Q.In a sample of 400 population from a village, 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village.

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

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:

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

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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). …

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

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

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

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