Question 441291
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b> 
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