Two schools entered their pupils for the Mathematics Olympiad examination and the numbers of pupils falling within each grade are shown in the table below. Use a X² test at 5% level of significance to determine whether proportion of pupils in the three grades are independent of the school.

• $H_0$: The proportion of pupils in the three grades is independent of the school. (There is no association between school and grade distribution).
• $H_1$: The proportion of pupils in the three grades is not independent of the school. (There is an association between school and grade distribution).

The formula for expected frequency is $E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$.

• $E_{\text{School X, Grade A}} = \frac{60 \times 38}{120} = 19$
• $E_{\text{School X, Grade B}} = \frac{60 \times 32}{120} = 16$
• $E_{\text{School X, Grade C}} = \frac{60 \times 50}{120} = 25$
• $E_{\text{School Y, Grade A}} = \frac{60 \times 38}{120} = 19$
• $E_{\text{School Y, Grade B}} = \frac{60 \times 32}{120} = 16$
• $E_{\text{School Y, Grade C}} = \frac{60 \times 50}{120} = 25$

Expected Frequencies Table:

The formula is $\chi^2 = \sum \frac{(O-E)^2}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency.

$df = (\text{number of rows} - 1) \times (\text{number of columns} - 1)$
$df = (2 - 1) \times (3 - 1) = 1 \times 2 = 2$

At a 5% level of significance ($\alpha = 0.05$) and $df = 2$, the critical value from the $\chi^2$ distribution table is 5.991.

Since the calculated $\chi^2$ value (6.7212) is greater than the critical value (5.991), we reject the null hypothesis ($H_0$).

Conclusion: There is sufficient evidence at the 5% significance level to conclude that the proportion of pupils in the three grades is not independent of the school.