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Unformatted text preview: STAT102 Midterm Solution Spring12 1. a. Hypothesis: H : C P ,H a : C < P Test statistic: t = 5- . 27 q 54 . 17( 1 10 + 1 11 ) = 1 . 64 ,s 2 p = (10- 1)8 . 7 2 + (11- 1)5 . 9 2 10 + 11- 2 = 54 . 17 P-value: (form calculator) P ( t 19 > 1 . 64) = 0 . 0587 Critical value: t . 05 , 19 = 1 . 729 Decision rule: Do not reject H . b. Test statistis: t = 5- . 27 q 8 . 7 2 10 + 5 . 9 2 11 = 1 . 61 P-value: P ( t 16 > 1 . 61) = 0 . 063 Critical value: t . 05 , 16 = 1 . 746 Decision rule: Do not reject H . c. No. Sally need not be a typical subject, and so her individual response need not be like that of the subjects in the study. This study only can not answer her question. [Prof. Browns 2nd answer: If Sally can be considered as typical in the sense that she is a random draw from the same population as that used in the study, then, Yes, we can make an estimate of how much her blood pressure would change. We can estimate that it would decrease by 5, since that was the mean decrease in the random sample of the study. We can even use the usual CI for this decrease, i.e. the decrease of a random subject given the drug is 5 t . 025 , 11 8 . 7. (Note that I did not divide by 10.) If we know anything further about Sally then this may not be a suitable answer. In particular if we know her current blood pressure this may not be suitable since we are not given the evidence from the sample about the correlation between current blood pressure and the effect of the drug.]current blood pressure and the effect of the drug....
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This note was uploaded on 04/04/2012 for the course STAT 102 taught by Professor Shaman during the Spring '08 term at UPenn.
- Spring '08