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4.3 Chi-squared tests

Syllabus
9231–2028–2029
Topic
4.3
Level
AS

Learning objectives

4.3.1Fitted distributions• fit a theoretical distribution, as prescribed by a given hypothesis, to given data Questions will not involve lengthy calculations.4.3.2Goodness-of-fit test• use a χ2-test, with the appropriate number of degrees of freedom, to carry out the corresponding goodness of fit analysis Classes should be combined so that each expected frequency is at least 5.4.3.3Independence test• use a χ2-test, with the appropriate number of degrees of freedom, for independence in a contingency table. Yates' correction is not required. Where appropriate, either rows or columns should be combined so that the expected frequency in each cell is at least 5.

Fit a prescribed distribution by turning its probabilities into expected counts

Under the stated hypothesis, identify the theoretical distribution and estimate only the parameters the question leaves unknown. Calculate each class probability from that fitted model, then multiply by the total frequency N: expected count E=Np. Preserve exhaustive classes, including a final tail such as X at least 4.

ForafittedPoissonmodel,estimateFor a fitted Poisson model, estimate\hat\lambda=\bar x=\frac{\sum xf}{N},thenthenP(X=r)=e^{-\hat\lambda}\frac{\hat\lambda^r}{r!},\qquad E_r=N P(X=r).AtailclasshasA tail class hasE_{\ge k}=N\left(1-\sum_{r=0}^{k-1}P(X=r)\right).

If $N=100$ and the fitted value is $\hat\lambda=1.2$, then the expected frequency for $X=0$ is100e^{-1.2}=30.12,and for $X=1$ it is100e^{-1.2}(1.2)=36.14.Continuewithunroundedprobabilitiesbeforecombininganysmalltailclasses.Continue with unrounded probabilities before combining any small tail classes.

Fitting does not mean copying observed proportions into expected counts. Matching an estimated mean does not establish good fit; it only supplies the model probabilities that the chi-squared analysis will assess.

Goodness of fit uses final combined classes and parameter-adjusted degrees of freedom

Set H0 to the prescribed distribution and H1 to not that distribution. Before calculating the statistic, combine adjacent classes so every final expected frequency is at least 5; combine their observed and expected counts together.

For the final $k$ classes,X^2=\sum\frac{(O-E)^2}{E}.If $p$ distribution parameters were estimated from these data,\text{df}=k-1-p.Compareintheuppertail:alargestatisticisevidenceagainstthefittedmodel.Compare in the upper tail: a large statistic is evidence against the fitted model.

With final observed counts $(20,30,50)$ and expected counts $(25,25,50)$,X^2=\frac{25}{25}+\frac{25}{25}+0=2.00.If no parameter was estimated, df $=3-1=2$; since $2.00<5.991$ at 5%, do not reject H0.

Count classes after combining and subtract every parameter estimated from the same data. A non-significant result means the data are compatible with the model at that level; it does not prove the model true.

Independence predicts each cell from its row and column margins

For an $r\times c$ contingency table, $H_0$ states that the two categorical variables are independent. The expected count in cell $(i,j)$ isE_{ij}=\frac{(\text{row }i\text{ total})(\text{column }j\text{ total})}{\text{grand total}},and df $=(r-1)(c-1)$.

ForobservedcountsFor observed counts\begin{pmatrix}30&20\10&40\end{pmatrix},rowtotalsare50,50andcolumntotals40,60,soexpectedcountsarerow totals are 50,50 and column totals 40,60, so expected counts are\begin{pmatrix}20&30\20&30\end{pmatrix}.ThusThusX^2=\frac{100}{20}+\frac{100}{30}+\frac{100}{20}+\frac{100}{30}=16.67.

Here df=1 and 16.67 exceeds the 5% critical value 3.841, so reject independence and conclude that there is evidence of an association. Every final expected cell must be at least 5; where needed, combine meaningful rows or columns before recalculating margins and df.

Use counts, not percentages, and do not apply Yates' correction in this syllabus. Association is not causation, and independence is not the same as mutually exclusive categories.