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Let's walk through the steps to solve this chi-square test problem based on the provided table and details.

### Step 1: Hypotheses
We are performing a chi-square test of independence, which helps to determine if there's a relationship between two categorical variables: the method used and whether or not the person quit smoking.

- **Null Hypothesis (H₀):** The methods for helping people quit smoking are independent (i.e., the method used has no effect on quitting smoking).
- **Alternative Hypothesis (Hₐ):** The methods for helping people quit smoking are dependent (i.e., the method used does affect quitting smoking).

### Step 2: Chi-square test statistic
We can use the chi-square formula for this contingency table:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
Where:
- \(O\) is the observed frequency from the table.
- \(E\) is the expected frequency, calculated by:
  \[
  E = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}}
  \]
The table contains the following information:

|            | Group 1 (Patch) | Group 2 (Medication) | Group 3 (No Intervention) | Row Total |
|------------|-----------------|----------------------|---------------------------|-----------|
| **Did Not Quit** | 52              | 54                   | 51                        | 157       |
| **Quit**        | 13              | 9                    | 4                         | 26        |
| **Column Total**| 65              | 63                   | 55                        | 183       |

We calculate the expected frequencies for each cell:
For example, for Group 1 (Did Not Quit):
\[
E = \frac{(65 \times 157)}{183} \approx 55.74
\]

We calculate this for each cell, then substitute into the chi-square formula to get the test statistic.

### Step 3: p-value
Once we have the test statistic (\(\chi^2\)), we can compare it against the chi-square distribution with degrees of freedom:
\[
df = (r - 1) \times (c - 1)
\]
Where:
- \(r = 2\) (number of rows: Quit and Did Not Quit)
- \(c = 3\) (number of groups)

Thus, \(df = (2-1) \times (3-1) = 2\).

We then find the p-value corresponding to the computed \(\chi^2\) value using the chi-square distribution table or a calculator.

### Step 4: Decision
If the p-value is less than the significance level (\(\alpha = 0.05\)), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

### Step 5: Conclusion
Based on the decision, we conclude whether there is sufficient evidence to suggest that the methods for helping people quit smoking are different.

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Would you like help with the calculations for the chi-square statistic and p-value, or do you want further explanation on any of these steps?

### Follow-up questions:
1. How do you calculate the expected frequencies for each cell?
2. Can you walk through how to apply the chi-square formula for this problem?
3. What does the p-value represent in this context?
4. How do you interpret the decision to reject or not reject the null hypothesis?
5. What happens if the significance level (\(\alpha\)) is changed from 0.05?

**Tip:** Always ensure the degrees of freedom are correctly calculated in a chi-square test, as they impact your final p-value.

A researcher wanted to determine which method would help smokers quit smoking. He divided a group of 183 smokers into three groups: Group 1 is given a patch, Group 2 is given oral medication, and Group 3 has no intervention. The results of the study are displayed in a table, and the task is to test the researcher's claim using a chi-square test with a significance level of α = 0.05.

Learn how to perform a chi-square test to determine if there is a difference in the effectiveness of smoking cessation methods: patch, medication, or no intervention. We calculate the chi-square statistic, test the null hypothesis, and interpret the results using a significance level of α = 0.05.

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