Question

Previous 21 22 23 24 25 26 27 28 Next Question 28 of 28 (6 points) View problem in a pop-up 8.6 Section Exercise 2 1,2 The average one-way airfare from Pittsburgh to Washington D.C. is $236. A random sample of 24 one-way fares during a particular month had a mean of $213 with a standard deviation of $43. At a = 0.01, is there sufficient evidence to conclude a difference from the stated mean? Source: www.fedstats.gov Part 1 out of 7 State the hypotheses and identify the claim with correct hypothesis. Ho: (select) (select) H1: (select) - $ (select) This hypothesis test is a (select) test. NEXT CHECK

Answer 1

H0: $\mu$ = $236

H1:   $\mu$ not = $236

This hypothesis test is a one sample t test or two tailed test

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 236 Ha: u 7236 This corresponds to a two-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 = 0.01, and the critical value for a two- tailed test is te = 2.807. The rejection region for this two-tailed test is R = {t:t > 2.807} (3) Test Statistics The t-statistic is computed as follows: X-MD s/n 213-236 43/24 = -2.62 (4) Decision about the null hypothesis Since it is observed that t = 2.62 <te = 2.807, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0153, and since p=0.0153 > 0.01, 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 different than 236, at the 0.01 significance level.