Question

1.The average number of years of post secondary education of employees in an industry is 1.5. A company claims that this average is higher for its employees. A random sample of 16 of its employees has a mean of 2.1 years of post secondary education with a standard deviation of 0.6 years.

a.Find a 95% confidence interval for the mean number years of post secondary education for the company’s employees. How does this compare with the industry value?

b.Find a 95% confidence interval for the standard deviation of number years of post secondary education for the company’s employees.

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 16- 1 ) = 2.131
2.1 ± t(0.05/2, 16 -1) * 0.6/√(16)
Lower Limit = 2.1 - t(0.05/2, 16 -1) 0.6/√(16)
Lower Limit = 1.7804
Upper Limit = 2.1 + t(0.05/2, 16 -1) 0.6/√(16)
Upper Limit = 2.4196
95% Confidence interval is ( 1.7804 , 2.4196 )

Part b)

(0.05/2, 16 - 1 ) = 27.4884
(1 - 0.05/2, 16 - 1) ) = 6.2621
Lower Limit = 0.4432
Upper Limit = 0.9286
95% Confidence interval is ( 0.4432 , 0.9286 )

( 0.4432 < σ < 0.9286 )