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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

Expert verified
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

01

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.
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-sample t-test
A two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In this context, it helps us decide whether male and female basketball players have different mean approval ratings for unsportsmanlike play.
Here’s how it works:
  • The test compares the sample means of two groups—in this case, male and female players.
  • It uses sample data to infer information about population parameters (i.e., the actual mean approval ratings in all high school basketball players).
  • The formula involves calculating a t-score, which tells us how far away the sample mean difference is from the hypothesized difference (0.5, in our case).
This test assumes that the two samples are independent, the data follows a normal distribution, and the variance in the two groups is approximately equal.
Applying the given data:
  • Mean for males (\(ar{X}_m = 2.76\)) and females (\(ar{X}_f = 2.02\)).
  • Standard deviations and sizes are used to calculate the t-score, which shows if the observed mean difference is statistically significant.
Null and Alternative Hypotheses
In hypothesis testing, it's essential to start by formulating two contrasting statements called the null and alternative hypotheses.
The **null hypothesis** (\(H0\)) assumes that there is no effect or difference. In this exercise, it proposes that the difference between the male and female approval rating means is less than or equal to 0.5. This is written as:\[\mu_m - \mu_f \le 0.5\]Where \(\mu_m\) and \(\mu_f\) stand for the mean approval scores for male and female players, respectively.
The **alternative hypothesis** (\(Ha\)) is what you want to prove. It opposes \(H0\), suggesting that the difference in mean ratings is greater than 0.5, expressed as:\[\mu_m - \mu_f > 0.5\]These hypotheses help determine the direction of the test. Here, we're doing a one-tailed test as we are checking if the male ratings exceed the female ratings by more than 0.5. Defining these hypotheses correctly ensures clarity in the testing process.
Significance Level
The significance level, often denoted as \(\alpha\), is a threshold that helps decide whether to reject the null hypothesis. It’s a probability measure that represents the risk of making a type I error—falsely rejecting a true null hypothesis.
In this exercise, \(\alpha = 0.05\), meaning there's a 5% risk of concluding that a difference exists when there's none. Choosing \(\alpha\) depends on how much risk you're willing to take. A common choice in many scientific disciplines is 0.05.
Once you perform the t-test, the resulting t-score is compared to a critical value from the t-distribution, which depends on \(\alpha\) and degrees of freedom (here, \(DF = 121\)). If your calculated t-score exceeds this critical value, you reject \(H0\).
Thus, the significance level is pivotal in hypothesis testing as it balances the risk of error and the need for evidence against the null hypothesis.

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