Question

The average score for the statistics course is 39%. A professor wants to see if the average score for students who are given colored pens on the first day of class is larger. The scores for the 16 randomly selected students who were given the colored pens are shown below. Assume that the distribution of the population is normal.

44, 34, 32, 48, 42, 33, 40, 41, 50, 36, 50, 51, 43, 48, 47, 45

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What can be concluded at the = 0.10 level of the significance?

1. For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
2. The null and alternative hypotheses would be:

H0:H0:  ? p μ  Select an answer ≠ ≤ ≥ =       

H1:H1:  ? p μ  Select an answer = ≠ > <    

3. The test statistic ? t z  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? ≤ >  αα
6. Based on this, we should Select an answer accept reject fail to reject  the null hypothesis.
7. Thus, the conclusion is that ...

Answer 1

The provided sample mean is 7 = 42.75 and the sample standard deviation is 6.3087, and the sample size is n = 16. S = .(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M = 39 Ha: u > 39 This corresponds to a right-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 = 0.10, and the critical value for a right-tailed test is te 1.341. The rejection region for this right-tailed test is R=tit > 1.341 (3) Test Statistics The t-statistic is computed as follows: Х – по t= 42.75 - 39 6.3087/116 2.378 s/n (4) Decision about the null hypothesis Since it is observed that t = 2.378 >te = 1.341, it is then concluded that the null hypothesis is rejected. = 0.0156 < Using the P-value approach: The p-value is p = 0.0156, and since p 0.10, 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 39, at the 0.10 significance level. Confidence Interval The 90% confidence interval is 39.985 < u < 45.515. Graphically T-Test Results: t-stats = 2.378, p-value =0.0156 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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