HW 7 Solution

HW 7 Solution - STAT 350 Assignment 7 Solutions (45 points)...

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STAT 350 Assignment 7 Solutions (45 points) 22. (4) In this problem, we have 507 n and ˆ 142 / 507 .28008 p . For a two-sided 99% confidence interval, we use * 2.575 z : ˆˆ (1 ) ˆ * p p n pz .28008(1 .28008) 507 .28008 2.575  .28008 2.575(.0199) (.228, .331)  25. (a) (3) Following the same format used for most confidence intervals, i.e., statistic (critical value) (standard error), an interval estimate for 1 - 2 is: 112 2 12 ) ) () p pp p z nn   . (b) (4) The response rate for no-incentive sample: p 1 = 75/110 = .6818, while the return rate for the incentive sample is p 2 = 66/98 = .6735. Using z = 1.96 (for a confidence level of 95%), a two-sided confidence interval for the true (i.e., population) difference in response rates 1 - 2 is: (.6818 - .6735) (1.96) 98 ) 6735 . 1 )( 6735 (. 110 ) 6818 . 1 ( 6818 . = .0083 .1273 =(-.119, .1356). The fact that this interval contains 0 as a plausible value of 1 - 2 means that it is plausible that the two proportions are equal. Therefore, including incentives in the questionnaires does not appear to have a significant effect on the response rate. (c) (4) Let i p ~ denote the sample proportion by adding 1 success and 1 failure to the i th sample. We calculate ) 2 /( ) 1 ( ~ n x p i , where x is the number of successes (or failures, whichever is desired) in the sample. Then: 67857 . ) 2 110 /( ) 1 75 ( ~ 1 p and . 67 . ) 2 98 /( ) 1 66 ( ~ 2 p
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HW 7 Solution - STAT 350 Assignment 7 Solutions (45 points)...

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