Newbold Stat8 Ism 09 Ge

Copyright 2013 Pearson Education 9-1

Chapter 9:

Hypothesis Tests of a Single Population

9.1

0 1: .2; : .2;p pH H 9.2

0 :H No change in interest rates is warranted

1 :H Reduce interest rates to stimulate the economy 9.3

0 : A Bp pH : There is no difference in the percentage of underfilled cereal packages

1 : A Bp pH : Lower percentage after the change 9.4

a. Motorist group perspective: H: Increasing the motorway speed limit would be safe. H: It would not be safe. b. Road Safety group perspective: H: Increasing the motorway speed limit would not be safe. H: It would be safe.

9.5 H: 5 minutes H: > 5 minutes

9.6 0 : B GH T T No difference in the total number of votes between Bush and Gore

1 : B GH T T Gore with more votes 9.7 A random sample is obtained from a population with a variance of 625 and the sample mean is

computed. Test the null hypothesis 0 : 100H versus the alternative 1 : 100H . 2 = 625 Compute the critical value cx and state the decision rule

a. n = 25. Reject 0H if 0cx x z n = 100 +1.645(25)/ 25 = 108.225

9-2 Statistics for Business & Economics, 7th edition

Copyright 2013 Pearson Education

b. n = 16. Reject 0H if 0cx x z n = 100 +1.645(25)/ 16 = 110.28125 c. n = 44. Reject 0H if 0cx x z n = 100 +1.645(25)/ 44 = 106.1998 d. n = 32 Reject 0H if 0cx x z n = 100 +1.645(25)/ 32 = 107.26994 9.8 A random sample of n = 25 is obtained from a population with a variance 2 and the

sample mean is computed. Test the null hypothesis 0 : 100H versus the alternative 1 : 100H with alpha = .05. Compute the critical value cx and state the decision rule

a. 2 = 225. Reject 0H if 0cx x z n = 100 +1.645(15)/ 25 = 104.935 b. 2 = 900. Reject 0H if 0cx x z n = 100 +1.645(30)/ 25 = 109.87 c. 2 = 400. Reject 0H if 0cx x z n = 100 +1.645(20)/ 25 = 106.58 d. 2 = 600. Reject 0H if 0cx x z n = 100 +1.645(24.4949)/ 25 = 108.0588

9.9 A random sample is obtained from a population with variance = 400 and the sample mean is computed to be 70. Consider the null hypothesis 0 : 80H versus the alternative 1 : 80H . Compute the p-value a. n = 25. 0xz

n

= 70 8020 25

= -2.50. ( 2.50)pp value P z = .0062

b. n = 16. 0xzn

= 70 8020 16

= -2.00. ( 2.00)pp value P z = .0228

c. n = 44. 0xzn

= 70 8020 44

= -3.32. ( 3.32)pp value P z = .0004

d. n = 32. 0xzn

= 70 8020 32

= -2.83. ( 2.83)pp value P z = .0023

9.10 A random sample of n = 25, variance = 2 and the sample mean is = 70. Consider the

null hypothesis 0 : 80H versus the alternative 1 : 80H . Compute the p-value

a. 2 = 225. 0xzn

= 70 8015 25

= -3.33. ( 3.33)pp value P z = .0004

b. 2 = 900. 0xzn

= 70 8030 25

= -1.67. ( 1.67)pp value P z = .0475

Chapter 9: Hypothesis Tests of a Single Population 9-3


c. 2 = 400. 0xzn

= 70 8020 25

= -2.50. ( 2.50)pp value P z = .0062

d. 2 = 600. 0xzn

= 70 8024.4949 25

= -2.04. ( 2.04)pp value P z = .0207

9.11 0 : 16H ; 1 : 16H ; reject 0H if Z.10 < -1.28 15.84 16

.4 16Z = -1.6, therefore, Reject 0H at the 10% level.

9.12 0 : 50H ; 1 : 50H ; reject 0H if Z.10 < -1.28 48.2 50

3 9Z = -1.8, therefore, Reject 0H at the 10% level.

9.13 a. 0 : 3H ; 1 : 3H ; reject 0H if Z.05 > 1.645 3.07 3

.4 64Z = 1.4, therefore, Do Not Reject 0H at the 5% level. b. p-value = 1 FZ(1.4) = 1 - .9192 = .0808 c. the p-value would be higher the graph should show that the p-value now

corresponds to the area in both of the tails of the distribution whereas before it was the area in one of the tails.

d. A one-sided alternative is more appropriate since we are not interested in detecting possible low levels of impurity, only high levels of impurity.



9.14 Test 0 : 100H ; 1 : 100H , using n = 25 and alpha = .05

a. 106, 15x s . Reject if 0 1,nx ts n

106 100

15 25 = 2.00. Since 2.00 is greater than

the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.

b. 104, 10x s . Reject if 0 1,nx ts n

, 104 100

10 25 = 2.00. Since 2.00 is greater than

the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.

c. Assuming a one-tailed test, 95, 10x s . Reject if 0 1,nx ts n

, 95 100

10 25 = -2.50.

Since -2.50 is less than the critical value of 1.711, there is insufficient evidence to reject the null hypothesis.

d. Assuming a one-tailed upper test, 92, 18x s . Reject if 0 1,nx ts n

, 92 100

18 25 =

-2.22. Since -2.22 is less than the critical value of 1.711, there is insufficient evidence to reject the null hypothesis.

9.15 Test 0 : 100H ; 1 : 100H , using n = 36 and alpha = .05



a. 106, 15x s . Reject if 0 1,nx ts n

, 106 10015 36

= 2.40. Since 2.40 is greater

than -1.690, there is insufficient evidence to reject the null hypothesis.

b. 104, 10x s . Reject if 0 1,nx ts n

, 104 100

10 36 = 2.40. Since 2.40 is greater

than -1.690, there is insufficient evidence to reject the null hypothesis.

c. 95, 10x s . Reject if 0 1,nx ts n

, 95 100

10 36 = -3.00. Since -3.00 is less than

the critical value of -1.690, there is sufficient evidence to reject the null hypothesis.

d. 92, 18x s . Reject if 0 1,nx ts n

, 92 100

18 36 = -2.67. Since -2.67 is less than

the critical value of -1.690, there is sufficient evidence to reject the null hypothesis. 9.16 0 1: 3; : 3;H H

2.4 31.8 100

t = -3.33, p-value is < .005. Reject 0H at any common level of alpha

9.17 0 1: 4; : 4;H H reject 0H if 2.576 > t1561,.005 > -2.576 4.27 4

1.32 1562t = 8.08, p-value is < .010. Reject 0H at any common level of alpha.

9.18 0 1: 0; : 0;H H

.078 0.201 76

t = 3.38, p-value is < .010. Reject 0H at any common level of alpha 9.19 0 1: 3; : 3;H H reject 0H if t171,.01 > 2.326

3.31 3.7 172

t = 5.81, p-value is < .01. Reject 0H at any common level of alpha. 9.20 0 1: 0; : 0;H H

2.91 011.33 170

t = -3.35, p-value is < .005. Reject 0H at any common level of alpha. 9.21 0 1: 125.32; : 125.32;H H reject 0H if |t15, .05/2 | > 2.131

131.78 125.3225.4 16

t = 1.017, p-value is > .200. Do not reject 0H at the .05 level.



9.22 a. No, the 95% confidence level provides for 2.5% of the area in either tail. This does not

correspond to a one-tailed hypothesis test with an alpha of 5% which has 5% of the area in one of the tails.

b. Yes. 9.23 0 1: 10; : 10;H H

8.82 102.4013 10

t = -1.554, p-value is between .100 and .050. Do not reject 0H at common levels of alpha.

9.24 0 1: 20; : 20;H H reject 0H if |t8, .05/2 | > 2.306 20.3556 20

.6126 9t = 1.741, therefore, do not reject 0H at the 5% level

9.25 0 1: 78.5; : 78.5;H H reject 0H if |t7, .10/2 | > 1.895 74.5 78.5

6.2335 8t = -1.815, therefore, do not reject 0H at the 10% level

9.26 H: 75; H: < 75; reject H if t, . < - 1.711 t = 70.2 75 = - 2.857, therefore reject H at the 5% level. 8.4 / 25 . 9.27 a. 0 1: 400; : 400;H H

381.35 40048.60 15

t = -1.486, p-value = .0797, therefore, reject 0H at alpha levels greater than 7.97%

b. Yes, with a larger sample size, the standard error would be smaller and hence, the calculated value of t would be larger. This would yield a smaller p-value and hence the companys claim could be rejected at a lower significance level than part a.

9.28 A random sample is obtained to test the null hypothesis of the proportion of women who

said yes to a new shoe model. 0 1: .25; : .25;p pH H . What value of the sample proportion is required to reject the null hypothesis with alpha = .03?



a. n = 400. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 +1.88 (.25)(1 .25) / 400 = .2907

b. n = 225. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 +1.88 (.25)(1 .25) / 225 = .30427

c. n = 625. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 +1.88 (.25)(1 .25) / 625 = .28256

d. n = 900. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 +1.88 (.25)(1 .25) / 900 = .2771

9.29 A random sample is obtained to test the null hypothesis of the proportion of women who

would purchase an existing shoe model. 0 1: .25; : .25;p pH H . What value of the sample proportion is required to reject the null hypothesis with alpha = .05? a. n = 400. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 1.645

(.25)(1 .25) / 400 = .2144

b. n = 225. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 1.645 (.25)(1 .25) / 225 = .2025

c. n = 625. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 1.645 (.25)(1 .25) / 625 = .2215

d. n = 900. Reject 0H if 0 0 0 (1 ) /cp p p z p p n = .25 1.645 (.25)(1 .25) / 900 = .22626

9.30 0 1: .25; : .25;p pH H .2908 .25

(.25)(.75) / 361z = 1.79, p-value = 1 FZ(1.79) = 1 - .9633 = .0367

Therefore, reject 0H at alpha greater than 3.67% 9.31

H: p 0.25; H: p < 0.25 reject H if Z < - 1.645 n = 360 p = 69/360 = 0.192 0.192 0.25 = -2.542. (i.e. < -1.645). Therefore reject H at the 5% level. 0.25(1-0.25)/360



9.32 0 1: .5; : .5;p pH H .45 .5

(.5)(.5) /160z = -1.26, p-value = 2[1 FZ(1.26)] = 2[1 .8962] = .2076

The probability of finding a random sample with a sample proportion this far or further from .5 if the null hypothesis is really true is .2076

9.33 0 1: .5; : .5;p pH H reject 0H if |z.10/2 | > 1.645 .5226 .5

(.5)(.5) /199z = .64, p-value = 2[1 FZ(.64)] = 2[1 .7389] = .5222

Therefore, do not reject 0H at the 10% alpha level. The p-value shows the probability of finding a random sample with a sample proportion this far or farther from .5 if the null hypothesis is really true is .5222

9.34 0 1: .5; : .5;p pH H .56 .5

(.5)(.5) / 50z = .85, p-value = 1 FZ(.85) = 1 .8023 = .1977

Therefore, reject 0H at alpha levels in excess of 19.77% 9.35 0 1: .75; : .75;p pH H .686 .75

(.25)(.75) /172z = -1.94, p-value = 1 FZ(1.94) = 1 .9738 = .0262

Therefore, reject 0H at alpha levels in excess of 2.62% 9.36 0 1: .75; : .75;p pH H .6931 .75

(.75)(.25) / 202z = -1.87, p-value = 1 FZ(1.87) = 1 .9693 = .0307

Therefore, reject 0H at alpha levels in excess of 3.07%



9.37 Compute the probability of Type II error and the power for the following

a. 5.10 . *( |cP x x = *( 5.041| 5.10)P x = 5.041 5.10

.1 16P z

= P(z -2.36) = .0091. Power = 1 .0091 = .9909

b. 5.03 . *( |cP x x = *( 5.041| 5.03)P x = 5.041 5.03

.1 16P z

= P(z .44) = .6700. Power = 1 .6700 = .3300

c. 5.15 . *( |cP x x = *( 5.041| 5.15)P x = 5.041 5.15

.1 16P z

= P(z -4.36) = .0000. Power = 1 .0000 = 1.0000

d. 5.07 . *( |cP x x = *( 5.041| 5.07)P x = 5.041 5.07

.1 16P z

= P(z -1.16) = .1230 . Power = 1 .1230 = .8770



9.38 What is the probability of Type II error if the actual proportion is

a. .52P . *(.46 .54 | )P p p p = * *

* * * *

.46 .54(1 ) (1 )

p pP zp p p p

n n

= .46 .52 .54 .52.52(1 .52) .52(1 .52)

600 600

P z

= P(-2.94 z .98) = .4984 + .3365 = .8349

b. .58P . *(.46 .54 | )P p p p = .46 .58 .54 .58.58(1 .58) .58(1 .58)

600 600

P z

= P(-5.96 z -1.99) = .5000 .4767 = .0233

c. .53P . *(.46 .54 | )P p p p = .46 .53 .54 .53.53(1 .53) .53(1 .53)

600 600

P z

= P(-3.44 z .49) = .4997 + .1879 = .6876

d. .48P . *(.46 .54 | )P p p p = .46 .48 .54 .48.48(1 .48) .48(1 .48)

600 600

P z

= P(-.98 z 2.94) = .3365 + .4984 = .8349

e. .43P . *(.46 .54 | )P p p p = .46 .43 .54 .43.43(1 .43) .43(1 .43)

600 600

P z

= P(1.48 z 5.44) = .5000 .4306 = .0694



9.39 a. 0H is rejected when 50

3 9X < -1.28 or when X < 48.2. Given an X = 48.2 hours,

the decision is to reject the null hypothesis.

b. The power of the test = 1 - = 1 P(Z > 48.2 493 9

) = 1 P(Z > -.80) = .2119

9.40 a. 0H is rejected when 3.4 64X > 1.645 or when X > 3.082. Since the sample mean is

3.07% which is less than the critical value, the decision is do not reject the null hypothesis.

b. The = P(Z < 3.082 3.1.4 64

) = 1 FZ(.36) = .3594. Power of the test = 1 - = .6406

9.41 a. 0H is rejected when 2.5758 < 4

1.32 1562X < 2.5758 or when 3.914 < X < 4.086.

Since the sample mean was 4.27, which is greater than the upper critical value, the decision is to reject the null hypothesis.

b. = P( 3.914 3.951.32 1562

< Z < 4.086 3.951.32 1562

) = P(-1.08 > Z > 4.07) = .8599

9.42 0H is rejected when .5

.25 / 802p < -1.28 or when p < .477

The power of the test = 1 - = 1 P(Z > .477 .45(.45)(.55) / 802

) = 1-P(Z > 1.54) = .9382



9.43 a. H: p 0.25; H: p < 0.25 reject H if Z < - 1.645 n = 340 p = 61/340 = 0.179

Reject when: P 0.25 < -1.645 0.25(1-0.25)/340 0.179 0.25 = -3.024. (i.e. < -1.645). Therefore reject H at the 5% level. 0.25(1-0.25)/340 Or when: p < 0.25 + 1.6450.25(1-0.25)/340 = 0.25 0.0386 = 0.2114

b. 0.2114 0.2 = 0.53 Probability = 1 - 0.7019 = 0.2981(29.8%) (0.8)(0.2)/340



9.44 a. 0H is rejected when 1.645 > .5.25 /199p > 1.645 or when .442 > p > .558. Since

the sample proportion is .5226 which is within the critical values. The decision is that there is insufficient evidence to reject the null hypothesis.

b. = P( .442 .6(.6)(.4) /199

< Z < .558 .6(.6)(.4) /199

) = 1-P(-4.55 < Z < -1.21) = .1131

9.45 a. 30.8 32(3 36

P Z ) = P(Z < -2.4) = 0.0082

b. 30.8 32(3 9

P Z ) = P(Z < -1.2) =0.1151.

The larger probability of a Type I error is due to the smaller sample size which increases the standard error of the mean.

c. 30.8 31(3 36

P Z ) = P(Z > -.4) =0.6554

9.46 a. .14 .09((.09)(.91) /100

P Z ) = P(Z > 1.75 ) = 0.0401

b. .14 .09((.09)(.91) / 400

P Z ) = P(Z > 3.49 ) = 0.0002 . The smaller probability of a

Type I error is due to the larger sample size which lowers the standard error of the mean.



c. .14 .20(

(.2)(.8) /100P Z ) = P(Z < -1.5) = .0668

d. i) lower, ii) higher

9.47 a. 2 20 1: 100; : 100;H H 2

22

( 1) 24(165)100

n s

= 39.6,

2 2(24,.025) (24,.010)39.36, 42.98 Therefore, reject 0H at the 2.5% level but not at the 1% level of significance.

b. 2 20 1: 100; : 100;H H 2

22

( 1) 28(165)100

n s

= 46.2,

2 2(28,.025) (28,.010)44.46, 48.28

Therefore, reject 0H at the 2.5% level but not at the 1% level of significance.

c. 2 20 1: 100; : 100;H H 2

22

( 1) 24(159)100

n s

= 38.16,

2 2(24,.050) (24,.025)36.42, 39.36 Therefore, reject

0H at the 5% level but not at the 2.5% level of significance.

d. 2 20 1: 100; : 100;H H

22

2

( 1) 37(67)100

n s

= 24.79,

2 2(37,.100) (37,.05)48.36, 52.19

Therefore, do not reject 0H at any common level of significance.



9.48 2 20 1: 500; : 500;H H reject 0H if 2(7,.10) > 12.02

22

2

( 1) 7(933.982)500

n s

= 13.0757, Therefore, reject 0H at the 10% level 9.49 a. s2 = 5.1556 b. 2 20 1: 2.25; : 2.25;H H reject 0H if 2(9,.05) > 16.92 2 9(5.1556)

2.25 = 20.6224. Reject 0H at the 5% level

9.50 2 20 1: 300; : 300;H H 2 29(480)

300 = 46.4, p-value = .0214. Reject 0H at the 5% level

9.51 The hypothesis test assumes that the population values are normally distributed

0 1: 2.0; : 2.0;H H reject 0H if 2(19,.05) > 30.14

22

2

19(2.36)(2)

= 26.4556. Do not reject 0H at the 5% level 9.52 0 1: 18.2; : 18.2;H H

22

2

24(15.3)(18.2)

= 16.961.

Do not reject 0H at the 10% level since 2 >15.66 = 2(24,.10) 9.53 a. The null hypothesis is the statement that is assumed to be true unless there is sufficient

evidence to suggest that the null hypothesis can be rejected. The alternative hypothesis is the statement that will be accepted if there is sufficient evidence to reject the null hypothesis

b. A simple hypothesis assumes a specific value for the population parameter that is being tested. A composite hypothesis assumes a range of values for the population parameter.

c. One sided alternatives can be either a one-tailed upper (> greater than) or a one-tailed lower (< less than) statement about the population parameter. Two sided alternatives are made up of both greater than or less than statements and are written as ( not equal to).

d. A Type I error is falsely rejecting the null hypothesis. To make a Type I error, the truth must be that the null hypothesis is really true and yet you conclude to reject the null and accept the alternative. A Type II error is falsely not rejecting the null hypothesis when in fact the null hypothesis is false. To make a Type II error, the null hypothesis must be false (the alternative is true) and yet you conclude to not reject the null hypothesis.



e. Significance level is the chosen level of significance that establishes the probability of a making a Type I error. This is represented by alpha. The power of the test, 1 , is the ability of the hypothesis test to identify correctly a false null hypothesis and reject it.

9.54 The p-value indicates the likelihood of getting the sample result at least as far away from

the hypothesized value as the one that was found, assuming that the distribution is really centered on the null hypothesis. The smaller the p-value, the stronger the evidence against the null hypothesis.

9.55 a. 45, 10.5409X s

b. 0 1: 40; : 40;H H reject 0H if t(9,.05) > 1.833 45 40

10.5409 10t = 1.50, therefore, do not reject 0H at the 5% level

9.56 a. False. The significance level is the probability of making a Type I error falsely

rejecting the null hypothesis when in fact the null is true. b. True

c. True d. False. The power of the test is the ability of the test to correctly reject a false null

hypothesis. e. False. The rejection region is farther away from the hypothesized value at the 1%

level than it is at the 5% level. Therefore, it is still possible to reject at the 5% level but not at the 1% level.

f. True g. False. The p-value tells the strength of the evidence against the null hypothesis.

9.57 a. 333 / 9 37; 312 8xX s = 6.245

0 1: 40; : 40;H H reject 0H if t8,.05 < -1.86 37 40

6.245 9t = -1.44, therefore, do not reject 0H at the 5% level



9.58 a. 776 800(120 100)

P Z ) = P(Z < -2) = .0228

b. 776 740( )120 100

P Z = P(Z > 3) = .0014

c. i) smaller ii) smaller d. i) smaller ii) larger

9.59 a. 0 1: .25; : .25;p pH H reject 0H if z.05 < -1.645 .215 .25

(.25)(.75) / 545z = -1.90, therefore, reject 0H at the 5% level

b. 0H is rejected when .25(.25)(.75) / 545p < -1.645 or when p < .2195

i) power = 1 P(Z > .2195 .2(.2)(.8) / 545

) = 1 P(Z > 1.14) = .8729

ii) power = 1 P(Z > .2195 .25(.25)(.75) / 545

) = 1 P(Z > -1.64) = .0505

iii) power = 1 P(Z > .2195 .3(.3)(.7) / 545

) = 1 P(Z > -4.1) = .0000

9.60 0 1: .5; : .5;p pH H .4808 .5

(.5)(.5) /104z = -.39, p-value = 2[1-FZ(.39)] = 2[1-.6517] = .6966

Therefore, reject 0H at levels in excess of 69.66% 9.61 H: p = 0.5; H: p > 0.5

n = 95

p = 54/95 = 0.568

Z = 0.568 0.5 = 1.33, 0.5(1- 0.5)/95

p-value = 0.0918, therefore reject H at alpha levels in excess of 9.18%

9.62 0 1: .25; : .25;p pH H reject 0H if z.05 > 1.645



.3333 .25(.25)(.75) /150

z = 2.356, therefore, reject 0H at the 5% level 9.63 0 1: .2; : .2;p pH H .2746 .2

(.2)(.8) /142z = 2.22, p-value = 1-FZ(2.22) = 1-.9868 = .0132

Therefore, reject 0H at levels in excess of 1.32% 9.64 Cost Model where W = Total Cost: W = 1,000 + 5X

1,000 5(400) 3,000W

2 2 125(5) (625) 15,625, 125, 2525

W W W

0H : W 3000; 1: 3000;H W Using the test statistic criteria: (3050 3000)/25 = 2.00 which yields a p-value of .0228, therefore, reject 0H at the .05 level. Using the sample statistic criteria: 3,000 (25)(1.645) 3041.1critX , 3,050calcX , since 3,050calcX > 3041.1critX , therefore, reject 0H at the .05 level.

9.65 0 1: 39, : 39H H

4040 39 1.19

5071

t . Probability of X = 40 given that is 39 is .1170. Therefore, the

Vice Presidents claim is not very strong.



9.66 Per capita consumption of fruits and vegetables x = 172.79; s = 19.254

0 1: 170; : 170;H H . Reject if 0 1,nx ts n

172.79 17019.254 3108

= 8.075 Since 8.075 is greater than the critical value of 1.645,

there is sufficient evidence to reject the null hypothesis. Per capita consumption of snack foods x = 114.11; s = 9.541


114.11 1149.541 3108

= 0.66. Since 0.66 is greater than the critical value of -1.645, there is no

sufficient evidence to reject the null hypothesis. Per capita consumption of soft drinks x = 66.81; s = 7.5


66.81 657.5 3108

= 13.487. Since 13.487 is greater than the critical value of -1.645,

there is no sufficient evidence to reject the null hypothesis. Per capita consumption of meat x = 70.38; s = 12.694


70.38 7012.694 3108

= 1.69. Since 1.69 is greater than the critical value of 1.645,

there is sufficient evidence to reject the null hypothesis.



9.67 Obesity rates of adults in the U.S. population. x = 28.29; s = 3.625


28.29 283.625 3140


there is sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in the U.S. population x = 14.19; s = 3.716


14.19 133.716 2691


there is sufficient evidence to reject the null hypothesis. 9.68

0 1: 40, : 40; 49.73 42.86XH H reject 0H One-Sample T: Salmon Weight Test of mu = 40 vs mu > 40 Variable N Mean StDev SE Mean Salmon Weigh 39 49.73 10.60 1.70 Variable 95.0% Lower Bound T P Salmon Weigh 46.86 5.73 0.000

At the .05 level of significance we have strong enough evidence to reject Ho that the true mean weight of salmon is no different than 40 in favor of Ha that the true mean weight is significantly greater than 40.

( )crit crit xX Ho t S : 40 + 1.686(1.70) = 42.8662

Population mean for = .50 (power=.50): tcrit = 0.0: 42.8662 + 0.0(1.70) = 42.8662 Population mean for = .25 (power=.75): tcrit = .681: 42.8662 + .681(1.70) = 44.0239 Population mean for = .10 (power=.90): tcrit = 1.28: 42.8662 + 1.28(1.70) = 45.0422 Population mean for = .05 (power=.95): tcrit = 1.645: 42.8662 + 1.645(1.70) = 45.6627



40 41 42 43 44 45 46

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

PopMean

Pow

er

Power curve

For beta = .50 .25 .10 and .05

9.69 a. 0 1: 1.6; : 1.6;H H reject 0H if |z.05|> 1.645

1.615 1.6.05 16

z = 1.20, p-value =2[1-FZ(1.2)]= 0.2302.

Do not reject 0H at the 10% level b. 0 1: .05; : .05;H H reject 0H if 2(15,.10) 22.31

2

22

15(.086)(.05)

= 44.376. Reject 0H at the 10% level



9.70 a. Assume that the population is normally distributed One-Sample T: Grams: Test of mu = 5 vs mu not = 5 Variable N Mean StDev SE Mean Grams:11-34 12 4.9725 0.0936 0.0270 Variable 95.0% CI T P Grams:11-34 ( 4.9130, 5.0320) -1.02 0.331

4.9725; .0936x s , 0 1: 5; : 5;H H reject 0H if |t(11, .025| > 2.201 4.9725 5.0936 12

t = -1.018. Do not reject 0H at the 5% level b. Assume that the population is normally distributed

0 1: .025; : .025;H H reject 0H if 2(11,.05) 19.68

22

2

11(.0936)(.025)

= 154.19. Therefore, reject 0H at the 5% level 9.71 333 / 9 37; 312000 8x s = 197.484

0 1: 6; : 6;H H reject 0H if 2(8,.10) 13.36

22

2

8(197.484)(6)

= 8666.651 . reject 0H at the 10% level 9.72 Obesity rates of adults in the U.S. population. x = 28.29; s = 3.625


28.29 283.625 3140


there is sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in the U.S. population

x = 14.19; s = 3.716


14.19 133.716 2691





9.73 Obesity rates of adults in California x = 23.34; s = 3.224


23.34 283.224 58

= -11.00. Since -11.00 is smaller than the critical value of 1.645,

there is no sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in California

x = 16.18; s = 2.535


16.18 132.535 57



Obesity rates of adults in Michigan. x = 29.49; s = 1.397


29.49 281.397 83


there is sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in Michigan

x = 14.02; s = 2.533


14.02 132.533 82



Obesity rates of adults in Minnesota. x = 27.24; s = 0.907


27.24 280.907 87


there is no sufficient evidence to reject the null hypothesis.



Low-income preschool obesity rate in Minnesota x = 12.51; s = 2.919


12.51 132.919 87


there is no sufficient evidence to reject the null hypothesis.

Obesity rates of adults in Florida. x = 26.82; s = 4.054


26.82 284.054 67


there is no sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in Florida x = 12.51; s = 2.919


13.53 134.338 66

= 0.993. Since 0.993 is smaller than the critical value of 1.645,

there is no sufficient evidence to reject the null hypothesis. 9.74 Mean weights of Men in the first interview x = 27.98; s = 5.468


27.98 305.468 2108

= -16.985. Since -16.985 is smaller than the critical value of -1.645,

there is sufficient evidence to reject the null hypothesis. Mean weights of Men in the second interview x = 28.07; s = 5.447


28.07 305.447 1925





There is no difference in the results obtained from the first and second Interviews for men. Mean weights of Women in the first interview x = 28.93; s = 7.022


28.93 307.022 2274


there is sufficient evidence to reject the null hypothesis. Mean weights of Women in the second interview x = 29.02; s = 7.071


29.02 307.071 2132


there is sufficient evidence to reject the null hypothesis. There is no difference in the results obtained from the first and second Interviews for women. 9.75 Mean weights of Immigrants in the first interview x = 27.63; s = 5.272


27.63 305.272 870


there is sufficient evidence to reject the null hypothesis. Mean weights of Immigrants in the second interview x = 27.75; s = 5.317


27.75 305.317 787


there is sufficient evidence to reject the null hypothesis. There is no difference in the results obtained from the first and second Interviews for immigrants



9.76 Mean weights of White people in the first interview x = 27.85; s = 6.065


27.85 306.065 2357


there is sufficient evidence to reject the null hypothesis. Mean weights of White people in the second interview x = 27.92; s = 6.117


27.92 306.117 2213


there is sufficient evidence to reject the null hypothesis. There is no difference in the results obtained from the first and second Interviews for White people. 9.77 Mean weights of Hispanic people in the first interview x = 28.83; s = 5.566


28.83 305.566 999


there is sufficient evidence to reject the null hypothesis. Mean weights of Hispanic people in the second interview x = 28.90; s = 5.603


28.90 305.603 934


there is sufficient evidence to reject the null hypothesis. There is no difference in the results obtained from the first and second Interviews for Hispanic people. 9.78 Mean weights of people who have been diagnosed with high blood pressure in the



first interview x = 30.15; s = 6.613


30.15 306.613 1522


there is no sufficient evidence to reject the null hypothesis. Mean weights of people who have been diagnosed with high blood pressure in the second interview x = 30.29; s = 6.651


30.29 306.651 1420


there is no sufficient evidence to reject the null hypothesis. There is no difference in the results obtained from the first and second Interviews for people diagnosed with high blood pressure.

Newbold Stat8 Ism 09 Ge

Documents

Transcript of Newbold Stat8 Ism 09 Ge