It is fairly straightforward to calculate the variance of the paired t-test:


But how can you calculate this variance if you have at each of the two measurements a number of measurements per patient (to exclude measurement error).

A real life example: a patient his fat is weighted by 4 doctors at the start of the trial, and by 4 different doctors at the ten of the trial.

Thank you!


1 Answer 1


Assuming 4 measurements per patient at each of two time points, let's define

$\bar{D} = \displaystyle\sum_{j=1}^4 (\bar{Y_{2j}}-\bar{Y_{1j}})/4 $


$Var(\bar{D}) = [ \displaystyle\sum_{j=1}^4 Var(\bar{Y_{1j}}) + \displaystyle\sum_{j=1}^4 Var(\bar{Y_{2j}}) + \displaystyle\sum_{j=1}^4\displaystyle\sum_{k\neq j}Cov(\bar{Y_{1j}}, \bar{Y_{1k}}) + \displaystyle\sum_{j=1}^4\displaystyle\sum_{k\neq j}Cov(\bar{Y_{2j}}, \bar{Y_{2k}}) - \displaystyle\sum_{j=1}^4\displaystyle\sum_{k=1}^4 Cov(\bar{Y_{2j}}, \bar{Y_{1k}})]/4$

This approach is analogous to the approach you laid out for the paired sample t-test, but even the paired sample t-test is simplified to

$Var(\bar{D}) = \sigma_d^2/n$ where $\sigma_d^2 = var(Y_{i2}-Y_{i1}) = var(D_i)$

A natural estimator for $\sigma_d^2$ is $\frac{1}{n-1}\sum(D_i-\bar{D})^2$

In your case, it might be easier to make the simplification that the measurement for the $i^{th}$ person at the $k^{th}$ time is the average across the 4 doctors at each time point so that $Y_{ik} = \displaystyle\sum_{j=1}^4Y_{ijk}/4$. Then, proceed as usual with the paired sample t-test.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.