Respond to each item completely, and provide all evidence of your calculations to receive full credit. A sample of $n=9$ college students took the GRE before and after a GRE prep course. For each student, the difference between the first score and the second score was measured. The results showed that the GRE scores were, on average, higher after the course with $M_{D}=17$ points and $s^{2}=324$. On the basis of this sample data, can you conclude that there was a significant difference in GRE scores after taking the training course? Click the link for Tables B1 and B2 $\downarrow$ 1. Perform a null hypothesis test using a two-tailed test and $a=.05$. Be sure to complete and label all 4 steps in the hypothesis testing.process. (8 points) 2. Calculate the effect size using Cohen's d if a significant effect exists. (2 points) 3. Calculate $r^{2}$ if a significant effect exists. (2 points) Edit View Insert Format Tools Table

Step 1 ：State the null hypothesis and the alternative hypothesis. The null hypothesis is that there is no significant difference in GRE scores after taking the training course, denoted as \(H0: \mu_D = 0\). The alternative hypothesis is that there is a significant difference in GRE scores after taking the training course, denoted as \(H1: \mu_D \neq 0\).

Step 2 ：Identify a test statistic. Since the standard deviation is known and the sample size is small, a t-test for dependent samples is used.

Step 3 ：Determine the critical region. For a two-tailed test with \(\alpha = .05\) and degrees of freedom \(df = n - 1 = 9 - 1 = 8\), the critical values from a t-distribution table are -2.306 and +2.306.

Step 4 ：Calculate the test statistic and make a decision. The test statistic t is calculated as \(t = (MD - \mu_D) / (sD / \sqrt{n}) = (17 - 0) / (18 / \sqrt{9}) = 17 / 6 = 2.83\). Since the calculated t (2.83) is greater than the critical value (2.306), the null hypothesis is rejected. Therefore, \(\boxed{\text{There is a significant difference in GRE scores after taking the training course.}}\)

Step 5 ：Calculate Cohen's d to measure the effect size. Cohen's d is calculated as \(d = MD / sD = 17 / 18 = 0.94\). According to Cohen's conventions, an effect size of 0.94 is considered large.

Step 6 ：Calculate R-squared to measure the proportion of the variance in the dependent variable that is predictable from the independent variable. R-squared is calculated as \(r^2 = t^2 / (t^2 + df) = 2.83^2 / (2.83^2 + 8) = 7.9889 / (7.9889 + 8) = 7.9889 / 15.9889 = 0.5\). Therefore, \(\boxed{50\%\text{ of the variance in the GRE scores can be explained by the training course.}}\)