The table lists survey results obtained from a random sample of different crime victims. Use a 0.01 significance level to test the claim that the type of crime is independent of whether the criminal was a stranger.
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To solve this problem, we need to perform a **Chi-Square Test for Independence**. This statistical test helps determine whether there is a significant association between two categorical variables—in this case, the type of crime and whether the criminal was a stranger.

### Steps to Solve:

1. **State the Hypotheses:**
   - **Null Hypothesis (H₀):** The type of crime is independent of whether the criminal was a stranger.
   - **Alternative Hypothesis (H₁):** The type of crime is not independent of whether the criminal was a stranger.

2. **Prepare the Data:**
   - If the table was provided, the data from the survey would be organized into a contingency table, showing the frequency of crimes where the criminal was a stranger or not, for each type of crime.

3. **Calculate the Expected Frequencies:**
   - For each cell in the contingency table, calculate the expected frequency using the formula:
     \[
     E = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}}
     \]

4. **Compute the Chi-Square Statistic:**
   - Use the formula:
     \[
     \chi^2 = \sum \frac{(O - E)^2}{E}
     \]
     Where:
     - \( O \) is the observed frequency.
     - \( E \) is the expected frequency.

5. **Determine the Degrees of Freedom (df):**
   - The degrees of freedom for a Chi-Square test in a contingency table is calculated by:
     \[
     df = (r - 1) \times (c - 1)
     \]
     Where \( r \) is the number of rows, and \( c \) is the number of columns.

6. **Find the Critical Value:**
   - Using the 0.01 significance level and the appropriate degrees of freedom, locate the critical value from the Chi-Square distribution table.

7. **Compare and Make a Decision:**
   - Compare the computed Chi-Square statistic with the critical value.
   - If the Chi-Square statistic is greater than the critical value, reject the null hypothesis.
   - Otherwise, do not reject the null hypothesis.

8. **Conclusion:**
   - Based on the comparison, conclude whether the type of crime is independent of whether the criminal was a stranger.

If you have the data or table available, I can assist you with the calculations.

Do you want more details on any of these steps? Also, here are five related questions you might consider next:

1. How do you calculate expected frequencies in a contingency table?
2. What are the assumptions of the Chi-Square test for independence?
3. How does the choice of significance level (e.g., 0.01) affect the results of a hypothesis test?
4. What is the difference between a Chi-Square test for independence and a Chi-Square test for goodness-of-fit?
5. How do you interpret a p-value in the context of a Chi-Square test?

**Tip:** Always ensure that the expected frequencies in a Chi-Square test are sufficiently large (typically at least 5) to meet the test's validity assumptions.

Explore how to test the claim of independence between crime types and whether the criminal was a stranger using Chi-Square analysis at a 0.01 significance level. Learn the steps and formulas required with a detailed example.

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