Question

5.  The reported average cost per workbook at a large college is $27.50.   A  professor claims that the actual average cost per workbook is higher than $27.50. A sample of 40 different workbooks has an average cost of $29.95. The population standard deviation is known to be $5.50. Can the null hypothesis be rejected at alpha = 0.05? For question #5 (multiple choice), tell whether this a right, left, or two- tailed test. For question #6, identify the null hypothesis from choices given. For question #7, give just your calculated z- value.  For question #8, convert your answer from #7 into a P-value, and place this value in the blank space provided. For question #9, answer the true/false question about the rejection of the null hypothesis.

5. Multiple choice question:  right-tailed, left-tailed, or two-tailed test

6. Multiple choice question:   enter your answer for the null hypothesis.

7. Fill in the blank, enter your calculated z-value

8. Fill in the blank with your answer from #7 converted into a P-value ,given to the ten thousandths place

9. True or False: The null hypothesis will be rejected based on the data given.

Answer 1

The provided sample mean is 7 = 29.95 and the known population standard deviation is o = 5.5, and the sample size is n = = 40. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ = 27.5 Ha: u > 27.5 This corresponds to a right-tailed test, for which a z-test for one mean, with known 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 right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R= {z: z > 1.64} (3) Test Statistics The Z-statistic is computed as follows: X - μo z = 29.95 – 27.5 5.5/V40 2.817 oln (4) Decision about the null hypothesis 2.817 > Zc = 1.64, it is then concluded that the null Since it is observed that z = hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0024, and since p = 0.0024 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is greater than 27.5, at the 0.05 significance level.

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