Math Problem Statement

Randomly selected deaths from car crashes were obtained, and the results are obtained in the accompanying table. Use a 0.05 significance level to test the claim that car crash fatalities occur with equal frequency on the different days of the week. How might the results be explained? Why does there appear to be an exceptionally large number of car crash fatalities on Saturday?

Solution

To solve this hypothesis test, we will:

1. Identify the hypotheses:

  • Null hypothesis (H0H_0): Car crash fatalities occur with equal frequency on all days of the week.
  • Alternative hypothesis (H1H_1): Car crash fatalities do not occur with equal frequency on all days of the week.

2. Perform the Chi-Square Goodness-of-Fit Test:

We are given the observed frequencies of fatalities for each day, and we assume that the expected frequencies would be equal if H0H_0 is true. To calculate the test statistic (χ2\chi^2), we need: χ2=(OE)2E\chi^2 = \sum \frac{(O - E)^2}{E} Where:

  • OO: Observed frequency
  • EE: Expected frequency

Observed Frequencies (OO):

  • Sun: 129
  • Mon: 110
  • Tues: 103
  • Wed: 100
  • Thurs: 107
  • Fri: 137
  • Sat: 163

Expected Frequency (EE):

Assuming equal frequencies, the expected frequency for each day is: E=Total number of fatalities7E = \frac{\text{Total number of fatalities}}{7} E=129+110+103+100+107+137+1637=8497=121.29E = \frac{129 + 110 + 103 + 100 + 107 + 137 + 163}{7} = \frac{849}{7} = 121.29

3. Calculate the Chi-Square Test Statistic (χ2\chi^2):

We calculate (OE)2/E(O - E)^2 / E for each day and sum these values:

χ2=(129121.29)2121.29+(110121.29)2121.29++(163121.29)2121.29\chi^2 = \frac{(129 - 121.29)^2}{121.29} + \frac{(110 - 121.29)^2}{121.29} + \ldots + \frac{(163 - 121.29)^2}{121.29}

4. Determine the p-value:

The degrees of freedom (dfdf) for this test is: df=Number of categories1=71=6df = \text{Number of categories} - 1 = 7 - 1 = 6

Using the chi-square distribution table or calculator, we compare the test statistic with the critical value for df=6df = 6 and α=0.05\alpha = 0.05, or we calculate the p-value directly.

5. Conclusion:

  • If the p-value is less than 0.05, reject H0H_0, meaning there is sufficient evidence to suggest that fatalities do not occur with equal frequency.
  • If the p-value is greater than 0.05, fail to reject H0H_0, meaning there is insufficient evidence to suggest a difference.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Chi-Square Goodness-of-Fit Test
Probability and Statistics

Formulas

Chi-Square Statistic: χ² = Σ((O - E)² / E)
Degrees of Freedom: df = Number of Categories - 1

Theorems

Chi-Square Distribution

Suitable Grade Level

College/University Statistics