Step 1 :State the appropriate null and alternative hypotheses. The appropriate null and alternative hypotheses are \(\mathrm{H}_{0}: \mu=24\) versus \(\mathrm{H}_{1}: \mu>24\).
Step 2 :Verify that the requirements to perform the test using the t-distribution are satisfied. The requirements are: The sample size is larger than 30, the students' test scores were independent of one another, the students were randomly sampled, a boxplot of the sample data shows no outliers, and the sample data come from a population that is approximately normal. In this case, the sample size is 200, which is larger than 30. The students were randomly sampled. We don't have information about the independence of the test scores, the presence of outliers, or the normality of the population, but we can assume these conditions are met.
Step 3 :Calculate the t-statistic using the formula \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, \(\bar{x} = 24.6\), \(\mu = 24\), \(s = 3.3\), and \(n = 200\). The calculated t-statistic is approximately 2.57.
Step 4 :Calculate the p-value. The p-value is the probability of observing a t-statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, the p-value is approximately 0.0054.
Step 5 :Compare the p-value to the significance level. The common significance level is 0.05. If the p-value is less than the significance level, reject the null hypothesis. In this case, the p-value (0.0054) is less than the significance level (0.05), so we reject the null hypothesis.
Step 6 :Interpret the results. Rejecting the null hypothesis suggests that the students who completed the core curriculum are ready for college-level mathematics, as they are scoring above 24 on the mathematics portion of the exam. Therefore, the final answer is \(\boxed{\text{Reject } H_{0}}\).