Question

Speeding Tickets A police chief records the gender of the drivers who receive speeding tickets. Test the claim at \(\alpha=0.05\) that the gender of the ticketed drivers is random. $$ \begin{array}{llllllllll} \text { M } & \text { M } & \text { M } & \text { F } & \text { F } & \text { M } & \text { F } & \text { M } & \text { F } & \text { M } \\ \text { M } & \text { F } & \text { M } & \text { M } & \text { M } & \text { F } & \text { M } & \text { M } & \text { F } & \text { F } \\ \text { F } & \text { M } & \text { M } & \text { F } & \text { M } & \text { M } & \text { F } & \text { M } & \text { M } & \text { M } \\ \text { M } & \text { M } & \text { F } & \text { M } & \text { F } & \text { F } & \text { F } & \text { M } & \text { M } & \text { M } \\ \text { F } & \text { F } & \text { M } & \text { F } & \text { F } & \text { F } & \text { M } & \text { M } & \text { M } & \text { M } \end{array} $$

Short answer

The gender of the ticketed drivers is random at the 0.05 significance level.

Step-by-step solution

Define the Hypotheses

First, we need to establish the null and alternative hypotheses.- Null Hypothesis \(H_0\): The gender of the ticketed drivers is random.- Alternative Hypothesis \(H_1\): The gender of the ticketed drivers is not random.

Conduct a Runs Test

A runs test can determine whether there is randomness in a categorical data sequence. A run is a sequence of similar items, followed by a different item. Count the total number of runs in the sequence. Here is the sequence of tickets: \(M, M, M, F, F, M, F, M, F, M, M, F, M, M, M, F, M, M, F, F, F, M, M, F, M, M, M, M, M, F, M, F, F, F, M, M, M, F, M, F, F, F, M, M, M, M, M\). Counting the runs, we have 22 runs.

Calculate Expected Number of Runs

To calculate the expected number of runs \(E(R)\):\[ E(R) = \frac{2n_1n_2}{n_1 + n_2} + 1 \]where \(n_1\) is the number of males (M), and \(n_2\) is the number of females (F). Count the M's and F's: \(n_1 = 27\), \(n_2 = 23\). Substitute these into the formula:\[ E(R) = \frac{2 \times 27 \times 23}{27 + 23} + 1 \approx 24.14 \]

Calculate the Standard Deviation of Runs

The standard deviation of the number of runs \(\sigma(R)\) is calculated by:\[ \sigma(R) = \sqrt{\frac{2n_1n_2(2n_1n_2 - n_1 - n_2)}{(n_1 + n_2)^2(n_1 + n_2 - 1)}} \]\[ \sigma(R) = \sqrt{\frac{2 \times 27 \times 23 (2 \times 27 \times 23 - 27 - 23)}{(27 + 23)^2(27 + 23 - 1)}} \approx 3.32 \]

Calculate the Z-Score

The Z-score is calculated to determine how far away the observed run count is from the expected run count.\[ Z = \frac{R - E(R)}{\sigma(R)} \]\[ Z = \frac{22 - 24.14}{3.32} \approx -0.64 \]

Decision Based on Z and Alpha

Check the Z-score against the critical Z-values at \(\alpha = 0.05\) for a two-tailed test, where the critical values are approximately \(\pm1.96\). Since \(-0.64\) is within \(-1.96\) and \(1.96\), we do not reject the null hypothesis at the 5% significance level.

Key concepts

Null Hypothesis

When conducting any statistical test, it's crucial to start with formulating a null hypothesis, denoted as \(H_0\). The null hypothesis is a statement of no effect or no difference, meaning it assumes any kind of observed variance in the data is entirely due to chance. In our case, regarding the gender of drivers receiving speeding tickets, the null hypothesis was established as: "The gender of the ticketed drivers is random."

This setup provides a baseline we can test against to see if there's enough evidence to suggest otherwise. We approach the problem with the presumption that there is no real pattern unless we find statistical proof to reject this hypothesis.

Alternative Hypothesis

The alternative hypothesis, denoted as \(H_1\), stands in contrast to the null hypothesis. It reflects what you might expect if there is actually an effect or a pattern in your data. Here, the alternative hypothesis was stated as: "The gender of the ticketed drivers is not random."

Formulating the alternative hypothesis is essential because it helps clarify what kind of results would lead you to reject the null hypothesis. The goal of conducting a test is often to determine whether evidence exists to support this alternative view. By observing the data, if it significantly deviates from the expected randomness, we might reject the null in favor of the alternative.

Z-Score

The Z-score is a measure of how far, and in what direction, a data point differs from the mean of the population. In statistical testing, it helps determine the position of a score in a normal distribution and quantifies the distance from the mean measured in terms of standard deviation.

For our runs test, the Z-score calculation formula is:

• \( Z = \frac{R - E(R)}{\sigma(R)} \)

where \(R\) is the observed number of runs, \(E(R)\) is the expected number of runs, and \(\sigma(R)\) is the standard deviation of the runs. A Z-score tells us how many standard deviations an element is from the mean.

In our example, the Z-score was calculated to be approximately \(-0.64\), which is needed to make a decision on the hypothesis.

Significance Level

The significance level of a test, denoted \(\alpha\), is the threshold at which you determine whether an observed effect is statistically significant. This value often signifies the probability of rejecting the null hypothesis when it is actually true, commonly set at 0.05 for a 5% risk.

For this exercise, the significance level is \(\alpha = 0.05\), and a two-tailed test was performed. This means that the test checks for statistically significant differences in either direction (both below and above the mean), considering critical Z-values of \(\pm 1.96\).

If the computed Z-score falls beyond this range, it indicates a need to reject the null hypothesis in favor of the alternative. Since the Z-score of \(-0.64\) falls within the range, there's insufficient evidence to reject the null hypothesis, suggesting that the sequence of ticketed drivers' genders is indeed random.