Question

A local insurance company is analyzing the cost of the two body shops their customers tend to use in hopes to save money. They randomly select 10 bills from their files of claim payments made in the last year to see if one company tends to be cheaper than the other. The data are shown below.

Crash-tastic: 800 1100 1500 1800 1900 1420 2100 2300 1400 1500

Mr. Fix It: 1700 1230 1220 1600 1120 1560 920 1800 1500 1400

Part 1 of 2: 5 pts Name the appropriate test we would use to determine if there is a significant difference in repair costs (2 pts). BE SPECIFIC! Explain why you made the choice you did (3 pts)!

Part 2 of 2: 5 pts Check each of the four appropriate conditions in detail (1 pt each) and indicate whether you can proceed with the test. If you cannot proceed, explain why, sighting the condition that wasn't met and why.

Answer 1

Hypothesis testing is more appropriate here. T-test can be used to determine the significance of the two groups.

The two groups are Crash-tastic and Mr. Fix it and both are independent.

Hence, t-test can be used to test the significant difference in repair costs.

B)

The t-test takes a sample from each of the two sets and establishes the problem statement by assuming a null hypothesis that the two means are equal after that certain values are calculated and compared against the standard values, and the assumed null hypothesis is accepted or rejected accordingly.

If the null hypothesis qualifies to be rejected, it indicates that data readings are strong and are not by chance.

This is 2 - independent sample 2-sided hypothesis test. Since n<30 so we can use t-test here.

State the hypothesis

Null Hypothesis

H₀= There is no significant difference in repair costs.

Alternate Hypothesis

Ha= There is a significant difference in repair costs.

Now Using Excel>Data Analysis>t-test two sample assuming equal variances.

Here is the output:

t-Test: Two-Sample Assuming Equal Variances
	Crash tastic	Mr. Fix it
Mean	1582	1405
Variance	205462.2	77272.22
Observations	10	10
Pooled Variance	141367.2
Hypothesized Mean Difference	0
df	18
t Stat	1.05265
P(T<=t) one-tail	0.153214
t Critical one-tail	1.734064
P(T<=t) two-tail	0.306428
t Critical two-tail	2.100922

Conclusion:

Now we get that p>0.05.

Thus, we cannot reject H₀.

Thus, we conclude that there is no significant difference in repair costs.