document.write( "Question 445470: 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 and 99 percent confidence intervals for the population mean. \n" ); document.write( "
Algebra.Com's Answer #306844 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi
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document.write( "The average years of employee experience at a company with 10,000 employees is
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document.write( " normally distributed, with a standard deviation of 12 years.
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document.write( "If a sample of 50 employees indicates a mean age of 38, 
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document.write( "calculate 90 percent confidence intervals for the population mean.
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document.write( "ME = 1.645[38/sqrt(50)] = 8.8402		
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document.write( "CI: \"38-8.8402+%3C+u+%3C+38+%2B+8.8402\"	
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document.write( "99 percent confidence intervals for the population mean.
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document.write( "ME =2.575[38/sqrt(50)] = 13.8381
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document.write( "CI: \"38-13.8381+%3C+u+%3C+38+%2B+13.8381\"	\r
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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( "92%	0.8	4%	z <-1.751	z >+1.751
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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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