This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 05/31/2011 for the course STAT 200 taught by Professor Andyregards during the Spring '11 term at World College.
 Spring '11
 AndyRegards

Click to edit the document details