Answer to Question #152046 in Statistics and Probability for JG

Question

A statistician studies the average age of professors in a local college in Georgia.
a) Based on a random sample of 25 professors, we found the average age is 54, with standard deviation of 6.3. Construct a 98% confidence interval for the population’s average age.
b) Based on another random sample of size 32, we found the average age is 53, assuming standard deviation of all professors in this college is 4.9. Construct an 83% confidence interval for the population’s average age and state the confidence statement.

Answer 1

a) n=25

μ=54

SD=6.3

Since population standard deviation is unknown and also sample size is less than 30 so we will use t statistics with

$df=n-1 \\ =25-1 \\ =24$

Now 98% confidence interval is given by:

$μ±t_{0.01} \times \frac{SD}{\sqrt{n}} = 54 ± 2.49 \times \frac{6.3}{\sqrt{25}} \\ = 54 ± 3.14$

Interval (50.86, 57.14)

b) n=32

μ=53

SD=4.9

We have to create 83% confidence interval for population mean since population SD is known so we will use Z statistics.

Now 83% confidence interval is given by:

$μ±Z_{0.085} \times \frac{SD}{\sqrt{25}} = 53 ± 1.37 \times \frac{4.9}{\sqrt{32}} \\ = 53 ± 1.19$

Interval (51.81, 54.19)