# A random sample of 16 life insurance policy holders showed that the mean value of their life...

## Question:

A random sample of 16 life insurance policy holders showed that the mean value of their life insurance policies is $200,000 with a standard deviation of$50,000. Assume that the values of life insurance policies for all such policyholders have an approximately normal distribution. The 99% confidence interval for the mean value of all life insurance policies is:

a. $174,823.50 to$225,176.50

b. $163,162.50 to$236,837.50

c. $145,249.25 to$254,750.75

d. $157,482.50 to$242,517.50

## Confidence Interval for Population Mean:

The confidence interval is a type of interval estimation that gives a range within which the true population mean is most likely to be contained at a given level of significance. By providing a range, a confidence interval captures the error made during estimation, which enables it to address the disadvantage of point estimates, which slightly differ from the true population mean.

The correct answer is (b). $163,162.50 to$236,837.50

We can get the following values from the question:

{eq}n=16\\\bar x=200,000\\s=50,000\\CI=99\% {/eq}

Since population standard deviation is unknown and sample size is small, we'll use student t distribution. Use the following formula to construct the confidence interval of the population mean:

{eq}\displaystyle P(\bar x\pm t_{\frac{\alpha}{2}}\times \frac{s}{\sqrt{n}}) {/eq}

Find the critical value t:

• Find the level of significance that correspond to 99% level of confidence:

{eq}\begin{align*} \alpha&=1-CI\\&=1-0.99\\&=0.01\\\displaystyle \frac{\alpha}{2}&=0.005 \end{align*} {/eq}

• Calculate the degrees of freedom:

{eq}\begin{align*} Df&=n-1\\&=16-1\\&=15 \end{align*} {/eq}

• Use the software or t table to determine the critical value t:

{eq}t_{0.005,df=15}=2.947 {/eq}

Now construct the confidence interval:

{eq}\displaystyle P(200,000\pm (2.947)(\frac{50,000}{\sqrt{16}}))\\P(200,000\pm 36,837.5)\\P($163,162.50\le \mu\le$236,837.50) {/eq}