https://file.aiquickdraw.com/mathbot/file/1728376857636c597be82.jpg

To solve this hypothesis testing problem based on the data provided, we will walk through each step carefully.

### a) **Determine the null and alternative hypotheses.**

- **Null Hypothesis (\( H_0 \))**: Smoking habits and seat belt usage are independent.
- **Alternative Hypothesis (\( H_a \))**: Smoking habits and seat belt usage are dependent.

Thus, the second option in the image is correct.

---

### b) **Determine the test statistic (\( \chi^2 \))**.

This involves calculating the Chi-Square statistic for the given contingency table. Here's a step-by-step guide:

1. **Observed Frequencies (O)**: From the table,
   - \( O_{11} = 67 \), \( O_{12} = 448 \)
   - \( O_{21} = 327 \), \( O_{22} = 2187 \)

2. **Expected Frequencies (E)**: 
   Expected frequency for each cell is calculated as:
   \[
   E_{ij} = \frac{(Row \ total)(Column \ total)}{Grand \ total}
   \]

   - Grand total = \( 67 + 448 + 327 + 2187 = 3029 \)
   - Row totals: 
     - Smoke = \( 67 + 448 = 515 \)
     - Do not smoke = \( 327 + 2187 = 2514 \)
   - Column totals: 
     - No seat belt = \( 67 + 327 = 394 \)
     - Seat belt = \( 448 + 2187 = 2635 \)

   Now we calculate expected frequencies:

   - \( E_{11} = \frac{515 \times 394}{3029} \approx 67.01 \)
   - \( E_{12} = \frac{515 \times 2635}{3029} \approx 447.99 \)
   - \( E_{21} = \frac{2514 \times 394}{3029} \approx 326.99 \)
   - \( E_{22} = \frac{2514 \times 2635}{3029} \approx 2187.01 \)

3. **Calculate \( \chi^2 \) statistic**:
   \[
   \chi^2 = \sum \frac{(O - E)^2}{E}
   \]
   Applying this formula:

   \[
   \chi^2 = \frac{(67 - 67.01)^2}{67.01} + \frac{(448 - 447.99)^2}{447.99} + \frac{(327 - 326.99)^2}{326.99} + \frac{(2187 - 2187.01)^2}{2187.01}
   \]

   \[
   \chi^2 \approx 0.0001 + 0.000002 + 0.000003 + 0.0000004
   \]

   \[
   \chi^2 \approx 0.0001
   \]

   **Rounded to two decimal places:**
   \[
   \chi^2 = 0.00
   \]

---

### c) **Determine the p-value**.

Using a Chi-Square distribution table for degrees of freedom:

- Degrees of freedom (\( df \)) = \( (2 - 1)(2 - 1) = 1 \)
- With \( \chi^2 \approx 0.00 \), the **p-value** is approximately **1.0000** (very high).

---

### d) **Make a decision**.

- Since the p-value is much larger than the significance level \( \alpha = 0.05 \), we **fail to reject the null hypothesis**.

---

### e) **Make a conclusion**.

- There is **not sufficient evidence** to support the claim that smoking habits and seat belt usage are related.

---

Do you want more details or have any questions? Here are 5 related questions for further exploration:

1. How would the test statistic change if one of the observed frequencies was slightly different?
2. What does it mean when two variables are independent in a Chi-Square test?
3. Why is the degrees of freedom calculated as \( (r-1)(c-1) \)?
4. How does increasing the sample size affect the Chi-Square statistic and p-value?
5. Could we use a different statistical test for this data, and if so, why?

**Tip**: Always ensure that your expected frequencies are greater than 5 in each cell for the Chi-Square test to be valid.

A researcher wants to determine whether smoking habits and seat belt usage are related. The data provided in a contingency table includes the following: Smoke (No Seat Belt = 67, Seat Belt = 448), Do not Smoke (No Seat Belt = 327, Seat Belt = 2,187). The significance level is α = 0.05.

Test whether smoking habits and seat belt usage are independent using a Chi-Square Test of Independence. The data includes 67 smokers without seat belts, 448 smokers with seat belts, 327 non-smokers without seat belts, and 2,187 non-smokers with seat belts, with a significance level of 0.05.

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