Answer to Question #349257 in Statistics and Probability for GAB

Question

2. Consider the population consisting of the values (1, 2, 8). List all possible samples of size 2 which can be drawn without replacement from the population. Find the following:

a. Population mean

b. Population variance

c. Population standard deviation

d. Mean of the samples and mean of the sampling distribution of mean

e. Variance of the sampling distribution of means

f. Standard deviation of the sampling distribution of means

Answer 1

a. We have population values 1,2,8, population size N=3 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{1+2+8}{3}=\dfrac{11}{3}$

b.Variance of population 

c.

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{27}(64+25+169)=\dfrac{86}{9}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{86}{9}}\approx3.0912$$

Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{3}C_2=3.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,2 & 1.5 \\ \hdashline 2 & 1,8 & 4.5 \\ \hdashline 3 & 2,8 & 5 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 1.5 & 1/3 & 3/6 & 9/12 \\ \hdashline 4.5 & 1/3 & 9/6 & 81/12 \\ \hdashline 5 & 1/3 & 10/6 & 100/12 \\ \end{array}$$

d. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{11}{3}=\mu$$

e. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{190}{12}-(\dfrac{11}{3})^2=\dfrac{43}{18}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

f.

$$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{\dfrac{43}{18}}\approx1.5456$$