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: 	

	

99 percent confidence intervals for the population mean.

ME =2.575[38/sqrt(50)] = 13.8381

CI: 	



	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

 

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