Let's solve this step-by-step:

We are testing for independence between the row variable and column variable using a chi-square test of independence. To calculate the chi-square statistic, we need to:

1. Determine row and column totals.
2. Compute expected counts for each cell.
3. Calculate the Pearson residuals (observed vs expected).
4. Find the chi-square test statistic.

Step 1: Observed Table and Totals

From the table:

• Grand Total: 217

Step 2: Expected Counts

The expected count for each cell is calculated as:

$E_{ij} = \frac{\text{(Row Total)} \times \text{(Column Total)}}{\text{Grand Total}}$

We calculate expected counts for all 6 cells:

For Row A:

• $E_{A,X} = \frac{118 \times 61}{217} \approx 33.17$
• $E_{A,Y} = \frac{118 \times 116}{217} \approx 63.07$
• $E_{A,Z} = \frac{118 \times 40}{217} \approx 21.76$

For Row B:

• $E_{B,X} = \frac{99 \times 61}{217} \approx 27.83$
• $E_{B,Y} = \frac{99 \times 116}{217} \approx 52.93$
• $E_{B,Z} = \frac{99 \times 40}{217} \approx 18.22$

The expected counts table is:

Step 3: Pearson Residuals

The Pearson residual for each cell is:

$R_{ij} = \frac{O_{ij} - E_{ij}}{\sqrt{E_{ij}}}$

Using observed $O$ and expected $E$:

Row A:

• $R_{A,X} = \frac{34 - 33.17}{\sqrt{33.17}} \approx 0.14$
• $R_{A,Y} = \frac{63 - 63.07}{\sqrt{63.07}} \approx -0.01$
• $R_{A,Z} = \frac{21 - 21.76}{\sqrt{21.76}} \approx -0.52$

Row B:

• $R_{B,X} = \frac{27 - 27.83}{\sqrt{27.83}} \approx -0.16$
• $R_{B,Y} = \frac{53 - 52.93}{\sqrt{52.93}} \approx 0.01$
• $R_{B,Z} = \frac{19 - 18.22}{\sqrt{18.22}} \approx 0.58$

The Pearson residuals table is:

Step 4: Chi-Square Test Statistic

The chi-square statistic is:

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

Calculating for all cells:

Row A:

• $\frac{(34 - 33.17)^2}{33.17} \approx 0.03$
• $\frac{(63 - 63.07)^2}{63.07} \approx 0.00$
• $\frac{(21 - 21.76)^2}{21.76} \approx 0.03$

Row B:

• $\frac{(27 - 27.83)^2}{27.83} \approx 0.02$
• $\frac{(53 - 52.93)^2}{52.93} \approx 0.00$
• $\frac{(19 - 18.22)^2}{18.22} \approx 0.03$

Summing these:

$\chi^2 = 0.03 + 0.00 + 0.03 + 0.02 + 0.00 + 0.03 = 0.11$

Final Answer:

The chi-square test statistic is approximately:

$\chi^2 = 0.11$

Do you need help interpreting the result or further explanation? Let me know!

Related Questions:

1. How do you interpret a chi-square test result for independence?
2. What are the degrees of freedom for this chi-square test?
3. How would you calculate the p-value for this test statistic?
4. What assumptions must be met to use the chi-square test?
5. How can you determine if the association between variables is significant?

Tip: The chi-square test assumes expected counts in all cells are sufficiently large (typically $E > 5$). Always check this assumption!