Answer to Question #335066 in Statistics and Probability for Ye nnnn

Question

Activity 2. Construct Me!

A population consists of the five numbers 2, 3, 6, 8 and 11.

Consider samples of size 2 that can be drawn from this population.

a. List all the possible samples and the corresponding mean.

Sample

Mean

19

b. Construct the sampling distribution of the sample means.

Sample Mean

Probability

Frequency

c. Draw a histogram of the sampling distribution of the

meets.

Answer 1

a. We have population values 2, 3, 6, 8 and 11 population size N=5 and sample size n=2.

Mean of population $(\mu)$ = 

$$\dfrac{2+3+6+8+11}{5}=6$$

Variance of population 

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

$$=\dfrac{16+9+0+4+25}{5}=10.8$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{10.8}\approx3.2863$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_2=10.$

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

b.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 5/2 & 1/10 & 5/20 & 25/40\\ \hdashline 8/2 & 1/10 & 8/20 & 64/40\\ \hdashline 9/2 & 1/10 & 9/20 & 81/40\\ \hdashline 10/2 & 1/10 & 10/20 & 100/40\\ \hdashline 11/2 & 1/10 & 11/20 & 121/40\\ \hdashline 13/2 & 1/10 & 13/20 & 169/40\\ \hdashline 14/2 & 2/10 & 28/20 & 392/40\\ \hdashline 17/2 & 1/10 & 17/20 & 289/40\\ \hdashline 19/2 & 1/10 & 19/20 & 361/40\\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1602}{40}-(6)^2=\dfrac{162}{40}=\dfrac{81}{20}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{81}{20}}\approx2.01246$$

c.