document.write( "Question 1200394: Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 372 with 96 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.\r
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Algebra.Com's Answer #834540 by Theo(13342)\"\" \"About 
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sample size is 372 with 96 successes.
\n" ); document.write( "p = 96 / 372
\n" ); document.write( "q = 1 - 96 / 372 = 276 / 372
\n" ); document.write( "standard error = sqrt((96/372 * 276/372) / 372) .0226869309.
\n" ); document.write( "critical z at 90% two tailed confidence intervall = plus or minus z = 1.644853626.
\n" ); document.write( "when z = - 1.644853626, z = (x - p) / s becomes:
\n" ); document.write( "-1.644853626 = (x - 96/372) / .0226869309.
\n" ); document.write( "solve for x to get:
\n" ); document.write( "x = -1.64485 * .0226869309 + 96/372 = .2207478357.
\n" ); document.write( "when z = 1.644853626:
\n" ); document.write( "x = 1.64485 * .0226869309 + 96/372 .2953811966.
\n" ); document.write( "your interval at 90% confidence interval is .221 to .295.
\n" ); document.write( "in interval notation, that would be (.221,.295), i believe.
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