In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion.


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?

Verified step by step guidance

1. Step 1: Define the null and alternative hypotheses. The null hypothesis (H₀) states that the tire selections follow a uniform distribution, meaning each tire is equally likely to be selected. The alternative hypothesis (H₁) states that the tire selections do not follow a uniform distribution.
2. Step 2: Calculate the expected frequency for each tire under the assumption of a uniform distribution. Since there are 41 students and 4 tires, the expected frequency for each tire is calculated as: $\frac{41}{4}$.
3. Step 3: Use the observed frequencies from the table (11 for Left Front, 15 for Right Front, 8 for Left Rear, and 6 for Right Rear) and the expected frequencies to compute the test statistic. The test statistic for a chi-square goodness-of-fit test is calculated using the formula: $\sum \frac{{(}^O}{E}$, where O represents the observed frequency and E represents the expected frequency.
4. Step 4: Determine the degrees of freedom for the chi-square test. The degrees of freedom are calculated as: $\mathrm{df}=k-1$, where k is the number of categories (in this case, 4 tires).
5. Step 5: Compare the calculated test statistic to the critical value from the chi-square distribution table at a significance level of 0.05, or use the P-value approach. If the test statistic exceeds the critical value or if the P-value is less than 0.05, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Interpret the result in the context of the problem to determine the likelihood of four students identifying the same tire when they really didn’t have a flat.

Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (H0) that represents no effect or no difference, and an alternative hypothesis (H1) that indicates the presence of an effect. The test assesses the evidence against H0 using a test statistic and a significance level, leading to a conclusion about whether to reject or fail to reject H0.

Uniform Distribution

A uniform distribution is a type of probability distribution where all outcomes are equally likely. In the context of this question, testing for a uniform distribution means evaluating whether the selection of tires by the students is evenly distributed across the available options. If the results significantly deviate from what would be expected under a uniform distribution, it suggests that the selections are not random and may indicate a bias or pattern.

P-value

The P-value is a measure that helps determine the strength of the evidence against the null hypothesis in hypothesis testing. It represents the probability of observing the test results, or something more extreme, assuming that the null hypothesis is true. A smaller P-value indicates stronger evidence against H0, and if it is less than the predetermined significance level (e.g., 0.05), the null hypothesis is rejected, suggesting that the observed data is unlikely under the null hypothesis.