Answer: \(\boxed{\chi^2=11.7}\), so there is evidence of an association between method of voting and gender at the \(1\%\) significance level.

We test

\[H_0:\text{ method of voting and gender are independent},\]

against

\[H_1:\text{ there is an association between method of voting and gender}.\]

The row totals are \(60\) and \(40\). The column totals are

\[15,\quad 33,\quad 52,\]

and the total number of members is \(100\).

The expected values are found using

\[E=\frac{(\text{row total})(\text{column total})}{\text{grand total}}.\]

For males, the expected values are

\[\frac{60(15)}{100}=9,\qquad \frac{60(33)}{100}=19.8,\qquad \frac{60(52)}{100}=31.2.\]

For females, the expected values are

\[\frac{40(15)}{100}=6,\qquad \frac{40(33)}{100}=13.2,\qquad \frac{40(52)}{100}=20.8.\]

Now calculate

\[\chi^2=\sum\frac{(O-E)^2}{E}.\]

So

\[\chi^2=\frac{(10-9)^2}{9}+\frac{(12-19.8)^2}{19.8}+\frac{(38-31.2)^2}{31.2}+\frac{(5-6)^2}{6}+\frac{(21-13.2)^2}{13.2}+\frac{(14-20.8)^2}{20.8}.\]

This gives

\[\chi^2=0.111+3.073+1.482+0.167+4.610+2.223=11.7\quad\text{to 3 significant figures}.\]

The degrees of freedom are

\[(2-1)(3-1)=2.\]

At the \(1\%\) significance level, the critical value is

\[\chi^2_{2,0.99}=9.21.\]

Since

\[11.7>9.21,\]

we reject \(H_0\).

Therefore there is evidence at the \(1\%\) level of an association between method of voting and gender.