Question

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000.a. Based on this sample information, develop a 90% confidence interval for the population mean yearly premium.b. How large a sample is needed to find the population mean within $250 at 99% confidence?

a.         Between $10,592 and $11,366, found by____________________

            b.         107, found by {(2.576 ´1000)/250}²= _____________________

Answer 1

(a)
SE = s/

= 1000/ = 223.6086

= 0.10

ndf = 20 - 1 = 19

From Table, critical values of t are given by

1.7291.

Confidence interval is given by:

t SE

= 10979 (1.7291 X 223.6086)

= 10979 386.6416

=(10592 , 11366)

(b)

Here,

= 0.01

From Table, critical values of

Z = 2.576

= 1000

e = 250

Substituting, we get: