homework_11_ans_stud

# homework_11_ans_stud - MTH 1210 Biostatistics Apr 1809...

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MTH 1210 Biostatistics Apr 18’09 Homework #11 (answers) ( Sampling Distribution of the Sample Proportion ) Use the handout # 12 and MS Excel program Three_distributions.xls, sheet 3 (see the textbook pgs 148, 149, 183) Below: is a sample proportion for an arbitrary random sample; p is a population proportion; it is assumed below that the sample size n is large enough and the sampling distribution of is approximately normally distributed with the mean µ = p ; 0 is the given value of with respect to which a probability of interest is estimated; * is the value of for a particular random sample under consideration. σ  is a standard deviation of the sampling distribution of the sample proportion. The sample size n is treated as "large enough" if n > 20 and if both np and n(1 – p ) are greater than 5 . 5.5.1 . Smith et al. (A-5) performed a retrospective analysis of data on 782 eligible patients admitted with myocardial infarction to a 46-bed cardiac service faculty. Of these patients, 248 ( 32 percent) reported a past myocardial infarction. Use 0.32 as the population proportion. Suppose 50 subjects are chosen at random from the population, what is the probability that over 40 percent would report previous myocardial infarction? np = 16 > 5 n(1 – p ) = 34 > 5 p = 0.32 σ  = [p(1 – p)/n] 1/2 = 0.065970 σ 2 = 0.004352 n = 50 0 = 0.4 z 0 = ( 0 – p)/σ  = 1.212678 P( 0 ) = f(Z ≤ z 0 ) = 0.887374 P( > 0 ) = f(Z > z 0 ) = 1 – f(Z ≤ z 0 ) = 0.112626 ANS: 0.113 5.5.2 . In the study cited in the problem 5.5.1,

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## This note was uploaded on 11/24/2011 for the course MATH 3000 taught by Professor Kzaer during the Spring '05 term at St. Johns College MD.

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homework_11_ans_stud - MTH 1210 Biostatistics Apr 1809...

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