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Unformatted text preview: AMS572.01 Final Exam Fall, 2010 Name ___________________________ID ________________Signature____________________ AMS Major? ______ Instruction: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please enter Yes or No for AMS Major. Please provide complete solutions for full credit. The exam goes from 2:15  4:45pm. Good luck! 1. ( for all ) The following data set from a study by the wellknown chemist and Nobel Laureate Linus Pauling gives the incidence of cold among 279 French skiers who were randomized to the Vitamin C and Placebo groups. Group Cold Yes No Vitamin C 17 122 Placebo 31 109 (a) Construct a 95% confidence interval for the difference between the two incidence rates; (b) Please test whether the incidence rates for the Placebo group is significantly higher than that of the Vitamin C group at the 5% level of significance. Please report the pvalue of your test. (c) Please write up the entire SAS program necessary to answer question raised in (b), including the data step. Answer: (a) VC: 139 , 122 . 122 17 17 1 1 = + = n p ; Placebo: 140 , 221 . 109 31 31 2 1 = + = n p ; The 100(1)% confidence interval for (p 1 p 2 ) is ( 29 ( 29 ( 29 ( 29  + + + 2 2 2 1 1 1 2 2 1 2 2 2 1 1 1 2 2 1 1 1 , 1 1 n p p n p p Z p p n p p n p p Z p p After plugging in Z 0.025 = 1.96 etc., we found the 95% CI to be [0.187, 0.011] (b) This is problem 9.12 in our text book. (*It is also OK, in fact better, if they used the pooled proportion in the denominator. *It is also OK if they did a 2sided test.) 1 (c) SAS code: Data cold; Input group $ outcome $ count; Datalines ; VC yes 17 VC no 122 Placebo yes 31 Placebo no 109 ; Run ; Proc freq data =cold; Tables group*outcome/ chisq ; Weight count; Run ; 2. ( for all ) People at high risk of sudden cardiac death can be identified using the change in a signal averaged electrocardiogram before and after prescribed activities. The current method is about 80% accurate. The method was modified, hoping to improve its accuracy. The new method is tested on 50 people and gave correct results on 46 patients. (a) Is this convincing evidence that the new method is more accurate? Please test at =.05. (b) If the new method actually has 90% accuracy, what power does a sample of 50 have to demonstrate that the new method is better at =.05? (c) How many patients should be tested in order for this power to be at least 0.75? Answer: This is problems 9.7 & 9.8 in our text book. 2 3. ( for all ) A classic tale involves four carpooling students who missed a test and gave as an excuse of a flat tire. On the makeup test, the professor asked the students to identify the particular tire that went flat. If they really did not have a flat tire, would they be able to identify the same tire? To mimic this situation, 40 other students were asked to identify the tire they would select. The data are:students were asked to identify the tire they would select....
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This note was uploaded on 01/31/2011 for the course AMS 572 taught by Professor Weizhu during the Fall '10 term at SUNY Stony Brook.
 Fall '10
 WeiZhu

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