S3.4 - Goodness of fit and contingency tables

Learning objectives

S3.4.1null and alternative hypothesesThe null and alternative hypotheses.; Applications to include the discrete uniform, binomial, n (O − E)2 Normal, Poisson and continuous uniform (rectangular) The use of ∑ i i as an distributions.; Lengthy calculations will not be required.; E i=1 i approximate χ2 statistic.S3.4.2Degrees of freedomDegrees of freedom.; Students will be expected to determine the degrees of freedom when one or more parameters are estimated from the data.; Cells should be combined when E < 5.; Yates’ i correction is not required.

A chi-squared test compares observed frequencies $O_i$ with frequencies $E_i$ expected under a null model. Large discrepancies produce a large statistic and evidence against $H_0$ ; the alternative says the stated model or independence claim does not hold.

For goodness of fit, write $H_0$ : the named distribution is a suitable model, and $H_1$ : it is not suitable. Required models include discrete uniform, binomial, Normal, Poisson and continuous uniform (rectangular). For a contingency table, write $H_0$ : the two named variables are independent (no association), and $H_1$ : they are associated.

\chi^2=\sum_i\frac{(O_i-E_i)^2}{E_i}

For goodness of fit, calculate $E_i=Np_i$ from the null distribution. For an $r\times c$ contingency table, $E_{ij}=(\text{row total})(\text{column total})/N$ . Compare the statistic with the upper-tail critical value for the chosen significance level and degrees of freedom.

For 50 observations in five equally likely categories, suppose $O=(8,12,9,11,10)$ and $E=(10,10,10,10,10)$ . Then $\chi^2=(4+4+1+1+0)/10=1.0$ . With 4 degrees of freedom, $1.0<9.488$ at 5%, so do not reject the discrete-uniform model.

A small statistic does not prove the model or independence; it means the data do not provide sufficient evidence against $H_0$ . Use frequencies, not raw measurements or percentages, and do not describe association as correlation.

Degrees of freedom select the correct chi-squared reference distribution. Count categories after any required combining, then subtract constraints created by the total frequency and by parameters estimated from the same data.

\nu=k-1-p\quad\text{for goodness of fit}

Here $k$ is the final number of cells and $p$ is the number of distribution parameters estimated from the sample. If every parameter is specified in advance, $p=0$ . For example, six final Poisson cells with the mean estimated from the data give $\nu=6-1-1=4$ . If combining reduces the table to five cells, use $\nu=3$ .

\nu=(r-1)(c-1)\quad\text{for an }r\times c\text{ contingency table}

The chi-squared approximation requires adequate expected frequencies. When $E_i<5$ , combine suitable neighbouring or tail categories before calculating the final statistic; recompute each combined $O$ and $E$ , and base $k$ and the degrees of freedom on the combined table.

A $3\times4$ contingency table has $\nu=(3-1)(4-1)=6$ . Its expected frequencies still come from row total $\times$ column total $/N$ ; the observed counts do not determine $\nu$ .

Subtract only parameters estimated from these data, not parameters supplied by the null model. Do not use the original number of cells after combining, and do not apply Yates' correction; it is not required in this specification.

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—S3.5 - Regression and correlationRead

S3.4 - Goodness of fit and contingency tables

Learning objectives

Use a chi-squared test for fit or association

Choose chi-squared degrees of freedom correctly

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