answers 3 - 1. C. In a matched pairs study, subjects are...

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1. C. In a matched pairs study, subjects are matched in pairs and the outcomes are compared within each matched pair. Key word: twins. 2. D. To analyze the data, we first subtract the exam score in Class 1 from that in Class 2 to obtain the difference for each pair of twin. Here D µ is the mean difference. The null hypothesis says that no difference occurs, and a H says that exam scores are higher in Class 1. When you subtract class1- class2, if class 1 scores are higher, D >0. 3. A. In a matched pairs analysis, we assume the population of difference has a normal distribution, so the t procedures can be applied here. And the t distribution is with n-1 (12-1=11) degrees of freedom. 4. C. Use the rule p-value <alpha, reject o H . Our reasonable alpha levels are .10,.05, and .01. We reject H 0 at all these levels. So, we can conclude the new method is better than the old method. 5. B. The goal of inference is to compare the responses (score mean) in two groups and the responses(score mean) in each group are independent of those in the other group. 6. C. Because H a expresses the effect we hope to find evidence for, so the H a here should be: µ 1 −µ 2 > 0. (the mean exam scores in Class 1 is higher than that in Class 2). If 1 is greater than 2 , we could get a value greater than zero. And then set up H 0 : µ 1 −µ 2 = 0 as the statement that the hoped-for effect is not present. 7. B. By hand, we use the conservative degrees of freedom, the smaller of 1 1 n and 1 2 n . Here, 1 n =12, 2 n =11, so we choose DF= 1 2 n =11-1=10. 8. A. It follows the t (n-1=10) distribution. For 99% confidence interval, Table D gives t*= 3.169. 9. B. From the CI, we are 95% confidence that the difference in the proportion is between -.163 and -.02. So we can say that females have a higher proportion of convictions for DUI than males. Since p1-p2 is negative, the proportion of group 2 (females) must be bigger than the proportion of group 1.
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10. B. From the confidence interval, we know the p1-p2 should be less than 0, which means we should reject the null. Besides, we always reject the null when the p-value is smaller than the alpha. So, the p-value is less than .05 here. 11. C. When we want to compare two proportions, we should have assumptions that the sample should be randomly selected and the sample size should be large enough. 12. A. From the CI, we are 95% confidence that the difference in the average typing rate ( freshmen µ - seniors ) is between –11.2 and -.9 . So, we can conclude that freshmen have a lower average typing rate than seniors. Since the results are negative, the 2 nd group must be higher seniors. Seniors are faster. 13. A. The p-value is small, so reject H 0 . So I is correct. II is not true because if the endpoints of the CI are given, use the term confidence, not probability. III is also wrong because the definition of the p-value is the probability that you would see a result this extreme if the null were true.
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This note was uploaded on 04/25/2010 for the course STA 2023 taught by Professor Ripol during the Spring '08 term at University of Florida.

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answers 3 - 1. C. In a matched pairs study, subjects are...

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