Goodness of fit tests and contingency tablesEdexcel A-Level Further Maths: Revision notes

A goodness of fit test checks whether observed frequencies are consistent with a proposed model. State the hypotheses in words: $\mathrm{H}_0$: the model fits (for example, the die is fair, or the data follow $\mathrm{B}(4,p)$); $\mathrm{H}_1$: the model does not fit. The test statistic is $X^2=\sum\frac{(O_i-E_i)^2}{E_i},$ where $O_i$ are the observed frequencies and $E_i$ the expected frequencies if $\mathrm{H}_0$ is true. $X^2$ is approximately $\chi^2_\nu$ distributed. A large value means the data are far from the model, so the test is always one-tailed (upper tail).

Find each $E_i$ as total $\times$ probability. For a discrete uniform model with $k$ outcomes, $E_i=\frac{n}{k}$. For a binomial model, $E_i=n\,\mathrm{P}(X=i)$ using $\mathrm{B}(N,p)$; for Poisson, $E_i=n\,\mathrm{P}(X=i)$ using $\mathrm{Po}(\lambda)$, with the last cell as 'r or more' so the probabilities sum to 1. The $\chi^2$ approximation is only reliable if every $E_i\geq5$. If a cell has $E_i<5$, combine it with a neighbouring cell (adding both $O$ and $E$) and count the combined cell as one. For example, expected frequencies $37.97, 41.60, 17.09, 3.12, 0.21$ become $37.97, 41.60, 20.43$.

The degrees of freedom are $\nu=\text{(number of cells after combining)}-1-(\text{number of parameters estimated from the data})$. The '$-1$' is because the total is fixed. If a parameter is given in the question (a fair die, $p=\frac12$, $\lambda=1.5$) nothing more is subtracted. If it is estimated from the sample, subtract one for each estimate: $\hat p=\frac{\bar x}{N}$ for a binomial, $\hat\lambda=\bar x$ for a Poisson. Examples: a die has $\nu=6-1=5$; $\mathrm{B}(4,p)$ with $p$ estimated and 3 cells left has $\nu=3-1-1=1$.

Discrete uniform: $\mathrm{H}_0$: each outcome is equally likely; $E_i=\frac nk$; no parameters estimated, so $\nu=k-1$. Binomial $\mathrm{B}(N,p)$: if $p$ is not given, estimate it from the sample mean, $\hat p=\frac{\bar x}{N}$, then $\nu=\text{cells}-2$. If $p$ is given, $\nu=\text{cells}-1$. Poisson $\mathrm{Po}(\lambda)$: if $\lambda$ is not given, use $\hat\lambda=\bar x$; the final cell 'r or more' has $E=n\left(1-\sum_{i<r}\mathrm{P}(X=i)\right)$. When a last cell like '4 or more' is used, you need the total of the observations in that cell to find $\bar x$.

A contingency table classifies each observation by two variables. The test asks whether they are independent: $\mathrm{H}_0$: there is no association between the variables (they are independent); $\mathrm{H}_1$: there is an association. Under $\mathrm{H}_0$, $E=\frac{\text{row total}\times\text{column total}}{\text{grand total}},$ and $\nu=(r-1)(c-1)$ for $r$ rows and $c$ columns. No parameters are estimated separately (the formula for $\nu$ already allows for the totals). Combine cells with $E<5$ by merging rows or columns, and recalculate $\nu$ from the new table. For a $2\times2$ table $\nu=1$.

Compare $X^2$ with the critical value from the $\chi^2_\nu$ table at the given significance level (for example $\nu=5$ at 5% gives $11.070$; $\nu=2$ gives $5.991$; $\nu=1$ gives $3.841$). If $X^2\geq$ critical value, reject $\mathrm{H}_0$. Alternatively, a calculator gives the p-value $\mathrm{P}(\chi^2_\nu\geq X^2)$: reject $\mathrm{H}_0$ if it is below the significance level. Worked example: a die rolled 120 times gives $E=20$ for each face, $X^2=\frac{94}{20}=4.7$, $\nu=5$, critical value $11.070$. Since $4.7<11.070$, there is no evidence that the die is unfair. Always write the conclusion in context.