document.write( "Question 126658This question is from textbook Elementary Statistics
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document.write( ": In a survey of 1002 people, 701 said that they voted in a recent presidental election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.\r
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document.write( "a. Find a 99% confidence interval estimate of the proportion of people who say that they voted.\r
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document.write( "b. Are the results consistent with the actual voter turnout of 61% why or why not? \n" );
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Algebra.Com's Answer #92825 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! In a survey of 1002 people, 701 said that they voted in a recent presidental election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote. \n" ); document.write( "------------------- \n" ); document.write( "a. Find a 99% confidence interval estimate of the proportion of people who say that they voted.\r \n" ); document.write( "\n" ); document.write( "p-hat = x/n = 701/1002 = 0.6996 \n" ); document.write( "E=z*sqrt[(p-hat)(q-hat)/n] \n" ); document.write( "= 2.575*sqrt[(0.6996)(0.3004)/1002] = 0.037292 \n" ); document.write( "--------- \n" ); document.write( "99% C.I.: (0.6996-0.03729,0.6996+0.0729) \n" ); document.write( "= (0.662308,0.736892)\r \n" ); document.write( "\n" ); document.write( "b. Are the results consistent with the actual voter turnout of 61% why or why not? \n" ); document.write( "No, because 61% does not fall in the confidence interval \n" ); document.write( "range of values for p.\r \n" ); document.write( "\n" ); document.write( "============================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |