Question

(1 point) A confidence interval for the true mean of the annual medical expenses of a middle-class American family is given as ($ 739, $ 781). If this interval is based on interviews with 100 families and a standard deviation of $ 114 is assumed.
(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) For the given Confidence interval ($ 739, $ 781) the mean is calculated as:

Sample Mean, M = (739+781)/2= $ 760

Because the Confidence interval is computed as:

M ± Z(s_M)

b) The confidence level is computed using Z score which is computed using Margin of error formula as:

Margin of Error= Z(s_M)

And Standard error= S/sqrt{n}

where S= Standard deviation and n= no of samples.

And Margin of error from Confidence interval is computed as:

Upper Limit= Mean+ Margin of error(E)

781=760+ E

E= 21

Thus, using the margin of error formula:

Using the Z table shown below The Confidence level at Z =1.8421 is 0.9342 that is 93.42%.