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Question 4 of 7 (1 point) Attempt 1 of Unlimited View question in a popup 8.3 Section Exercise 16 (p-value College tuition: The mean annual tuition and fees in the 2013-2014 academic year for a sample of 22 private colleges in California was $33,750 with a standard deviation of $7400. A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is less than $35,000? Use the 0.05 level of significance and the P-value method with the TI-84 Plus calculator Part: 0/4 Part 1 of 4 State the appropriate null and alternate hypotheses. HO H: This hypothesis test is a (Choose one) v test. Save For Later Submit Assignment

Answer 1

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The provided sample mean is X = 33750 and the sample standard deviation is s= 7400, and the sample size is n = 22. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u 235000 Ha: u < 35000 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is te = -1.721. The rejection region for this left-tailed test is R=t:t< -1.721 (3) Test Statistics The t-statistic is computed as follows: X-Mo t= | sim 33750 - 35000 7400//55 --0.792

(4) Decision about the null hypothesis Since it is observed that t = -0.792 > t = -1.721, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.2185, and since p=0.2185 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 35000, at the 0.05 significance level.