An insurance company specializing in automobile insurance, wishes to compare the

Question

An insurance company specializing in automobile insurance, wishes to compare the repair costs of moderately damaged cars (repair costs between $ 700 and $ 1400) at two garages. The company is trying to determine whether mean costs of repairs at garage 2 is higher than garage 1 The repair cost for each car at both garages is given (Cost Estimates in Hundreds of Dollars) a. At 0.05 level of significance, is there an evidence that mean cost of repairs at garage 2 is higher than garage 1. b. Determine and interpret the meaning of p value in (a) c. Construct and interpret a 95% confidence interval estimate of mean difference in repair costs at garage 1 and garage 2 d. What assumption is necessary about the population distribution in order to perform the test in (a)?

Explanation / Answer

a. Follow the steps to determineif there is evidence that mean cost of repairs at garage 2 is higher than garage 1.

Hypotheses: H0:mu1-mu2=0 (there is no difference in population mean cost of repair for garage 1 and 2)

H1:mu1-mu2<0 (population mean cost of repair at garage 2 is higher than garage 1)

Assume groups (garage 1 and 2) are independent, cars are randomly assigned to either of the garage, and repair cost for both the garage are roughly symmetrical. The assumptions are reasonably met, thus use Student's t model to perform a 2-sample t test.

The sample size are equal (n1=n2=7), the sample standard deviation (using calculator) for cost of repair at garage 1 and 2 are 1.25 and 1.51. Thus, s1/s2=1.25/1.51=0.83<2, therefore, use pooled standard deviation.

Test statistic:

Polled standard devaition, sp=sqrt[{(n1-1)s1^2+(n2-1)s2^2}/(n1+n2-2)], where, n represent sample size, s is sampel standard deviation.

sp=sqrt[{(7-1)1.25^2+(7-1)1.51^2}/(7+7-2)]=1.3859

Test statistic, t=(x1bar-x2bar)/sp sqrt(1/n1+1/n2), where, xbar is sample mean and the values are computed using calculator.

=(9.33-10.13)/1.3859 sqrt(1/7+1/7)

=-1.08

p value at 12 degrees of freedom [df=n1+n2-2] is 0.151.

Rejection rule: Per rejection rule, reject null hypothesis if p value is less than alpha=0.05. Here, p value is not less than 0.05, therefore, fail to reject null hypothesis.

Conclusion: There is insufficiet sample evidence to suggest that mean cost of repairs at garage 2 is higehr than garage 1.

b. The p value is 0.151 implies that assuming the null hypothesis to be true, one is likely to obtain a t value of -1.08 or higher.

c. The 95% c.i of mean difference in repair costs at garage 1 and 2 is as follows:

(x1bar-x2bar)+-talpha/2 (sp)sqrt(1/n1+1/n2), where, t alpha/2 is t critical at 0.025 (alpha=0.05, alpha/2=0.025) and 12 degrees of freedom.

=(1.25-1.51)+-2.179(1.3859)sqrt(1/7+1/7)

=(-2.414, 0.814)

d. The population distribution of repair cost at garage 1 and 2 are normally distributed. Violation of normality assumption would not allow one to use two-sample t test for the difference between two means.

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