Question

Motor Vehicle Fatalities The table below lists motor vehicle fatalities by day of the week for a recent year (based on data from the Insurance Institute for Highway Safety). Use a 0.01 significance level to test the claim that auto fatalities occur on the different days of the week with the same frequency. Provide an explanation for the results.

Day	Sun.	Mon.	Tues.	Wed.	Thurs.	Fri.	Sat.
Frequency	5304	4002	4082	4010	4268	5068	5985

Short answer

There is enough evidence to conclude thatmotor vehicle fatalities do not occur equally frequently on all days of the week.

The number of auto fatalities is higher on weekends than on weekdays because drinking activity increases on weekends (since individuals have more free time) and as a result, people tend to drive under the influence of alcohol which leads to accidents.

Step-by-step solution

Given information

Data are given on the frequency of auto fatalities on different days of the week.

The claim is to test that auto fatalities occur on different days of the week with the same frequency at the significance level 0.01.

Hypotheses

The null hypothesis is as follows:

Motor vehicle fatalities occur with equal frequency on all days of the week.

The alternative hypothesis is as follows:

Motor vehicle fatalities do not occur with equal frequency on all days of the week.

Observed and expected frequencies

Let 1, 2, …..,6 represent the different days of the week starting from Sunday and ending on Saturday.

The observed frequencies of fatalities are tabulated as shown:

\({O_1}\)=5304

\({O_2}\)=4002

\({O_3}\)=4082

\({O_4}\)=4010

\({O_5}\)=4268

\({O_6}\)=5068

\({O_7}\)=5985

It is given that fatalities are believed to occur with the same frequency on different days.

Therefore, the expected frequency for each of the 7 days is the same and is equal to:

\(E = \frac{n}{k}\)

Where,

n is the total frequency

k is the number of days.

The value of n is computed as shown:

\(\begin{aligned}{c}n = 5304 + 4002 + .... + 5985\\ = 32719\end{aligned}\)

The value of k is equal to 7.

Thus, the expected frequency is equal to:

\(\begin{aligned}{c}E = \frac{n}{k}\\ = \frac{{32719}}{7}\\ = 4674.143\end{aligned}\)

Test statistic

The test statistic value is computed as shown:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\; \sim {\chi ^2}_{\left( {k - 1} \right)}\\ = \frac{{{{\left( {5304 - 4674.143} \right)}^2}}}{{4674.143}} + \frac{{{{\left( {4002 - 4674.143} \right)}^2}}}{{4674.143}} + ....... + \frac{{{{\left( {5985 - 4674.143} \right)}^2}}}{{4674.143}}\\ = 787.018\end{aligned}\]

Thus, \({\chi ^2} = 787.018\).

Critical value, p-value and conclusion of the test

The degrees of freedom are computed as shown:

\(\begin{aligned}{c}df = k - 1\\ = 7 - 1\\ = 6\end{aligned}\)

The critical value of\({\chi ^2}\)with 6 degrees of freedom and\(\alpha = 0.01\)for a right-tailed test is equal to 16.8119.

Referring to the chi-square table, the p-value is obtained using the degrees of freedom and test statistic, which is equal to 0.000.

Since the value of the test statistic is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

Thus, there is enough evidence to conclude thatmotor vehicle fatalities do not occur equally frequently on all days of the week.

Explanation for the Conclusion

The number of auto fatalities is higher on weekends as compared to weekdays because drinking activity increases on weekends (as people have more free time) and thus, people tend to drive under the influence of alcohol which leads to accidents.