document.write( "Question 1158231: Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the p-value? a 0.3461
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document.write( "b. 0.1539
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document.write( "c. 0.3078
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document.write( "d. 0.0100 \n" );
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Algebra.Com's Answer #781259 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! assuming randomness, this is a one-sample proportion \n" ); document.write( "z=(p hat-p)/sqrt (p*(1-p)/n) \n" ); document.write( "=-0.05/sqrt(.41*.59/100) \n" ); document.write( "=0.3078 \n" ); document.write( "C \n" ); document.write( "Note: I interpret this as a two way test. Nowhere did it say \"at least 41%\" but only 41%. If it were a one way test, the p-value would be half of this. \n" ); document.write( " |