Question

6.13

A firm has a generous but rather complicated policy concerning end-of-year bonuses for its lower-level managerial personnel. The policy’s key factor is a subjective judgment of “contribu- tion to corporate goals.” A personnel officer took samples of 24 female and 36 male managers to see whether there was any difference in bonuses, expressed as a percentage of yearly salary. The data are listed here:

Gender Bonus Percentage

F 9.2 7.7 11.9 6.2 9.0 8.4 6.9 7.6 7.4

F 8.0 9.9 6.7 8.4 9.3 9.1 8.7 9.2 9.1

F 8.4 9.6 7.7 9.0 9.0 8.4

M 10.4 8.9 11.7 12.0 8.7 9.4 9.8 9.0 9.2

M 9.7 9.1 8.8 7.9 9.9 10.0 10.1 9.0 11.4

M 8.7 9.6 9.2 9.7 8.9 9.2 9.4 9.7 8.9

M 9.3 10.4 11.9 9.0 12.0 9.6 9.2 9.9 9.0

a. What are the populations of interest in this study?

b. Is there significant evidence that the mean bonus percentage for males is more than five units larger than the mean bonus percentage for females? Use a 5 .05.

c. What is the level of significance of your test?

d. Estimate the difference in the mean bonus percentages for males and females using a 95% confidence interval.

Step 1

a

The population of interest in this study are lower-level managerial employees (both males and females) of the firm.

b.

We are comparing two group means to identify if there is any discrimination in doling out bonuses. We will hence use a two sample T-test to get to the bottom of this.

The null hypothesis is that the difference in means of two sample is 5. Here, 5 is the null-hypothesized difference which we will use to calculate our test statistic.

Standard Error in difference of means = [ Variance of Males' Bonus percentages /n1 + Variance of Females' bonus percentages/n2] ^ 0.5

= [ Variance of Males' Bonus percentages /36 + Variance of Females' bonus percentages/24] ^ 0.5

= [1.007714/36 + 1.413623/24 ]^ 0.5

= 0.294776 ~ 0.295

T-statistic = Difference in sample means - Null-hypothesized difference in means/ Standard Error in the difference of sample means

= (9.68 - 8.53 - 5) / 0.295

= 1.15 -5/ 0.295

= -13.0608

Negative value of test statistic indicates that the difference in sample means is far less than the hypothesized difference.

Step 2

Degrees of freedom in this case will be lesser of the n1 - 1 and n2-1 ie ie Min ( 35,23) = 23

Now that we know df = 23 and alpha = 0.05

We also know that this is a one-tailed test as hypothesis says difference is larger than 5. It would be two-tailed if the hypothesis had said larger or smaller but it didn't. So we read the number under t(0.05) /one-tail /df= 23 to get our critical value.

Critical value = 1.714

Since our test-statistic is more extreme in the opposite direction of the critical value, we reject the null hypothesis. There is no significant evidence that the mean bonus percentage for males is more than five units larger than the mean bonus percentage for females

Step 3

c

Level of significance is 0.05 which is the probability of rejecting a null hypothesis ev...

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