Question

A confidence interval for the true mean of the annual medical expenses of a middle-class American family is given as ($ 736, $ 779). If this interval is based on interviews with 100 families and a standard deviation of $ 128 is assumed. Suppose all annual medical expenses of middle-class American families follow an approximately normal distribution. (a) What is the sample mean of annual medical expenses ? (b) What is the confidence level of the interval estimate (as a decimal) ?

Answer 1

a)
lower limit = mean - ME = 736
upper limit = mean + ME = 779

sample mean = (779 + 736)/2 = 757.5

b)
ME = 779 - 757.5 = 21.5

std. dev., s = 128 and n = 100
ME = z * s/sqrt(n)
z = 21.5 * sqrt(100)/128
z = 1.6797

P(z > 1.6797) = 0.0465
Confidence level = 100 - 2*0.0465*100 = 90.70