This is a Chi-Square Goodness-of-Fit Test problem to determine if the observed crash distribution matches the expected distribution of crashes based on the age distribution.

Steps to Solve:

1. State the Null and Alternative Hypotheses:

   • $H_0$: The observed crash distribution conforms to the expected age distribution.
   • $H_a$: The observed crash distribution does not conform to the expected age distribution.

2. Calculate the Expected Frequencies: Multiply the total number of drivers (300) by the expected percentages for each age group:

   • Under 25: $300 \times 0.16 = 48$
   • 25-44: $300 \times 0.44 = 132$
   • 45-64: $300 \times 0.27 = 81$
   • Over 64: $300 \times 0.13 = 39$

3. Compute the Test Statistic: The test statistic for a Chi-Square Goodness-of-Fit test is: $\chi^2 = \sum \frac{(O - E)^2}{E}$ where $O$ is the observed frequency, and $E$ is the expected frequency. Calculate for each group:

   • Under 25: $\frac{(116 - 48)^2}{48} = \frac{68^2}{48} = 96.17$
   • 25-44: $\frac{(70 - 132)^2}{132} = \frac{(-62)^2}{132} = 29.12$
   • 45-64: $\frac{(46 - 81)^2}{81} = \frac{(-35)^2}{81} = 15.12$
   • Over 64: $\frac{(68 - 39)^2}{39} = \frac{29^2}{39} = 21.56$

   Sum these values to find $\chi^2$: $\chi^2 = 96.17 + 29.12 + 15.12 + 21.56 = 161.97$

4. Find the Critical Value: Degrees of freedom (df) = Number of categories - 1 = 4 - 1 = 3. At a significance level of 0.025, the critical value from the Chi-Square table is approximately $11.345$.

5. Make a Decision:

   • If $\chi^2 > \text{critical value}$, reject $H_0$.
   • Here, $161.97 > 11.345$, so we reject $H_0$.

Conclusion:

• The conclusion is B: There is sufficient evidence to warrant the rejection of the claim that the distribution of crashes conforms to the distribution of ages.

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Do you want further clarification or a breakdown of the calculations? Here are 5 related questions:

1. How is the Chi-Square test used in hypothesis testing?
2. Can you explain why degrees of freedom matter in this test?
3. How do we interpret the critical value in a Chi-Square test?
4. What other scenarios would use a Chi-Square Goodness-of-Fit Test?
5. How does significance level affect the test result?

Tip:

Always check if the expected frequencies are all greater than 5. If not, the Chi-Square test may not be valid, and you might need to combine categories or use a different test.