An education organization claims that the mean SAT scores for male athletes and male non-athletes at a college are different. A random sample of 26 male athletes at the college has a mean SAT score of 1189 and a standard deviation of 218. A random sample of 18 male non-athletes at the college has a mean SAT score of 1376 and a standard deviation of 186. At α=0.05, can you support the organization’s claim? Interpret the decision in the context of the original claim. Assume the populations are normally distributed and the population variances are equal.

Verified step by step guidance

1. Identify the hypotheses for the test. The null hypothesis \(H_0\) states that the mean SAT scores for male athletes and male non-athletes are equal: \(\mu_1 = \mu_2\). The alternative hypothesis \(H_a\) states that the means are different: \(\mu_1 \neq \mu_2\).
2. Since the population variances are assumed equal and the samples are independent, use the two-sample t-test for equal variances. Calculate the pooled standard deviation \(s_p\) using the formula: \[s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\] where \(n_1\) and \(n_2\) are the sample sizes, and \(s_1\) and \(s_2\) are the sample standard deviations.
3. Calculate the test statistic \(t\) using the formula: \[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means.
4. Determine the degrees of freedom for the test, which is \(df = n_1 + n_2 - 2\). Then, find the critical t-value(s) from the t-distribution table for a two-tailed test at significance level \(\alpha = 0.05\).
5. Compare the absolute value of the calculated test statistic \(|t|\) to the critical t-value. If \(|t|\) is greater than the critical value, reject the null hypothesis \(H_0\) and conclude there is sufficient evidence to support the claim that the mean SAT scores differ. Otherwise, do not reject \(H_0\). Interpret this decision in the context of the original claim about the difference in mean SAT scores.

Key Concepts

Two-Sample t-Test for Means

This test compares the means of two independent groups to determine if there is a statistically significant difference between them. It uses sample means, standard deviations, and sizes to calculate a t-statistic, which is then compared to a critical value based on the chosen significance level (α). Here, it helps test if male athletes and non-athletes have different mean SAT scores.

Assumption of Equal Population Variances

When performing a two-sample t-test, assuming equal variances means the variability in both groups is similar. This allows pooling the sample variances to get a more accurate estimate of the common variance, which affects the calculation of the test statistic and degrees of freedom. This assumption simplifies the test and is crucial for valid results.

Significance Level and Hypothesis Testing

The significance level (α = 0.05) defines the threshold for rejecting the null hypothesis, which states there is no difference between group means. If the test statistic falls in the critical region, we reject the null and support the claim of a difference. Interpreting the decision involves relating statistical results back to the real-world context of SAT scores.