simple random sample

A sample $S$ of size $n$ from a population $U$ of size $N$ is calleda simple random sample if

1. it is a sample without replacement, and

2. the probability of picking this sample is equal to theprobability of picking any other sample of size $n$ from the samepopulation $U$.

From the first part of the definition, there are $\left(\genfrac{}{}{0pt}{}{N}{n}\right)$samples of $n$ items from a population of $N$ items. From thesecond part of the definition, the probability of any sample of size$n$ in $U$ is a constant. Therefore, the probability of picking aparticular simple random sample of size $n$ from a population ofsize $N$ is ${\left(\genfrac{}{}{0pt}{}{N}{n}\right)}^{-1}$.

Remarks Suppose $x_1,x_2,\mathrm{\dots },x_n$ are valuesrepresenting the items sampled in a simple random sample of size$n$.

• The sample mean $\overline{x}=\frac{1}{n}{\sum }_{i=1}^n{x}_i$ is anunbiased estimator of the true population mean $\mu$.

• The sample variance $s^2=\frac{1}{n-1}{\sum }_{i=1}^n{(x_i-\overline{x})}^2$ is an unbiased estimator of$S^2$, where $(\frac{N-1}{N})S^2={\sigma }^2$ is the true variance ofthe population given by

  $${\sigma }^2:=\frac{1}{N}\sum _{i=1}^N{(x_i-\overline{x})}^2.$$

• The variance of the sample mean $\overline{x}$ from the truemean $\mu$ is

  $$\left(\frac{N-n}{nN}\right)S^2.$$

  The larger the sample size,the smaller the deviation from the true population mean. When$n=1$, the variance is the same as the true population variance.