document.write( "Question 366127: A sample of 89 golfers showed that their average score on a particular golf course was 90.98 with a standard deviation of 6.47.
\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 98% confidence interval of the mean score for all 89 golfers.
\n" ); document.write( "(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 89.
\n" ); document.write( "(C) Which confidence interval is larger and why?
\n" ); document.write( "

Algebra.Com's Answer #260923 by stanbon(75887)\"\" \"About 
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A sample of 89 golfers showed that their average score on a particular golf course was 90.98 with a standard deviation of 6.47.
\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( "-------------------
\n" ); document.write( "x=bar = 90.98\r
\n" ); document.write( "\n" ); document.write( "(A) Find the 98% confidence interval of the mean score for all 89 golfers.
\n" ); document.write( "ME = 2.3263[6.47/sqrt(89)] = 1.5955
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\n" ); document.write( "CI: 90.98-1.5955 < u < 90.98+1.5955
\n" ); document.write( "CI: 89.3845 < u < 92.5755
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\n" ); document.write( "\n" ); document.write( "(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 89.
\n" ); document.write( "ME = 2.3263*6.47/sqrt(135) = 1.2954
\n" ); document.write( "---
\n" ); document.write( "CI: 90.89-1.2954 < u < 90.89+1.2954
\n" ); document.write( "CI: 89.5946 < u < 92.1854
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\n" ); document.write( "(C) Which confidence interval is larger and why?
\n" ); document.write( "The width of a CI is always 2*ME
\n" ); document.write( "Since the ME is larger for n = 89, its CI is larger.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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