It is possible to compute population size \(M_g\) from sampling frame. The standard deviation of \(g\)-th stratum is
$$S_g^2 =\frac{1}{M_g-1} \sum\limits_{k=1}^{M_g} \left(y_{gk}-\bar{Y}_g \right)^2= \frac{1}{M_g-1} \sum\limits_{k=1}^{M_g} y_{gk}^2 - \frac{M_g}{M_g-1}\bar{Y}_g^2$$
\(\sum\limits_{k=1}^{M_g} y_{gk} ^2\) and \(\bar{Y}_g^2\) have to be estimated to estimate \(S_g^2\). Estimate of \(\sum\limits_{k=1}^{M_g} y_{gk}^2\) is \(\sum\limits_{h=1}^{H} \frac{N_h}{n_h} \sum\limits_{i=1}^{n_h} y_{gi}^2 z_{hi}\), where
\( z_{hi} = \left\{
\begin{array}{ll}
0, & h_i \notin \theta_g \\
1, & h_i \in \theta_g
\end{array}
\right. \)
, \(\theta_g\) is the index group of successfully surveyed units belonging to \(g\)-th stratum. #'Estimate of \(\bar{Y}_g^2\)
is
$$\hat{\bar{Y}}_g^2=\left( \hat{\bar{Y}}_g \right)^2-\hat{Var} \left(\hat{\bar{Y}} \right)$$
$$\hat{\bar{Y}}_g =\frac{\hat{Y}_g}{M_g}= \frac{1}{M_g} \sum\limits_{h=1}^{H} \frac{N_h}{n_h} \sum\limits_{i=1}^{n_h} y_{hi} z_{hi}$$
So the estimate of \(S_g^2\) is
\(s_g^2=\frac{1}{M_g-1} \sum\limits_{h=1}^{H} \frac{N_h}{n_h} \sum\limits_{i=1}^{n_h} y_{hi}^2 z_{hi} -\)
\(-\frac{M_g}{M_g-1} \left( \left( \frac{1}{M_g} \sum\limits_{h=1}^{H} \frac{N_h}{n_h} \sum\limits_{i=1}^{n_h} y_{hi} z_{hi} \right)^2 - \frac{1}{M_g^2} \sum\limits_{h=1}^{H} N_h^2 \left(\frac{1}{n_h} - \frac{1}{N_h}\right) \frac{1}{n_h-1} \sum\limits_{i=1}^{n_h} \left(y_{hi} z_{hi} - \frac{1}{n_h} \sum\limits_{t=1}^{n_h} y_{ht} z_{ht} \right)^2 \right)\)
Two conditions have to realize to estimate \(S_g^2: n_h>1, \forall g\) and \(\theta_g \ne 0, \forall g.\)
Variance of \(\hat{Y}\) is
$$ Var\left( \hat{Y} \right) = \sum\limits_{g=1}^{G} M_g^2 \left( \frac{1}{m_g} - \frac{1}{M_g} \right) S_g^2 $$
Estimate of \(\hat{Var}\left( \hat{Y} \right)\) is
$$\hat{Var}\left( \hat{Y} \right) = \sum\limits_{g=1}^{G} M_g^2 \left( \frac{1}{m_g} - \frac{1}{M_g} \right)s_g^2$$