Question

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores. Diet Vegan Omnivore Sample Size Sample Mean Sample SD 88 5.20 1.05 93 5.65 1.10 Calculate a 99% CI for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores. (Use Hvegan - Momnivore. Round your answers to three decimal places.) (mmol/l) Interpret the interval. O We cannot draw a conclusion from the given information. O We are 99% confident that the true average cholesterol level for vegans is greater than that of omnivores by an amount outside the confidence interval. O We are 99% confident that the true average cholesterol level for vegans is less than that of omnivores by an amount within the confidence interval. O We are 99% confident that the true average cholesterol level for vegans is greater than that of omnivores by an amount within the confidence interval.

Answer 1

Friday

31/07/2020

Given that Standard error = = v(51^2/n1)+((s2^2/n2) = v(1.052/88 ) -1.112/93) = 0.1598 Therefore, the 99 % confidence interval of the difference in population mean is = X1 = 5.2 s1 = 1.05 n1 = 88 And CI= 99 X2 = 5.65 s2 = 1.1 n2 = 93 At 99% confidence levelt is = a = 1-9996 = 1-0.99 = 0.01 Therefore, a/2 = 0.005 Lower limite = (1-2)-E = (5.2-5.65)-0.4161 = -0.866 Here, Point estimate = = X1-22 = 5.2-5.65 = -0.45 Upper limite = (1-x2)+E = (5.2-5.65) +0.4161 = -0.034 Here, tc = = t0.005,179 = 2.604 Degree of freedom is Interpretation Option C 2 + 12 df + ng (12-1) = 0.0007 /0.00000364 = 178.985 = 179 min-1 Now, Margin of error = E= tcstd error = 2.604*0.1598 = 0.4161

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