Question

The article "The Relationship of Task and Ego Orientation to Sportsmanship Attitudes and the Perceived Legitimacy of Injurious Acts" (Research Quarterly for Exercise and Sport [1991]: \(79-87\) ) examined the extent of approval of unsporting play and cheating. High school basketball players completed a questionnaire that was used to arrive at an approval score, with higher scores indicating greater approval. A random sample of 56 male players resulted in a mean approval rating for unsportsmanlike play of \(2.76\), whereas the mean for a random sample of 67 female players was \(2.02 .\) Suppose that the two sample standard deviations were \(.44\) for males and \(.41\) for females. Is it reasonable to conclude that the mean approval rating is higher for male players than for female players by more than .5? Use \(\alpha=.05\).

Short answer

To provide a short answer, we have to finish the t-test by comparing our computed t-score with the critical value. If t-score > \(t_{critical}\), the hypothesis that the mean approval for unsportsmanlike play is higher for males by more than 0.5 is upheld.

Step-by-step solution

Formulate the hypothesis

The null hypothesis (\(H0\)) would be that the difference between the two means is less than or equal to 0.5. The alternative hypothesis (\(Ha\)) is that the difference between the two means is more than 0.5. So, \(H0: \mu_m - \mu_f ≤ 0.5\) and \(Ha: \mu_m - \mu_f > 0.5\) where \(\mu_m\) and \(\mu_f\) are the mean approval ratings for males and females, respectively.

Perform the t-test

We need to apply the two-sample t-test for difference of means formula: \(t = \frac{(\bar{X}_m - \bar{X}_f) - (D0)}{\sqrt{\frac{s^2_m}{n_m} + \frac{s^2_f}{n_f}}}\), where \(\bar{X}_m = 2.76\) (mean of males), \(\bar{X}_f = 2.02\) (mean of females), \(D0 = 0.5\) (claimed difference), \(s_m = 0.44\) (standard deviation of males), \(n_m = 56\) (number of males), \(s_f = 0.41\) (standard deviation of females) and \(n_f = 67\) (number of females). Substituting these values into the formula gives us the t-score.

Determine the critical value and Conclusion

Check the t-table for a t-critical value with \(\alpha = 0.05\) for a one-tailed test and degrees of freedom \(DF = n_m + n_f - 2 = 56 + 67 - 2 = 121\). If the computed t-score is greater than \(t_{critical}\), then reject the null hypothesis and conclude that the mean approval rating is higher for male players than for female players by more than 0.5.