Q.A quality-control department finds that the population standard deviation of the fill-weight of a certain packaged product is 20 gm. If a random sample of 100 packets is drawn, find the standard error of the sample mean.

$S.E.(\bar{X}) = \dfrac{\sigma}{\sqrt{n}}$.

The sample mean $\bar{X}$ has variance $\dfrac{\sigma^2}{n}$, where $\sigma^2$ is the population variance and $n$ is the sample size. Its standard error is the square root of that variance:

$S.E.(\bar{X}) = \sqrt{\frac{\sigma^2}{n}} = \frac{\sigma}{\sqrt{n}}.$

The standard error shrinks as the sample size $n$ increases, reflecting that larger samples give more precise estimates of the mean.

Errors that occur in sampling are grouped into two broad categories — sampling errors and non-sampling errors; option (b).

Classification.

1. Sampling errors — arise because only a part of the population is studied (differences between the sample estimate and the true population value).
2. Non-sampling errors — arise from data collection, measurement, recording, tabulation, etc., and can occur even in a complete census.

Hence there are two types of errors in sampling.

$S.E.(\bar{x}) = \dfrac{\sigma}{\sqrt{n}}$.

If a population has standard deviation $\sigma$ and samples of size $n$ are drawn, the sample means $\bar{x}$ vary around the population mean. Their standard deviation — the standard error of the mean — is

$S.E.(\bar{x}) = \frac{\sigma}{\sqrt{n}}.$

It decreases as the sample size $n$ increases, showing that larger samples give more precise estimates of the mean.