Answer to Question #341213 in Statistics and Probability for Xchely

Question

A group of students got the following scores in a test 6,9,12,15, and 21. Consider samples of size 3 that can be drawn from this population.

A. List all the possible samples and the corresponding mea.

B. Construct the sampling distribution of the sample means

Answer 1

We have population values 6,9,12,15,21, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{6+9+12+15+21}{5}=12.6$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{5}(43.56+12.96+0.36$$

$$+5.76+70.56)=26.64$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{26.64}\approx5.1614$$

A. Select a random sample of size 3 without replacement. We have a sample distribution of sample mean.

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 6,9,12 & 9 \\ \hdashline 2 & 6,9,15 & 10 \\ \hdashline 3 & 6,9,21 & 12 \\ \hdashline 4 & 6,12,15 & 11 \\ \hdashline 5 & 6,12,21 & 13 \\ \hdashline 6 & 6,15,21 & 14 \\ \hdashline 7 & 9,12,15 & 12 \\ \hdashline 8 & 9,12,21 & 14 \\ \hdashline 9 & 9,15,21 & 15 \\ \hdashline 10 & 12,15,21 & 16 \\ \hdashline \end{array}$$

B.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 9 & 1/10 & 9/10 & 81/10 \\ \hdashline 10 & 1/10 & 10/10 & 100/10 \\ \hdashline 11 & 1/10 & 11/10 & 121/10 \\ \hdashline 12 & 2/10 & 24/10 & 288/10 \\ \hdashline 13 & 1/10 & 13/10 & 169/10 \\ \hdashline 14 & 2/10 & 28/10 & 392/10 \\ \hdashline 15 & 1/10 & 15/10 & 225/10 \\ \hdashline 16 & 1/10 & 16/10 & 256/10 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{126}{10}=12.6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1632}{10}-(12.6)^2=4.44= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{4.44}\approx2.1071$$