document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "a. Based on this sample information, develop a 90 percent confidence interval for the
\n" ); document.write( "population mean yearly premium.\r
\n" ); document.write( "\n" ); document.write( "b. How large a sample is needed to find the population mean within $250 at 99 percent
\n" ); document.write( "confidence?
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Algebra.Com's Answer #256397 by stanbon(75887)\"\" \"About 
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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.
\n" ); document.write( "a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium.
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\n" ); document.write( "x-bar = 10979
\n" ); document.write( "ME = 1.645*1000/sqrt(20) = 367.83
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\n" ); document.write( "90% CI: 10979-367.82 < u < 10979+367.83
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\n" ); document.write( "b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?
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\n" ); document.write( "n = [zs/E]^2 = [2.5758*1000/250]^2 = 107 when rounded up
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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