A particular article presents results of a survey of adults

A particular article presents results of a survey of adults with diabetes. The average body mass index (BMI) in a sample of 1,558 men was 30.3, with a standard deviation of 0.5. The average BMI in a sample of 1,922 women was 31.1 with a standard deviation of 0.2. Find a 99% confidence interval for the difference in mean BMI between men and women with diabetes. Round your final answers to four decimal places. The confidence interval is ( , ).

Solution

Given a=1-0.99=0.01, Z(0.005) = 2.58 (from standard normal table)

So the lower bound is

(xbar1-xbar2) -Z*sqrt(s1^2/n1+s2^2/n2)

=(30.3-31.1)-2.58*sqrt(0.5^2/1558+0.2^2/1922)

=-0.8347

So the upper bound is

(xbar1-xbar2) +Z*sqrt(s1^2/n1+s2^2/n2)

=(30.3-31.1)+2.58*sqrt(0.5^2/1558+0.2^2/1922)

=-0.7653