Week 3 Assignment

Week 3 Assignment 7.11 A. The sort should be approximately manakin because the central intend theorem. The hear size could be considered larger-than-life so the sampling dispersion will be normal. B. The compressed is the same as the people inculpate, 20. The standard deviation is 4/sqrt(64) = 4/8 = 0.5 C. z = (xbar - mu)/(sigma/?n) z = (21 - 20)/(4/?64) z = 1 / 0.5 z = 2 p = 0.9772 p = 1 - 0.9772 = 0.0228 D. z = (xbar - mu)/(sigma/?n) z = (19.385 - 20)/(4/?64) z = -0.615 / 0.5 z = -1.23 p = 0.1093 7.30 A. z = (phat - p)/sqrt[p (1-p)/n] z = (0.32 - 0.3) / sqrt [0.3(1 - 0.3)/1011] z = 0.02 / sqrt (0.00020772) z = 1.421 p = 0.9223 Since we fate greater than, p = 1 - 0.9223 = 0.0777 B. Maybe, but we did not hang an Alpha level rather beginning. 8.8 A. if ? = 0.05 then CI (95%) for the mean is 5.46±z(0.025)2.47/?100 = (4.976,5.944) if ? = 0.

01 then CI (99%) for the mean is 5.46±z(0.005)2.47/?100 = (4.824,6.096) B. Yes, because the focal ratio pose (5.944) < 6 C. No, because the upper limit (6.096) >6 D. We are 95% confident(p) that the mean is less than 6 8.38 95% confidence interval: p +/- z * sqrt [p(1 - p)/n] 0.5571 +/- 1.96 * sqrt [( 0.5571 * 0.4429)/350] 0.5571 +/- 1.96 * sqrt [0.00070496] 0.5571 +/- 0.05204 (0.5051, 0.6092) Since 0.48 is not at bottom the confidence interval, we flowerpot be 95% received that the true(a) proportion is above 0.48If you trust to get a large-minded of the mark essay, order it on our website: Ordercustompaper.com

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