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.
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x-bar = 10979
ME = 1.645*1000/sqrt(20) = 367.83
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90% CI: 10979-367.82 < u < 10979+367.83
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b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?
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n = [zs/E]^2 = [2.5758*1000/250]^2 = 107 when rounded up
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Cheers,
Stan H.
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