document.write( "Question 623123: Could someone please assist in helping me with this?...\r
\n" ); document.write( "\n" ); document.write( "A sample of 142 golfers showed that their average score on a particular golf course was 90.46 with a standard deviation of 3.84.
\n" ); document.write( "Answer each of the following (show all work
\n" ); document.write( "and state the final answer to at least two decimal places.):
\n" ); document.write( "(A) Find the 95% confidence interval of the mean score for all 142 golfers.
\n" ); document.write( "(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 120 golfers instead of a sample of 142.
\n" ); document.write( "(C) Which confidence interval is larger and why? \n" ); document.write( "

Algebra.Com's Answer #391971 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "(A) Find the 95% confidence interval of the mean score for all 142 golfers.
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document.write( "ME = 1.96[3.84/sqrt(142)]   and   90.46 - ME <  < 90.46 + ME
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document.write( "(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 120 golfers 
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document.write( "ME = 1.96[3.84/sqrt(120)]  and   90.46 - ME <  < 90.46 + ME
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document.write( "(C) Which confidence interval is larger and why?  for the smaller sample as then ME is larger
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