document.write( "Question 933936: A poll asked for people's opinion on whether closing local newspapers would hurt civic life; 430 of 1001 respondents said it would hurt civic life a lot.\r
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document.write( "A. Find the proportion of the respondents who said closing local papers would hurt civic life a lot.\r
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document.write( "(round to three decimal places as needed)\r
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document.write( "B. Find a 95% confidence interval for the population proportion who believed closing newspaper would hurt civic life a lot. Assume the poll used a simple random sample (SRS) (In fact, it used random sampling, but a more complex method than SRS\r
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document.write( "(, ) round to three decimal places as needed\r
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document.write( "C. Find an 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot.\r
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document.write( "(, ) round to three decimal places as needed\r
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document.write( "D. Which interval is wider?\r
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document.write( "Thank you \n" );
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Algebra.Com's Answer #567133 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A poll asked for people's opinion on whether closing local newspapers would hurt civic life; 430 of 1001 respondents said it would hurt civic life a lot. \n" ); document.write( " A. Find the proportion of the respondents who said closing local papers would hurt civic life a lot. (round to three decimal places as needed) \n" ); document.write( "Ans: 430/1001 = 0.430 or 430 out of 1000 \n" ); document.write( " ------------------------------ \n" ); document.write( "B. Find a 95% confidence interval for the population proportion who believed closing newspaper would hurt civic life a lot. Assume the poll used a simple random sample (SRS) (In fact, it used random sampling, but a more complex method than SRS (, ) round to three decimal places as needed \n" ); document.write( "--- \n" ); document.write( "p-hat = 0.43 \n" ); document.write( "ME = 1.96*sqrt[0.43*0.57/1001] = 0.031 \n" ); document.write( "------ \n" ); document.write( "95% CI:: 0.43-0.031 < p < 0.43+0.031 \n" ); document.write( "----------------------------------------- \n" ); document.write( "C. Find an 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot. \n" ); document.write( " (, ) round to three decimal places as needed \n" ); document.write( "ME:: 1.2820*sqrt[0.43*0.57/1001] = 0.02 \n" ); document.write( "---- \n" ); document.write( "80% CI: 0.43-0.02 < p < 0.43+0.02 \n" ); document.write( "----- \n" ); document.write( "D. Which interval is wider? \n" ); document.write( "95% CI is wider than 80% CI. \n" ); document.write( "The more confidence you want, the wider the CI must be. \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "---------------- \n" ); document.write( " \n" ); document.write( " |