document.write( "Question 1053391: Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a 90% confidence interval. Express the answer in the form ą E and round to the nearest ten-thousandth.\r
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Algebra.Com's Answer #668627 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!
ME = $\"z%2Asqrt%28%28p%281-p%29%29%2Fn%29\"$
\n" ); document.write( "z = 1.645 and p = 2250/4000 = .5625
\n" ); document.write( "ME = $\"1.645%2Asqrt%28%28.5625%28.4375%29%29%2F4000%29\"$ round to the nearest ten-thousandth.
\n" ); document.write( "2250 ą ME
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\n" ); document.write( " = CI z = value
\n" ); document.write( "90% z =1.645
\n" ); document.write( "92% z = 1.751
\n" ); document.write( "95% z = 1.96
\n" ); document.write( "98% z = 2.326
\n" ); document.write( "99% z = 2.576
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