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# McNemar’s test for paired proportions

## What is McNemar’s Test?

## The Null Hypothesis

## How Does the Test Work?

## Test Assumptions

## What if Assumptions Aren’t Met?

## Performing McNemar’s Test

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

## References

- What is McNemar’s Test?
- The Null Hypothesis
- How Does the Test Work?
- Test Assumptions
- What if Assumptions Aren’t Met?
- Performing McNemar’s Test
- References

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

- 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 prevalence in the population remains consistent under both conditions

- 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

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

- 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]

- 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.

(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.

- Oxford handbook of medical statistics (Janet Peacock Philip J Peacock).
- Bland M. An introduction to medical statistics. 3rd ed. Oxford: Oxford University Press, 2000.
- 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.