Answer to Question #348157 in Statistics and Probability for Lyn

Question

A group of students got the following scores in an achievement test:9,12,15,18,21 and 24 . consider the sample of size 3 that can be drawn from this population. Construct a sampling distribution of the resulting means and find the probability

Answer 1

We have population values 9,12,15,18,21, 24, population size N=6 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{9+12+15+18+21+24}{6}=16.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{6}(56.25+20.25+2.25$$ $$+2.25+20.25+56.25)=26.25$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{26.25}\approx5.1235$$

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 $^{N}C_n=^{6}C_3=20.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 9,12,15 & 12 \\ \hdashline 2 & 9,12,18 & 13 \\ \hdashline 3 & 9,12,21 & 14 \\ \hdashline 4 & 9,12,24 & 15 \\ \hdashline 5 & 9,15,18 & 14 \\ \hdashline 6 & 9,15,21 & 15 \\ \hdashline 7 & 9,15, 24 & 16 \\ \hdashline 8 & 9,18,21 & 16 \\ \hdashline 9 & 9,18,24 & 17 \\ \hdashline 10 & 9, 21,24 & 18 \\ \hdashline 11 & 12,15,18 & 15 \\ \hdashline 12 & 12, 15,21 & 16 \\ \hdashline 13 & 12, 15, 24 & 17 \\ \hdashline 14 & 12,18,21 & 17 \\ \hdashline 15 & 12,18,24 & 18 \\ \hdashline 16 & 12, 21,24 & 19 \\ \hdashline 17 & 15, 18,21 & 18 \\ \hdashline 18 & 15, 18,24 & 19 \\ \hdashline 19 & 15, 21,24 & 20 \\ \hdashline 20 & 18, 21,24 & 21 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 12 & 1/20 & 12/20 & 144/20 \\ \hdashline 13 & 1/20 & 13/20 & 169/20 \\ \hdashline 14 & 2/20 & 28/20 & 392/20 \\ \hdashline 15 & 3/20 & 45/20 & 675/20 \\ \hdashline 16 & 3/20 & 48/20 & 768/20 \\ \hdashline 17 & 3/20 & 51/20 & 867/20 \\ \hdashline 18 & 3/20 & 54/20 & 972/20 \\ \hdashline 19 & 2/20 & 38/20 & 722/20 \\ \hdashline 20 & 1/20 & 20/20 & 400/20 \\ \hdashline 21 & 1/20 & 21/20 & 441/20 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{330}{20}=16.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{5550}{20}-(16.5)^2=5.25= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{5.25}\approx2.2913$$