McNemar's test statistic is given by:

enter image description here

Yates's correction defines something like this:

enter image description here

However, the correction for the McNemar's test statistic requires 1 (not 0.5) be subtracted, which is not the same as other chi-square tests. Why?


1 Answer 1


The uncorrected McNemar statistic is the square of a standardized* difference in two counts; the continuity correction is $\frac12$ to both, but it's $-\frac12$ for the larger of the two and $+\frac12$ for the smaller of the two[1].

* under the null

As a result, when you take $|B-C|$ the total of the two continuity corrections is always $-1$.

(This post may be of some possible interest as well.)

Edit in response to question in comment

You need to think about what is "as, or more extreme" here.

The calculation effectively conditions on the discordant-pair total $B+C=N_{d}$ and does an approximation to the binomial test that $p_B=0.5$. Imagine you have observed say $b=4$ and $c=12$:

enter image description here

However, the continuity correction reasoning itself doesn't rely on the binomial -- it only relies on the fact that we want to look at the cases that are at least as extreme as $(4,12)$, which are $b=4,3,2,1,0$ plus the "other tail" values $12,13,14,15,16$. Then the usual continuity correction reasoning would say you want the approximating normal from 4.5 down and from 11.5 up.

So more generaly we're always taking both values toward the middle, $(b+c)/2$ by 0.5, which means the larger values reduce by 0.5 and the smaller values increase by 0.5. This reasoning follows through to the squared form used in the chi-squared version.

[1] Edwards, A.L. (1948),
"Note on the correction for continuity in testing the significance of the difference between correlated proportions."
Psychometrika, 13:3 (Sept), pp185-187.

  • $\begingroup$ But why is it $+0.5$ for the smaller of the two? Isn't Yates' idea based on that symmetric binomial rv are better approximated by normal when shifted back by 0.5, so shouldn't the corrections cancels out? Edwards' paper doesn't seem to explained this. Thanks. $\endgroup$
    – Francis
    Commented Jun 17, 2016 at 4:16
  • 1
    $\begingroup$ @Francis see my update which should motivate it. $\endgroup$
    – Glen_b
    Commented Jun 17, 2016 at 4:54

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