document.write( "Question 628652: Problem 6)
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document.write( "A government survey conducted to estimate the mean price of houses in a metropolitan area is designed to have a margin of error of $10,000. Pilot studies suggest that the population standard deviation is $70,000. Estimate the minimum sample size needed to estimate the population mean with the stated accuracy. \r
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document.write( "Problem 7)
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document.write( "A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 490 college students showed that 33% of them had, or intended to, cheat on examinations. Find the margin of error for the 95% confidence interval.
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Algebra.Com's Answer #395774 by John10(297)![]() ![]() You can put this solution on YOUR website! Problem 7)\r \n" ); document.write( "\n" ); document.write( "A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 490 college students showed that 33% of them had, or intended to, cheat on examinations. Find the margin of error for the 95% confidence interval.\r \n" ); document.write( "\n" ); document.write( "______________________________________________\r \n" ); document.write( "\n" ); document.write( "We have p = 0.33 and q = 1- 0.33 = 0.67 , n = 490, the critical value is 1.96 (95% confidence interval)\r \n" ); document.write( "\n" ); document.write( "standard deviation = sqrt(0.33 * 0.67 * 490) = 10.41\r \n" ); document.write( "\n" ); document.write( "E = (zs)/sqrt(n)= (1.96 * 10.41)/sqrt(490) = 0.92\r \n" ); document.write( "\n" ); document.write( "Hope it helps:)\r \n" ); document.write( "\n" ); document.write( "John10 \n" ); document.write( " \n" ); document.write( " |