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


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.

Answer and Explanation:

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}

Learn more about this topic:

Using the t Distribution to Find Confidence Intervals

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 6

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