document.write( "Question 612153: a sample of 85 golfers showed that their average score on a particular golfers course was 91.31 with a standard deviation of 5.37. show all work.\r
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document.write( "A) find the 99% confidence interval of the mean score for all 85 golfers.\r
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document.write( "B) find the 99% confidence interval of the mean score for all golfers if this is a sample of 120 golfers instead of a sample of 85.\r
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document.write( "C) which onfidence interval is smaller and why? \n" );
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Algebra.Com's Answer #385350 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! a sample of 85 golfers showed that their average score on a particular golfers course was 91.31 with a standard deviation of 5.37. show all work.\r \n" ); document.write( "\n" ); document.write( "A) find the 99% confidence interval of the mean score for all 85 golfers. \n" ); document.write( "(89.775,92.845) \n" ); document.write( "------------------------\r \n" ); document.write( "\n" ); document.write( "B) find the 99% confidence interval of the mean score for all golfers if this is a sample of 120 golfers instead of a sample of 85. \n" ); document.write( "(90.027,92.593) \n" ); document.write( "------------------------\r \n" ); document.write( "\n" ); document.write( "C) which onfidence interval is smaller and why? \n" ); document.write( "Interval B is slightly smaller. \n" ); document.write( "ME = t*s/sqrt(n) \n" ); document.write( "Notice that ME and sqrt(n) are indirectly related. \n" ); document.write( "As n gets larger, ME gets smaller. \n" ); document.write( "=============================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "================== \n" ); document.write( " |