{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

420Hw01ans - STAT 420(10 points(due Friday September 5 by...

This preview shows pages 1–4. Sign up to view the full content.

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 σ = \$1,500. If a random sample of 25 junior executives yields an average salary of \$16,400, what is the 95-percent confidence interval for μ , the average salary of all junior executives. σ = \$1,500, n = 25, X = \$16,400. n z ° 2 X α ± 95% confidence, α = 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 - ± α 95% confidence, α = 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 )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 σ = 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 0 : μ = 20 and H a : μ = 19.8. Hint: Treat this problem as a left-tailed 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 α = 0.025. Hint: Recall: Power = P ( Reject H 0 | H 0 is not true ) = P ( Reject H 0 | μ = 19.8 ) . Find the rejection region first and state it in terms of the sample mean X . n = 20. α = 0.025. Rejection Region: Z = n ° ± 0 X - < – z α . Z = 20 25 . 0 20 X - < 1.96 . 20 25 . 0 96 . 1 20 X - < = 19.89 . Power = P ( X < 19.89 | μ = 19.8 ) = ° ° ° ± ² ³ ³ ³ ´ µ - < 20 25 . 0 8 . 19 89 . 19 Z P = P ( Z < 1.61 ) = 0.9463 .
b) If a sample of n = 20 items has mean weight 19.9, what conclusion is drawn about μ at the 2.5-percent significance level? Compute the P value for this case, and interpret it.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 10

420Hw01ans - STAT 420(10 points(due Friday September 5 by...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online