Statistics question?
A random sample of 31 women were put on a diet for 2 months. The mean Weight Loss was found with 4.1 kg a standard deviation of 1.3 kg. (A) Construct a confidence interval of 92% of the actual average weight loss in a population of women like those who put a diet. Been taken where appropriate. (B) Interpret the confidence interval in part (a). (c) What sample size is required to produce a 98% interval confidence for the true mean weight loss with a margin of error of 0.25 kg?
(a). According to the central limit theorem, since the size of our sample of n = 31 is large (the thumb of rule is our sample size n is at least equal to or greater than 30), then the sampling distribution of the sample of all means possible weights of women-x-bar will follow the normal distribution z. This remains the case that despite our original population distribution of weights of all women are not normally distributed. Our level of confidence = 92%, ie 100 (1-alpha)% = 92% of 1-alpha = 0.92 Alpha = 0.08 A 92% confidence interval for the true mean weight loss = [x-bar (+ -) z (alpha / 2) * (s / √ n)] = [4.1 (+ -) z (0.08 / 2) * (1.3 / √ 31) ] = [4.1 (+ -) Z0.04 *) 1.3 / √ 31)] = [4.1 (+ -) 1.75 * (1.3 / √ 31)] = [3.7,4.5] (b). The interval indicates that we are above 92% confident that the actual average weight of women must be one of the values lie between 3.7 kg and 4.5 kg. (c). Let n1 be the required sample size. n1 = [Z (alpha / 2) * S / B] ^ 2 n1 = [Z (0.08 / 2) * 1.3/0.25] ^ 2 n1 = [1.75 * 1.3/0.25] ^ 2 n1 = 83 Therefore, a random sample of size n = 83 is necessary to obtain a confidence interval 98% to the actual average weight loss with a margin of error of 0.25 kg. Hope this helps.
