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)\"\" \"About 
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( "
\n" );