When is Yates' correction applied?

For a 2\times2 contingency table, where \nu=1.

Why is Yates' correction needed?

Observed frequencies are discrete but the chi-squared distribution is continuous, so the uncorrected statistic is too large.

Does Yates' correction raise or lower the statistic?

It lowers it, making the test more cautious.

What is the modulus used for in the corrected formula?

Every difference is made positive, so 0.5 always reduces the size of the difference.

In a 2\times2 table, how do the four |O-E| values compare?

They are all equal, so calculate once.

Corrected statistic 2.90, \nu=1, 5\% level: conclusion?

2.90<3.841, so do not reject H0.

Yates' correctionAQA A-Level Further Maths: Flashcards

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What is Yates' correction?

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What is Yates' correction?: A continuity correction: subtract $0.5$ from each $|O-E|$ before squaring.
State the corrected test statistic.: $\sum\frac{(|O_i-E_i|-0.5)^2}{E_i}$
When is Yates' correction applied?: For a $2\times2$ contingency table, where $\nu=1$ .
Why is Yates' correction needed?: Observed frequencies are discrete but the chi-squared distribution is continuous, so the uncorrected statistic is too large.
Does Yates' correction raise or lower the statistic?: It lowers it, making the test more cautious.
Degrees of freedom for a $2\times2$ table?: $(2-1)(2-1)=1$
$5\%$ critical value for $\nu=1$ ?: $3.841$
$1\%$ critical value for $\nu=1$ ?: $6.635$
What is the modulus used for in the corrected formula?: Every difference is made positive, so $0.5$ always reduces the size of the difference.
In a $2\times2$ table, how do the four $|O-E|$ values compare?: They are all equal, so calculate once.
Does the rule that every $E>5$ still apply?: Yes.
Corrected statistic $2.90$ , $\nu=1$ , $5\%$ level: conclusion?: $2.90<3.841$ , so do not reject $H_0$ .

Exam questions on Yates' correction

A gardener compares germination for 60 seeds given fertiliser and 60 seeds given none. Of the seeds given fertiliser, 42 germinate. Of the seeds given none, 30 germinate. A chi-squared test for association between treatment and germination is to be carried out.
State the degrees of freedom of the test statistic and explain why Yates' correction is applied.2 marks
A survey of 50 people records which of two products, A or B, each prefers. Of the 30 men, 20 prefer A and 10 prefer B. Of the 20 women, 6 prefer A and 14 prefer B.
Calculate the value of the test statistic with Yates' correction applied.2 marks
A trial compares two treatments for a skin condition. Of 20 patients given treatment A, 14 were cured. Of 20 patients given treatment B, 8 were cured.
Calculate the value of the test statistic with Yates' correction applied.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).