The Chi-Square Family, Small Expected Counts and the Link to z

What the chi-square distribution looks like

The $\chi^2$ statistic is a sum of squares, so it is never negative and its distribution is right-skewed, starting at $0$. There is one curve for each $df$; the mean of the curve equals $df$, and as $df$ grows the curve becomes more symmetric and shifts right. Because only large values of $\chi^2$ signal disagreement with $H_0$, the P-value is always the area in the upper tail: on a calculator, $\chi^2\text{cdf}(\text{statistic},\ \infty,\ df)$. Example: $\chi^2=4{,}0$ with $df=1$ gives P $\approx0{,}046$; the same $4{,}0$ with $df=3$ gives P $\approx0{,}26$ — the statistic means nothing without its $df$.

When an expected count is below 5

The $\chi^2$ approximation breaks down when expected counts are small. If a category has expected count under $5$, the standard remedy is to combine it with a neighbouring, logically similar category (e.g. "5 or more" absences), recompute the expected counts, and reduce $df$ accordingly because there are now fewer categories. If combining is not sensible, report that the condition fails and that the P-value is unreliable — that sentence earns the point.

Chi-square and z on a 2×2 table

For a $2\times2$ table the chi-square test of homogeneity and the two-sided two-proportion $z$ test are the same test: $\chi^2=z^2$ and the P-values agree exactly. The $z=2{,}0$ from the drug-versus-placebo study becomes $\chi^2=4{,}0$ with $df=1$ and the same P-value $0{,}046$. Use $z$ when the alternative is one-sided or when you need a confidence interval for the difference; chi-square cannot do either.