Question

The article "Workaholism in Organizations: Gender Differences" (Sex Roles [1999]: \(333-346\) ) gave the following data on 1996 income (in Canadian dollars) for random samples of male and female MBA graduates from a particular Canadian business school: \begin{tabular}{lccc} & \(\boldsymbol{n}\) & \(\boldsymbol{x}\) & \(\boldsymbol{s}\) \\ \hline Males & 258 & \(\$ 133,442\) & \(\$ 131,090\) \\ Females & 233 & \(\$ 105,156\) & \(\$ 98,525\) \\ \hline \end{tabular} a. For what significance levels would you conclude that the mean salary of female MBA graduates of this business school is above \(\$ 100,000\) ? b. Is there convincing evidence that the mean salary for female MBA graduates of this business school is lower than the mean salary for the male graduates?

Short answer

The solutions to these questions depend on the computed p-values, which may be found using a standard distribution table or statistical software. For a, any significance level greater than this p-value would lead us to conclude that the mean salary of female MBA graduates is above $100,000. On part b, if the p-value is less than the significance level, there is convincing evidence that the mean salary of female MBA graduates of this business school is less than the mean salary of male MBA graduates.

Step-by-step solution

Formulate the Hypothesis for Part a

The null hypothesis (\(H_0\)) can be formulated as: The mean salary of female MBA graduates of this business school is equal to $100,000. The alternative hypothesis (\(H_1\)) is: The mean salary of female MBA graduates of this business school is above $100,000. Mathematically, \(H_0: \mu = \$100,000\) and \(H_1: \mu > \$100,000\).

Statistical Calculation for Part a

Compute the test statistic (z) using the formula: \[z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\] We plug the values from the exercise into this equation, where \(\bar{x} = \$105,156\), \(\mu_0 = \$100,000\), \(s = \$98,525\), and \(n = 233\). Calculate the z-score and find the corresponding p-value from the standard normal distribution table or via computation.

Decision for Part a

Using the p-value, compare it with the standard significance level (usually 0.05). If the p-value is less than 0.05, we reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis. The significance levels at which we would conclude that the mean salary of female MBA graduates is above $100,000 are those greater than the computed p-value.

Formulate the Hypothesis for Part b

The null hypothesis (\(H_0\)) is that the mean salary for female MBA graduates is equal to that of the male graduates. The alternative hypothesis (\(H_1\)) is that the mean salary for female MBA graduates is lower than that of the male graduates. In mathematical notation, \(H_0: \mu_F = \mu_M\) and \(H_1: \mu_F < \mu_M\).

Statistical Calculation for Part b

Compute the test statistic using the formula for the hypothesis test of difference between two means (since the standard deviations are known, we will use z-test): \[z = \frac{\bar{x}_F - \bar{x}_M} {\sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}}}\] where \(\bar{x}_F = \$105,156\), \(\bar{x}_M = \$133,442\), \(s_F = \$98,525\), \(s_M = \$131,090\), \(n_F = 233\), and \(n_M = 258\). After finding the z-score, compute the p-value.

Decision for Part b

Using the p-value, compare it with the standard significance level (usually 0.05). If the p-value is less than the level of significance, we reject the null hypothesis in favor of the alternative hypothesis. If the p-value is larger than the level of significance, we fail to reject the null hypothesis.