Question

A study is conducted in a company that employs 800 engineers. A random sample of 50 engineers reveals that the average sample age is 34.3 years. Historically, the population standard deviation of the age of the company’s engineers is approximately 8 years. Construct a 98% confidence interval to estimate the average age of all the engineers in this company. 

Answer 1

$n=800 \\ \bar{x}=34.3 \\ \sigma= 8$

Two-sided confidence interval:

$CI = (\bar{x} - \frac{Z_c \times \sigma}{\sqrt{n}}, \bar{x} + \frac{Z_c \times \sigma}{\sqrt{n}})$

For 98 % confidence $α= 1 -0.98 = 0.02$

$\frac{α}{2} = \frac{0.02}{2} = 0.01 \\ Z_c = Z_{0.01} = 2.33 \\ CI = (34.3 - \frac{2.33 \times 8}{\sqrt{800}}, 34.3 + \frac{2.33 \times 8}{\sqrt{800}}) \\ = (34.3 -2.64, 34.3 +2.64) \\ = (31.66, 36.94)$