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A researcher wanted to determine whether pedestrian deaths were uniformly distributed over the days of the week. She randomly selected 300 pedestrian deaths, recorded the day of the week on which the death occurred, and obtained the following results (the data are based on information obtained from the Insurance Institute for Highway Safety) $$ \begin{array}{lc|lc} \begin{array}{l} \text { Day of } \\ \text { the Week } \end{array} & \text { Frequency } & \begin{array}{l} \text { Day of } \\ \text { the Week } \end{array} & \text { Frequency } \\ \hline \text { Sunday } & 39 & \text { Thursday } & 41 \\ \hline \text { Monday } & 40 & \text { Friday } & 49 \\ \hline \text { Tuesday } & 30 & \text { Saturday } & 61 \\ \hline \text { Wednesday } & 40 & & \\ \hline \end{array} $$ Test the belief that the day of the week on which a fatality happens involving a pedestrian occurs with equal frequency at the \(\alpha=0.05\) level of significance.

Step by step solution

01

State the hypotheses

Formulate the null and alternative hypotheses. The null hypothesis (H0) states that pedestrian deaths are uniformly distributed over the days of the week. The alternative hypothesis (H1) is that pedestrian deaths are not uniformly distributed: - Null Hypothesis (H0): Pedestrian deaths are uniformly distributed. - Alternative Hypothesis (H1): Pedestrian deaths are not uniformly distributed.

02

Determine the expected frequencies

If pedestrian deaths are uniformly distributed, each day of the week should have the same number of deaths. Calculate the expected frequency by dividing the total number of deaths by the number of days: \[ \text{Expected frequency per day} = \frac{300}{7} \approx 42.86 \]

03

Compute the test statistic

Use the chi-square test statistic formula to compare observed and expected frequencies: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency.day of the week results: \[ \text{Sunday:} \chi^2 = \frac{(39 - 42.86)^2}{42.86} \approx 0.35 \] \[ \text{Monday:} \chi^2 = \frac{(40 - 42.86)^2}{42.86} \approx 0.19 \] \[ \text{Tuesday:} \chi^2 = \frac{(30 - 42.86)^2}{42.86} \approx 3.87 \] \[ \text{Wednesday:} \chi^2 = \frac{(40 - 42.86)^2}{42.86} \approx 0.19 \] \[ \text{Thursday:} \chi^2 = \frac{(41 - 42.86)^2}{42.86} \approx 0.08 \] \[ \text{Friday:} \chi^2 = \frac{(49 - 42.86)^2}{42.86} \approx 0.89 \] \[ \text{Saturday:} \chi^2 = \frac{(61 - 42.86)^2}{42.86} \approx 7.33 \] Summing these values yields: \[ \chi^2 = 0.35 + 0.19 + 3.87 + 0.19 + 0.08 + 0.89 + 7.33 \approx 12.90 \]

04

Determine the degrees of freedom and critical value

The degrees of freedom (df) for this test is determined by the number of categories minus 1: \[ df = 7 - 1 = 6 \] Using the chi-square distribution table and 鈲� = 0.05, the critical chi-square value for 6 degrees of freedom is approximately 12.59.

05

Compare the test statistic and critical value

Compare the calculated test statistic \(\chi^2\) to the critical value from the chi-square distribution table: \[ \text{Calculated } \chi^2 \approx 12.90 \] \[ \text{Critical value} = 12.59 \] Since 12.90 > 12.59, we reject the null hypothesis.

06

State the conclusion

Based on the comparison, there is sufficient evidence to reject the null hypothesis at the 0.05 level of significance. Therefore, we conclude that pedestrian deaths are not uniformly distributed over the days of the week.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions or inferences about population parameters based on sample data. It begins with formulating two hypotheses: the null hypothesis (H0) and the alternative hypothesis (H1).
The null hypothesis represents a statement of no effect or no difference, assuming any observed effect is due to sampling variability. In contrast, the alternative hypothesis suggests there is a significant effect or difference.
In our exercise, the null hypothesis (H0) is that pedestrian deaths are uniformly distributed over the days of the week. The alternative hypothesis (H1) is that pedestrian deaths are not uniformly distributed. This helps us determine if the observed frequencies of deaths differ significantly from what we'd expect if distribution was uniform.

Expected Frequency

Expected frequency refers to the frequency we would expect in each category if the null hypothesis is true.
In a chi-square test, expected frequencies are used to see how observed data compare to what we would expect based on our null hypothesis.
For our exercise, if pedestrian deaths are uniformly distributed, each day should have an equal number of deaths. We can calculate the expected frequency by dividing the total number of deaths by the number of days:
\ \ \ \[ \frac{300}{7} \approx 42.86 \] \
This means if the deaths were uniformly distributed, we'd expect around 42.86 deaths each day of the week.

Degrees of Freedom

Degrees of freedom (df) is a crucial concept in statistics that describes the number of values in the final calculation of a statistic that are free to vary.
In the context of the chi-square test, degrees of freedom are determined by the number of categories minus one. Essentially, it tells us how many categories are involved in the test.
For our exercise, we have seven days in a week. So, the degrees of freedom would be: \ \ \ \[ df = 7 - 1 = 6 \] \ A higher number of degrees of freedom generally means a more flexible model and affects the critical value we use to determine significance.

Level of Significance

The level of significance, denoted by \ \( \text{\( \alpha \)} \), is a threshold set by the researcher before conducting the test. It represents the probability of rejecting the null hypothesis when it is actually true, essentially controlling the risk of a Type I error.
For our exercise, the level of significance is given as 0.05 or 5%. This means we are willing to accept a 5% risk of concluding that pedestrian deaths are not uniformly distributed when they actually are.
The critical value from the chi-square distribution table at this level of significance and with 6 degrees of freedom helps us to decide whether to reject the null hypothesis. In this case, the critical value is 12.59. If our calculated chi-square value exceeds this, we reject the null hypothesis.