Chi-Square Tests of Independence: Expected Counts and Inference

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Chi-square statistic for independence

The sum across all cells of each squared observed-minus-expected difference divided by its expected count: $\chi^2=\sum \frac{(O-E)^2}{E}$.

Degrees of freedom for an independence test

For a table with $r$ rows and $c$ columns, $df=(r-1)(c-1)$.

Why is the independence test right-tailed?

Large differences between observed and expected counts produce large chi-square statistics, providing evidence against independence.

Decision rule using a p-value

Reject the null hypothesis when the p-value is at most the significance level $\alpha$; otherwise, fail to reject it.

What does a test of independence establish about causation?

Evidence of association alone does not establish that one factor causes the other.

Quiz questions

1. Which expression gives the chi-square statistic for a test of independence?Answer: $\sum \frac{(O-E)^2}{E}$ across all cells
2. A contingency table has 4 rows and 3 columns. How many degrees of freedom does its chi-square test of independence have?Answer: 6
3. Why does a chi-square test of independence use a right-tailed rejection region?Answer: Large observed–expected discrepancies produce large chi-square values.
4. For a test of independence with significance level $\alpha=0.05$, the p-value is 0.03. What decision should be made?Answer: Reject the null hypothesis of independence.
5. A test of independence finds evidence that two factors are associated. What conclusion about causation is justified by that result alone?Answer: The result alone does not establish that either factor causes the other.