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.
### (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.
