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Unformatted text preview: is about normal with mean of p and standard error SE( ) = . Example. Suppose the proportion of all college students who have used marijuana in the past 6 months is p = .40. For a class of size N = 200, representative of all college students on use of marijuana, what is the chance that the proportion of students who have used mj in the past 6 months is less than .32 (or 32%)? Solution. The mean of the sample proportion is p and the standard error of is SE( ) = . For this marijuana example, we are given that p = .4. We then determine SE( ) = = = = 0.0346 So, the sample proportion is about normal with mean p = .40 and SE( ) = 0.0346. The zscore for .32 is z = (.32  .40) / 0.0346 = 2.31. Then using Standard Normal Table Prob( < .32) = Prob(Z <. 2.31) = 0.0104....
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 Spring '11
 AndyRegards
 Normal Distribution, Standard Deviation, Standard Error, sample proportion

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