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The sample size tends to result in a smaller sampling error when a sample means is used to estimate a population mean.

For SRSWR (simple random sampling with replacements) S.D. of $\stackrel{¯}{x}={\mathrm{σ}}_{\stackrel{¯}{x}}^{-}$

$\frac{\mathrm{σ}}{\sqrt{n}}$.

i.e. sample size is inversely proportional to the sequence root. When a result, as the sample size $n$grows, the S.D. of the sample mean ${\mathrm{σ}}_{\stackrel{¯}{x}}$lowers. That is, if the standard deviation ${\mathrm{σ}}_{\stackrel{¯}{x}}$falls, the values of all potential sample means will cluster around the mean of $\overline{x}$, i.e. the population mean $\mathrm{μ}$, and the concentration of sample means around $\mathrm{μ}$will grow. As a result, a sample drawn from all feasible samples is more likely to have a sample mean that is close to the population mean than a sample drawn from a lower sample size. As a result, as the sample size grows, the sampling error lowers.

For AWAWOR (without replacement)

${\mathrm{σ}}_{\stackrel{¯}{x}}=\sqrt{\frac{N-n}{N-1}}·\frac{\mathrm{σ}}{\sqrt{n}}$

$=\sqrt{\frac{N-n}{n}}·\frac{\mathrm{σ}}{\sqrt{N-1}}$

$=\sqrt{\frac{N}{n}}-1·\frac{\mathrm{σ}}{\sqrt{N-1}}$

Also here if we increase $\mathrm{n}$, the value of $\frac{N}{n}$decreases and hence the value of $\sqrt{\frac{N}{n}}-1$decreases.