Question

NorthCo is a well established company with thousands of employees and they want to estimate the average age of their workers. A sample of 123 employees was taken. Assuming the population has a standard deviation of 8.5, we want to find a 90% confidence interval for the population mean.

If the mean of the sample is 25.3, what is the confidence interval estimate?
For full marks your answer should be accurate to at least two decimal places.

Confidence Interval:

Answer 1

Solution:

Given that,

= 25.3

s = 8.5

n = 123

Degrees of freedom = df = n - 1 = 123 - 1 = 122

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,122 = 1.657

Margin of error = E = t_/2,df * (s /n)

= 1.657 * (8.5 / 123)

= 1.27

The 90% confidence interval estimate of the population mean is,

- E < < + E

25.3 - 1.27 < < 25.3 +1.27

24.03 < < 26.57

(24.03, 26.57)