The necessary value for z12 was determined to be 1.960. Recall the given information. Sample 1 Sample 2 n = 400 n2 = 300 P1 = 0.56 P2 = 0.39 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. lower bound = P1 - P2 - Z«/2 V P;(1 – P1) , P2(1 – P2) n2 0.56(1 – 0.56). 0.39(1 – 0.39) 300 = 0.56 - 0.39 – 400 %3D Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P;(1 – P1) , P2(1 - P2) upper bound = P1- P2 + Z«/2\V n1 n2 0.56(1 – 0.56) 0.39(1 - 0.39) = 0.56 - 0.39 + 400 300 Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lower bound of to an upper bound of
The necessary value for z12 was determined to be 1.960. Recall the given information. Sample 1 Sample 2 n = 400 n2 = 300 P1 = 0.56 P2 = 0.39 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. lower bound = P1 - P2 - Z«/2 V P;(1 – P1) , P2(1 – P2) n2 0.56(1 – 0.56). 0.39(1 – 0.39) 300 = 0.56 - 0.39 – 400 %3D Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P;(1 – P1) , P2(1 - P2) upper bound = P1- P2 + Z«/2\V n1 n2 0.56(1 – 0.56) 0.39(1 - 0.39) = 0.56 - 0.39 + 400 300 Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lower bound of to an upper bound of
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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