Question 445470
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Hi
The average years of employee experience at a company with 10,000 employees is
 normally distributed, with a standard deviation of 12 years.
If a sample of 50 employees indicates a mean age of 38, 
calculate 90 percent confidence intervals for the population mean.
ME = 1.645[38/sqrt(50)] = 8.8402		
CI: {{{38-8.8402 < u < 38 + 8.8402}}}	
	
99 percent confidence intervals for the population mean.
ME =2.575[38/sqrt(50)] = 13.8381
CI: {{{38-13.8381 < u < 38 + 13.8381}}}	

	a	a/2	crtical regions	
90%	0.1	5%	z <-1.645	z >+1.645
92%	0.8	4%	z <-1.751	z >+1.751
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