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

$Var(\overline{D})=Var(\overline{Y_1})+Var(\overline{Y_2})-2*Covar[Var(\overline{Y_1})*Var(\overline{Y_2})]$

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!

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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 $

Then

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

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.

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