SOLUTION: 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

Question 359163: 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 percent confidence interval for the
population mean yearly premium.
b. How large a sample is needed to find the population mean within $250 at 99 percent
confidence?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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 percent confidence interval for the population mean yearly premium.
---
x-bar = 10979
ME = 1.645*1000/sqrt(20) = 367.83
---
90% CI: 10979-367.82 < u < 10979+367.83
----------------------------------------------
b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?
---
n = [zs/E]^2 = [2.5758*1000/250]^2 = 107 when rounded up
================
Cheers,
Stan H.

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