How do I get the sample deviation from population deviation when sampling is done without replacement from a finite population? The deriving step by step was needed. enter image description here
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Do it yourself, adjust formula to $\sigma_{\bar{x}}^2 = \frac{1}{N(N-1)} \sum_{i=1}^{N} (x_i - \mu)^2$ to account for the dependence between the values in the sample, do few simple steps and you'll get $SE(\bar{x}) = \frac{\sigma}{\sqrt{N(N-1)}} \cdot \sqrt{\frac{N-n}{N-1}}$ – rumathe Feb 16 '24 at 01:24