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This problem involves testing whether different methods of quitting smoking (patch, oral medication, and no intervention) are equally effective. We will use a Chi-Square test for independence.

Let's break down each part of the question:

### a. Determine the null and alternative hypotheses:

The correct hypotheses would be:
- \( H_0 \): The methods for helping people quit smoking are the same.
- \( H_a \): The methods for helping people quit smoking follow a different distribution.

So, the second set of hypotheses is correct.

### b. Determine the test statistic:

We need to calculate the Chi-Square test statistic (\( \chi^2 \)).

1. **Observed values (O):**
   \[
   \begin{array}{|c|c|c|c|}
   \hline
   & \text{Group 1 (Patch)} & \text{Group 2 (Oral Medication)} & \text{Group 3 (No Intervention)} \\
   \hline
   \text{Did Not Quit} & 52 & 54 & 51 \\
   \hline
   \text{Quit} & 13 & 9 & 4 \\
   \hline
   \end{array}
   \]

2. **Total values**:
   \[
   \text{Total} = 183, \text{where } N_{quit} = 13+9+4 = 26, N_{not \, quit} = 52+54+51 = 157
   \]

3. **Expected values (E)** for each cell:
   \[
   E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}
   \]

After computing these, the test statistic \( \chi^2 \) can be calculated using the formula:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]

Let me calculate this for you.The Chi-Square test statistic \( \chi^2 \) is approximately \( 3.96 \).

### c. Determine the p-value:

Next, we need to compute the p-value. Since this is a Chi-Square test, the degrees of freedom (df) can be calculated as:
\[
df = (r - 1) \times (c - 1)
\]
where \( r \) is the number of rows (2: "Did Not Quit" and "Quit") and \( c \) is the number of columns (3: Groups 1, 2, 3).

Let me calculate the p-value for you.The p-value is approximately \( 0.1381 \).

### d. Make a decision:

Since the p-value \( 0.1381 \) is greater 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 the methods for helping people quit smoking are different.

---

Do you need more details or have any questions? Here are some related questions to explore:

1. What is the impact of using a different significance level?
2. How would the results change if there were more participants in each group?
3. Can this test be used for other types of interventions, like different types of therapies?
4. What assumptions are made in the Chi-Square test?
5. How can effect size be measured in this context?

**Tip:** The Chi-Square test is most accurate when all expected values are greater than 5.

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: Group 1 - 52 did not quit, 13 quit; Group 2 - 54 did not quit, 9 quit; Group 3 - 51 did not quit, 4 quit. Test the researcher’s claim using a significance level of α = 0.05.

Explore how to test the effectiveness of different smoking cessation methods (patch, oral medication, no intervention) using the Chi-Square Test. Includes a full solution for calculating the test statistic, p-value, and making a decision based on a significance level of α = 0.05.

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