Download Math 243 Final Exam Solutions: Statistical Analysis and Hypothesis Testing and more Exams Probability and Statistics in PDF only on Docsity! Math 243 Final Exam Solutions Chris Phan, Summer 2004 Multiple Choice Questions Question Form A (Yellow) Form B (Blue) 1 C B 2 B C 3 D C 4 C D 5 C B 6 A C 7 D A 8 D C For help with question 7, please see p. 238. Open Response The questions on each form are nearly identical, except in a different order. The numbers below refer to the question number on Form A (Yellow). The question number for Form B (Blue) is given in parentheses. 9. (12.) This is a matched pairs design, so we do a one-sample t-test on the differences within each of the pairs. (We can’t do a two-sample t-test, because the selection of volunteers in the second population is influenced by the selection of volunteers in the first population.) Suppose we do drug − placebo (the other way is analogous). Then our hypothesis are H0 : µ = 0 and Ha : µ < 0. We obtain t = −0.5937, to which the calculator gives a P -value of P = 0.2837. Thus, as P 6< α = .05, we fail to reject the null hypothesis and conclude that there is not significant evidence, at the 5% level, that those taking the drug in each pair suffered from hiccups for less time than those taking the placebo. (If you are doing this by hand, you will compare t to t∗ = −1.833, the α = .05 left-tailed critical value from the t distribution with 9 degrees of freedom. You accept H0 since t 6< t∗.) 10. (11.) Since we know σ = 0.5, we use a z-test. The hypotheses are H0 : µ = 2000 and Ha : µ < 2000. We obtain z = −0.4, which our calculator gives a P -value of P = 0.3446. Thus, as P 6< α = .05, we fail to reject the null hypothesis and conclude that there is not significant evidence, at the 5% level, that the mean volume of bottles produced at the company is less than 2000 ml. (If you are doing this by hand, you will compare z to z∗ = −1.645, the α = .05 left-tailed critical value for the Standard Normal distribution. You accept H0 since z 6< z∗.) 11. (9.) Here, we use a one-proportion z-test. The hypotheses are H0 : p = 0.5 and Ha : p > 0.5. We obtain p̂ = .5078, which has a z-score of z = 0.5. The calculator gives this a P -value of P = 0.3085, and so we fail to reject the null hypothesis as P 6< α = .05. Thus, there is not significant evidence, at the 5% level, that over half of voters currently plan to vote for the measure. (By hand, you would compare z to z∗ = 1.645, the right-tailed α = .05 critical value for the Standard Normal distribution.You accept H0 since z 6> z∗.) 12. (10.) Here, we use a two-proportion z-test. Suppose we consider the students with the discount to be Popu- lation 1, and the students without the discount Population 2. Then, our hypotheses are H0 : p1 = p2 and Ha : p1 < p2. We obtain p̂1 = .0207, p̂2 = .0366, p̂ = .0313, which yields z = −2.3777 and a P -value of P = .0087. Thus, as P < α = .01, we reject the null hypothesis and conclude that, at the 1% level, there is significant evidence that the proportion of students enjoying the discount who have been at fault in a serious accident is less than the proportion among students who do not enjoy the discount. (If you are doing this by hand, you compare z to z∗ = −2.326, the α = .01 left-tailed critical value for the Standard Normal distribution. You reject H0 as z < z∗.) 1
