# Degree of freedom in t test involving weighted mean

Suppose I have repeated measurements of the same sample done by different people. Each person has different number of repeats. Here are the measurement data:

Person #1: 2, 3, 2

Person #2: 3, 3, 6

Person #3: 2, 5, 6, 4

Person #4: 2, 3, 4, 5

Person #5: 3, 2

I need to do a t test for each person to evaluate how consistent his results are compared to all the other people's results. (I'm not interested in knowing if "person" is a significant factor in the ANOVA setting.) I kind of have to do it this way, but if you have other better alternatives to evaluate the consistency for each one of them against the rest of the group, I would love to hear them.

The t test is:
$t_i=\displaystyle\frac{\bar{x}_i-\bar{\bar{x}}_{-i}}{\sqrt{s_i^2/n_i+(SE\ of\ \bar{\bar{x}}_{-i})^2}}$
where $\bar{x}_i$ is the mean of the $i$-th person's data with the sample size being $n_i$, and $\bar{\bar{x}}_{-i}$ is the weighted mean of the other persons' data $\Sigma_{j\neq i}n_j\bar{x}_j/\Sigma_{j\neq i}n_j$.

This post shows me how to get the standard error of the weighted mean. But it leaves me clueless about how to get the degree of freedom for this t test because it seems to me the Welch-Satterthwaite effective degree of freedom can no longer be directly used.

• Because you have correlated values within person, this is probably not the place for a weighted mean. You might envision this as a random effects model, with random per-subject intercepts. – Frank Harrell Jun 13 '17 at 22:19
• @FrankHarrell I don't understand why the values are correlated within person. Each person tests the same sample repeatedly and in blind fashion multiple times. The data on my post are not real data, I just made up some numbers. – hooyeh Jun 14 '17 at 2:31