Question

7. Use the following collection of 30 test scores: 100, 65, 67, 61, 62, 70, 75, 73, 88, 77, 83, 79, 95, 79, 80, 102, 86, 87, 87, 91, 87, 89, 92, 90, 99, 87, 72, 93, 79, 52 a. Construct a frequency distribution table using 6 classes: include classes, frequencies, class marks, class boundaries, relative frequencies and relative percentages. b. Find the mean to the nearest tenth and sample standard deviation to the nearest tenth. c. Use the results from part b to determine the Z-score associated with a data value of 79. 8. In a recent poll on education 64% of all college students stay up all night studying for final exams. A professor at a local college says this is too low and wants to test this claim. He surveys 5,000 college students and finds that 3250 have stayed up all night studying for final exams. Which statement can we conclude. Use a -5% a. State the null and alternative hypothesis b. Determine and draw the hypothesis testing model (label everything) c. Determine the critical value(s) and test statistic d. Determine the conclusion
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Answer 1

8.

The following information is provided: The sample size is N = 5000, the number of favorable cases is X = 3250, and the sample proportion is p 0.65, and the significance level is a = 0.05 - X N 3250 5000 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p = 0.64 Ha: p > 0.64 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to 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: z = = P – po V po(1 – po)/n 0.65 – 0.64 10.64(1 – 0.64)/5000 1.473 (4) Decision about the null hypothesis 1.64, it is then concluded that the null hypothesis is Since it is observed that z = 1.473 < Zc = not rejected. Using the P-value approach: The p-value is p = 0.0704, and since p = 0.0704 > 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 proportion p is greater than Po, at the a = = 0.05 significance level. Confidence Interval The 95% confidence interval for p is: 0.637 < p < 0.663. Graphically Z-Test Results: z-stats = 1.473, p-value =0.0704 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
Thus there is no sufficient evidence to conclude that the proportion is greater than 64%.

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