hw8_sol - Statistics 528 - Homework 8 Solutions 6.7 (a)

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Statistics 528 - Homework 8 Solutions 6.7 (a) (123.8-1.645(10/sqrt(15)), 123.8+1.645(10/sqrt(15)))=(119.6, 128.0) (b) (123.8-1.96(10/sqrt(15)), 123.8+1.96(10/sqrt(15)))=(118.7, 128.9) (c) (123.8-2.576(10/sqrt(15)), 123.8+2.576(10/sqrt(15)))=(117.1, 130.5) (d) As the confidence level increases, the margin of error gets bigger. 6.10 (a) (123.8-1.96(10/sqrt(50)), 123.8+1.96(10/sqrt(50)))=(121.03, 126.57) (b) The margin of error for the sample of 50 plots is smaller. More data provide more accurate estimate. (c) In both cases, more data result in a narrower interval due to the decrease in the standard deviation of the estimate. 6.16 n= (1.96*8000/500) 2 =983.45. n=984 is needed. 6.22 (a) From Table D, P(Z>=2.326)=0.01, so z .01 =2.326. The middle 98% of values of lie between -2.326 and 2.326 under the standard normal curve. A 98% confidence interval is (10.0023-2.326(0.0002/sqrt(5)), 10.0023+2.326(0.0002/sqrt(5)))=(10.0021, 10.0025)g (b) n= (2.326*0.0002/0.0001) 2 =21.64, so use n=22. =21....
View Full Document

Page1 / 2

hw8_sol - Statistics 528 - Homework 8 Solutions 6.7 (a)

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online