A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is 534 per year with a standard deviation of 63.

Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a 99% confidence interval for the population mean, μ.

Answer :

Given :

Sample mean $(x\overline{)}=534$

Sample standard deviation $(s)=63$

Sample size $(n)=25$

Confidence interval level $=99$

Solution :

The significance level is calculated as:

$$\alpha =1–0.99=0.01$$

The degree of freedom is calculated as:

$$\text{df}=n–1=25–1=24$$

The critical value is calculated as:

$$t_c=t_{\frac{\alpha }{2},\text{df}}=t_{\frac{0.01}{2},24}=2.797$$

The confidence interval is calculated as:

$$CI=\overline{x}\pm t_c\times \frac{s}{\sqrt{n}}=534\pm 2.797\times \frac{63}{\sqrt{25}}=(498.76,569.24)$$