Question

A confidence interval for the true mean of the annual medical expenses of a middle-class American family is given as ($ 745, $ 782). If this interval is based on interviews with 90 families and a standard deviation of $ 124 is assumed. Suppose all annual medical expenses of middle-class American families follow an approximately normal distribution.

What is the sample mean of annual medical expenses ?

(b) What is the confidence level of the interval estimate (as a decimal) ?

Answer 1

The confidence interval for mean is: 1 Zertica 745, 782) The upper limit is: 782 The lower limit is: 745. Calculate the sample mean of annual medical expenses. Lower limit + Upper limit Mean= 745+ 782 2 = 763.5 Hence, the sample mean is: 763.5 b) Calculate the confidence level. The margin of error is, ME - Upper – Lower _782 – 745 2 = 18.5

ME=2...( → Zerita MENN Zcritic a= = 124 critical = 1.4154 Now, calculate the confidence level. (USe MS Excel) Confidence level =(=NORMSDIST(1.4154)) =0.921524 92% Hence, the required confidence level is: 92%.