Flat Tire and Missed Class A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn’t have a flat tire?

Tire	Left Front	Right Front	Left Rear	Right Rear
Number Selected	11	15	8	16

Forty-onestudents are required to select one of the four tires to examine whether the four students who missed the test actually had a flat tire. Of 41, one was not included as he used a spare tire.

Let O denote the observed frequencies of students who guess the four tires.

Let E denote the frequencies of students who are expected to guess each of the four tires according to the uniform distribution.

Thus, if the students are uniformly distributed, the frequency corresponding to each tire is computed below:

\(\begin{aligned}{c}E = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{students}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{tires}}}}\\ = \frac{{40}}{4}\\ = 10\end{aligned}\)

Assume the students are randomly selected, and as allthe expected values are greater than 5, the test requirements are fulfilled.

The null hypothesis for conducting the given test is as follows:

The number of students whoselect the four tires followsa uniform distribution. Thus, the students cannot identify the correct tire.

The alternative hypothesis for conducting the given test is as follows:

The number of students whoselect the four tires does not follow a uniform distribution. Thus, the students can identify the correct tire.

The table given below shows the necessary calculations.

The value of the test statistic can be given as follows:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 0.1 + 2.5 + 0.4 + 1.6\\ = 4.6\end{aligned}\)

Thus,\({\chi ^2} = 4.6\).

Let k be the number of types of tire categories, which is4.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 4 - 1\\ = 3\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with three degrees of freedom is equal to 7.815.

Moreover, the p-value is equal to 0.203.

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to be rejected.

There is enough evidence to support the statement that thenumber of students that select the four tires followsa uniform distribution. Thus, the students cannot identify the correct tire.

Since the null hypothesis is rejected, it can be said that an equal number of students identify each of the four tires as the flat tire.

Itindicates that the likelihood of the four students identifying the same tire as the flat tire is relativelylow when they did not havea flat tire in reality.