Question

It is a common belief that SUVs are safer than cars. In a collision, does an SUV sustain less damage (cost of repair) than a car?

SUV into Car SUV Damage ($) Car Damage ($)
Honda CR-V into Honda Civic $1,721 $1,274
Toyota RAV4 into Toyota Corolla $1,434 $2,327
Hyundai Tuscon into Kia Forte $850 $3,223
Volkswagon Tiguan into Volkswagon Golf $2,329 $2,058
Jeep Patriot into Dodge Caliber $1,415 $3,095
Ford Escape into Ford Focus $1,470 $3,386
Nissan Rogue into Nissan Sentra

$2,884 $4,560
Test the claim above at the α=0.1 level of significance. Note: A normal probability plot of difference in car damages between SUVs and other cars indicates the population could be normal and a boxplot indicated no outliers.

1) what is the test statistic?

Answer 1

The test statistic for a paired t-test on SUV versus car collision damage costs is calculated using the mean difference in repair costs, the hypothesized difference (usually zero for no difference), the sample standard deviation, and the sample size. This value is then compared to the critical t-value at a 0.1 significance level to determine if SUVs sustain statistically less damage than cars.

Step-by-step explanation:

Testing the claim that SUVs sustain :

The question requires testing the claim that SUVs sustain less damage than cars in a collision using the provided damage cost data for various SUVs and cars. To test this claim statistically, we calculate the mean differences in damage costs between SUVs and cars. Given the normality and no outliers from the plots, we can proceed with a paired t-test to find out if there is a significant difference in costs.

After calculating mean differences, we use the t-test formula to get the test statistic: t = (mean difference - hypothesized difference) / (sample standard deviation / sqrt(sample size)) For hypothesis testing, we compare this t-value to the critical t-value from the t-distribution table at the α=0.1 significance level with the appropriate degrees of freedom. If the computed t-value is greater than the critical t-value, we reject the null hypothesis, which states there is no difference in damage costs.