In: Statistics and Probability
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:
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 = t0.05,122 = 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)