Q.Using the observed table and the expected frequencies from Worked Example 4, compute the chi-square statistic and comment on whether sex and product preference appear to be associated.

With every expected frequency equal to 25, the chi-square statistic sums $(O-E)^2/E$ over all four cells.

Each cell contributes $(\pm 5)^2 / 25 = 25/25 = 1$, over 4 cells: …

Step 1 — Recall observed $O$ and expected $E$. From Worked Examples 4:

Step 2 — Add the last column.

Step 3 — Interpret. Since $\chi^2 = 4 > 0$, the observed table departs from what independence would predict, so there is some association between sex and product preference — men leaned towards 'Liked' while women leaned towards 'Disliked'. (Whether this association is large enough to be declared statistically significant would need a formal test using degrees of freedom, which is beyond this chapter's scope.) …

For a $2\times2$ contingency table with cells $\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}$, the direct formula $\chi^2 = \dfrac{N(ad-bc)^2}{(a+b)(c+d)(a+c)(b+d)}$ gives the same value as summing $(O-E)^2/E$, and avoids comp …

Dividing by $O$ instead of $E$ in each term is a frequent error — the denominator of every term is the expected frequency. Reporting $\chi^2$ as $0$ because 'the positive and negative $O-E$ values cancel' shows the squaring …