document.write( "Question 1159480: A national magazine claims that the average college student watches less television that general public. The national average is 30 hours per week, with a standard deviation of 1.5 hours. A sample of 30 college student has a mean of 26 hours. a=.05 \n" ); document.write( "
Algebra.Com's Answer #782526 by Boreal(15235)\"\" \"About 
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Assuming random sample, normality
\n" ); document.write( "probability of this occurring (could also be a hypothesis test, but what is wanted is not stated).
\n" ); document.write( "z=(x-mean)/sigma/sqrt(n)
\n" ); document.write( "=(26-30)/1.5/sqrt (30)
\n" ); document.write( "=-4*sqrt(30)/1.5\r
\n" ); document.write( "\n" ); document.write( "=-14.61
\n" ); document.write( "This has a p-value of <<0.0001, in this case essentially 0.\r
\n" ); document.write( "\n" ); document.write( "If the hypothesis is that the two are the same, it is rejected. ]
\n" ); document.write( "If the hypothesis is that college students watch more or the same, and the alternative is that they don't, that hypothesis would also be rejected.
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