[DATA] Pedestrian Deaths 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). 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.

Verified step by step guidance

1. Step 1: State the hypotheses for the chi-square goodness-of-fit test. The null hypothesis \(H_0\) is that pedestrian deaths are uniformly distributed across the seven days of the week, meaning each day has the same expected frequency. The alternative hypothesis \(H_a\) is that the deaths are not uniformly distributed.
2. Step 2: Calculate the expected frequency for each day under the null hypothesis. Since there are 300 total deaths and 7 days, the expected frequency for each day is \(\frac{300}{7}\).
3. Step 3: Use the observed frequencies given for each day: Sunday (39), Monday (40), Tuesday (30), Wednesday (40), Thursday (41), Friday (49), and Saturday (61).
4. Step 4: Compute the chi-square test statistic using the formula: \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\) where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each day.
5. Step 5: Determine the degrees of freedom, which is the number of categories minus 1, so \(df = 7 - 1 = 6\). Then, compare the calculated chi-square statistic to the critical value from the chi-square distribution table at \(\alpha = 0.05\) and \(df = 6\) to decide whether to reject or fail to reject the null hypothesis.

Key Concepts

Chi-Square Goodness-of-Fit Test

This test determines if observed categorical data fit an expected distribution. Here, it checks whether pedestrian deaths are equally likely on each day of the week. The test compares observed frequencies to expected frequencies under the assumption of uniform distribution.

Expected Frequency Calculation

Expected frequency is the count we would expect in each category if the null hypothesis is true. For equal distribution across seven days, it is total deaths divided by seven. This value is used to compare against observed frequencies in the chi-square test.

Significance Level and Hypothesis Testing

The significance level (alpha = 0.05) sets the threshold for rejecting the null hypothesis. If the chi-square test statistic exceeds the critical value at this level, we reject the belief that deaths are uniformly distributed across days.