document.write( "Question 1197599: A random sample of 800 home owners in a particular city found 232 home owners who had a swimming pool in their backyard. Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard. Express your results to the nearest hundredth of a percent.
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Algebra.Com's Answer #830921 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Hi \r\n" );
document.write( "  p(had pool) = 232/800 = .29\r\n" );
document.write( "95% CI   ⇒   z = 1.96   |  2-sided  Invnorm( .95 + .05/2) = Invnorm(.975)\r\n" );
document.write( "ME = \"z%2Asqrt%28%28p%281-p%29%29%2Fn%29\"\r\n" );
document.write( "Plug and Play\r\n" );
document.write( "ME = \"1.96%2Asqrt%28%28.29%28.71%29%29%2F800%29\" \r\n" );
document.write( ".29 - ME < \"mu\" < .29 + 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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