Answer to Question #261912 in Statistics and Probability for Alex

Question

a population consists of three numbers (2,5,9). Consider all possible samples of size 2 which can be drawn from the population. Find the variance of the sampling distribution of the sample means.

Answer 1

There are $\binom{N}{n}=N!/(N-n)!*n!$ possible ways of drawing a sample of size $n$ from a population size of $N$.

In this question, $N=3$ and $n=2$. Therefore, number of ways to draw samples of size $n=2$ is,

$\binom{3}{2}=3!/2!=3$

The samples are,

(2,5),(2,9),(5,9)

The population mean$(\mu)$ is given as,

$\mu=\sum(x)/N=(2+5+9)/3=16/3=5.3333$

The population variance$(\sigma^2)$ is given as,

$\sigma^2=\displaystyle\sum^2_{i=1}(x_i-\mu)^2/N=24.6667/3=8.2222$

Define the variance of the sampling distribution of the sample means by $\sigma^2_m$ then it is given by,

$\sigma^2_m=\sigma^2/n=8.22222/2=4.1111(4\space dp)$

Thus, the variance of the sampling distribution of the sample means is $\sigma^2_m=4.1111$