I have several measurements from the same population spanning over the course of several years, each year with its own mean and standard error (based on the same replicates at same location each time), and I want to calculate the grand mean and grand standard error. The grand mean seems fairly straight-forward (average the means?) but I am not sure if I can calculate the grand SE averaging annual SE values. For instance:
means <- (35,23,38,40)
se <- (2,4,7,6)
grand_mean <- mean(means)
grand_se <- mean(se) ?
Thanks
Best Answer
Suppose we have observations taken across $m$ different years, $\{ X_{1j} \}_{j=1}^{n_1}, \{ X_{2j} \}_{j=1}^{n_2}, \ldots , \{ X_{mj} \}_{j=1}^{n_m}$ where $\text{Var}(X_{i1}) = \sigma^2_i$ for $1 \leq i \leq m$ and everything is assumed independent. Then we can directly calculate the variance of the overall mean,
$$ \begin{align} \text{Var} \left ( \frac{\sum_{i=1}^{m} \sum_{j=1}^{n_i} X_{ij}}{ \sum_{i=1}^{m} n_i} \right ) &= \frac{ \sum_{i=1}^{m} \sum_{j=1}^{n_i} \sigma^2_i }{ \left ( \sum_{i=1}^{m} n_i \right )^2 } \\ &= \frac{ \sum_{i=1}^{m} n_i \sigma^2_i }{\left ( \sum_{i=1}^{m} n_i \right )^2} \end{align} $$
You could then estimate this quantity by replacing each $\sigma^2_i$ with $\hat{\sigma}^2_i \equiv \sum_{j=1}^{n_i} (x_{ij} - \bar{x}_i)^2 / (n_i - 1)$, the sample variance within year $i$. However, it's not clear what you'd be estimating with the overall average if the means also vary with time.