Chi squared tests

FS1-6 Chi squared tests — coverage pack

1 specification leaves · notes, questions, answers and worked methods

FS1-6.1 · Goodness of fit tests and contingency tables. The null and alternative hypotheses. Degrees of freedom.

For goodness of fit, $H_0$ states that the proposed distribution fits; for a contingency table, $H_0$ states that the two classifications are independent.
Use $\chi^2=\sum\dfrac{(O-E)^2}{E}$ . For $k$ goodness-of-fit groups, $df=k-1-m$ when $m$ parameters are estimated from the data; for an $r\times c$ table, $df=(r-1)(c-1)$ .
In a contingency table, each expected frequency is $E=(\text{row total})(\text{column total})/(\text{grand total})$ .
A common error is to leave an expected frequency below $5$ ; adjacent or sensible categories should be combined before calculating the statistic and degrees of freedom.

Tier 1 · Easy

1. A goodness-of-fit test has four categories with model probabilities $0.1,0.2,0.3,0.4$ and a sample size of $120$ . No parameter is estimated. Find the expected frequencies and the degrees of freedom.[3 marks]
Answer
- Expected frequencies $12,24,36,48$
- $df=3$
Method: Multiply $120$ by each model probability to obtain $12,24,36,48$ . With $k=4$ groups and no estimated parameter, $df=k-1=3$ .

Tier 2 · Standard

1. After fitting a one-parameter model from the same data, five ordered categories have observed frequencies $41,33,17,7,2$ and expected frequencies $40,32,18,7,3$ . Carry out a goodness-of-fit test at the $5\%$ level of significance. The critical value for $2$ degrees of freedom is $5.991$ .[8 marks]
Answer
- Combine the final two categories, giving observed $41,33,17,9$ and expected $40,32,18,10$
- $H_0$ : the fitted model is suitable; $H_1$ : it is not suitable
- $\chi^2=0.2118$
- $df=2$
- Do not reject $H_0$ ; there is insufficient evidence that the model does not fit
Method: The final expected frequency is $3<5$ , so combine it with the adjacent fourth category. This gives $O=(41,33,17,9)$ and $E=(40,32,18,10)$ . Then $\chi^2=\dfrac{1^2}{40}+\dfrac{1^2}{32}+\dfrac{(-1)^2}{18}+\dfrac{(-1)^2}{10}=0.211806\ldots$ . After combining there are $k=4$ groups, and one parameter was estimated, so $df=k-1-1=2$ . Since $0.2118<5.991$ , do not reject $H_0$ .

Tier 3 · Hard

1. The observed counts in a $3\times4$ contingency table are $\begin{array}{c|rrrr}&A&B&C&D\\ \hline R&24&18&20&18\\ S&16&22&20&22\\ T&10&20&30&20\end{array}$ . Test independence at the $5\%$ level of significance. The critical value for $6$ degrees of freedom is $12.592$ .[10 marks]
Answer
- $H_0$ : row and column classifications are independent; $H_1$ : they are associated
- Expected row in each case: $(16.667,20,23.333,20)$
- $\chi^2=9.577$
- $df=6$
- Do not reject $H_0$ ; there is insufficient evidence of an association
Method: Each row total is $80$ , the column totals are $50,60,70,60$ , and the grand total is $240$ . Thus every expected row is $(80/240)(50,60,70,60)=(16.6667,20,23.3333,20)$ . Summing $(O-E)^2/E$ over all $12$ cells gives $\chi^2=9.5771429\ldots$ . The degrees of freedom are $(3-1)(4-1)=6$ . Since $9.577<12.592$ , do not reject independence.