Answer to Question #343288 in Statistics and Probability for Ayen

Question

1. A population consists of six values (6, 9, 12, 15, 18, and 21). (35 points) a. Select a random sample of size 3. Explain the random sampling that you used. (3 pts.) b. How many possible samples can be drawn? (3 pls.)

c. list all possible samples and compute the mean of each sample. (10 pls.)

d. Construct a frequency distribution of the sample means obtained in step 2 including

f: P(x): P(): *P(): EP(x): Ex-P(F) and 2 P(F). (13 pts.)

e. Determine the mean, voriance and standard deviation of the sample mean. (6 pts.)

Answer 1

1.

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

Mean of population $(\mu)$ = $\dfrac{6+9+12+15+18+21}{6}=13.5$

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.

b. The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_3=20.$

c.

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

d.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 9 & 1/20 & 9/20 & 81/20 \\ \hdashline 10 & 1/20 & 10/20 & 100/20 \\ \hdashline 11 & 2/20 & 22/20 & 242/20 \\ \hdashline 12 & 3/20 & 36/20 & 432/20 \\ \hdashline 13 & 3/20 & 39/20 & 507/20 \\ \hdashline 14 & 3/20 & 42/20 & 588/20 \\ \hdashline 15 & 3/20 & 45/20 & 675/20 \\ \hdashline 16 & 2/20 & 32/20 & 512/20 \\ \hdashline 17 & 1/20 & 17/20 & 289/20 \\ \hdashline 18 & 1/20 & 18/20 & 324/20 \\ \hdashline \end{array}$$

e. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{270}{20}=13.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{3750}{20}-(13.5)^2=5.25= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{5.25}\approx2.2913$$