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