Question

Aalbedte ADA Artboede Normal AaBbce Albede AaBb No Spacing Heading Heading 2 Sty as Para Die LMU E2300, Chap 8 HW Part II: Name: 1. One of the major problems faced by an auto insurance company is the filing of fraudulent claims. An insurance company carefully investigated 1000 auto claims filed with it and found 108 of them to be fraudulent. Construct a 96% confidence interval for the corresponding population proportion. Show all work in the space below. b. Write a full sentence interpreting the numbers you have calculated, i.e. what do these numbers stand for? _(narrower/wider) c. If you used a 90% Confidence Level instead, the Cl would be Answer without doing any calculations.

Answer 1

a)

Sample proportion is -

$\widehat{p} = \frac{x}{n} = \frac{108}{1000} = 0.108$

The critical value of test statistic is -

$Z_{\frac{\alpha}{2}} = Z_{ \frac{0.04}{2}} = Z_{0.02} = 2.054$

The confidence interval is given by -

$\widehat{p} \pm Z_{\frac{\alpha}{2}} \left ( \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}} \right )$

So, we get -

$CI = 0.108 \pm 2.054\left ( \sqrt{\frac{0.108 \times 0.892}{1000}} \right )$

$\\ So,\:\: 0.108 - 0.0202 < P < 0.108 + 0.0202 \\ \\ \Rightarrow 0.0878 < P < 0.1282$

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b)

This confidence interval indicates that if we perform random sampling of 1000 auto claim several times, then 96% of times, we would get the estimate of population proportion of fraud claims to be within 0.0878 and 0.1282.

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c)

If we use a 90% confidence interval, the width of interval would be narrower because we are reducing the confidence level. This would reduce the value of critical z-statistic and thus margin of error.