Question

A nutritionist wants to compare the mean protein content of grilled chicken sandwiches from Arby’s and McDonald’s. You randomly select several chicken sandwiches from both restaurants and find the following:

Arby’s finds from a sample of 15 sandwiches, a mean protein content of 29g with a sample standard deviation of 1.5g.
McDonald’s finds from a sample of 20 sandwiches, a mean protein content of 25g with a sample standard deviation of 2g.

Find a 95% confidence interval for the true difference in mean protein contents.

Answer 1

21.5 7 Solution: - Given, u = 15, X = 29 S, - 20 X 2= 25 X = 2 ok At 95% Confidence Tudenvalid, 1- 0.95 = 0.05 0.05 0.025 A 2 2 df=uth -2 = 15+ 20-2 = 33. t 2.035 L. dr * Spa (1,-1) 5²+ (12-DS uth -2 (15-1) 1.5² + ( 20-1) 2² 15t 20-2 1.805 Se - Sp + ㅗ и, I 1.805 I Uz ㅗ + 0.616 LIS 20 The 95% Confidence interval 4-M is, Ý, - Xq - tx se < M-42 (X, - Xqt txse 29 - 25 - 2.035 X 0.616 29+ 25 + 2.035 X 0.616 . ( 2.746, 5.254)