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The image provided shows a contingency table with observed values for two categorical variables, followed by a hypothesis test to check for independence between the row and column variables. The hypothesis is tested at a significance level of 0.05.

Here’s how we proceed:

### Step 1: Hypotheses
- \( H_0 \) (Null Hypothesis): Row and column variables are independent.
- \( H_1 \) (Alternative Hypothesis): Row and column variables are not independent.

### Step 2: Observed Table
The contingency table is presented as follows:

\[
\begin{array}{|c|c|c|c|}
\hline
  & C_1 & C_2 & \text{Total} \\
\hline
R_1 & 15 & 7 & 22 \\
R_2 & 37 & 43 & 80 \\
\hline
\text{Total} & 52 & 50 & 102 \\
\hline
\end{array}
\]

### Step 3: Expected Frequencies
We calculate the expected values for each cell using the formula:

\[
E_{ij} = \frac{( \text{Row Total} ) ( \text{Column Total} )}{ \text{Grand Total} }
\]

For example, for cell \( (R_1, C_1) \):
\[
E_{11} = \frac{22 \times 52}{102} \approx 11.22
\]

Similarly, we calculate for other cells:
\[
E_{12} = \frac{22 \times 50}{102} \approx 10.78
\]
\[
E_{21} = \frac{80 \times 52}{102} \approx 40.78
\]
\[
E_{22} = \frac{80 \times 50}{102} \approx 39.22
\]

### Step 4: Chi-Square Test Statistic
The chi-square test statistic is calculated using the formula:

\[
\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
\]

Where \( O_{ij} \) are the observed frequencies and \( E_{ij} \) are the expected frequencies. Using the observed and expected values:

For \( (R_1, C_1) \):
\[
\frac{(15 - 11.22)^2}{11.22} \approx 1.29
\]
For \( (R_1, C_2) \):
\[
\frac{(7 - 10.78)^2}{10.78} \approx 1.32
\]
For \( (R_2, C_1) \):
\[
\frac{(37 - 40.78)^2}{40.78} \approx 0.35
\]
For \( (R_2, C_2) \):
\[
\frac{(43 - 39.22)^2}{39.22} \approx 0.36
\]

Summing these gives:

\[
\chi^2 = 1.29 + 1.32 + 0.35 + 0.36 = 3.32
\]

### Step 5: Degrees of Freedom
The degrees of freedom \( df \) are calculated as:

\[
df = ( \text{Number of Rows} - 1) \times (\text{Number of Columns} - 1) = (2-1) \times (2-1) = 1
\]

### Step 6: Critical Value and Conclusion
At a significance level \( \alpha = 0.05 \), with 1 degree of freedom, the critical value from the chi-square distribution table is approximately \( 3.841 \).

Since the test statistic \( \chi^2 = 3.32 \) is less than the critical value \( 3.841 \), we **fail to reject the null hypothesis**. This suggests that there is not enough evidence to conclude that the row and column variables are dependent.

### Final Answer:
We fail to reject the null hypothesis, so the row and column variables appear to be independent at the 0.05 significance level.

---

Do you want further clarification on any of the steps, or need additional details?

Here are 5 related questions for deeper understanding:
1. How are the expected frequencies calculated in a contingency table?
2. What does it mean if we reject or fail to reject the null hypothesis in a chi-square test?
3. How do we choose the significance level for hypothesis testing?
4. How can the degrees of freedom impact the critical value in a chi-square test?
5. What are other types of tests used for categorical data analysis?

**Tip**: Always ensure that the sample size is sufficiently large when applying the chi-square test, as small sample sizes can lead to inaccurate results.

Determine whether the row and column variables are independent using the Chi-square test at a significance level of 0.05. The contingency table provided has the following observed values: R1-C1: 15, R1-C2: 7, R2-C1: 37, R2-C2: 43, with totals 52 (C1), 50 (C2), and 102 (Total).

This problem involves performing a Chi-square test for independence on a 2x2 contingency table with observed values. The solution involves calculating expected frequencies, test statistic, and comparing with the critical value at a 0.05 significance level. Suitable for college-level statistics or econometrics students.

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