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To test the hypothesis of association between appendectomy and diabetes, a chi-square test of independence is typically used. Based on the data provided, the observed frequencies are as follows:

|                | Appendectomy Present | Appendectomy Absent | Total |
|----------------|----------------------|---------------------|-------|
| **Diabetes**   | 8                    | 48                  | 56    |
| **No Diabetes**| 34                   | 161                 | 195   |
| **Total**      | 42                   | 209                 | 251   |

The null hypothesis (\( H_0 \)) and alternative hypothesis (\( H_1 \)) are formulated as follows:

- \( H_0 \): There is no association between appendectomy and diabetes (they are independent).
- \( H_1 \): There is an association between appendectomy and diabetes (they are not independent).

### Steps for Chi-Square Test:
1. **Calculate Expected Frequencies**:
   Expected frequency for each cell = \(\frac{{\text{Row total} \times \text{Column total}}}{\text{Grand total}}\).

2. **Chi-Square Statistic**:
   \[
   \chi^2 = \sum \frac{{(O_i - E_i)^2}}{E_i}
   \]
   where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency.

3. **Degrees of Freedom**:
   Degrees of freedom = \( (r - 1) \times (c - 1) \), where \( r \) is the number of rows and \( c \) is the number of columns. Here, \( df = (2 - 1) \times (2 - 1) = 1 \).

4. **Compare \( \chi^2 \) value** to the critical value from the chi-square distribution table at a significance level (usually \( \alpha = 0.05 \)).

I will now calculate the expected frequencies and the chi-square statistic.The expected frequencies and the chi-square statistic are as follows:

- **Expected frequencies**:
  - Diabetes with appendectomy: 9.37
  - Diabetes without appendectomy: 46.63
  - No diabetes with appendectomy: 32.63
  - No diabetes without appendectomy: 162.37

- **Chi-square statistic**: 0.31
- **Degrees of freedom**: 1

Next, we compare this chi-square statistic to the critical value from the chi-square distribution with 1 degree of freedom at a significance level (e.g., \( \alpha = 0.05 \)).

The critical value for \( \alpha = 0.05 \) and 1 degree of freedom is 3.841. Since the calculated chi-square value (0.31) is less than the critical value, we **fail to reject the null hypothesis**.

### Conclusion:
There is not enough evidence to suggest an association between appendectomy and diabetes.

Would you like further details on the steps or calculations? Here are some related questions for practice:

1. What assumptions must be satisfied for a chi-square test to be valid?
2. How would you interpret the result if the chi-square statistic had been larger than the critical value?
3. Can the chi-square test be used for small sample sizes? If not, which alternative test could be used?
4. What is the significance level (\( \alpha \)) and how does it affect the conclusion of the test?
5. How would the chi-square test differ if there were more than two categories for diabetes and appendectomy?

**Tip**: Ensure that expected frequencies are sufficiently large (usually at least 5) to apply the chi-square test accurately.

Test the hypothesis of association between appendectomy and diabetes.

Test the association between appendectomy and diabetes using a chi-square test of independence. The observed frequencies include diabetes cases with and without appendectomy. Learn how to calculate expected frequencies, the chi-square statistic, and interpret the results.

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