Question

A population consists of 4 number 3, 7, 11, 15 of sample size n=2 which can be drawn without replacement from the population.

a. Population mean

b. Population variance

c.population standard deviation

d. Mean of the sampling distribution of the sample means

e. Variance of the sampling distribution of the sample means

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

Answer 1

We have population values 3,7,11,15, population size N=4 and sample size n=2.

a. Mean of population $(\mu)$ = $\dfrac{3+7+11+15}{4}=9$

b. Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{36+4+4+36}{4}=20$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{20}\approx4.472136$$

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 $^{N}C_n=^{4}C_2=6.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 3,7 & 5 \\ \hdashline 2 & 3,11 & 7 \\ \hdashline 3 & 3,15 & 9 \\ \hdashline 4 & 7, 11& 9 \\ \hdashline 5 & 7,15 & 11 \\ \hdashline 6 & 11,15 & 13 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 5 & 1/6 & 5/6 & 25/6 \\ \hdashline 7 & 1/10 & 7/6 & 49/6 \\ \hdashline 9 & 2/6 & 18/6 & 162/6 \\ \hdashline 11 & 1/6 & 11/6 & 121/6 \\ \hdashline 13 & 1/6 & 13/6 & 169/6 \\ \hdashline \end{array}$$

d. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{54}{6}=9=\mu$$

e. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{526}{6}-(9)^2=\dfrac{20}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

f.

$$\sigma_{\bar{X}}=\sqrt{\dfrac{20}{3}}\approx2.581989$$