Answer to Question #217588 in Statistics and Probability for Sanju

Question

A population consists of four numbers 2, 3, 4, 5. Consider all possible distinct

samples of size two with replacement. Find (i) the population mean (ii) the

population standard deviation (iii) the sampling distribution of means (iv)

standard deviation of the sampling distribution of means

Answer 1

(i)

$$mean=\mu=\dfrac{2+3+4+5}{4}=3.5$$

(ii)

$$Variance=\sigma^2$$

$$=\dfrac{(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2}{4}$$

$$=1.25$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{1.25}=\dfrac{\sqrt{5}}{2}$$

(iii) We have population values $2, 3, 4, 5,$ population size $N=4,$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn with replacement is

$$N^n=4^2=16$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} No & Sample & Mean \\ \hline 1 & (2, 2) & 2 \\ \hdashline 2 & (2,3) & 2.5 \\ \hdashline 3 & (2,4) & 3\\ \hdashline 4 & (2,5) & 3.5 \\ \hdashline 5 & (3,2) & 2.5 \\ \hdashline 6 & (3, 3) & 3\\ \hdashline 7 & (3,4) & 3.5 \\ \hdashline 8 & (3, 5) & 4 \\ \hdashline 9 & (4,2) & 3 \\ \hdashline 10 & (4, 3) & 3.5 \\ \hdashline 11 & (4,4) & 4 \\ \hdashline 12 & (4, 5) & 4.5 \\ \hdashline 13 & (5,2) & 3.5 \\ \hdashline 14 & (5, 3) & 4 \\ \hdashline 15 & (5,4) & 4.5 \\ \hdashline 16 & (5, 5) & 5\\ \hdashline \end{array}$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} \bar{x} & f & f(\bar{x})& \bar{x}f(\bar{x})& \bar{x}^2f(\bar{x}) \\ \hline 2 & 1 & 1/16 & 2/16 & 4/16 \\ \hdashline 2.5 & 2 & 2/16 & 5/16 & 12.5/16 \\ \hdashline 3 & 3 & 3/16 & 9/16 &27/16 \\ \hdashline 3.5 & 4 & 4/16 & 14/16 & 49/16 \\ \hdashline 4 & 3 & 3/16 & 12/16 & 48/16 \\ \hdashline 4.5 & 2 & 2/16 & 9/16 & 40.5/16 \\ \hdashline 5 & 1 & 1/16 & 5/16 & 25/16 \\ \hdashline Total &1 6 & 1 & 3.5 & 103/8\\ \hdashline \end{array}$$

$$E(\bar{X})=\sum\bar{x}f(\bar{x})=3.5$$

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

$$=\dfrac{103}{8}-\dfrac{49}{4}=\dfrac{5}{8}$$

$$E(\bar{X})=3.5=\mu$$

(iv)

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

$$=\dfrac{103}{8}-\dfrac{49}{4}=\dfrac{5}{8}$$

$$Var(\bar{X})=\sigma_{\bar{x}}^2=\dfrac{5}{8}=\dfrac{\sigma^2}{n}=\dfrac{5}{4(2)}$$

$$\sigma_{\bar{x}}=\dfrac{\sqrt{10}}{4}=\dfrac{\sigma}{\sqrt{n}}$$