document.write( "Question 851681: the average local cell phone call length was reported to be 2.27 minutes. A random sample of 20 phone calls showed an average of 2.98 minutes with a standard deviation of 0.98 minute. At 0.05 level of significance can it be concluded that the average differs from the population average? \n" ); document.write( "
Algebra.Com's Answer #512928 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! the average local cell phone call length was reported to be 2.27 minutes. A random sample of 20 phone calls showed an average of 2.98 minutes with a standard deviation of 0.98 minute. At 0.05 level of significance can it be concluded that the average differs from the population average? \n" ); document.write( "----- \n" ); document.write( "Ho: p = 2.27 (claim) \n" ); document.write( "Ha: p # 2.27 \n" ); document.write( "-------------------- \n" ); document.write( "x-bar = 2.98 \n" ); document.write( "---- \n" ); document.write( "t(2.98) = (2.98-2.27)/[0.98/sqrt(20)] = 3.24 \n" ); document.write( "------ \n" ); document.write( "p-value = 2*P(t > 3.24 when df = 19) = 2*tcdf(3.24,100) = 0.0051 \n" ); document.write( "--------- \n" ); document.write( "Conclusion: Since the p-value is less tahn 5% reject Ho. \n" ); document.write( "The test results do not support the claim at the 5% \n" ); document.write( "level of significance. \n" ); document.write( "======== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "================ \n" ); document.write( " |