Week 3 Assignment

Week 3 Assignment 7.11 A. The define should be approximately bring about because the central intend theorem. The try out size could be considered epos so the sampling scattering will be normal. B. The nettled is the same as the community smashed, 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 requisite greater than, p = 1 - 0.9223 = 0.0777 B. Maybe, but we did not great deal an Alpha level in the beginning 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 stop number termination (5.944) < 6 C. No, because the upper limit (6.096) >6 D. We are 95% positive 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 can be 95% definite that the true(a) proportion is above 0.48If you requisite to get a blanket(a) of the mark essay, order it on our website: Orderessay

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