Anystate Auto Insurance Company took a random sample of 384 insurance claims paid out during a...

Question:

Anystate Auto Insurance Company took a random sample of {eq}384 {/eq} insurance claims paid out during a {eq}1 {/eq}-year period. The average claim paid was {eq}\$1,550 {/eq}. Assume {eq}\sigma = \$234 {/eq}.

A) Find a {eq}0.90 {/eq} confidence interval for the mean claim payment.

B) Find a {eq}0.99 {/eq} confidence interval for the mean claim payment.

Confidence Interval for a Mean:

The confidence interval is a type of interval estimation that gives a range of values likely to be the true population mean. The upper and lower bounds of the interval occur plus and minus margin of error from the best point estimate (P-hat).

Answer and Explanation:

a).

Given that;

{eq}n=384\\\bar x=1550\\\sigma=234 {/eq}

Use equation below to construct .90 confidence interval for the mean:

{eq}\displaystyle \left(\bar X\pm Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\right) {/eq}

Find critical value that corresponds to 90% level of confidence:

{eq}\displaystyle \frac{\alpha}{2}=\frac{1-0.90}{2}=0.05\\Z_{0.05}=\pm 1.645 {/eq}

Plug in values into the formula and solve for upper and lower bounds:

{eq}\displaystyle \left(1550\pm 1.645\times \frac{234}{\sqrt{384}}\right)\\(1550\pm 19.6)\\(1530.4, 1569.6) {/eq}


b).

Repeat the same procedure as above:

{eq}\displaystyle \frac{\alpha}{2}=\frac{1-0.99}{2}=0.005\\z_{0.005}=\pm 2.58\\\displaystyle \left(1550\pm 2.58\times \frac{234}{\sqrt{384}}\right)\\(1550\pm 30.8)\\(1519.2, 1580.8) {/eq}


Learn more about this topic:

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Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
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