document.write( "Question 285475: The health of employees is monitored by periodically weighing them in. A sample of 54
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document.write( "employees has a mean weight of 175.2 lb. Assuming that is known to be 121.2 lb, use a
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document.write( "0.10 significance level to test the claim that the population mean of all such employees
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document.write( "weights is less than 200 lb.
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document.write( "Perform a hypothesis test \n" );
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Algebra.Com's Answer #207057 by stanbon(75887) ![]() You can put this solution on YOUR website! The health of employees is monitored by periodically weighing them in. \n" ); document.write( "A sample of 54 employees has a mean weight of 175.2 lb. \n" ); document.write( "Assuming that is known to be 121.2 lb, use a \n" ); document.write( "0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. \n" ); document.write( "Perform a hypothesis test \n" ); document.write( "-------------- \n" ); document.write( "Ho: u >= 200 \n" ); document.write( "Ha: u < 200 (claim) \n" ); document.write( "------------------------- \n" ); document.write( "sample mean = 175.2; sample standard deviation = 121.2 \n" ); document.write( "----------- \n" ); document.write( "test statistic: t(175.2) = (175.2-200)/[121.2/sqrt(54)] = -1.5036.. \n" ); document.write( "p-value: P(t<-1.5036 with df= 53) = tcdf(-10,-1,5036,53) = 0.0693 \n" ); document.write( "----- \n" ); document.write( "Conclusion: Since the p-value is less than 10% reject Ho. \n" ); document.write( "The test results support the claim. \n" ); document.write( "======================================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |