document.write( "Question 740412: Test the claim that the mean time Americans spend watching TV per day is 4.4 hours at the 0.10 level of significance. A random sample of 80 people had a mean time of 2.225 hours viewing per day and a standard deviation of 1.088 hours.\r
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document.write( "I have a hard time determining whether to use a z test or t test. Here I think it is a z test because the sample is over 30. \n" );
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Algebra.Com's Answer #451619 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Test the claim that thee mean time Americans spend watching TV per day is 4.4 hours at the 0.10 level of significance. A random sample of 80 people had a mean time of 2.225 hours viewing per day and a standard deviation of 1.088 hours. \n" ); document.write( "I have a hard time determining whether to use a z test or t test. Here I think it is a z test because the sample is over 30. \n" ); document.write( "------ \n" ); document.write( "Some texts use a t-test for ALL means testing. Others use the t-test \n" ); document.write( "only for samples <= 30. You'll have to check your text to see which \n" ); document.write( "way to go, or do it both ways. \n" ); document.write( "------- \n" ); document.write( "Ho: u = 4.4(claim) \n" ); document.write( "Ha: u # 4.40 \n" ); document.write( "======= \n" ); document.write( "z(2.225) = (2.225-4.4)/1.088 = -1.9991 \n" ); document.write( "---- \n" ); document.write( "p-value = 2*P(z < -1.9991) = 2*normalcdf(-100,-1.9991) = 0.04560 \n" ); document.write( "----- \n" ); document.write( "Since the p-value is < 10%, reject Ho. \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( " |