Step 1 :State the null and alternative hypotheses: \(\mathrm{H}_{0}: \mu=516, \mathrm{H}_{1}: \mu > 516\)
Step 2 :Calculate the test statistic using the formula for a one-sample t-test: \(t_{0} = \frac{x_{\text{bar}} - \mu}{s / \sqrt{n}}\)
Step 3 :Substitute the given values into the formula: \(t_{0} = \frac{521 - 516}{116 / \sqrt{2000}} \approx 1.93\)
Step 4 :Calculate the p-value using the t-distribution and the degrees of freedom (df = n - 1 = 1999): p-value \approx 0.027
Step 5 :Compare the p-value to the level of significance: 0.027 < 0.10
Step 6 :Since the p-value is less than the level of significance, reject the null hypothesis
Step 7 :Conclude that the mean score after the review course is significantly higher than the original mean score
Step 8 :Final Answer: The test statistic is approximately \(\boxed{1.93}\) and the p-value is approximately \(\boxed{0.027}\). Since the p-value is less than the level of significance (0.10), we reject the null hypothesis and conclude that the mean score after the review course is significantly higher than the original mean score.