Question

The average final exam score for the statistics course is 75%. A professor wants to see if the average final exam score for students who are given colored pens on the first day of class is lower. The final exam scores for the 15 randomly selected students who were given the colored pens are shown below. Assume that the distribution of the population is normal. 63, 81, 82, 65, 75, 49. 86, 75, 56, 62, 72, 83, 81, 66, 48 What can be concluded at the the a = 0.05 level of significance level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: u 75 H: NE 75 c. The test statistic (please show your answer to 3 decimal places.) d. The p-value = 1 (Please show your answer to 4 decimal places.) e. The p-value is ? va f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... O The data suggest that the population mean final exam score for students who are given colored pens at the beginning of class is not significantly lower than 75 at a = 0.05, so there is statistically insignificant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is lower than 75. O The didht suggest the populaton mean is significantly lower than 75 at a = 0.05, so there is statistically significant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is lower than 75. O The data suggest the population mean is not significantly lower than 75 at a = 0.05, so there is statistically insignificant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is equal to 75.

Answer 1

Standard Deviation, s: 12.356606098995 Count, N: 15 Sum, Ex: 1044 Mean, X 69.6 Variance, s2: 152.68571428571 Steps 1 N S = (Xi – T), N 1 i=1 S2 = E(xi - 7)2 N-1 (63 - 69.6)2 + ... + (48 - 69.6)2 15-1 2137.6 14 = 152.68571428571 S = 7152.68571428571 = 12.356606098995

The provided sample mean is X = 69.6 and the sample standard deviation is 12.3566, and the sample size is n = 15. S = (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 75 Ha: u < 75 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 0.05, and the Based on the information provided, the significance level is a = critical value for a left-tailed test is te = -1.761. The rejection region for this left-tailed test is R=t:t< -1.761 (3) Test Statistics The t-statistic is computed as follows: Х – по t= 69.6 – 75 12.3566/V15 -1.693 s/n (4) Decision about the null hypothesis Since it is observed that t = -1.693 > tc = -1.761, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0563, and since p= 0.0563 > 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 75, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 62.757 < p < 76.443. Graphically. T-Test Results: t-stats = -1.693, p-value =0.0563 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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