Question

A random sample of 15 employees was selected. The average age in the sample was 31 years with a variance of 49 years. Assuming ages are normally distributed, the 98% confidence interval for the population average age is _____.

Answer 1

Answer:

Assuming ages are normally distributed, the 98% confidence interval for the population average age is [26.3, 35.7].

Step-by-step explanation:

We have to construct a 98% confidence interval for the mean.

The information we have is:

- Sample mean: 31

- Variance: 49

- Sample size: 15

- The age is normally distributed.

We know that the degrees of freedom are

$k=n-1=15-1=14$

Then, the t-value for a 98% CI is t=2.625 (according to the t-table).

The standard deviation can be estimated from the variance as:

$s=\sqrt {s^2}=\sqrt{49}=7$

The margin of error is:

$E=t_{14}*s/\sqrt{n}\\\\E=2.625*7/\sqrt{15}=18.375/3.873=4.7$

Then, the CI can be constructed as:

$M-t*s/\sqrt{n}\leq\mu\leq M+t*s/\sqrt{n}\\\\31-4.7\leq \mu \leq 31+4.7\\\\26.3\leq \mu \leq 35.7$