document.write( "Question 1174171: 5. Suppose that a market research firm is hired to estimate the percent of adults living in
\n" ); document.write( "a large city who have cell phones. Five hundred randomly selected adult residents in
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\n" ); document.write( "sampled, 415 responded yes - they own cell phones. Using a 95% confidence level,
\n" ); document.write( "compute a confidence interval estimate for the true proportion of adult residents of this
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Algebra.Com's Answer #799572 by ewatrrr(24785)\"\" \"About 
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document.write( "    x̄  = 415/500 = .83   n = 500\r\n" );
document.write( " Using a 95% confidence level, compute a confidence interval estimate\r\n" );
document.write( " for the true proportion of adult residents of this city who have cell phones. \r\n" );
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document.write( "   ME = \"z%2Asqrt%28%28p%281-p%29%29%2Fn%29\"\r\n" );
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document.write( "   ME = \"1.96%2Asqrt%28%28%28.83%29%28.17%29%29%2F500%29%29\" = .0006\r\n" );
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document.write( "    .83 - .0006 < \"mu\" < .83 + .0006\r\n" );
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document.write( "Wish You the Best in your Studies.\r\n" );
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document.write( "CI	z = value\r\n" );
document.write( "90%	z =1.645\r\n" );
document.write( "92%	z = 1.751\r\n" );
document.write( "95%	z = 1.96\r\n" );
document.write( "98%	z = 2.326\r\n" );
document.write( "99%	z = 2.576\r\n" );
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