document.write( "Question 1197598: A recent survey showed that among 775 randomly selected subjects who completed 4 years of college, 162 smoke and 613 do not smoke. Determine a 95% confidence interval for the true proportion of the given population that smokes.
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Algebra.Com's Answer #830920 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi \r\n" );
document.write( "  p(smoke) = 162/775 = .209\r\n" );
document.write( "95% CI   ⇒   z = 1.96   |  2-sided  Invnorm( .95 + .05/2) = Invnorm(.975)\r\n" );
document.write( "ME = \r\n" );
document.write( "ME = \r\n" );
document.write( ".209 - ME <  < .209 + ME\r\n" );
document.write( "Will let You finish it Up\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
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