document.write( "Question 1199185: A random sample of 225 adults was given an IQ test. It was found that 117 of them scored higher than 100. Based on this, compute a 99% confidence interval for the proportion of all adults whose IQ score is greater than 100. Then find the lower limit and upper limit of the 99% confidence interval.
\n" ); document.write( "Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. \n" ); document.write( "
Algebra.Com's Answer #832942 by Theo(13342)\"\" \"About 
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sample size is 225.
\n" ); document.write( "117 of them scored higher than 100.
\n" ); document.write( "your sample mean proportion is 117/225 = .52
\n" ); document.write( "let p = .52
\n" ); document.write( "q = 1 - p = .48
\n" ); document.write( "let n = 225
\n" ); document.write( "standard error = sqrt(p * q / n) = sqrt(.52 * .48 / 225) = .0333
\n" ); document.write( "99% confidence interval z-score = plus or minus 2.5758
\n" ); document.write( "z-score formula = (x - m) / s
\n" ); document.write( "x is the raw score.
\n" ); document.write( "m is the mean
\n" ); document.write( "s is the standard error.\r
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\n" ); document.write( "\n" ); document.write( "on the low side of the confidence interval, your formula becomes:
\n" ); document.write( "-2.5758 = (x - .52) / .0333
\n" ); document.write( "solve for x to get x = -2.5758 * .0333 + .52 = .4342\r
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\n" ); document.write( "\n" ); document.write( "on the high side of the confidence interval, your formula becomes:
\n" ); document.write( "2.5758 = (x - .52) / .0333
\n" ); document.write( "solve for x to get x = 2.5758 * .0333 + .52 = .6058\r
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\n" ); document.write( "\n" ); document.write( "your 99% confidence interval is from .4342 to .6058
\n" ); document.write( "round to 2 decimal places to get .43 to .61
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