91Ó°ÊÓ

Consider the given question,

A finite population of size N.

The sample size is1.

Standard deviation of sample mean $\overline{x}$is given below,

role="math" localid="1652639300769" ${\mathrm{σ}}_{\overline{x}}=\left\{\begin{array}{l}\sqrt{\frac{N-n}{N-1}.\frac{\mathrm{σ}}{\sqrt{n}}}\mathrm{When}\mathrm{sampling}\mathrm{is}\mathrm{done}\mathrm{without}\mathrm{replacement}......\left(i\right)\\ \frac{\mathrm{σ}}{\sqrt{n}}\mathrm{When}\mathrm{sampling}\mathrm{is}\mathrm{done}\mathrm{with}\mathrm{replacement}......\left(ii\right)\end{array}\right.$

For $n=1$,

For case (i), role="math" localid="1652639382759" ${\mathrm{σ}}_{\overline{x}}=\frac{N-n}{N-1}.\frac{\mathrm{σ}}{\sqrt{1}}$

For case (ii), ${\mathrm{σ}}_{\overline{x}}=\frac{\mathrm{σ}}{\sqrt{1}}$

Therefore, we can say case (i) is equal to case (ii).

As for sample of size 1, there is no difference between sampling with replacement and sampling without replacement. As we draw only one observation from the population there is no chance of repetition of same population unit in the sample and hence sampling with or without replacement does not matter in this case.

So, the expressions of standard deviation of sample mean for sample of size 1 are identical for both the cases.

As if we draw a sample of size Nwhich is the size equal to population size, without replacement, all the population units occur in the sample exactly once, i.e, the sample is nothing but the actual population. Hence, the only possible sample mean is equal to the population mean, i.e, the deviation of sample mean from its mean is 0. Hence the standard deviation of sample means is equal to 0.

For sample size $n=N$and when sampling is done without replacement,

$$\begin{gathered} {\mathrm{σ}}_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}.\frac{\mathrm{σ}}{\sqrt{n}} \\ =\sqrt{\frac{0}{N-1}}.\frac{\mathrm{σ}}{\sqrt{n}} \\ =0 \end{gathered}$$