Answer to Question #304216 in Statistics and Probability for Janjan

Question

A population consists of the four numbers 1, 2, and 4. List all the possible samples of size n=2 which can be drawn replacement from the population. Find the following

a.) Population means

b.) Population Variance

c.) Population Standard Deviation

d.) Mean of the sampling distribution of sample means

e.) Variance of the sampling distribution of sample means

f.) Standard deviation of the sampling distribution of sample means

Answer 1

We have population values $1,2,4,5$ population size $N=4,$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn with replacement is $4^2=16.$

a)

$$mean=\mu=\dfrac{1+2+4+5}{4}=3$$

b)

$$Variance=\sigma^2=\dfrac{1}{4}((1-3)^2+(2-3)^2$$

$$+(4-3)^2+(5-3)^2)=2.5$$

c)

$$\sigma=\sqrt{\sigma^2}\approx1.581139$$

d)

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

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} \bar{x} & f & f(\bar{x})& \bar{x}f(\bar{x})& \bar{x}^2f(\bar{x}) \\ \hline 1& 1 & 1/16 & 1/16 & 2/32 \\ \hdashline 1.5 & 2 & 2/16 & 3/16 & 9/32 \\ \hdashline 2 & 1 & 1/16 & 2/16 & 8/32 \\ \hdashline 2.5 & 2 & 2/16 & 5/16 & 25/32 \\ \hdashline 3 & 4 & 4/16 & 12/16 & 72/32 \\ \hdashline 3.5 & 2 & 2/16 & 7/16 & 49/32 \\ \hdashline 4 & 1 & 1/16 & 4/16 & 32/32 \\ \hdashline 4.5 & 2 & 2/16 & 9/16 & 81/32 \\ \hdashline 5 & 1 & 1/16 & 5/16 & 50/32 \\ \hdashline Total & 16 & 1 & 3 & 41/4 \\ \hdashline \end{array}$$

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{x}f(\bar{x})=5$$

e)

$$Var(\bar{X})=\sigma_{\bar{X}}^2=\sum\bar{x}^2f(\bar{x})-(\sum\bar{x}f(\bar{x}))^2$$ $$=\dfrac{41}{4}-3^2=\dfrac{5}{4}$$

f)

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{5}{4}}=\dfrac{\sqrt{5}}{2}\approx1.118034$$

Check

$$\mu_{\bar{X}}=E(\bar{X})=5=\mu$$

$$Var(\bar{X})=\dfrac{5}{4}=\dfrac{\sigma^2}{n}=\dfrac{2.5}{2}$$