Well, under i.i.d. assumptions for the $x_i$ the variance of the weighted sample mean relative to the population variance would have a factor $\left(\sum_iw_i^2\right)/\left(\sum_iw_i\right)^2$, which reduces to $1/n$ if all of the weights are equal. And there would be a comparable substitution for the "$1-1/n$" factor on the unbiased sample variance as well. See here for example.
For explanation, we have $\bar{x}=\frac{1}{W}\sum_iw_ix_i$, where $W=\sum_iw_i$.
So then using $\mathbb{V}[\,]$ to denote variance, we have
\begin{align}
\mathbb{V}[\bar{x}] &= \mathbb{V}\left[\tfrac{1}{W}\sum_iw_ix_i\right] \\
&=\sum_i\mathbb{V}\left[\frac{\,w_i\,}{W}\,x_i\right] \\
&=\sum_i\left[\left(\frac{\,w_i}{W}\right)^2\,\mathbb{V}[x_i]\right] \\
&=\mathbb{V}[x]\frac{\sum_iw_i^2}{W^2} \\
\end{align}
Recall that weighted mean is
$$
\bar x = \frac{\sum_i w_i x_i} {\sum_i w_i}
$$
with $w_i \ge 0$, where arithmetic mean is just a special case with $w_i = 1/n$
$$
\bar x = \frac{\sum_i 1/n \; x_i} {\sum_i 1/n} = \frac{\sum_i 1/n \; x_i} {1} = \sum_i 1/n \; x_i = 1/n \sum_i x_i
$$
If you want to weight by inverse of standard deviations $w_i = 1/\sigma_i$
$$
\sum_i \frac{\phi_i}{\sigma_i} \bigg/ \sum_i \frac{1}{\sigma_i}
$$
If you want to correct additionally for the sample sizes, use standard error instead of standard deviation as a weight.
If you include $N$ in normalization, the result is on completely different scale (compare to using arithmetic mean) as compared to scaling by the weights alone.
> set.seed(42)
> k <- 10
> n <- 100 * k
> grp <- rep(1:k, length.out=n)
> x <- rnorm(n)
> phi <- as.vector(by(x, grp, mean))
> sigma <- as.vector(by(x, grp, sd))
> mean(x)
[1] -0.02582443
> mean(phi)
[1] -0.02582443
> sum(phi/sigma) / sum(1/sigma)
[1] -0.02440608
> sum(phi/sigma) / (n*sum(1/sigma))
[1] -2.440608e-05
Best Answer
The geometric mean and the mean whether weighted or unweighted are different parameters of a distribution. So the question of which parameter to estimate depends on which aspect of the distribution you are interested in. For a normal distribution the mean and variance are the natural parameters and it seems that it would make more sense to estimate the population mean.
Now consider a variable Y=exp(X) where X has a normal distribution. Y is said to have a lognormal distribution. Consider the sample geometric mean for a sample of size n, Y$_1$, Y$_2$,...,Y$_n$.
G$_m$ = Π (Y$_i$)$^1$$^/$$^n$ is the sample geometric mean for the geometric mean parameter of the distribution of Y. ln(G$_m$)=Σln(Y$_i$)/n. Since ln(Y)=X the log of the geometric mean is the sample mean for the corresponding normal random variables X$_i$. So for a lognormal distribution the geometric mean may be the natural parameter to estimate since the log of it is the same mean for normal random variables.