Week 3 Assignment

Week 3 Assignment 7.11 A. The function should be approximately run because the central intend theorem. The ideal size could be considered epic so the sampling statistical distribution will be normal. B. The unshakable is the same as the tribe compressed, 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 call for greater than, p = 1 - 0.9223 = 0.0777 B. Maybe, but we did not decorate an Alpha level earlier 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 velocity specify (5.944) < 6 C. No, because the upper limit (6.096) >6 D. We are 95% assured 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 squeeze out be 95% accredited that the true(a) proportion is above 0.48If you motivation to get a huge of the mark essay, order it on our website: Orderessay

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