Lesson_05_examples

# Lesson_05_examples - What should be the approximate mean...

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1 Suppose that State College police department want to estimate the true proportion of all underage drinking among all PSU students. They have come to the statistics department for help with this study. To make the estimation, we randomly asked n = 400 PSU freshmen and sophomores whether they have ever drank underage before. a) Suppose 200 of the 400 students answered with “Yes.” And 40 of 400 answered with a big fat “YES”. Then the sample proportion (also called the point estimate) p ˆ =(200+40)/400 = 0.60 b) What is the standard error of sample proportion (p), i.e. SE(p ˆ )? SE( p ˆ )= n ) p ˆ 1 ( p ˆ = 400 ) 4 . 0 ( 6 . = 0.02449 c) Verify that normal approximation methods would be appropriate for this study. n* p ˆ =400*.6=240>10 n*(1- p ˆ )=400*.4=160 > 10, so YES 2 Suppose the amount of beer that each PSU students drink per week have approximately a normal distribution with mean =40 ounces and standard deviation =16 ounces. a. Suppose n = 16 students were randomly selected from main campus.
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Unformatted text preview: What should be the approximate mean and standard deviation of the distribution of the mean amount of beer of these 16 students? Mean will still be 16, i.e. X =16, also known as expected amount beer is 16 SE( X ) = 16 16 n =16/4=4 b. For these 16 students, there is about a 68% chance that the mean amount of beer will be between ___ and ___. 16 16 * 1 40 =(40-4, 40+4)=(36, 44). c. If the sample mean these 16 students was 48 ouches, use the Standard Normal Table to find P( X >48). P( X > 48) = P ) 00 . 2 ( 16 16 40 48 Z P Z P n X Z = 1-P(Z<2.00)=1-0.9772=0.0228 d. Write an interpretation for this outcome in part ii. There is approximately a 2.3% chance that a sample of 16 students would have a mean of more than 48 ouches....
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## This note was uploaded on 10/08/2010 for the course STAT 200 taught by Professor Barroso,joaor during the Fall '08 term at Penn State.

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