Question

Given the population: 1, 3, 4, 6, and 8; and suppose samples of size 3 are drawn from this population:

1. What is the mean and standard deviation of the population?

2. How many different samples of size n=3 can be drawn from the population? List them with their corresponding means

3. Construct the sampling distribution of the sample means.

4. What is the mean of the sampling distribution of the sample means?

5. What is the standard deviation of the sampling distribution of the sample means?

Answer 1

1. We have population values 1, 3, 4, 6, 8, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{1+3+4+6+8}{5}=4.4$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{5}(11.56+1.96+0.16$$

$$+2.56+12.96)=5.84$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{5.84}\approx2.4166$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,3,4 & 8/3 \\ \hdashline 2 & 1,3,6 & 10/3 \\ \hdashline 3 & 1,3,8 & 12/3 \\ \hdashline 4 & 1,4,6 & 11/3 \\ \hdashline 5 & 1,4,8 & 13/3 \\ \hdashline 6 & 1,6,8 & 15/3 \\ \hdashline 7 & 3,4,6 & 13/3 \\ \hdashline 8 & 3,4,8 & 15/3 \\ \hdashline 9 & 3,6,8 & 17/3 \\ \hdashline 10 & 4,6,8 & 18/3 \\ \hdashline \end{array}$$

3.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 8/3 & 1/10 & 8/30 & 64/90 \\ \hdashline 10/3 & 1/10& 10/30 & 100/90 \\ \hdashline 11/3 & 1/10 & 11/30 & 121/90 \\ \hdashline 12/3 & 1/10 & 12/30 & 144/90 \\ \hdashline 13/3 & 2/10 & 26/30 & 338/90 \\ \hdashline 15/3 & 2/10 & 30/30 & 450/90 \\ \hdashline 17/3 & 1/10 & 17/30 & 289/90 \\ \hdashline 18/3 & 1/10 & 18/30 & 324/90 \\ \hdashline \end{array}$$

4. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=4.4=\mu$$

5. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1830}{90}-(4.4)^2=\dfrac{2.92}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{2.92}{3}}\approx0.9866$$