The **mcnemar test calculator** is designed for analyzing **paired nominal data**, particularly in contexts such as **before-and-after studies** or **matched-pair designs**.

By comparing the frequencies of

discordant pairs, theMcNemar testoffers insights into whether observed changes are statistically significant or merely due to chance.

## McNemar Test Examples Chart

Study Type | Before (No) | Before (Yes) | After (No) | After (Yes) | McNemar’s Chi-square | p-value |
---|---|---|---|---|---|---|

Smoking Cessation | 65 | 35 | 80 | 20 | 15.0 | 0.0001 |

Diet Adherence | 40 | 60 | 55 | 45 | 7.5 | 0.0062 |

Medication Response | 70 | 30 | 75 | 25 | 1.0 | 0.3173 |

Exercise Habits | 55 | 45 | 30 | 70 | 25.0 | <0.0001 |

Sleep Quality | 50 | 50 | 60 | 40 | 2.5 | 0.1138 |

## McNemar Test Formula

The **McNemar Test** formula is:

`chi^2 = (b - c)^2 / (b + c)`

Where:

bis the number of subjects who changed from negative to positive.cis the number of subjects who changed from positive to negative.

In a study involving 100 participants testing a new medication:

- 20 improved (changed from symptomatic to asymptomatic).
- 10 worsened (changed from asymptomatic to symptomatic).

Calculating the test statistic:

`chi^2 = (20 - 10)^2 / (20 + 10) = 100 / 30 = 3.33`

**With 1 degree of freedom, if this chi^2 value exceeds the critical value (3.84 for alpha = 0.05), we reject the null hypothesis, concluding that there is a significant difference.**

## How to Calculate the McNemar Test

To perform the McNemar test:

- Create a
**2×2 contingency table**of before and after results. - Identify the
**discordant pairs**(b and c). - Apply the formula: chi^2 = (b – c)^2 / (b + c).
- Compare the result to the chi-square distribution with 1 degree of freedom.

A study evaluates the effectiveness of a new teaching method.

Before:

- 60 students passed.
- 40 students failed.

After:

- 75 students passed.
- 25 students failed.

Before\After | Pass | Fail |
---|---|---|

Pass | 55 | 5 |

Fail | 20 | 20 |

Here, b (failed before, passed after) = 20 and c (passed before, failed after) = 5.

Calculating the test statistic:

`chi^2 = (20 - 5)^2 / (20 + 5) = 225 / 25 = 9`

**With chi^2 = 9 and 1 degree of freedom, the p-value is approximately 0.0027, which is less than the typical significance level of 0.05. This indicates that the new teaching method had a statistically significant effect on student performance.**