Newbold Stat8 Ism 09 Ge

Chapter 9: Hypothesis Tests of a Single Population

9.1 9.2

9.3

H : p  .2; H : p  .2; 0

H H

1

0

:

No change in interest rates is warranted

1

:

Reduce interest rates to stimulate the economy

H :p 0

A



p : There is no difference in the percentage of underfilled cereal B

packages

H :p 1

A



p : Lower percentage after the change B

9.4 a. Motorist group perspective: H₀: Increasing the motorway speed limit would be safe. H₁: It would not be safe. b. Road Safety group perspective: H₀: Increasing the motorway speed limit would not be safe. H₁: It would be safe. 9.5 H₀:   5 minutes H₁:  > 5 minutes

9.6

H : T  T No difference in the total number of votes between Bush and Gore H : T T Gore with more votes 0

1

B

G

B

G

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 H 0 :   100 versus the alternative H 1 :   100 . 2 = 625 Compute the critical value xc and state the decision rule a. n = 25. Reject 108.225

H

0

if x  xc  0  z 

n = 100 +1.645(25)/

Copyright © 2013 Pearson Education

9-1

25 =

9-2

th

Statistics for Business & Economics, 7 edition

b. n = 16. Reject 110.28125 c. n = 44. Reject 106.1998 d. n = 32 Reject

H

0

if x  xc  0  z 

n = 100 +1.645(25)/ 16 =

H

0

if x  xc  0  z 

n = 100 +1.645(25)/ 44 =

H

if x  xc  0  z 

0

n = 100 +1.645(25)/ 32 = 107.26994

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 H 0 :   100 versus the alternative

9.8

H

1

:   100 with alpha = .05. Compute the critical value xc and state the decision rule

a.  2 = 225. Reject

0

if x  xc  0  z 

n = 100 +1.645(15)/ 25 = 104.935

b.  2

0

if x  xc  0  z 

n = 100 +1.645(30)/ 25 = 109.87

0

if x  xc  0  z 

n = 100 +1.645(20)/ 25 = 106.58

0

if x  xc  0  z 

n = 100 +1.645(24.4949)/ 25 = 108.0588

c. 

2

d. 

2

9.9

H = 900. Reject H = 400. Reject H = 600. Reject H

A random sample is obtained from a population with variance = 400 and the sample mean is computed to be 70. Consider the null hypothesis H 0 :   80 versus the alternative

H : 1

 80 . Compute the p-value

x  0  n x  0 b. n = 16. z   n x  0 c. n = 44. z   n x  0 d. n = 32. z   n a. n = 25. z 

9.10

70  80 20 25 70  80 = 20 16 70  80 = 20 44 70  80 = 20 32

=

= -2.50. p  value  P( z p  2.50) = .0062

= -2.00. p  value  P( z p  2.00) = .0228 = -3.32. p  value  P( z p  3.32) = .0004

= -2.83. p  value  P( z p  2.83) = .0023

A random sample of n = 25, variance =  2 and the sample mean is = 70. Consider the null hypothesis H 0 :   80 versus the alternative H 1 :   80 . Compute the pvalue

x  0 70  80 = = -3.33. p  value  P ( z p  3.33) = .0004  n 15 25 x  0 70  80 = = -1.67. p  value  P( z p  1.67) = .0475 b.  2 = 900. z   n 30 25

a.  2 = 225. z 



9-3

x  0 70  80 = = -2.50. p  value  P( z p  2.50) = .0062 20 25  n x  0 70  80 = = -2.04. p  value  P( z p  2.04) = .0207 d.  2 = 600. z  24.4949 25  n c.  2 = 400. z 

9.11

H : 0

Z

9.12

9.13

0

a.

H : 1

 16 ; reject

H

15.84  16 = -1.6, therefore, Reject .4 16

H : Z

 16 ;

 50 ;

H : 1

 50 ; reject

H : 0

 3;

H : 1

 3 ; reject

H

H

48.2  50 = -1.8, therefore, Reject 3 9

if Z.10 < -1.28

0

H

at the 10% level.

if Z.10 < -1.28

0

H

0

0

0

at the 10% level.

if Z.05 > 1.645

3.07  3 = 1.4, therefore, Do Not Reject H 0 at the 5% level. .4 64 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.

Z


9-4

9.14

th


Test

H : 0

 100 ;

H : 1

 100 , using n = 25 and alpha = .05

x  0

106  100 = 2.00. Since 2.00 is greater than s n 15 25 the critical value of 1.711, there is sufficient evidence to reject the null hypothesis. x  0 104  100 = 2.00. Since 2.00 is greater than b. x  104, s  10 . Reject if  tn 1, , s n 10 25 the critical value of 1.711, there is sufficient evidence to reject the null hypothesis. x  0 95  100 = -2.50. c. Assuming a one-tailed test, x  95, s  10 . Reject if  tn 1, , s n 10 25 Since -2.50 is less than the critical value of 1.711, there is insufficient evidence to reject the null hypothesis. x  0 92  100 = d. Assuming a one-tailed upper test, x  92, s  18 . Reject if  tn 1, , s n 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. a.

9.15

x  106, s  15 . Reject if

Test

H : 0

 100 ;

H : 1

 tn 1,

 100 , using n = 36 and alpha = .05



9-5

x  0 106  100 = 2.40. Since 2.40 is greater  tn 1,  , 15 36 s n than -1.690, there is insufficient evidence to reject the null hypothesis. x  0 104  100 = 2.40. Since 2.40 is greater b. x  104, s  10 . Reject if  tn 1, , s n 10 36 than -1.690, there is insufficient evidence to reject the null hypothesis. x  0 95  100 = -3.00. Since -3.00 is less than c. x  95, s  10 . Reject if  tn 1, , s n 10 36 the critical value of -1.690, there is sufficient evidence to reject the null hypothesis. x  0 92  100 = -2.67. Since -2.67 is less than d. x  92, s  18 . Reject if  tn 1, , 18 36 s n the critical value of -1.690, there is sufficient evidence to reject the null hypothesis. a. x  106, s  15 . Reject if

9.16

H : 0

t

9.17

9.18

0

0

9.20

0

9.21

0

0

0

at any common level of alpha

if 2.576 > t1561,.005 > -2.576

H

at any common level of alpha.

0

 0; H 1 :   0;

 3; H 1 :   3; reject

H

0

H

0

at any common level of alpha

if t171,.01 > 2.326

H

0


 0; H 1 :   0;

2.91  0 = -3.35, p-value is < .005. Reject 11.33 170

H : t

0

3.31  3 = 5.81, p-value is < .01. Reject .7 172

H : t

H

.078  0 = 3.38, p-value is < .010. Reject .201 76

H : t

 4; H 1 :   4; reject

H

4.27  4 = 8.08, p-value is < .010. Reject 1.32 1562

H : t

9.19

2.4  3 = -3.33, p-value is < .005. Reject 1.8 100

H : t

 3; H 1 :   3;

 125.32; H 1 :   125.32; reject

H

0

H

0


if |t15, .05/2 | > 2.131

131.78  125.32 = 1.017, p-value is > .200. Do not reject 25.4 16


H

0

at the .05 level.

9-6

th


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

H : 0

 10; H 1 :   10;

8.82  10 = -1.554, p-value is between .100 and .050. Do not reject 2.4013 10 common levels of alpha. t

9.24

H : 0

t 9.25

H

0

0

 78.5; H 1 :   78.5; reject

0

at

if |t8, .05/2 | > 2.306

20.3556  20 = 1.741, therefore, do not reject .6126 9

H : t

 20; H 1 :   20; reject

H

H

0

H

at the 5% level

0

if |t7, .10/2 | > 1.895

74.5  78.5 = -1.815, therefore, do not reject 6.2335 8

H

0

at the 10% level

9.26 H₀:   75; H₁:  < 75; reject H₀ if t₂₄, .₀₅ < - 1.711 t = 70.2 – 75 = - 2.857, therefore reject H₀ at the 5% level. 8.4 / √25 .

9.27

a.

H : 0

 400; H 1 :   400;

381.35  400 = -1.486, p-value = .0797, therefore, reject H 0 at alpha levels 48.60 15 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 company’s claim could be rejected at a lower significance level than part a. t

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. H 0 : p  .25; H 1 : p  .25; . What value of the sample proportion is required to reject the null hypothesis with alpha = .03?



a. n = 400. Reject

H

0

if pˆ  pˆ c  p0  z

9-7

p0 (1  p0 ) / n = .25 +1.88

(.25)(1  .25) / 400 = .2907 b. n = 225. Reject

H

0


p0 (1  p0 ) / n = .25 +1.88

(.25)(1  .25) / 225 = .30427

c. n = 625. Reject

H

0


p0 (1  p0 ) / n = .25 +1.88

(.25)(1  .25) / 625 = .28256 d. n = 900. Reject

H

0


p0 (1  p0 ) / 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. H 0 : p  .25; H 1 : p  .25; . What value of the sample proportion is required to reject the null hypothesis with alpha = .05? a. n = 400. Reject H 0 if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645 (.25)(1  .25) / 400 = .2144

b. n = 225. Reject

H

0

if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645

(.25)(1  .25) / 225 = .2025 c. n = 625. Reject

H

0

if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645

(.25)(1  .25) / 625 = .2215

d. n = 900. Reject

H

0

if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645

(.25)(1  .25) / 900 = .22626

9.30

H : p  .25; H : p  .25; 0

1

.2908  .25 = 1.79, p-value = 1 – FZ(1.79) = 1 - .9633 = .0367 z (.25)(.75) / 361 Therefore, reject H 0 at alpha greater than 3.67% 9.31 H₀: p  0.25; H₁: p < 0.25 reject H₀ if Z < - 1.645 n = 360 p = 69/360 = 0.192 0.192 – 0.25 = -2.542. (i.e. < -1.645). Therefore reject H₀ at the 5% level. √0.25(1-0.25)/360


9-8 9.32

th


H : p  .5; H : p  .5; 0

1

.45  .5 = -1.26, p-value = 2[1 – FZ(1.26)] = 2[1 – .8962] = .2076 (.5)(.5) /160 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 z

9.33

H : p  .5; H : p  .5; reject H 0

1

0

if |z.10/2 | > 1.645

.5226  .5 = .64, p-value = 2[1 – FZ(.64)] = 2[1 – .7389] = .5222 (.5)(.5) /199 Therefore, do not reject H 0 at the 10% alpha level. The p-value shows the probability z

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

H : p  .5; H : p  .5; 0

1

.56  .5 = .85, p-value = 1 – FZ(.85) = 1 – .8023 = .1977 z (.5)(.5) / 50 Therefore, reject H 0 at alpha levels in excess of 19.77% 9.35

H : p  .75; H : p  .75; 0

1

.686  .75 = -1.94, p-value = 1 – FZ(1.94) = 1 – .9738 = .0262 z (.25)(.75) /172 Therefore, reject H 0 at alpha levels in excess of 2.62% 9.36

H : p  .75; H : p  .75; 0

1

.6931  .75 = -1.87, p-value = 1 – FZ(1.87) = 1 – .9693 = .0307 z (.75)(.25) / 202 Therefore, reject H 0 at alpha levels in excess of 3.07%



9-9

9.37 Compute the probability of Type II error and the power for the following  5.041  5.10  a.   5.10 .   P( x  xc |    * = P ( x  5.041|  *  5.10) = P  z   .1 16   = P(z ≤ -2.36) = .0091. Power = 1 – .0091 = .9909  5.041  5.03  b.   5.03 .   P ( x  xc |    * = P ( x  5.041|  *  5.03) = P  z   .1 16   = P(z ≤ .44) = .6700. Power = 1 – .6700 = .3300  5.041  5.15  c.   5.15 .   P( x  xc |    * = P ( x  5.041|  *  5.15) = P  z   .1 16   = P(z ≤ -4.36) = .0000. Power = 1 – .0000 = 1.0000  5.041  5.07  d.   5.07 .   P( x  xc |    * = P ( x  5.041|  *  5.07) = P  z   .1 16   = P(z ≤ -1.16) = .1230 . Power = 1 – .1230 = .8770


9-10 9.38

th


What is the probability of Type II error if the actual proportion is     * * .46  p .54  p  a. P  .52 .   P(.46  pˆ  .54 | p  p* ) = P  z  p* (1  p* ) p* (1  p* )    n n      .46  .52 .54  .52   = P(-2.94 ≤ z ≤ .98) = .4984 + .3365 = .8349 = P z .52(1  .52)   .52(1  .52)   600 600    .46  .58 .54  .58   b. P  .58 .   P(.46  pˆ  .54 | p  p* ) = P  z .58(1  .58)   .58(1  .58)   600 600 = P(-5.96 ≤ z ≤ -1.99) = .5000 – .4767 = .0233    .46  .53 .54  .53   c. P  .53 .   P(.46  pˆ  .54 | p  p* ) = P  z .53(1  .53)   .53(1  .53)   600 600 = P(-3.44 ≤ z ≤ .49) = .4997 + .1879 = .6876    .46  .48 .54  .48   d. P  .48 .   P(.46  pˆ  .54 | p  p* ) = P  z .48(1  .48)   .48(1  .48)   600 600 = P(-.98 ≤ z ≤ 2.94) = .3365 + .4984 = .8349    .46  .43 .54  .43   e. P  .43 .   P(.46  pˆ  .54 | p  p* ) = P  z .43(1  .43)   .43(1  .43)   600 600 = P(1.48 ≤ z ≤5.44) = .5000 – .4306 = .0694



9.39

9.40

9.41

9.42

9-11

X  50 < -1.28 or when X < 48.2. Given an X = 48.2 hours, 3 9 the decision is to reject the null hypothesis. 48.2  49 ) = 1 – P(Z > -.80) = .2119 b. The power of the test = 1 -  = 1 – P(Z > 3 9

a.

H

0

is rejected when

X 3 > 1.645 or when X > 3.082. Since the sample mean is .4 64 3.07% which is less than the critical value, the decision is do not reject the null hypothesis. 3.082  3.1 ) = 1 – FZ(.36) = .3594. Power of the test = 1 -  = .6406 b. The  = P(Z < .4 64 a.

H

0

is rejected when

X 4 < 2.5758 or when 3.914 < X < 4.086. 1.32 1562 Since the sample mean was 4.27, which is greater than the upper critical value, the decision is to reject the null hypothesis. 3.914  3.95 4.086  3.95 Z > 4.07) = .8599 b.  = P( 1.32 1562 1.32 1562 a.

H

0

is rejected when –2.5758 <

p  .5 < -1.28 or when p < .477 .25 / 802 .477  .45 ) = 1-P(Z > 1.54) = .9382 The power of the test = 1 -  = 1 – P(Z > (.45)(.55) / 802

H

0

is rejected when


9-12

th


9.43 a. H₀: p  0.25; H₁: p < 0.25 reject H₀ if Z < - 1.645 n = 340 pˆ = 61/340 = 0.179 Reject when: P – 0.25 √0.25(1-0.25)/340

< -1.645

= -3.024. (i.e. < -1.645). Therefore reject H₀ at the 5% level. 0.179 – 0.25 √0.25(1-0.25)/340 Or when: pˆ < 0.25 + – 1.645√0.25(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

9-13

p  .5 > 1.645 or when .442 > p > .558. Since .25 /199 the sample proportion is .5226 which is within the critical values. The decision is that there is insufficient evidence to reject the null hypothesis. .442  .6 .558  .6 b.  = P( < Z< ) = 1-P(-4.55 < Z < -1.21) = .1131 (.6)(.4) /199 (.6)(.4) /199

a.

H

0

is rejected when –1.645 >

30.8  32 ) = P(Z < -2.4) = 0.0082 3 36 30.8  32 ) = P(Z < -1.2) =0.1151. b.   P( Z  3 9 The larger probability of a Type I error is due to the smaller sample size which increases the standard error of the mean. 30.8  31 c.   P ( Z  ) = P(Z > -.4) =0.6554 3 36

9.45

a.   P ( Z 

9.46

a.   P( Z 

.14  .09 ) = P(Z > 1.75 ) = 0.0401 (.09)(.91) /100 .14  .09 b.   P( Z  ) = P(Z > 3.49 ) = 0.0002 . The smaller probability of a (.09)(.91) / 400 Type I error is due to the larger sample size which lowers the standard error of the mean.


9-14

th


.14  .20 ) = P(Z < -1.5) = .0668 (.2)(.8) /100 d. i) lower, ii) higher

c.   P( Z 

9.47 a.

:   100; H 1 :   100;   2

H

0



2 (24,.025)

 39.36, 

2 (24,.010)

Therefore, reject b.

2 (28,.025)

 44.46, 

Therefore, reject

2 (24,.050)

 36.42, 

(24,.025)



(37,.100)

 48.36, 

H

0

2 (37,.05)

Therefore, do not reject



2



28(165) = 46.2, 100

(n  1) s 2



2



24(159) = 38.16, 100

 39.36

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

2 2 2 H 0 :   100; H 1 :   100;   2

(n  1) s 2

at the 2.5% level but not at the 1% level of significance.

2

Therefore, reject

24(165) = 39.6, 100

 48.28

(28,.010) 0



at the 2.5% level but not at the 1% level of significance.

0

2

H



2

 42.98

2 2 2 H 0 :   100; H 1 :   100;  



d.

H

(n  1) s 2

2 2 2 H 0 :   100; H 1 :   100;  



c.

2

2

(n  1) s 2



2



37(67) = 24.79, 100

 52.19

H

0

at any common level of significance.



9.48

H

0

:  2  500; H 1 :  2  500; reject

2  9.49

H

0



7(933.982) = 13.0757, Therefore, reject 500

9(5.1556) = 20.6224. Reject 2.25

H

0

H

0

H

0

at the 10% level

if  2(9,.05) > 16.92

at the 5% level

29(480) = 46.4, p-value = .0214. Reject 300

H

0

at the 5% level

The hypothesis test assumes that the population values are normally distributed 2 H 0 :   2.0; H 1 :   2.0; reject H 0 if  (19,.05) > 30.14 19(2.36) 2   = 26.4556. Do not reject (2) 2 2

9.52

if  2(7,.10) > 12.02

:  2  300; H 1 :  2  300;

2  9.51



2

0

a. s2 = 5.1556 b. H 0 :  2  2.25; H 1 :  2  2.25; reject

2  9.50

(n  1) s 2

H

9-15

H

0

H

0

at the 5% level

:   18.2; H 1 :   18.2;

24(15.3) 2 = 16.961. (18.2) 2 Do not reject H 0 at the 10% level since  2 >15.66 =  2 (24,.10)

2 

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.


9-16

th


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 – β, is the ability of the hypothesis test to identify correctly a false null hypothesis and reject it. 9.54

9.55

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. a. X  45, s  10.5409

H :

b. t

0

 40; H 1 :   40; reject

H

0

if t(9,.05) > 1.833

45  40 = 1.50, therefore, do not reject 10.5409 10

H

0

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. X  333 / 9  37; sx  312 8 = 6.245

H : 0

t

 40; H 1 :   40; reject

H

0

if t8,.05 < -1.86

37  40 = -1.44, therefore, do not reject 6.245 9

H

0

at the 5% level



776  800 ) = P(Z < -2) = .0228 120 100) 776  740 b.   P( Z  ) = P(Z > 3) = .0014 120 100 c. i) smaller ii) smaller d. i) smaller ii) larger

9.58

a.   P ( Z 

9.59

a.

H : p  .25; H : p  .25; reject H 0

1

0

if z.05 < -1.645

.215  .25 = -1.90, therefore, reject H 0 at the 5% level (.25)(.75) / 545 p  .25 < -1.645 or when p < .2195 b. H 0 is rejected when (.25)(.75) / 545 .2195  .2 power = 1 – P(Z > ) = 1 – P(Z > 1.14) = .8729 i) (.2)(.8) / 545 .2195  .25 power = 1 – P(Z > ) = 1 – P(Z > -1.64) = .0505 ii) (.25)(.75) / 545 .2195  .3 power = 1 – P(Z > ) = 1 – P(Z > -4.1) = .0000 iii) (.3)(.7) / 545 z

9.60

H : p  .5; H : p  .5; 0

1

.4808  .5 = -.39, p-value = 2[1-FZ(.39)] = 2[1-.6517] = .6966 z (.5)(.5) /104 Therefore, reject H 0 at levels in excess of 69.66% 9.61 H₀: p = 0.5; H₁: p > 0.5 n = 95 pˆ = 54/95 = 0.568

Z = 0.568 – 0.5 = 1.33, √0.5(1- 0.5)/95 p-value = 0.0918, therefore reject H₀ at alpha levels in excess of 9.18%

9.62

H : p  .25; H : p  .25; reject H 0

1

0

if z.05 > 1.645


9-17

9-18

th


.3333  .25 = 2.356, therefore, reject (.25)(.75) /150

z

9.63

H

0

at the 5% level

H : p  .2; H : p  .2; 0

1

.2746  .2 = 2.22, p-value = 1-FZ(2.22) = 1-.9868 = .0132 (.2)(.8) /142 Therefore, reject H 0 at levels in excess of 1.32% z

9.64 Cost Model where W = Total Cost: W = 1,000 + 5X W  1, 000  5(400)  3, 000 125  2W  (5)2 (625)  15, 625,  W  125,  W   25 25 H 0 : W  3000; H1: W  3000;

Using the test statistic criteria: (3050 – 3000)/25 = 2.00 which yields a p-value of .0228, therefore, reject H 0 at the .05 level. Using the sample statistic criteria: X crit  3, 000  (25)(1.645)  3041.1 , X calc  3, 050 , since X calc  3, 050 > X crit  3041.1 , therefore, reject

9.65

H

0

H

:   39, H 1 :   39

0

at the .05 level.

40  39  1.19 . Probability of X = 40 given that  is 39 is .1170. Therefore, the 50 71 Vice President’s claim is not very strong. t40 



9.66

9-19

Per capita consumption of fruits and vegetables x = 172.79; s = 19.254 x  H 0 :   170; H 1 :   170; . Reject if s n0  tn1, 172.79  170 = 8.075 Since 8.075 is greater than the critical value of 1.645, 19.254 3108 there is sufficient evidence to reject the null hypothesis. Per capita consumption of snack foods x = 114.11; s = 9.541

H : 0

 114; H 1 :   114; . Reject if

x  0  tn 1, s n

114.11  114 = 0.66. Since 0.66 is greater than the critical value of -1.645, there is no 9.541 3108 sufficient evidence to reject the null hypothesis. Per capita consumption of soft drinks x = 66.81; s = 7.5

H : 0

 65; H 1 :   65; . Reject if

x  0 s

n

 tn 1,

66.81  65 = 13.487. Since 13.487 is greater than the critical value of -1.645, 7.5 3108 there is no sufficient evidence to reject the null hypothesis. Per capita consumption of meat x = 70.38; s = 12.694

H : 0

 70; H 1 :   70; . Reject if

x  0 s

n

 tn 1,

70.38  70 = 1.69. Since 1.69 is greater than the critical value of 1.645, 12.694 3108 there is sufficient evidence to reject the null hypothesis.


9-20 9.67

th


Obesity rates of adults in the U.S. population. x = 28.29; s = 3.625 x  H 0 :   28; H 1 :   28; . Reject if s n0  tn1, 28.29  28 = 4.461. Since 4.461 is greater than the critical value of 1.645, 3.625 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 x  H 0 :   13; H 1 :   13; . Reject if s n0  tn1, 14.19  13 = 16.589. Since 16.589 is greater than the critical value of 1.645, 3.716 2691 there is sufficient evidence to reject the null hypothesis.

9.68

H

0

:   40, H 1 :   40; X  49.73  42.86 reject

H

0

One-Sample T: Salmon Weight Test of mu = 40 vs mu > 40 Variable Salmon Weigh Variable Salmon Weigh

N 39

Mean 49.73

95.0% Lower Bound 46.86

StDev 10.60 T 5.73

SE Mean 1.70 P 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.

X crit  Ho  tcrit ( S x ) : 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



Power curve For beta = .50 .25 .10 and .05 1.0 0.9 0.8

Power

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 40

41

42

43

44

45

PopMean

9.69

a.

H : 0

 1.6; H 1 :   1.6; reject

H

0

if |z.05|> 1.645

1.615  1.6 = 1.20, p-value =2[1-FZ(1.2)]= 0.2302. .05 16 Do not reject H 0 at the 10% level

z

b.

H

0

:   .05; H 1 :   .05; reject

15(.086) 2   = 44.376. Reject (.05) 2 2

H H

0

0

if  2(15,.10)  22.31

at the 10% level


46

9-21

9-22 9.70

th


a. Assume that the population is normally distributed One-Sample T: Grams: Test of mu = 5 vs mu not = 5 Variable N Mean Grams:11-34 12 4.9725 Variable Grams:11-34

x  4.9725; s  .0936 , t

(

StDev 0.0936

95.0% CI 4.9130, 5.0320)

H : 0

SE Mean 0.0270

T -1.02

P 0.331

 5; H 1 :   5; reject

4.9725  5 = -1.018. Do not reject .0936 12

H

0

H

0

if |t(11, .025| > 2.201

at the 5% level

b. Assume that the population is normally distributed 2 H 0 :   .025; H 1 :   .025; reject H 0 if  (11,.05)  19.68

2  9.71

H

0

at the 5% level

x  333 / 9  37; s  312000 8 = 197.484

H

0

:   6; H 1 :   6; reject

2  9.72

11(.0936) 2 = 154.19. Therefore, reject (.025) 2

H

0

if  2(8,.10)  13.36

8(197.484) 2 = 8666.651 . reject (6) 2

H

0

at the 10% level

Obesity rates of adults in the U.S. population. x = 28.29; s = 3.625 x  H 0 :   28; H 1 :   28; . Reject if s n0  tn1, 28.29  28 = 4.461. Since 4.461 is greater than the critical value of 1.645, 3.625 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 x  H 0 :   13; H 1 :   13; . Reject if s n0  tn1, 14.19  13 = 16.589. Since 16.589 is greater than the critical value of 1.645, 3.716 2691 there is sufficient evidence to reject the null hypothesis.



9.73

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

H : 0

 28; H 1 :   28; . Reject if

x  0  tn 1, s n

23.34  28 = -11.00. Since -11.00 is smaller than the critical value of 1.645, 3.224 58 there is no sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in California x = 16.18; s = 2.535 x  H 0 :   13; H 1 :   13; . Reject if s n0  tn1, 16.18  13 = 9.462. Since 9.462 is greater than the critical value of 1.645, 2.535 57 there is sufficient evidence to reject the null hypothesis. Obesity rates of adults in Michigan. x = 29.49; s = 1.397

H : 0

 28; H 1 :   28; . Reject if

x  0  tn 1, s n

29.49  28 = 9.743. Since 9.743 is greater than the critical value of 1.645, 1.397 83 there is sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in Michigan x = 14.02; s = 2.533 x  H 0 :   13; H 1 :   13; . Reject if s n0  tn1, 14.02  13 = 3.662. Since 3.662 is greater than the critical value of 1.645, 2.533 82 there is sufficient evidence to reject the null hypothesis. Obesity rates of adults in Minnesota. x = 27.24; s = 0.907

H : 0

 28; H 1 :   28; . Reject if

x  0  tn 1, s n

27.24  28 = -7.778. Since -7.778 is smaller than the critical value of 1.645, 0.907 87 there is no sufficient evidence to reject the null hypothesis.


9-23

9-24

th


Low-income preschool obesity rate in Minnesota x = 12.51; s = 2.919 x  H 0 :   13; H 1 :   13; . Reject if s n0  tn1, 12.51  13 = -1.576. Since -1.576 is smaller than the critical value of 1.645, 2.919 87 there is no sufficient evidence to reject the null hypothesis. Obesity rates of adults in Florida. x = 26.82; s = 4.054

H : 0

 28; H 1 :   28; . Reject if

x  0  tn 1, s n

26.82  28 = -2.39. Since -2.39 is smaller than the critical value of 1.645, 4.054 67 there is no sufficient evidence to reject the null hypothesis. Low-income preschool obesity rate in Florida x = 12.51; s = 2.919 x  H 0 :   13; H 1 :   13; . Reject if s n0  tn1, 13.53  13 = 0.993. Since 0.993 is smaller than the critical value of 1.645, 4.338 66 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 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 27.98  30 = -16.985. Since -16.985 is smaller than the critical value of -1.645, 5.468 2108 there is sufficient evidence to reject the null hypothesis. Mean weights of Men in the second interview x = 28.07; s = 5.447 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 28.07  30 = -15.562. Since -15.562 is smaller than the critical value of -1.645, 5.447 1925 there is sufficient evidence to reject the null hypothesis.



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 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 28.93  30 = -7.238. Since -7.238 is smaller than the critical value of -1.645, 7.022 2274 there is sufficient evidence to reject the null hypothesis. Mean weights of Women in the second interview x = 29.02; s = 7.071 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 29.02  30 = -6.385. Since -6.385 is smaller than the critical value of -1.645, 7.071 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 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 27.63  30 = -13.261. Since -13.261 is smaller than the critical value of -1.645, 5.272 870 there is sufficient evidence to reject the null hypothesis. Mean weights of Immigrants in the second interview x = 27.75; s = 5.317 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 27.75  30 = -11.856. Since -11.856 is smaller than the critical value of -1.645, 5.317 787 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


9-25

9-26 9.76

th


Mean weights of White people in the first interview x = 27.85; s = 6.065 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 27.85  30 = -17.18. Since -17.18 is smaller than the critical value of -1.645, 6.065 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 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 27.92  30 = -16.019. Since -16.019 is smaller than the critical value of -1.645, 6.117 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 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 28.83  30 = -6.622. Since -6.622 is smaller than the critical value of -1.645, 5.566 999 there is sufficient evidence to reject the null hypothesis. Mean weights of Hispanic people in the second interview x = 28.90; s = 5.603 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 28.90  30 = -5.984. Since -5.984 is smaller than the critical value of -1.645, 5.603 934 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 Copyright © 2013 Pearson Education


first interview x = 30.15; s = 6.613

H : 0

 30; H 1 :   30; . Reject if

x  0  tn 1, s n

30.15  30 = 0.913. Since 0.913 is greater than the critical value of -1.645, 6.613 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 x  H 0 :   30; H 1 :   30; . Reject if s n0  tn1, 30.29  30 = 1.656. Since 1.656 is greater than the critical value of -1.645, 6.651 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.


9-27

Newbold Stat8 Ism 09 Ge

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