document.write( "Question 441291: A sample of 106 golfers showed that their average score on a particular golf course was 87.98 with a standard deviation of 5.39.
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\n" ); document.write( "B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of 106.
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Algebra.Com's Answer #304614 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi
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document.write( "sample of 106 golfers showed that their average score on a particular golf
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document.write( " course was 87.98 with a standard deviation of 5.39.
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document.write( "A) Find the 90% confidence interval of the mean score for all 106 golfers.
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document.write( " ME = 1.645[5.39/sqrt(106)] = .8612
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document.write( " CI: 87.98-.8612 < u < 87.98+.8612
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document.write( " CI: 86.9988 < u < 88.8412 
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document.write( "B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of 106.
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document.write( "  ME = 1.645[5.39/sqrt(130)] = .7776
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document.write( " CI: 87.2024 < u < 88.7576
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document.write( "C) Which confidence interval is larger and why?
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document.write( " confidence interval is larger for the smaller sample size of 106 due to 
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document.write( "the fact the ME is larger\r
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document.write( "Summary of z values for various confidence intervals
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document.write( "	a	a/2	crtical regions	
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document.write( "90%	0.1	5%	z <-1.645	z >+1.645
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document.write( "95%	0.05	2.50%	z <-1.96	z >+1.96
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document.write( "98%	0.02	1%	z <-2.33	z >+2.33
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document.write( "99%	0.01	0.50%	z<-2.575	z >+2.575
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