document.write( "Question 1054795: 1. A regeneration survey has indicated that 75 of the 90 plots examined were stocked. \r
\n" ); document.write( "\n" ); document.write( "a) Find the 99% confidence limits for the fraction of the total area being stocked.
\n" ); document.write( "b) How large a sample is needed if we wish to be 95% confident that our sample proportion will be within 0.05 of the true fraction?\r
\n" ); document.write( "\n" ); document.write( "2. Another survey from another area indicated 75% stocking from 100 plots. Is this different from the area described in Question 1? Use 90% and 95% confidence limits.
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Algebra.Com's Answer #670024 by ewatrrr(24785)\"\" \"About 
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p = 75/90 = 5/6 and q = 1/6
\n" ); document.write( "99%
\n" ); document.write( "a. ME = \"z%2Asqrt%28%28p%281-p%29%29%2Fn%29\" = ± \"2.576%2Asqrt%28%28%285%2F6%29%281%2F6%29%29%2F90%29\"
\n" ); document.write( "5/6 - ME < \"mu\" < 5/6 + ME
\n" ); document.write( "b. n = \"%28z%2FME%29%5E2+%28p%281-p%29%29%29\" = \"%281.645%2F.05%29%5E2+%28%285%2F6%29%281%2F6%29%29%29\"
\n" ); document.write( "2. Yes: p = 75/100 = 3/4 and q = 1/4
\n" ); document.write( "follow steps in a above for z = 1.645 and z = 1.96
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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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