Suppose the mean IQ score of people in a certain country is 103 . Suppose the director of a college obtains a simple random sample of 38 students from that country and finds the mean IQ is 107.1 with a standard deviation of 13.8 Complete parts (a) through (d) below.
(a) Consider the hypotheses $\mathrm{H}_{0}, \mu=103$ versus $\mathrm{H}_{1} \mu> 103$. Explain what the director is testing. Perform the test at the $\alpha=0.01$ level of significance. Write a conclusion for the test.
Explain what the director is testing. Choose the correct answer below.
A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to 103.
B. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 103
C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than 103.
D. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually equal to 103.
Find the test statistic for this hypothesis test
1.83 (Round to two decimal places as needed)
Find the P-value for this hypothesis test
(Round to three decimal places as needed.)
Answer
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 \(\boxed{0.038}\).
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 \(\boxed{0.038}\).
