Hi

sample of 106 golfers showed that their average score on a particular golf

 course was 87.98 with a standard deviation of 5.39.

A) Find the 90% confidence interval of the mean score for all 106 golfers.

 ME = 1.645[5.39/sqrt(106)] = .8612

 CI: 87.98-.8612 < u < 87.98+.8612

 CI: 86.9988 < u < 88.8412 

B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of 106.

  ME = 1.645[5.39/sqrt(130)] = .7776

 CI: 87.2024 < u < 88.7576

C) Which confidence interval is larger and why?

 confidence interval is larger for the smaller sample size of 106 due to 

the fact the ME is larger



Summary of z values for various confidence intervals

	a	a/2	crtical regions	

90%	0.1	5%	z <-1.645	z >+1.645

95%	0.05	2.50%	z <-1.96	z >+1.96

98%	0.02	1%	z <-2.33	z >+2.33

99%	0.01	0.50%	z<-2.575	z >+2.575

 

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