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Unformatted text preview: STAT 420 Fall 2008 Homework #1 (10 points) (due Friday, September 5, by 3:00 p.m.) From the textbook: 3.5 The salary of junior executives in a large retailing firm is normally distributed with standard deviation = $1,500. If a random sample of 25 junior executives yields an average salary of $16,400, what is the 95percent confidence interval for , the average salary of all junior executives. = $1,500, n = 25, X = $16,400. n z & 2 X 95% confidence, = 0.05, z 0.025 = 1.960. 25 500 , 1 960 . 1 400 , 16 16,400 588 ( 15,812 , 16,988 ) 3.12 Redo Problem 3.5 without assuming that the population standard deviation is known. You are given s = $1,575. s = $1,575, n = 25, X = $16,400. ( ) n t n s 1 2 X 95% confidence, = 0.05, t 0.025 ( 24 ) = 2.064. 25 575 , 1 064 . 2 400 , 16 16,400 650.16 ( 15,749.84 , 17,050.16 ) 3.18 ( a ), ( b ) A cereal packaging machine is supposed to turn out boxes that contain 20 oz of cereal on the average. Past experience indicates that the standard deviation of the content weights of the packages turned out by the machine is = 0.25 oz. Boxes containing less than 19.8 oz of cereal, on the average, are considered unacceptable. Thus the manufacturer wishes to test H : = 20 and H a : = 19.8. Hint: Treat this problem as a lefttailed test since 19.8 < 20. a) If samples of size n = 20 are drawn from the current production to test the hypotheses, fund the power function of the test assuming a significance level of = 0.025. Hint: Recall: Power = P ( Reject H  H is not true ) = P ( Reject H  = 19.8 ). Find the rejection region first and state it in terms of the sample mean X . n = 20. = 0.025. Rejection Region: Z = n & X < z . Z = 20 25 . 20 X < 1.96. 20 25 . 96 . 1 20 X  < = 19.89. Power = P ( X < 19.89  = 19.8 ) = & & &  < 20 25 ....
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This note was uploaded on 05/09/2010 for the course MATH 420 taught by Professor Stepanov during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
 Stepanov
 Standard Deviation

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