Question

The Insurance Institute for Highway Safety collects safety information about all types of cars every year, and publishes a report of Top Safety Picks among all cars, makes, and models. Table 11.58 presents the number of Top Safety Picks in six-car categories for the two years 2009 and 2013. Analyze the table data to conclude whether the distribution of cars that earned the Top Safety Picks safety award has remained the same between 2009 and 2013. Derive your results at the 5%

significance level.

Short answer

Conclusion: There is sufficient evidence to insure that the distribution of cars remained same between years 2009 and$2013.$

Step-by-step solution

Given Information

We have to to conclude whether the distribution of cars that earned the Top Safety Picks safety award has remained the same between 2009 and 2013

Null Hypothesis

The null hypothesis is shown below: $H_0$ : The distribution of cars remained the same between the years 2009 and$2013.$

Alternative Hypothesis

Against the alternative hypothesis as shown below:

$H_a$: The distribution of cars remained different between the years 2009 and$2013.$

Degrees of Freedom

The degrees of freedom can be calculated by the formula given below:

$df=(numberofcolumn-1)$

Therefore

df=(6-1)

=5

Step 5  Distribution

From the above calculation, it is clear that the distribution for the test is ${\chi }_s^2$.

Expected Frequencies 

The observed value table is already given in the textbook. Calculate the expected frequencies by using the formula given below:

$E=\frac{\left(\text{row total}\right)\left(\text{column total}\right)}{\text{overall total}}$

All calculations can be done in an excel worksheet. Hence, the expected (E) value table is shown below:

Step 7  Test Statistic

$=\sum _{\left(ixj\right)}\frac{(O-E{)}^2}{E}$

The test statistic of the independence test is given below:

Test statistic

To calculate $\frac{(O-E{)}^2}{E}$apply the formula

$=\left((B4-B12{)}^2\right)/B12$

in cell B19 and drag the same formula up to cell G20. After that, take the total of columns total and rows total. The table of test statistics is shown below:

Hence the test statistic is 6.65

 p-Value

The$p$-value can be calculated in excel by using CHIDIST ( ) formula as shown below:

Hence the p- value is 0.25

Step 9  The Chi-square sketch

The Chi-square sketch is given below

Decision, reason and conclusion

i. Alpha:$0.05$

ii: Decision: Do not reject the null hypothesis$H_0$

iii. Reason for decision: Because $p-$value$>\alpha$

iv. Conclusion: There is sufficient evidence to insure that the distribution of cars remained same between years 2009 and$2013.$