STAT 1124 Assignment 7 Model Answers

# STAT 1124 - μ = 4.88 6 Question 15.28 Page 383 c z =(5.91-4.88(0.79/sqrt(49 = 9.03 7 Question 15.30 Page 383 b p = 2 P(z<-1.3 = 0.1936 8

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STAT 1124 – Section 2 Assignment 7 Due Thursday April 23, 2009 Model Answers 1. Question 14.24, Page 358 n = (0.258*0.79/0.2) 2 103.5 104 2. Question 14.30, Page 360 a. A 99% confidence interval is 137 ± 2.58*(65/sqrt(269)) = 137 ± 10.209 b. The sample is a simple random sample. 3. Question 14.32, Page 360 A 99% confidence interval is 248 ± 2.58*(65/sqrt(270)) = 137 ± 10.19 The outlier has a huge impact on the confidence interval 4. Question 14.34, Page 360 The interval refers to the mean NEAP score, not the individual scores, which will be much more variable. 5. Question 15.26, Page 383 a. H
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Unformatted text preview: : μ = 4.88 6. Question 15.28, Page 383 c. z = (5.91-4.88)/(0.79/sqrt(49)) = 9.03 7. Question 15.30, Page 383 b. p = 2 * P(z < -1.3) = 0.1936 8. Question 15.32, Page 383 Note: there is a problem with the question. All of the choices are incorrect. p = P(z > -1.3) = 1 - 0.0968 = 0.9032 9. Question 15.46, Page 384 The parameter μ = the mean difference (fortified minus unfortified) in blood folic acid levels H : μ = 0 H a : μ > 0 10. Question 15.52, Page 385 No: “ p=0.03 means that the evidence form the sample occurs with 3% probability (not very often) if H is true....
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## This note was uploaded on 07/14/2010 for the course STATS 1124 taught by Professor Michael during the Summer '09 term at Langara.

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