Question

answer please can simplify process

Two experimental diets, designed to add weight to malnourished adults in developing countries, are being tested. The following sample data for weight gain after 6-weeks on the diet (in kg) was obtained (assume normally distributed data): n = 8, mean = 5.80, S1 = 1.613 for Diet A and n2 = 9, mean = 7.27, S2 = 0.99 for Diet B. Suppose you decided to use the non-pooled t-test and present your results in a confidence interval. What would be the 90% confidence interval for the difference in the average weight gain under the two diets given that degrees of freedom are 11.36? O A. -1.47 +0.78 OB.-1.47 +0.90 O C. -1.47 +1.18 O D.-1.47 1.15 O E. - 1.47 1.00

Answer 1

nias 12=9 1 5-80 ܕ ، $2=7-27 5 = 1.613 S₂=0.99 df=12 (11:36) At qoy confidence level tx=11782 The margin of error (E)=td2 Jeht be ne E- = 1:782 x (1.61332 (o.aqj2 1.18 9 pointestinate (5,5%) = 5.80-7:27=11:47 The Confidence internal fã-2) I E=/-1.49+118 option