = ± 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 0.H At the 5% level, the evidence is sufficient to say that the drug has changed the mean pH level. The conclusion is the same as with the P-value method.

29. For a right-tailed test and α = 0.01, critical value is 0 2.412t = 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 0.H 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 α = 0.05, critical value is 0 1.676t = − 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 0.H 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 8.3 1. The value of p comes from H0

. Note that q = 1 – p.

2. p̂ pz

pqn

−= , where ˆ rp

n= is the sample proportion, p is the value from the null hypothesis, and 1q p= − .

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. (a) Yes, ( ) ( )30 0.5 15 5; 30 0.5 15 5np nq= = > = = > .

(b) 0 1: 0.50; : 0.50H p H p= ≠

(c) 12ˆ 0.4;30

p = = ( )

ˆ 0.4 0.5 1.100.5 0.5

30

p pzpqn

− −= = ≈ −

(d) ( ) ( )ˆvalue 2 2 1.10 0.2713pP P z z P z− = > = > − ≈

(e) Since value 0.2713 0.05,P α− = > = do not reject the null hypothesis. (f) The sample proportion based on 30 trials is not sufficiently different from 0.50 to justify rejecting the null hypothesis at the 5% level of significance. 6. (a) Yes, ( ) ( )60 0.18 10.8 5; 60 0.82 49.2 5np nq= = > = = > .

(b) 0 1: 0.18; : 0.18H p H p= >

(c) 18ˆ 0.3;60

p = = ( )

ˆ 0.3 0.18 2.420.18 0.82

60

p pzpqn

− −= = ≈

(d) ( ) ( )ˆvalue 2.42 0.0078pP P z z P z− = > = > ≈

(e) Since value 0.0078 0.01,P α− = < = reject the null hypothesis.

(f) The sample proportion based on 60 trials is sufficiently different from 0.18 to justify rejecting the null hypothesis at the 1% level of significance. 7. (i) (a)

01

0.01: 0.301: 0.301

H pH p

α ==<

(b) Use the standard normal distribution. The p̂ 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.

46ˆ 0.214215

rpn

= = ≈

0.301(0.699)

215

ˆ 0.214 0.301 2.78pqn

p pz − −= = ≈ −

(c) P value = P(z < −2.78) = 0.0027

(d) Since a P value of 0.0027 ≤ 0.01, we reject 0.H 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 0H to be false. However, because our sample data lead us to reject 0H and conclude that there are too few numbers with a leading digit 1, more investigation is merited. 8. (i) (a)

01

0.01: 0.301: 0.301

H pH p

α ==<

(b) Use the standard normal distribution. The p̂ 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.

92ˆ 0.404228

rpn

= = ≈

0.301(0.699)

228

ˆ 0.404 0.301 3.39pqn

p pz − −= = ≈

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

3.393.002.001.000.00-1.00-2.00-3.00-3.50

Standard Normal Distribution

(d) Since a P value of 0.0003 ≤ 0.01, we reject 0.H 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 0H to be false. However, because our data led us to reject 0H and conclude that there are “too many” numbers with a leading digit 1, more investigation is merited. 9. (a)

01

0.01: 0.70: 0.70

H pH p

α ==≠

(b) Use the standard normal distribution. The p̂ 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.

24ˆ 0.7532

rpn

= = ≈

0.70(0.30)

32

ˆ 0.75 0.70 0.62pqn

p pz − −= = ≈

(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 0.H 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%.

10. (a) 01

0.05: 0.67: 0.67

H pH p

α ==<


21ˆ 0.552638

rpn

= = =

0.67(0.33)

38

ˆ 0.5526 0.67 1.54pqn

p pz − −= = = −

(c) P value = P(z < −1.54) = 0.0618

(d) Since a P value of 0.0618 > 0.05, we fail to reject 0.H 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%. 11. (a)

01

0.01: 0.77: 0.77

H pH p

α ==<


15ˆ 0.555627

rpn

= = =

0.77(0.23)

27

ˆ 0.5556 0.77 2.65pqn

p pz − −= = = −

(c) P value = P(z < −2.65) = 0.0004

(d) Since a P value of 0.0004 ≤ 0.01, we reject 0.H 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. 12. (a)

01

0.05: 0.24: 0.24

H pH p

α ==≠


12ˆ 0.214356

rpn

= = ≈

0.24(0.76)

56

ˆ 0.2143 0.24 0.45pqn

p pz − −= = ≈ −

(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 0.H 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. 13. (a)

01

0.01: 0.50: 0.50

H pH p

α ==<

(b) Use the standard normal distribution. The p̂ 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.

10ˆ 0.294134

rpn

= = =

0.5(0.5)

34

ˆ 0.2941 0.50 2.40pqn

p pz − −= = = −

(c) P value = P(z < −2.40) = 0.0082

(d) Since a P value of 0.0082 ≤ 0.01, we reject 0.H 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. 14. (a)

01

0.05: 0.75: 0.75

H pH p

α ==≠


64ˆ 0.771183

rpn

= = =

0.75(0.25)

83

ˆ 0.7711 0.75 0.44pqn

p pz − −= = =

(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 0.H 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%. 15. (a)

01

0.01: 0.261: 0.261

H pH p

α ==≠


61ˆ 0.1924317

rpn

= = =

0.261(0.739)

317

ˆ 0.1924 0.261 2.78pqn

p pz − −= = ≈ −

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

(d) Since a P value of 0.0054 ≤ 0.01, we reject 0.H 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. 16. (a)

01

0.01: 0.214: 0.214

H pH p

α ==>


136ˆ 0.2759493

rpn

= = =

0.214(0.786)

493

ˆ 0.2759 0.214 3.35pqn

p pz − −= = =

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

4.003.352.001.000.00-1.00-2.00-3.00-4.00


(d) Since a P value of 0.0004 ≤ 0.01, we reject 0.H 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. 17. (a)

01

0.01: 0.47: 0.47

H pH p

α ==>


490ˆ 0.48711006

rpn

= = =

0.47(0.53)

1006

ˆ 0.4871 0.47 1.09pqn

p pz − −= = =

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

(d) Since a P of 0.1379 > 0.01, we fail to reject 0.H 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%. 18. (a)

01

0.05: 0.80: 0.80

H pH p

α ==<

(b) Use the standard normal distribution. The p̂ 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.

88ˆ 0.7652115

rpn

= = =

0.80(0.20)

115

ˆ 0.7652 0.80 0.93pqn

p pz − −= = = −

(c) P value = P(z < −0.93) = 0.1762

(d) Since a P value of 0.1762 > 0.05, we fail to reject 0.H 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%. 19. (a)

01

0.05: 0.092: 0.092

H pH p

α ==>


29ˆ 0.1480196

rpn

= = =

0.092(0.908)

196

ˆ 0.1480 0.092 2.71pqn

p pz − −= = =

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

(d) Since a P value of 0.0034 ≤ 0.05, we reject 0.H 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%. 20. (a)

01

0.01: 0.12: 0.12

H pH p

α ==<


16ˆ 0.0766209

rpn

= = =

0.12(0.88)

209

ˆ 0.0766 0.12 1.93pqn

p pz − −= = = −

(c) P value = P(z < −1.93) = 0.0268

(d) Since a P value of 0.0268 > 0.01, we fail to reject 0.H 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. 21. (a)

01

0.01: 0.82: 0.82

H pH p

α ==≠


56ˆ 0.767173

rpn

= = =

0.82(0.18)

73

ˆ 0.7671 0.82 1.18pqn

p pz − −= = = −

(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 0.H 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%. 22. For a two-tailed test and α = 0.01, critical values are 0 2.58.z± = ± 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 0.H Result is consistent with the P-value conclusion. 23. For a left-tailed test and α = 0.01, critical value is 0 2.33.z = − 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 0.H Result is consistent with the P-value conclusion. 24. For a two-tailed test and α = 0.01, critical value is 0 2.33.z = 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 0.H Result is consistent with the P-value conclusion. Section 8.4 1. Paired data are dependent.

2. Take the difference of corresponding paired data values. The formula for the test statistic is d

d

dt sn

µ−= , with

d being the mean of the differences from the sample, dµ being the mean difference from the population, ds being the standard deviation of the differences from the sample, and n being the sample size. 3. H0 0dµ =: . 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) Yes. The sample size is sufficiently large. Use a Student’s t with 1 35df n= − = .

(b) 0 1: 0; : 0d dH Hµ µ= ≠

(c) 0.8 0 2.42

36

d

d

dt sn

µ− −= = =

(d) Table 6 gives 0.02 value 0.05P< − < ; TI-84 gives value 0.0218P − = (e) Reject the null hypothesis as value 0.0218 0.05P α− = < = . (f) At the 5% level of significance, the sample mean of differences is sufficiently different from 0 that we conclude the population mean of the differences is not zero. 8. (a) Yes. The distribution of differences is mound-shaped and symmetric. Use a Student’s t with 1 19df n= − = .

(b) 0 1: 0; : 0d dH Hµ µ= >

(c) 2 0 1.7895

20

d

d

dt sn

µ− −= = =

(d) Table 6 gives 0.025 value 0.05P< − < ; TI-84 gives value 0.0448P − = (e) Do not reject the null hypothesis as value 0.0448 0.01P α− = > = . (f) At the 1% level of significance, the sample mean of differences is not sufficiently different from 0 to justify rejecting the null hypothesis. 9. (a)

01

0.05: 0: 0

dd

HH

αµµ

==≠

Since ≠ is in 1,H 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 8 d = B − A 3 −2 5 4 10 −15 6 7

2.25, 7.78dd s= =

7.788

2.25 0 0.818d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 8 − 1 = 7 From 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 0.H 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. 10. (a)

01

0.01: 0: 0

dd

HH

αµµ

==≠


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

0.37, 0.47dd s= =

0.477

0.37 0 2.08d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 7 − 1 = 6 From 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 0.H 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. 11. (a)

01

0.01: 0: 0

dd

HH

αµµ

==>

Since > is in 1,H 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, symmetric distribution.

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

12.6, 22.66dd s= =

22.665

12.6 0 1.243d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 5 − 1 = 4 From 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 0.H 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. 12. (a)

01

0.05: 0: 0

dd

HH

αµµ

==>


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

4.50, 4.1226dd s= =

4.122610

4.50 0 3.452d

ds

n

dt − −= = ≈

µ

(c) d.f. = n − 1 = 10 − 1 = 9 From 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.

4.0003.4522.0001.0000.000-1.000-2.000-3.000-4.000

Student's t Distribution, d.f. = 9

(d) Since the P-value interval is ≤ 0.05, we reject 0.H 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. 13. (a)

01

0.05: 0: 0

dd

HH

αµµ

==>


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

6.125, 9.83dd s= =

9.838

6.125 0 1.76d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 8 − 1 = 7 From 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 0.H 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.

14. (a) 01

0.01: 0: 0

dd

HH

αµµ

==≠


Pair 1 2 3 4 5 6 7 8 9 d = 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 16 −1.3 4.0 5.6 −7.6 0.9 4.1 6.5

1.10, 3.745dd s= =

3.74516

1.10 0 1.175d

ds

n

dt µ− −= = ≈

(c) d.f. = n − 1 = 16 − 1 = 15 From 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 0.H 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. 15. (a)

01

0.05: 0: 0

dd

HH

αµµ

==>


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

6, 21.5dd s= =

21.58

6 0 0.789d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 8 − 1 = 7 From 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 0.H 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. 16. (a)

01

0.05: 0: 0

dd

HH

αµµ

==>


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

6.125, 8.079dd s= =

8.0798

6.125 0 2.144d

ds

n

dt µ− −= = ≈


(d) Since the P-value interval is ≤ 0.05, we reject 0.H 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. 17. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.05: 0: 0

HH

αµµ

==>

Since > is in 1,H a right-tailed test is used. (b) Use the Student’s t distribution. The sample size is greater than 30. 2.472, 12.124dd s= =

12.12436

2.472 0 1.223d

ds

n

dt µ− −= = ≈


(d) Since the P-value interval > 0.05, we fail to reject 0.H 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. 18. (i) Use a calculator to verify. Rounded values are used in part (ii). (ii) (a)

01

0.05: 0: 0

HH

αµµ

==<

Since < is in 1,H a left-tailed test is used. (b) Use the Student’s t distribution. The sample size is greater than 30. 5.7391, 15.910dd s= − =

15.91046

5.7391 0 2.447d

ds

n

dt µ− − −= = ≈ −

(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 0.H 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. 19. (a)

01

0.05: 0: 0

dd

HH

αµµ

==>


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

2.0, 4.5dd s= =

4.59

2.0 0 1.33d

ds

n

dt µ− −= = =


(d) Since the P-value interval is > 0.05, we fail to reject 0.H 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. 20. (a)

01

0.05: 0: 0

dd

HH

αµµ

==>


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

0.4, 0.447dd s= =

0.4476

0.4 0 2.192d

ds

n

dt µ− −= = =


(d) Since the P-value interval is ≤ 0.05, we reject 0.H 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.

21. (a) 01

0.05: 0: 0

dd

HH

αµµ

==>


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

0.775, 1.0539dd s= =

1.05398

0.775 0 2.080d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 8 − 1 = 7 From 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 0.H 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. 22. (a) 2.25; 7.778dd s≈ ≈ ; t0.95 = 2.365 with d.f. = 7; E ≈ 2.365; 95% confidence interval for µd 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) α = 0.05; c = 1 –α = 0.95; H0: µd = 0; H1: µd ≠ 0; Since µd = 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 α = 0.05 in Problem 9.

23. For a two-tailed test with α = 0.05 and d.f. = 7, critical values are 0 2.365.t± = ± The sample test statistic t = 0.818 is between −2.365 and 2.365, so we do not reject 0.H This conclusion is the same as that reached by the P-value method. 24. For a right-tailed test with α = 0.01 and d.f. = 4, the critical value is 0 3.747.t = The sample test statistic t = 1.243 is to the left of 0 ,t so we do not reject 0.H This conclusion is the same as that reached by the P-value method.

Section 8.5 1. (a) 0H says the population means are equal.

(b) 1 2

1 2

1 2

x xz

n nσ σ−

=+

(c)

1 2

1 22 2

1 2

x xts sn n

−=

+

with df = smaller sample size – 1 or use Satterthwaite’s formula

2. Use the Student’s t distribution. 3. H0: μ1 = μ2 or H0: μ1 – μ2

= 0.

4. (a) 0H says the population proportions are equal.

(b) 1 2

1 2

ˆ ˆp pzpq pqn n

−=

+

5. The best estimate is 1 2

1 2

r rpn n+

=+

.

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) A Student’s t with 48df = . Samples are independent, the population standard deviations are not known, and the sample sizes are sufficiently large. (b) 0 1 2 1 1 2: ; :H Hµ µ µ µ= ≠

(c) 1 2 10 12 2;x x− = − = −

1 2

1 22 2 2 2

1 2

10 12 3.0373 449 64

x xts sn n

− −= = ≈ −

++

(d) Table 6 gives 0.001 value 0.01P< − < ; TI-84 gives value 0.0030P − = (e) Reject the null hypothesis as value 0.0030 0.01P α− = < = . (f) At the 1% level of significance, the sample evidence is sufficiently strong to reject the null hypothesis and conclude the population means are different. 10. (a) A Student’s t with 8df = . Samples are independent, the population standard deviations are not known, and the populations are mound-shaped and symmetric. (b) 0 1 2 1 1 2: ; :H Hµ µ µ µ= >

(c) 1 2 20 10 1;x x− = − =

1 2

1 22 2 2 2

1 2

20 19 0.8942 316 9

x xts sn n

− −= = ≈

++

(d) Table 6 gives 0.125 value 0.250P< − < ; TI-84 gives value 0.1943P − = (e) Do not reject the null hypothesis as value 0.1943 0.05P α− = > = . (f) At the 5% level of significance, the sample evidence is not sufficiently strong to reject the null hypothesis and conclude the population mean from the first population exceeds that of the second population. 11. (a) The standard normal distribution. Samples are independent, the population standard deviations are known, and the sample sizes are sufficiently large. (b) 0 1 2 1 1 2: ; :H Hµ µ µ µ= ≠

(c) 1 2 10 12 2;x x− = − = −

1 2

1 22 2 2 2

1 2

10 12 3.043 449 64

x xz

n nσ σ− −

= = ≈ −

++

(d) Table 5 gives value 0.0024P − = (e) Reject the null hypothesis as value 0.0024 0.01P α− = < = . (f) At the 1% level of significance, the sample evidence is sufficiently strong to reject the null hypothesis and conclude the population means are different. 12. (a) The standard normal distribution. Samples are independent, the population standard deviations are known, and the populations are normal. (b) 0 1 2 1 1 2: ; :H Hµ µ µ µ= >

(c) 1 2 20 10 1;x x− = − =

1 2

1 22 2 2 2

1 2

20 19 0.892 316 9

x xz

n nσ σ− −

= = ≈

++

(d) Table 5 gives value 0.1867P − = (e) Do not reject the null hypothesis as value 0.1867 0.05P α− = > = . (f) At the 5% level of significance, the sample evidence is not sufficiently strong to reject the null hypothesis and conclude the population mean from the first population exceeds that of the second population.

13. (a) 45 70 0.65775 100

p += ≈

+

(b) The standard normal distribution. ( )1 75 0.657 49.275 5n p = = >

( )2 100 0.657 65.7 5n p = = >

( )1 75 0.343 25.725 5n q = = >

( )2 100 0.343 34.3 5n q = = >

(c) 0 1 2:H p p= ; 1 1 2:H p p≠

(d) 1 245 70ˆ ˆ 0.175 100

p p− = − = − ; ( ) ( )

1 2

1 2

ˆ ˆ 0.6 0.7 1.380.657 0.343 0.657 0.343

75 100

p pzpq pqn n

− −= = ≈ −

+ +

(e) Table 5 gives value 0.1676P − =

(f) Do not reject the null hypothesis as value 0.1676 0.05P α− = > = . (g) At the 5% level of significance, the difference between the sample proportions is too small to justify rejecting the null hypothesis that the probabilities are equal.

14. (a) 60 156 0.36

200 400p += =

+

(b) The standard normal distribution. ( )1 200 0.36 72 5n p = = >

( )2 400 0.36 144 5n p = = >

( )1 200 0.64 128 5n q = = >

( )2 400 0.64 256 5n q = = >

(c) 0 1 2:H p p= ; 1 1 2:H p p<

(d) 1 260 156ˆ ˆ 0.09200 400

p p− = − = − ; ( ) ( )

1 2

1 2

ˆ ˆ 0.3 0.39 2.170.36 0.64 0.36 0.64

200 400

p pzpq pqn n

− −= = ≈ −

+ +

(e) Table 5 gives value 0.015P − = (f) Reject the null hypothesis as value 0.015 0.05P α− = < = . (g) At the 5% level of significance, the sample proportion from the first sample is sufficiently smaller than the sample proportion from the second sample to conclude the population proportion from the first population is smaller than the population proportion from the second population. 15. (a)

0 1 21 1 2

0.01: :

HH

αµ µµ µ

==>

(b) Use the standard normal distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are known. 1 2 2.8 2.1 0.7x x− = − =

2 2 2 21 21 2

1 2 1 2

0.5 0.710 10

( ) ( ) 0.7 0 2.57

n n

x xzσ σ

µ µ− − − −= = ≈

++

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

(d) Since P value = 0.0051 ≤ 0.01, we reject 0.H 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. 16. (a)

0 1 21 1 2

0.01: :

HH

αµ µµ µ

==≠

(b) Use the standard normal distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are known. 1 2 43 36 7x x− = − =

2 2 221 21 2

1 2 1 2

152112 14

( ) ( ) 7 0 0.96

n n

x xzσ σ

µ µ− − − −= = ≈

++

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

(d) Since P value = 0.3370 > 0.01, we fail to reject 0.H 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. 17. (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==≠

(b) Use the standard normal distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are known. 1 2 4.9 4.3 0.6x x− = − =

2 2 2 21 21 2

1 2 1 2

1.5 1.246 51

( ) ( ) 0.6 0 2.16

n n

x xzσ σ

µ µ− − − −= = ≈

++

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

(d) Since P value = 0.0308 ≤ 0.05, we reject 0.H 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. 18. (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==<

(b) Use the standard normal distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are known. 1 2 15.2 19.7 4.5x x− = − = −

2 2 2 21 21 2

1 2 1 2

7.2 5.232 35

( ) ( ) 4.5 0 2.91

n n

x xzσ σ

µ µ− − − − −= = ≈ −

++

(c) P value = P(z < −2.91) = 0.0018

(d) Since P value = 0.0018 ≤ 0.05, we reject 0.H 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. 19. (i) Use a calculator to verify. Use rounded results to compute t. (ii) (a)

0 1 21 1 2

0.01: :

HH

αµ µµ µ

==<

(b) Use the Student’s t distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are unknown.

1 2 3.51 3.87 0.36x x− = − = −

2 2 2 21 21 2

1 2 1 2

0.81 0.9410 12

( ) ( ) 0.36 0 0.965s sn n

x xt − − − − −= = ≈ −

++

µ µ

(c) Since 1 2 ,n n< 1. . 1 10 1 9d f n= − = − = 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 0.H 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. 20. (i) Use a calculator to verify. Use rounded results to compute t. (ii) (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==>

(b) Use the Student’s t distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are unknown. 1 2 109.50 99.36 10.14x x− = − =

2 2 2 21 21 2

1 2 1 2

15.41 11.5716 14

( ) ( ) 10.14 0 2.053s sn n

x xt − − − −= = ≈

++

µ µ

(c) Since 1 2 ,n n< 1. . 1 14 1 13d f n= − = − = . 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 0.H 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. 21. (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==≠

(b) Use the Student’s t distribution. Both sample sizes are large, 30,in ≥ and 1σ and 2σ are unknown. 1 2 344.5 354.2 9.7x x− = − = −

2 2 2 21 21 2

1 2 1 2

49.1 50.930 30

( ) ( ) 9.7 0 0.751s sn n

x xt − − − − −= = ≈ −

++

µ µ

(c) Since 1 2 ,n n= 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 0.H 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. 22. (a)

0 1 21 1 2

0.01: :

HH

αµ µµ µ

==>

(b) Use the Student’s t distribution. Both sample sizes are large, 30,in ≥ and 1σ and 2σ are unknown. 1 2 368.4 349.2 19.2x x− = − =

2 2 2 21 21 2

1 2 1 2

39.5 56.630 30

( ) ( ) 19.2 0 1.524s sn n

x xt − − − −= = ≈

++

µ µ

(c) Since 1 2 ,n n= 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 0.H 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. 23. (i) Use a calculator to verify. Use rounded results to compute t. (ii) (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==≠

(b) Use the Student’s t distribution. We assume that the population distributions for both are approximately normal and mound-shaped and that 1σ and 2σ are unknown. 1 2 4.75 3.93 0.82x x− = − =

2 2 2 21 21 2

1 2 1 2

2.82 2.4316 15

( ) ( ) 0.82 0 0.869s sn n

x xt − − − −= = ≈

++

µ µ

(c) Since 2 1,n n< 2. . 1 15 1 14.d f n= − = − = 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 0.H 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. 24. (i) Use a calculator to verify. Use rounded results to compute t. (ii) (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==>

(b) Use the Student’s t distribution. We assume the population distributions for both are approximately normal and mound-shaped and that 1σ and 2σ are unknown. 1 2 12.53 10.77 1.76x x− = − =

2 2 2 21 21 2

1 2 1 2

2.39 2.4014 16

( ) ( ) 1.76 0 2.008s sn n

x xt − − − −= = ≈

++

µ µ

(c) Since 1 2 ,n n< 1. . 1 14 1 13.d f n= − = − = From Table 6 in Appendix II, 2.008 falls between entries 1.771 and 2.160. Use one-tailed areas to find 0.025 < P value < 0.050. Using a TI-84, d.f. ≈ 27.50, and P value ≈ 0.0273.

(d) Since the P-value interval is ≤ 0.05, we reject 0.H 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. 25. (i) Use a calculator to verify. Use rounded results to compute t. (ii) (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==≠

(b) Use the Student’s t distribution. We assume the population distributions for both are approximately normal and mound-shaped and that 1σ and 2σ are unknown. 1 2 4.86 6.5 1.64x x− = − = −

2 2 2 21 21 2

1 2 1 2

3.18 2.887 8

( ) ( ) 1.64 0 1.041s sn n

x xt − − − − −= = ≈ −

++

µ µ

(c) Since 1 2 ,n n< 1. . 1 7 1 6.d f n= − = − = From Table 6 in Appendix II, 1.041 falls between entries 0.718 and 1.273. Use two-tailed areas to find 0.250 0.05, we fail to reject 0.H 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. 26. (i) Use a calculator to verify. Use rounded results to compute t. (ii) (a)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==<

(b) Use the Student’s t distribution. We assume the population distributions for both are approximately normal and mound-shaped and that 1σ and 2σ are unknown. 1 2 1.5x x− = −

2 2 2 21 21 2

1 2 1 2

2.38 2.787 8

( ) ( ) 1.5 0 1.126s sn n

x xt µ µ− − − − −= = ≈ −

++

(c) Since 1 2 ,n n< 1. . 1 7 1 6.d f n= − = − = From Table 6 in Appendix II, find 0.125 0.05, we fail to reject 0.H No, the data are not statistically significant. (e) At the 5% level of significance, the evidence is insufficient to indicate that the population mean time lost due to stressors is greater than that lost due to intimidators.

27. (a) ( )

( ) ( )

2 2 2 21 21 2

2 22 21 2

1 1 2 2

2 20.81 0.94

10 122 2 2 2

0.81 0.941 11 19 10 11 121 1

. . 19.96

s sn n

s sn n n n

d f

− −

+ + ≈ = ≈

++

(Note: Some software will truncate this to 19.) (b) In Problem 19, 1. . 1 10 1 9.d f n= − = − = The convention of using the smaller of 1 1n − and 2 1n − leads to a d.f. that is always less than or equal to that computed by Satterthwaite’s formula.

28. (a) 2 2 2 2

1 21 2

1 2

( 1) ( 1) 15(2.82 ) 14(2.43 ) 2.6392 16 15 2

n s n ss

n n− + − +

= = ≈+ − + −

1 2

1 21 1 1 1

16 15

4.75 3.93 0.8652.639n n

x xts

− −= = ≈

+ +

1 2. . 2 16 15 2 29d f n n= + − = + − = 0 1 2: H µ µ= 1 1 2: H µ µ≠ 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 0.H (b) Using unpooled sample standard deviation, t ≈ 0.869, d.f. = 14, and fail to reject 0.H 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.

29. (a) 0 1 21 1 2

0.05: :

H p pH p p

α ==≠

(b) Use the standard normal distribution. The number of trials is sufficiently large because 1 1 2, , ,n p n q n p and 2n q are each larger than 5.

1 2

1 2

59 56 0.2911220 175

r rpn n+ +

= = ≈+ +

1 1 0.2911 0.7089q p= − = − =

1 21 2

1 2

59 56ˆ ˆ0.268, 0.32220 175

r rp pn n

= = ≈ = = =

1 2ˆ ˆ 0.268 0.32 0.052p p− = − = −

1 2

1 20.2911(0.7089) 0.2911(0.7089)

220 175

ˆ ˆ 0.052 1.13pq pqn n

p pz − −= = ≈ −

+ +

(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 0.H 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. 30. (a)

0 1 21 1 2

0.05: :

H p pH p p

α ==<


1 2

1 2

21 22 0.244393 83

r rpn n+ +

= = =+ +

1 1 0.2443 0.7557q p= − = − =

1 21 2

1 2

21 22ˆ ˆ0.226, 0.26593 83

r rp pn n

= = ≈ = = ≈

1 2ˆ ˆ 0.226 0.265 0.039p p− = − = −

1 2

1 20.2443(0.7557) 0.2443(0.7557)

93 83

ˆ ˆ 0.039 0.61pq pqn n

p pz − −= = ≈ −

+ +

(c) P value = P(z < −0.61) = 0.2709

(d) Since the P value = 0.2709 > 0.05, we fail to reject 0.H No, the data are not statistically significant. (e) At the 5% level of significance, there is insufficient evidence to conclude that the population proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. 31. (a)

0 1 21 1 2

0.01: :

H p pH p p

α ==≠


1 2

1 2

12 7 0.0676153 128

r rpn n+ +

= = =+ +

1 1 0.0676 0.9324q p= − = − =

1 21 2

1 2

12 7ˆ ˆ0.0784, 0.0547153 128

r rp pn n

= = ≈ = = ≈

1 2ˆ ˆ 0.0784 0.0547 0.0237p p− = − =

1 2

1 20.0676(0.9324) 0.0676(0.9324)

153 128

ˆ ˆ 0.0237 0.79pq pqn n

p pz −= = ≈

+ +

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

(d) Since the P value = 0.4296 > 0.01, we fail to reject 0.H 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 high-school dropouts on Oahu is different from that of Sweetwater County.

32. (a) 0 1 21 1 2

0.05: :

H p pH p p

α ==<


1 2

1 2

141 125 0.5278288 216

r rpn n+ +

= = =+ +

1 1 0.5278 0.4722q p= − = − =

1 21 2

1 2

141 125ˆ ˆ0.4896, 0.5787288 216

r rp pn n

= = = = = =

1 2ˆ ˆ 0.4896 0.5787 0.0891p p− = − = −

1 2

1 20.5278(0.4722) 0.5278(0.4722)

288 216

ˆ ˆ 0.0891 1.98pq pqn n

p pz − −= = ≈ −

+ +

(c) P value = P(z < −1.98) = 0.0239

(d) Since the P value = 0.0239 < 0.05, we reject 0.H 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. 33. (a)

0 1 21 1 2

0.01: :

H p pH p p

α ==<


1 2

1 2

37 47 0.42100 100

r rpn n+ +

= = =+ +

1 1 0.42 0.58q p= − = − =

1 21 2

1 2

37 47ˆ ˆ0.37, 0.47100 100

r rp pn n

= = = = = =

1 2ˆ ˆ 0.37 0.47 0.10p p− = − = −

1 2

1 20.42(0.58) 0.42(0.58)

100 100

ˆ ˆ 0.10 1.43pq pqn n

p pz − −= = ≈ −

+ +

(c) P value = P(z < -1.43) ≈ 0.0764

(d) Since the P value = 0.0764 > 0.01, we fail to reject 0.H 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 adults believing in extraterrestrials who attended college is higher than the proportion who did not attend college. 34. (a)

0 1 21 1 2

0.05: :

H p pH p p

α ==>


1 2

1 2

45 36 0.669459 62

r rpn n+ +

= = ≈+ +

1 1 0.6694 0.3306q p= − = − =

1 21 2

1 2

45 36ˆ ˆ0.7627, 0.580659 62

r rp pn n

= = ≈ = = ≈

1 2ˆ ˆ 0.7627 0.5806 0.182p p− = − ≈

1 2

1 20.6694(0.3306) 0.6694(0.3306)

59 62

ˆ ˆ 0.182 2.13pq pqn n

p pz −= = ≈

+ +

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

(d) Since the P value = 0.0166 < 0.05, we reject 0.H Yes, the data are statistically significant.

(e) At the 5% level of significance, there is sufficient evidence to conclude that the population proportion of conservative voters who prefer art with fully clothed people is higher. 35. (a)

0 1 21 1 2

0.05: :

H p pH p p

α ==<


1 2

1 2

45 71 0.2189250 280

r rpn n+ +

= = ≈+ +

1 1 0.2189 0.7811q p= − = − =

1 21 2

1 2

45 71ˆ ˆ0.18, 0.2536250 280

r rp pn n

= = = = = ≈

1 2ˆ ˆ 0.18 0.2536 0.074p p− = − ≈ −

1 2

1 20.2189(0.7811) 0.2189(0.7811)

250 280

ˆ ˆ 0.074 2.04pq pqn n

p pz − −= = ≈ −

+ +

(c) P value = P(z < −2.04) = 0.0207

(d) Since the P value = 0.0207 < 0.05, we reject 0.H Yes, the data are statistically significant. (e) At the 5% level of significance, there is sufficient evidence to conclude that the population proportion of trusting people in Chicago is higher in the older group. 36. 0 1 2 1 1 2: ; : ;H Hµ µ µ µ= > for α = 0.01 and a right-tailed test, the critical value 0 2.33;z = sample test statistic z = 2.57 is in the critical region, reject 0.H This result is consistent with the P-value method. 37. 0 1 2 1 1 2: ; : ;H Hµ µ µ µ= < for d.f. = 9, α = 0.01 in the one-tail area row, the critical value 0 2.821.t = − Sample test statistic t = −0.965 is not in the critical region, fail to reject 0.H This result is consistent with the P-value method. 38. 0 1 2 1 1 2: ; : ;H p p H p p= ≠ for α = 0.05 and a two-tailed test, the critical values are 0 1.96;z± = ± sample test statistic z = −1.13 is not in the critical region, fail to reject 0.H This result is consistent with the P-value method.

Chapter Review Problems 1. 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)

01

0.05: 11.1: 11.1

HH

==≠

αµµ


0.636

10.8 11.1 3.00n

xz σµ− −

= = = −

(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 0.H 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)

01

0.05: 0.35: 0.35

H pH p

α ==>


39ˆ 0.481581

rpn

= = =

0.35(0.65)

81

ˆ 0.4815 0.35 2.48pqn

p pz − −= = =

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

(d) Since a P value of 0.0066 ≤ 0.05, we reject 0.H 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)

01

0.01: 0.8 A: 0.8 A

HH

αµµ

==>


0.419

1.4 0.8 4.390sn

xt µ− −= = ≈

(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.

5.004.393.002.001.000.00-1.00-2.00-3.00


(d) Since the entire P-value interval ≤ 0.01, we reject 0.H 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)

0 1 21 1 2

0.01: :

HH

αµ µµ µ

==>


2 2 2 21 21 2

1 2 1 2

2.1 2.012 12

( ) ( ) 2.5 2.986s sn n

x xt − − −= = ≈

++

µ µ

(c) Since 1 2 ,n n= 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 0.H 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)

01

0.01: 0.60: 0.60

H pH p

α ==<

(b) Use the standard normal distribution. The p̂ 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 than 5.

40ˆ 0.444490

rpn

= = =

0.60(0.40)

90

ˆ 0.4444 0.60 3.01pqn

p pz − −= = = −

(c) P value = P(z < −3.01) = 0.0013

(d) Since a P value of 0.0013 ≤ 0.01, we reject 0.H 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)

0 1 21 1 2

0.05: :

HH

αµ µµ µ

==≠

(b) Use the Student’s t distribution. We assume that both population distributions are approximately normal and that 1σ and 2σ are unknown. 1 2 53 62 9x x− = − = −

2 2 2 21 21 2

1 2 1 2

19 1551 60

( ) ( ) 9 0 2.735s sn n

x xt − − − − −= = ≈ −

++

µ µ

(c) Since 1 2 ,n n< 1. . 1 51 1 50.d f n= − = − = 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 0.H 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)

01

0.01: 40: 40

HH

αµµ

==>

Since > is in 1,H use a right-tailed test. (b) Use the standard normal distribution. We assume that x has a normal distribution with known standard deviation .σ Note that µ is given in the null hypothesis.

994

43.1 40 3.34n

xz σµ− −

= = ≈

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

4.003.342.001.000.00-1.00-2.00-3.00-4.00


(d) Since a P value of 0.0004 ≤ 0.01, we reject 0.H 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)

0 1 21 1 2

0.05: :

H p pH p p

α ==<


1 2

1 2

12 18 0.162288 97

r rpn n+ +

= = ≈+ +

1 1 0.1622 0.8378q p= − = − =

1 21 2

1 2

12 18ˆ ˆ0.1364, 0.185688 97

r rp pn n

= = ≈ = = ≈

1 2ˆ ˆ 0.1364 0.1856 0.0492p p− = − = −

1 2

1 20.1622(0.8378) 0.1622(0.8378)

88 97

ˆ ˆ 0.0492 0.91pq pqn n

p pz − −= = ≈ −

+ +

(c) P value = P(z < −0.91) = 0.1814

(d) Since the P value = 0.1814 > 0.05, we fail to reject 0.H 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) 0 1 21 1 2

0.05: :

HH

αµ µµ µ

==≠


2 2 2 21 21 2

1 2 1 2

0.8 0.955 51

( ) ( ) 0.3 0 1.808s sn n

x xt − − − −= = ≈

++

µ µ

(c) Since 2 1,n n< 2. . 1 51 1 50.d f n= − = − = 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 0.H 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)

01

0.05: 0.36: 0.36

H pH p

α ==<


33ˆ 0.275120

rpn

= = =

0.36(0.64)

120

ˆ 0.275 0.36 1.94pqn

p pz − −= = = −

(c) P value = P(z < −1.94) = 0.0262

(d) Since a P value of 0.0262 ≤ 0.05, we reject 0.H 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)

01

0.05: 7 oz: 7 oz

HH

αµµ

==≠


0.58

7.3 7 1.697sn

xt µ− −= = ≈

(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 0.H 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)

01

0.01: 0: 0

dd

HH

αµµ

==>

(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 F d = exp − cont 14 14 7 7 12 5

9.83, 3.97dd s= =

3.976

9.83 0 6.07d

ds

n

dt µ− −= = =

(c) d.f. = n − 1 = 6 − 1 = 5 From 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.

7.0006.0665.0004.0003.0002.0001.0000.000-1.000-2.000-3.000


(d) Since the P-value interval is ≤ 0.01, we reject 0.H 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)

01

0.05: 0: 0

dd

HH

αµµ

==<

(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 d = before − after −6.4 −7.2 −8.6 −3.8 1.3

4.94, 3.90dd s= − =

3.905

4.94 0 2.832d

ds

n

dt µ− − −= = = −


(d) Since the P-value interval is ≤ 0.05, we reject 0.H 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)

01

0.05: 48 months: 48 months

HH

==<

αµµ


8.6110

44.2 48 1.396sn

xt µ− −= = ≈ −


(d) Since the entire P-value interval > 0.05, we fail to reject 0.H 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.

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Chapter 8: Hypothesis Testing