document.write( "Question 765249: Approximately 19.4% of the U.S population age 5 years and above speaks a language other than English at home. In a large metropolitan area, it was found that 94 residents (age 5 years and above) out of 400 randomly selected residents spoke a language other than English at home. Use the z-test for for a proportion, and test the claim that the proportion for this metropolitan area is higher than the national proportion. Use alpha=0.01 and report the p-value.\r
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document.write( "Any help with this will be deeply appreciated. \n" );
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Algebra.Com's Answer #466099 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Approximately 19.4% of the U.S population age 5 years and above speaks a language other than English at home. In a large metropolitan area, it was found that 94 residents (age 5 years and above) out of 400 randomly selected residents spoke a language other than English at home. Use the z-test for for a proportion, and test the claim that the proportion for this metropolitan area is higher than the national proportion. Use alpha=0.01 and report the p-value. \n" ); document.write( "------ \n" ); document.write( "Ho: p = 0.194 \n" ); document.write( "Ha: p > 0.194 (claim) \n" ); document.write( "----- \n" ); document.write( "sample proportion = p-hat = 94/400 = 0.235 \n" ); document.write( "---- \n" ); document.write( "test stat:: z(0.235) = (0.235-0.194)/sqrt[0.194*0.806/400] = 2.0737 \n" ); document.write( "----- \n" ); document.write( "p-value = P(z > 2.0737) = invNorm(2.0737,100) = 0.0191 \n" ); document.write( "---- \n" ); document.write( "Conclusion: Since the p-value is greater tha 1% fail to reject Ho. \n" ); document.write( "---- \n" ); document.write( "The test results do not support the claim. \n" ); document.write( "====================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "====================== \n" ); document.write( " |