Question

Explain why increasing the sample size tends to result in a smaller sampling error when a sample means is used to estimate a population mean.

Short answer

When we raise $n$, the value of ${\sigma }_{\overline{x}}$ falls, resulting in a tiny sampling error.

Step-by-step solution

Given Information

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

Explanation

For SRSWR (simple random sampling with replacements) S.D. of $\overline{x}={\sigma }_{\overline{x}}^{-}$

$\frac{\sigma }{\sqrt{n}}$.

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 ${\sigma }_{\overline{x}}$lowers. That is, if the standard deviation ${\sigma }_{\overline{x}}$falls, the values of all potential sample means will cluster around the mean of $\overline{x}$, i.e. the population mean $\mu$, and the concentration of sample means around $\mu$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)

${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}·\frac{\sigma }{\sqrt{n}}$

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

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

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.