Question

Consider a population consisting of 2,5,6,8, and 10. Suppose samples of size 2 are drawn from this population.

Q1.Compute the mean and the standard deviation of the population.

Q2. List all samples of size 5 and compute the mean for each sample.

Q3. Construct the sampling distribution of the sample means.

Q4. Calculate the mean of the sampling distribution of the sample means.

Q5. Calculate the standard deviation of the sampling distribution of the sample means.

Answer 1

Q1. We have population values 2,5,6,8,10, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{2+5+6+8+10}{5}=6.2$

Variance of population 

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

$$=\dfrac{17.64+1.44+0.04+3.24+14.44}{5}=7.36$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{7.36}\approx2.712932$$

Q2. The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_3=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,5,6 & 13/3 \\ \hdashline 2 & 2,5,8 & 15/3 \\ \hdashline 3 & 2,5,10 & 17/3 \\ \hdashline 4 & 2,6,8 & 16/3 \\ \hdashline 5 & 2,6,10 & 18/3 \\ \hdashline 6 & 2,8,10 & 20/3 \\ \hdashline 7 & 5,6,8 & 19/3 \\ \hdashline 8 & 5,6,10 & 21/3 \\ \hdashline 9 & 5,8,10 & 23/3 \\ \hdashline 10 & 6,8,10 & 24/3 \\ \hdashline \end{array}$$

Q3.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 13/3 & 1/10 & 13/30 & 169/90 \\ \hdashline 15/3 & 1/10 & 15/30 & 225/90 \\ \hdashline 16/3 & 1/10 & 16/30 & 256/90 \\ \hdashline 17/3 & 1/10 & 17/30 & 289/90 \\ \hdashline 18/3 & 1/10 & 18/30 & 324/90 \\ \hdashline 19/3 & 1/10 & 19/30 & 361/90 \\ \hdashline 20/3 & 1/10 & 20/30 & 400/90 \\ \hdashline 21/3 & 1/10 & 21/30 & 441/90 \\ \hdashline 23/3 & 1/10 & 23/30 & 529/90 \\ \hdashline 24/3 & 1/10 & 24/30 & 576/90 \\ \hdashline \end{array}$$

Q4. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=6.2=\mu$$

Q5.The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{119}{3}-(6.2)^2=\dfrac{3.68}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{3.68}{3}}\approx1.107550$$