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McNemar’s test for paired proportions

This material is reproduced based on the examples from: Oxford handbook of medical statistics (Janet Peacock Philip J Peacock) [1].

What is McNemar’s Test?

  • A statistical method to assess the association between two paired proportions
  • Applicable for matched case-control studies or ‘before and after’ studies
  • Based on the chi-squared distribution with 1 degree of freedom
  • Generates a P value, estimates, and a confidence interval

See Bland, Chapter 13 [2].

The Null Hypothesis

  • The prevalence in the population remains consistent under both conditions

How Does the Test Work?

  • The test focuses on discordant pairs (yes/no, no/yes) where exposure differs
  • Concordant pairs (yes/yes, no/no) are disregarded since they don’t provide information on differences within pairs
  • Expected frequencies are calculated assuming no association (null hypothesis is true), meaning the frequencies are equal in both discordant pairs (yes/no, no/yes)
  • Observed frequencies are compared to expected values
  • If observed frequencies deviate significantly from expected values, this implies a real association
  • The test employs a formula based on the chi-squared distribution to calculate a P value

Test Assumptions

  • Requires a large sample
  • For the test to be valid, each expected frequency should be greater than 5

What if Assumptions Aren’t Met?

  • The P value might be too small, resulting in potentially false significant outcomes
  • If numbers are small but the rule of thumb is satisfied, apply the version of the test with a continuity correction (see Bland, Chapter 13) [2]

Performing McNemar’s Test

  • Always use frequencies, not percentages, for calculations
  • The test is typically carried out using a computer program – the calculations following Table 1 demonstrate how the test operates.

Table 1 a,b: Matched Case-Control Study of Asthma Death and Short-Acting B2 Agonist Use [3] (Two Presentations)

(a)

Died (case) Survived (control) No. of pairs Notation
No No 411 a
Yes No 69 b
No Yes 45 c
Yes Yes 7 d

(b) Results arranged as a 2x2 table

Died (case) Used β2 agonist Yes No Total
Survived Yes 411 45 456
(control) No 69 7 76
Total   480 52 532
  • Expected frequency = (b+c)/2 = (69+45)/2 = 57
  • Test statistic is: \(\frac{\sum_{\text{discordant cells}}(O - E)^2}{E} = \frac{(69 - 57)^2 + (45 - 57)^2}{57} = 5.05\)

This follows a chi-squared distribution with 1 degree of freedom and has P=0.031, indicating a relationship between the use of short-acting $\beta$2 agonist and death from asthma.

References

  1. Oxford handbook of medical statistics (Janet Peacock Philip J Peacock).
  2. Bland M. An introduction to medical statistics. 3rd ed. Oxford: Oxford University Press, 2000.
  3. Anderson HR, Ayres JG, Sturdy PM, Bland JM, Butland BK, Peckitt C et al. Bronchodilator treatment and deaths from asthma: case-control study. BMJ 2005; 330(7483):117.