document.write( "Question 1054795: 1. A regeneration survey has indicated that 75 of the 90 plots examined were stocked. \r
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document.write( "a) Find the 99% confidence limits for the fraction of the total area being stocked.
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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
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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. \n" );
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Algebra.Com's Answer #670024 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! p = 75/90 = 5/6 and q = 1/6 \n" ); document.write( "99% \n" ); document.write( "a. ME = \n" ); document.write( "5/6 - ME < \n" ); document.write( "b. n = \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 \n" ); document.write( "| \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 \n" ); document.write( " \n" ); document.write( " |