Step 1 :The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 103. This is a one-tailed t-test problem where we are testing if the sample mean is significantly greater than the population mean. The null hypothesis is that the population mean is 103 and the alternative hypothesis is that the population mean is greater than 103.
Step 2 :To find the test statistic, we can use the formula for the t-score which is \((\text{sample mean} - \text{population mean}) / (\text{sample standard deviation} / \sqrt{\text{sample size}})\).
Step 3 :To find the p-value, we can use the cumulative distribution function (CDF) of the t-distribution with degrees of freedom equal to sample size - 1. The p-value is the probability that a t-distributed random variable is greater than the observed t-score, which can be found as \(1 - \text{CDF(t-score)}\).
Step 4 :Let's calculate these values. The population mean (mu) is 103, the sample mean (x_bar) is 107.1, the sample standard deviation (s) is 13.8, and the sample size (n) is 38. The significance level (alpha) is 0.01.
Step 5 :The test statistic (t-score) is approximately 1.83 and the p-value is approximately 0.038.
Step 6 :Since the p-value is greater than the significance level (0.01), we fail to reject the null hypothesis. This means that we do not have sufficient evidence to conclude that the population mean IQ score is greater than 103.
Step 7 :The correct answer to the first question is B. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 103. The test statistic for this hypothesis test is approximately \(\boxed{1.83}\). The P-value for this hypothesis test is approximately \(\boxed{0.038}\).