Answer to Question #217609 in Statistics and Probability for Queen

Question

consider a group of n= 4 people with the following ages: 16, 18, 20 and 22. Consider samples of size n= 2 from the group. If X is the average age of the two people in a sample, find:
the mean and variance of the sampling distribution of X.
compare these values to the mean and variance of the population.

Answer 1

We have population values $16, 18, 20, 22,$ population size $N=4,$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{4}{2}=6$$

$$mean=\mu=\dfrac{16+18+20+22}{4}=19$$

$$Variance=\sigma^2$$

$$=\dfrac{(16-19)^2+(18-19)^2+(20-19)^2+(22-19)^2}{4}=5$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} No & Sample & Mean \\ \hline 1 & (16, 18) & 17 \\ \hdashline 2 & (16, 20) & 18 \\ \hdashline 3 & (16, 22) & 19 \\ \hdashline 4 & (18, 20) & 19 \\ \hdashline 5 & (18, 22) & 20 \\ \hdashline 6& (20, 22) & 21 \\ \hdashline \end{array}$$

The sampling distribution of the sample mean $\bar{x}$ and its mean and standard deviation are:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} \bar{x} & f & f(\bar{x})& \bar{x}f(\bar{x})& \bar{x}^2f(\bar{x}) \\ \hline 17 & 1 & 1/6 & 17/6 & 289/6 \\ \hdashline 18 & 1 & 1/6 & 18/6 & 324/6 \\ \hdashline 19 & 2 & 2/6 & 38/6 & 722/6 \\ \hdashline 20 & 1 & 1/6 & 20/6 & 400/6 \\ \hdashline 21 & 1 & 1/6 & 21/6 & 441/6 \\ \hdashline Total & 6 & 1 & 19 & 1088/3 \\ \hdashline \end{array}$$

$$E(\bar{X})=\sum\bar{x}f(\bar{x})=19$$

$$Var(\bar{X})=\sum\bar{x}^2f(\bar{x})-(\sum\bar{x}f(\bar{x}))^2$$

$$=\dfrac{1088}{3}-19^2=\dfrac{5}{3}$$

$$E(\bar{X})=19=\mu$$

$$Var(\bar{X})=\dfrac{5}{3}=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5}{2}(\dfrac{4-2}{4-1})$$