Question

Insurance adjustors are concerned that Dave’s Auto Repair is charging higher-than-necessary prices to their clients. To see if the estimates are unreasonably high, a sample of 10 damaged cars was taken to Dave’s and another repair shop and the estimates (in dollars) were recorded. Here are the results: Car 1 2 3 4 5 6 7 8 9 10 Dave’s 1410 1550 1250 1300 900 1520 1750 3600 2250 2840 Other 1250 1300 1250 1200 950 1575 1600 3380 2125 2600 a) For each car, subtract the estimate of the other garage from Dave’s estimate. Find the mean and standard deviation for the differences. You do not need to find the mean or sd by hand - use a calculator, Excel, or R. b) (12 points total) Test the null hypothesis that there is no difference between the estimates of the two garages. Show all the steps as demonstrated in class. c) Construct a 95% confidence interval for the difference in estimates.

Answer 1

(a)

H0: Null Hypothesis: = 0

HA: Alternative Hypothesis: 0

From the given data, values of d = Dave's - Other are got as follows:

Car	1	2	3	4	5	6	7	8	9	10
Dave's	1410	1550	1250	1300	900	1520	1750	3600	2250	2840
Other	1250	1300	1250	1200	950	1575	1600	3380	2125	2600
d = Dave's - Other	160	250	0	100	- 50	- 55	150	220	125	240

From d values, the following statistics are calculated:

n = 10

= 114

(b)

s_d = 114.4018

SE = s_d/

= 114.4018/

= 36.1770

Test statistic is given by:

t = 114/36.1770

= 3.1512

ndf = n - 1 = 10 - 1 = 9

= 0.05

From Table, critical values of t= 2.2622

Since the calculated value of t = 3.1532 is greater than critical value of t = 2.2622, the difference is significant. Reject null hypothesis.

Conclusion:

The data do not support the claim that there no difference between the estimates of the two garages.

(c)

Confidence interval:

114 (2.2622 X 36.1770)

= 114 81.8396

= ( 32.1604 ,195.8396)

Confidence Interval:

32.1604 < < 195.8396