# Statistics For Managers Using Microsoft Excel Global Edition 8th Edition By David – Test Bank

Chapter 12

CHAPTER 12

12.1 (a) For df = 1 and = 0.01, = 6.635.

(b)For df = 1 and = 0.005, = 7.879.

(c)For df = 1 and = 0.10, = 2.706.

12.2 (a) For df = 2 and = 0.05, = 5.991.

(b)For df = 2 and = 0.025, = 7.378.

(c)For df = 2 and = 0.005, = 10.597.

12.3 (a)-(b)

Observed Freq Expected Freq Observed Freq Expected Freq Total Obs, Row 1

20 20 30 30 50

chi-sq contrib=0chi-sq contrib=0

Observed Freq Expected Freq Observed Freq Expected Freq Total Obs, Row 2

30 30 45 45 75

chi-sq contrib=0chi-sq contrib=0

Total Obs, Col 1 Total Obs, Col 2 GRAND TOTAL

50 75 125

(c) = 0 + 0 + 0 + 0 = 0. Since < 3.841, it is not significant at the 5% level of significance.

12.4 (a)

Observed Freq Expected Freq Observed Freq Expected Freq Total Obs, Row 1

20 25 30 25 50

chi-sq contrib=1.00chi-sq contrib=1.00

Observed Freq Expected Freq Observed Freq Expected Freq Total Obs, Row 2

30 25 20 25 50

chi-sq contrib=1.00chi-sq contrib=1.00

Total Obs, Col 1 Total Obs, Col 2 GRAND TOTAL

50 50 100

(b)Decision rule: If > 3.841, reject H0.

Test statistic: = 1.00 + 1.00 + 1.00 + 1.00 = 4

Decision: Since = 4 is greater than the critical value of 3.841, it is significant at the 5% level of significance.

12.5 (a)

H0: H1: where population: 1 = males, 2 = females

Decision rule: df = 1. If > 6.635, reject H0.

Test statistic: = 6.0465

Decision: Since = 6.0465 is less than the upper critical bound of 6.6349, do not reject H0. There is not enough evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.01 level of significance.

(b)p-value = 0.0139. The probability of obtaining a test statistic of 6.0465 or larger when the null hypothesis is true is 0.0139.

12.5 (c) (a) H0: H1: where Populations: 1 = males, 2 = females

cont.

Decision: Since = 1.0623 is smaller than the upper critical bound of

6.635, do not reject H0. There is not enough evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.01 level of significance.

(b)p-value = 0.3027. The probability of obtaining a test statistic of 1.0623 or larger when the null hypothesis is true is 0.3027.

(d)The results of (a) – (c) are exactly the same as those of Problem 10.29. The in (a) and the in Problem 10.29 (a) satisfy the relationship that = and the p-value obtained in (b) is exactly the same as the p-value in Problem 10.29 (b). Similarly, the in (c) and the in Problem 10.29 (d) satisfy the relationship that = and the p-value obtained in (c) is exactly the same as the p-value in Problem 10.29 (d).

12.6(a)H0: H1:

(b)Decision rule: df = 1. If > 3.841, reject H0.

Test statistic: = 41.884

Decision: Reject H0. There is enough evidence of a difference between the free-to-air TV viewers and their cable TV counterparts in terms of their inclination to visit the website of the advertiser after viewing the advertisement on TV.

(c)Decision rule: df = 1. If > 6.635, reject H0.

Test statistic: = 41.884

Decision: Reject H0. There is enough evidence of a difference between the free-to-air TV viewers and their cable TV counterparts in terms of their inclination to visit the website of the advertiser after viewing the advertisement on TV.

(d)The decision to reject H0 remains unchanged despite ‘relaxing’ the value to 0.01.

12.7(a)Decision rule: df = 1. If > 3.841, reject H0.

Test statistic: = 0.074

Decision: Since = 0.074 > 3.841, do not reject H0. There is not enough evidence to conclude that there is gender bias in securing Series AVC funding.

(b) p-value = 0.79. The probability of obtaining a test statistic of 0.074 or larger when the null hypothesis is true is 0.79.

12.8 (a) H0: H1:

PHStat output with computation:

Decision rule: df = 1. If > 3.8415, reject H0.

Test statistic: = 11.5702

Decision: Since = 11.5702 is greater than the upper critical bound of 3.8415,

reject H0. There is enough evidence of a significant difference between organizations with 500 to 2,499 employees and organizations with 2,500+ employees with respect to the proportion that have employee recognition programs.

(b)p-value = 0.0007. The probability of obtaining a difference in proportions that gives rise to a test statistic above 11.5702 is 0.0007 if there is no difference in the proportions.

(c)The results of (a) and (b) are exactly the same as those of Problem 10.32. The in (a) and the ZSTAT in Problem 10.32 (a) satisfy the relationship that = 11.5702 = = and the p-value in Problem 10.32 (b) is exactly the same as the p-value obtained in (b).

12.9 (a) Decision rule: df = 1. If > 3.841, reject H0.

Test statistic: = 9.803

Decision: Since = 9.803 > 3.841, reject H0. There is enough evidence of a difference between the proportion of hoteliers and bankers who had a working weekend in the last quarter of 2015.

(b)p-value = 0.0017. The probability of obtaining a test statistic of 9.803 or larger when the null hypothesis is true is 0.0017.

12.10 (a) PHStat output with computation:

(b)Decision rule: df = 1. If > 3.841, reject H0.

Test statistic: = 0.216

Decision: Since = 0.216 < 3.841, do not reject H0. There is not enough evidence of a difference between the opinion on upselling potential of outlet managers of restaurants deploying digital menus and those not deploying digital menus.

(c)p-value = 0.642. The probability of obtaining a test statistic of 0.216 or larger when the null hypothesis is true is 0.642.

12.11 (a) df = (r – 1)(c – 1) = (2 – 1)(5 – 1) = 4

(b) = 9.488

(c) = 13.277

12.12 (a) The expected frequencies in the first row are 43.53, 63.12, and 78.35.

The expected frequencies in the second row are 56.47, 81.88, and 101.65.

Expected Frequencies

Column variable

Row variable C1 C2 C3 Total

R1 43.53 63.12 78.35 185

R2 56.47 81.88 101.65 240

Total 100 145 180 425

(b) = 10.612, which is significant at = 0.05 as it is greater than the critical value of 5.99.

12.13 (a) The expected frequencies in the first row are 25, 25, and 25.

The expected frequencies in the second row are 25, 25, and 25.

(b) = 4.000. The critical value with 2 degrees of freedom and = .05 is 5.991. The result is not deemed significant.

12.14 (a) It is evident from the below output that there is a statistically significant difference in the proportion of age groups downloading travel apps. There isn’t adequate evidence to support the null hypothesis that these proportions are equal.

(b)p-value = 0.029. The probability of obtaining a data set which gives rise to a test statistic of 9.017 or more is 0.029 if the null hypothesis is true.

12.14 (c) Excel output of the Marascuilo procedure:

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