Question

Example 8.26·#17 The average movie ticket in 2010 cost $789. A random sample of 15 movie tickets from the suburbs of a large U.S. city indicated that the mean cost was $11.09 with a standard deviation of $4.86. At the 0.01 level of significance, can it be concluded that the mean is higher than the national average? Givens n-and df- Solution Step 1 State the null and alternative hypothesis Ho : μ Step 2. 167 Step 3 stat-test→ #21 test Step 4 Step 5 At 0.01 there is...

Answer 1

The provided sample mean is X11.09 and the sample standard deviation is s- 4.86, and the sample size is n 15. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: μ-7.89 Ha: μ > 7.89 This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. Based on the information provided, the significance level is α 0.01, and the critical value for a right-tailed test is te2.624 The rejection region for this right-tailed test is R-t:t>2.624 (3) Test Statistics The t-statistic is computed as follows: 2.55 /Vn 4.86/V15 (4) Decision about the null hypothesis Since it is observed that t 2.55 < tc2.624, it is then concluded that the null hypothesis is not rejected

Using the P-value approach: The p-value is p = 0.0116, and since p 0.0116 > 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 greater than 7.89, at the 0.01 significance level. Confidence Interval The 99% confidence interval is 7.355 < < 14.825. Graphically T-Test Results: t-stats 2.55, p-value -0.0116 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 4.03.5 -3.0 2.5-2.01.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Hence,

At $\alpha$ = 0.01 there is not enough evidence to conclude that the mean is higher than the national average.