Insurance adjusters are concerned about the high estimates they are receiving from Jocko's Garage. To see if the estimates are unreasonably high, each of 10 damaged cars was taken to Jocko's and to another garage, and the estimates (in dollars) were recorded.

\begin{tabular}{|c|c|c|c|c|c|}
\hline Car & 1 & 2 & 3 & 4 & 5 \\
\hline Jocko's & 1410 & 1550 & 1250 & 1300 & 900 \\
\hline Other & 1250 & 1300 & 1250 & 1200 & 950 \\
\hline Car & 6 & 7 & 8 & 9 & 10 \\
\hline Jocko's & 1520 & 1750 & 3600 & 2250 & 2840 \\
\hline Other & 1575 & 1600 & 3380 & 2125 & 2600 \\
\hline
\end{tabular}

(a) For each car, subtract the estimate of the other garage from Jocko's estimate. Find the mean and the standard deviation for this difference.

(b) Test the null hypothesis that there is no difference between the estimates of the two garages. Be sure to specify the null and alternative hypotheses, and the test statistic with degrees of freedom.

(c) Construct a [tex]$95 \%$[/tex] confidence interval for the difference in estimates.

(d) The insurance company is considering seeking repayment from 1000 claims filed with Jocko's last year. Using your answer to part (c), what repayment would you recommend the insurance company seek?



Answer :

Sure, let's address each part of the question step by step:

### (a) Calculate the differences, mean difference, and standard deviation for these differences

For each car, subtract the other garage's estimate from Jocko's estimate:

[tex]\[ \begin{aligned} & \text{Car 1: } 1410 - 1250 = 160 \\ & \text{Car 2: } 1550 - 1300 = 250 \\ & \text{Car 3: } 1250 - 1250 = 0 \\ & \text{Car 4: } 1300 - 1200 = 100 \\ & \text{Car 5: } 900 - 950 = -50 \\ & \text{Car 6: } 1520 - 1575 = -55 \\ & \text{Car 7: } 1750 - 1600 = 150 \\ & \text{Car 8: } 3600 - 3380 = 220 \\ & \text{Car 9: } 2250 - 2125 = 125 \\ & \text{Car 10: } 2840 - 2600 = 240 \\ \end{aligned} \][/tex]

The differences are: [tex]\([160, 250, 0, 100, -50, -55, 150, 220, 125, 240]\)[/tex]

Next, calculate the mean difference:

[tex]\[ \text{Mean Difference} = \frac{160 + 250 + 0 + 100 - 50 - 55 + 150 + 220 + 125 + 240}{10} = 114.0 \][/tex]

Calculate the standard deviation of the differences:

[tex]\[ \text{Standard Deviation} = 114.4018 \][/tex]

### (b) Test the null hypothesis that there is no difference between the estimates of the two garages

Null Hypothesis ([tex]\(H_0\)[/tex]): The mean difference in estimates between Jocko's Garage and the other garage is 0. ([tex]\(\mu_d = 0\)[/tex])

Alternative Hypothesis ([tex]\(H_a\)[/tex]): The mean difference in estimates between Jocko's Garage and the other garage is not 0. ([tex]\(\mu_d \neq 0\)[/tex])

To test this hypothesis, we use the paired sample t-test.

Calculate the t-statistic and p-value:

[tex]\[ \text{t-statistic} = 3.1512 \][/tex]
[tex]\[ \text{p-value} = 0.011715 \][/tex]

Since the p-value is less than 0.05, we reject the null hypothesis at the 5% significance level, indicating that there is a statistically significant difference in the estimates between the two garages.

### (c) Construct a 95% confidence interval for the difference in estimates

We construct a 95% confidence interval for the mean difference.

[tex]\[ \text{95% Confidence Interval} = (32.1619, 195.8381) \][/tex]

### (d) Recommend repayment for 1000 claims

Using the 95% confidence interval, we can recommend repayment for 1000 claims. To calculate this, we multiply the bounds of the confidence interval by the number of claims:

[tex]\[ \text{Lower Bound Repayment (for 1000 claims)} = 1000 \times 32.1619 = 32161.86 \text{ dollars} \][/tex]
[tex]\[ \text{Upper Bound Repayment (for 1000 claims)} = 1000 \times 195.8381 = 195838.14 \text{ dollars} \][/tex]

So, the insurance company could seek repayment ranging from approximately [tex]\(32162\)[/tex] dollars to [tex]\(195838\)[/tex] dollars for 1000 claims based on the 95% confidence interval.