Chi-squared goodness of fit testsEdexcel International A Level Further Maths: Revision notes

A goodness of fit test asks whether observed data could reasonably have come from a stated model, such as a discrete or continuous uniform, binomial, Poisson or Normal distribution. $H_0$: the model fits the data (for example, 'the number of breakdowns follows a Poisson distribution'). $H_1$: the model does not fit the data. The test is always one-tailed: only a large value of the test statistic counts as evidence against the model. Write the hypotheses about the whole distribution, not about a single parameter.

For each cell, compare the observed frequency $O_i$ with the expected frequency $E_i$ given by the model: $X^2=\sum\frac{(O_i-E_i)^2}{E_i}.$ If $H_0$ is true, $X^2$ is approximately $\chi^2$ distributed. A small value means the data are close to the model; a large value means they are far from it. Under the model $E_i=n\times P(\text{cell } i)$, and the expected frequencies must add up to the total frequency $n$. Yates' correction is not required.

Discrete uniform (a fair die, $n$ throws): every cell has $E_i=\frac{n}{6}$. For example, 120 throws gives $E=20$ per face. Continuous uniform on $[a,b]$: a class of width $w$ has probability $\frac{w}{b-a}$. Waiting times uniform on 0 to 20 minutes in four classes of width 5 give $P=0.25$ each. Binomial $B(n,p)$: $E_i=N\times P(X=i)$. If $p$ is not given, estimate it as $\hat p=\frac{\text{sample mean}}{n}$. Poisson $Po(\lambda)$: $E_i=N\times P(X=i)$, and estimate $\lambda$ by the sample mean. Make the last cell '$r$ or more', with $E=N-\sum$ of the others. Normal $N(\mu,\sigma^2)$: $E_i=N\times P(\text{class})$, found by standardising with $Z=\frac{X-\mu}{\sigma}$. The two end classes extend to $\pm\infty$.

The $\chi^2$ approximation is only reliable if every expected frequency is at least 5. If any $E_i<5$, combine it with a neighbouring cell, adding both the expected and the observed frequencies, and repeat until every $E_i\ge5$. For a Poisson model the small cells are usually in the tail, so combine into '$r$ or more'. Check the combined value: with $E=3.45$ and $1.17$ the total $4.62$ is still below 5, so a third cell must be added. The number of cells $k$ used in the degrees of freedom is the number after combining.

$\nu=k-1-m,$ where $k$ is the number of cells after combining and $m$ is the number of parameters estimated from the data. One degree of freedom is lost because the total of the expected frequencies is fixed. No parameters estimated (a die, a uniform distribution with given limits, a fully specified binomial, Poisson or Normal): $\nu=k-1$. Binomial with $p$ estimated, or Poisson with $\lambda$ estimated: $\nu=k-2$. Normal with $\mu$ and $\sigma^2$ both estimated: $\nu=k-3$.

1. State $H_0$ and $H_1$.
2. Find the expected frequencies and combine cells where $E_i<5$.
3. Calculate $X^2=\sum\frac{(O_i-E_i)^2}{E_i}$.
4. State $\nu$ and the critical value of $\chi^2_\nu$ at the stated significance level, from the tables.
5. Reject $H_0$ if $X^2$ is greater than the critical value; otherwise there is insufficient evidence to reject it. Conclude in context. Example: a die gives frequencies 14, 25, 17, 22, 24, 18 in 120 throws. $E=20$, $X^2=\frac{94}{20}=4.7$, $\nu=5$ and the 5% critical value is $11.070$. Since $4.7<11.07$, there is no evidence that the die is unfair. The same statistic $\sum\frac{(O_i-E_i)^2}{E_i}$ is used to test contingency tables, covered later in this unit.