Families USA, a monthly magazine that discusses issues related to health and hea

Question

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 19 of its subscribers. It found that the annual health insurance premium s for a family with coverage through an employer averaged $11,100. The standard deviation of the sample was $920. a.based on this sample information, develope a 98% confidence interval for the population mean yearly premium (round answer to nearest whole dollar amount) b. How large of a sample is needed to find the population mean within $250 at 95 percent confidence?

Explanation / Answer

Answer:

a- Develop a 98% confidence interval for the population mean yearly premium

n= 19

x- bar = $11,100

standard deviation = $920

CI at 98% is = 2.33

Fomula when population SD is known =

                           X-bar + (plus, minus) z*S.D./sqrt (n)

= 2.33*920/ sqrt (19)

= 491.76

= $11,100 + 491.76

b- large of a sample is needed to find the population mean within $250 at 95 percent confidence will be :

n = (zs/e)2

z= 1.96(value at 95%)

s (S.D)= $920

e= 250

n= (1.96*920 / 250 ) 2

= 52.024

n = 52

so, 52 to samples is needed to find the population mean within $250 at 95% confidence level.

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