document.write( "Question 1197343: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If we want to estimate this proportion to within 3% at the 95% confidence level, how many randomly selected teenagers must we survey? \n" ); document.write( "
Algebra.Com's Answer #830570 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi \r\n" );
document.write( " 95% confident ⇒  z = 1.96, \r\n" );
document.write( "p = .50(Generally .5 is used if p is not given)  ME = .03\r\n" );
document.write( "How large of a sample size is required?\r\n" );
document.write( "n = \"%28z%2FME%29%5E2+%28p%281-p%29%29%29\"\r\n" );
document.write( "n = \"%281.96%2F.03%29%5E2+%28.5%28.5%29%29%29\" = 1067.11  0r sample Size = 1068\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" );
document.write( "Wish You the Best in your Studies.\r\n" );
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