How is the expected frequency of a cell calculated?

\frac{row total\timescolumn total}{grand total}

What convention applies to expected frequencies?

All expected frequencies should be greater than 5.

What do you do if an expected frequency is 5 or less?

Combine adjacent categories, then recalculate the degrees of freedom.

Is the chi-squared test for association one-tailed or two-tailed?

One-tailed (upper tail): reject H0 only for a large test statistic.

What does a large test statistic suggest?

Observed frequencies are far from those expected under independence, so there is evidence of association.

How do you find the sources of association?

Look for cells with large contributions \frac{(O-E)^2}{E} and compare O with E in context.

Contingency tables and the chi-squared testAQA A-Level Further Maths: Flashcards

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Write the null hypothesis for a chi-squared test on a contingency table.

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Write the null hypothesis for a chi-squared test on a contingency table.: There is no association between the two variables (they are independent).
How is the expected frequency of a cell calculated?: $\frac{\text{row total}\times\text{column total}}{\text{grand total}}$
State the test statistic.: $\sum\frac{(O_i-E_i)^2}{E_i}$
State the degrees of freedom for an $n\times m$ table.: $(n-1)(m-1)$
What convention applies to expected frequencies?: All expected frequencies should be greater than 5.
What do you do if an expected frequency is 5 or less?: Combine adjacent categories, then recalculate the degrees of freedom.
Is the chi-squared test for association one-tailed or two-tailed?: One-tailed (upper tail): reject $H_0$ only for a large test statistic.
What does a large test statistic suggest?: Observed frequencies are far from those expected under independence, so there is evidence of association.
How do you find the sources of association?: Look for cells with large contributions $\frac{(O-E)^2}{E}$ and compare $O$ with $E$ in context.
Critical value for $\nu=2$ at $5\%$ ?: $5.991$
Critical value for $\nu=4$ at $5\%$ ?: $9.488$
A $3\times4$ table: degrees of freedom?: $(3-1)(4-1)=6$

Exam questions on Contingency tables and the chi-squared test

A survey of 150 employees records how they travel to work and whether they live in town. Of the 40 who walk, 30 live in town. Of the 60 who take the bus, 25 live in town. Of the 50 who drive, 15 live in town. A chi-squared test for association is to be carried out.
State the null hypothesis and the number of degrees of freedom for the test.2 marks
Hospital records for 200 patients from three wards, A, B and C, give each patient's recovery outcome. Ward A has 70 patients: 48 recovered fully, 20 partially and 2 not at all. Ward B has 60 patients: 35 fully, 22 partially and 3 not at all. Ward C has 70 patients: 27 fully, 36 partially and 7 not at all.
Explain why two of the outcome categories must be combined before the test is carried out, and state the degrees of freedom after doing so.2 marks
A cafe owner records the drink chosen by 120 customers: 60 in the morning and 60 in the afternoon. In the morning, 18 chose tea, 30 chose coffee and 12 chose juice. In the afternoon, 24 chose tea, 14 chose coffee and 22 chose juice.
Find the expected frequency for each cell, assuming the drink is independent of the time of day, and calculate the value of the test statistic $\sum\frac{(O-E)^2}{E}$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).