The annual salaries (in thousands of dollars) of 8 men in middle
management at a given company are : 55.5, 64.8, 68.2, 70.2, 52.4, 56.8, 60.6, 72.5 while those for 6 women are : 56.2, 48.8, 58.4, 50.9, 60.2, 54.5.
Let X and Y denote the salaries of the men and women respectively.
Assuming normal distribution and equal standard deviation, test the null hypothesis
Hx = My
against the alternative hypothesis
Hx > Hy
at
5 percent level of significance.
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We have a hypothesis testing problem comparing salaries between men and women in middle management. We have 8 men's salaries and 6 women's salaries, all in thousands of dollars. We assume normal distributions with equal standard deviations. Our null hypothesis states that the mean salaries are equal, while the alternative hypothesis suggests men's mean salary is greater than women's. We'll test this at a 5 percent significance level.
Now let's calculate the sample statistics. For men, we have 8 observations with a sample mean of 62.56 thousand dollars and sample variance of 54.95. For women, we have 6 observations with a sample mean of 54.83 thousand dollars and sample variance of 20.77. Since we assume equal population variances, we calculate the pooled standard deviation using the formula shown. The pooled variance is 40.85, giving us a pooled standard deviation of 6.39.
Now we calculate the test statistic using the t-test formula. The numerator is the difference between sample means: 62.56 minus 54.83 equals 7.73. The denominator involves the pooled standard deviation times the square root of one over n1 plus one over n2, which equals 3.41. Therefore, our calculated t-statistic is 2.267. With 12 degrees of freedom and a 5 percent significance level, the critical value is 1.782. Since our calculated t-value of 2.267 exceeds the critical value, we can see this visually on the t-distribution.
Now we make our decision. Our decision rule is to reject the null hypothesis if the calculated t-value exceeds the critical value. Since 2.267 is greater than 1.782, we reject the null hypothesis. At the 5 percent significance level, we have sufficient evidence to conclude that the mean salary of men is greater than the mean salary of women. The p-value is approximately 0.021, which is less than 0.05, confirming our decision to reject the null hypothesis. The test statistic falls in the rejection region shown in red.
To summarize our hypothesis test: We tested whether men's mean salary exceeds women's mean salary using a two-sample t-test with pooled variance. Our calculated t-statistic of 2.267 exceeded the critical value of 1.782, leading us to reject the null hypothesis at the 5 percent significance level. We concluded that men's mean salary is significantly higher than women's in this company's middle management.