Answer to Question #348078 in Statistics and Probability for teya

Question

Using the data below, perform the following tasks

23 26 29 32 35

A. Construct a sampling distribution of the mean when n=3

B. Draw a histogram for the sample means

Answer 1

A.

We have population values 23, 26, 29, 32, 35, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{23+26+29+32+35}{5}=29$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{36+9+0+9+36}{5}=18$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{18}=3\sqrt{2}\approx4.24264$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 23,26,29 & 26 \\ \hdashline 2 & 23,26,32 & 27 \\ \hdashline 3 & 23,26,35 & 28\\ \hdashline 4 & 23,29,32 & 28 \\ \hdashline 5 & 23,29,35 & 29 \\ \hdashline 6 & 23,32,35 & 30 \\ \hdashline 7 & 26,29,32 & 29 \\ \hdashline 8 & 26,29,35 & 30 \\ \hdashline 9 & 26,32,35 & 31 \\ \hdashline 10 & 29,32,35 & 32 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 26 & 1/10 & 26/10 & 676/10 \\ \hdashline 27 & 1/10 & 27/10 & 729/10 \\ \hdashline 28 & 2/10 & 56/10 & 1568/10 \\ \hdashline 29 & 2/10 & 58/10 & 1682/10 \\ \hdashline 30 & 2/10 & 60/10 & 1800/10 \\ \hdashline 31 & 1/10 & 31/10 & 961/10 \\ \hdashline 32 & 1/10 & 32/10 & 1024/10 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=29=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$

$$=\dfrac{8440}{10}-(29)^2=3= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{3}\approx1.73205$$

B.