Answer to Question #344350 in Statistics and Probability for Maria

Question

Consider a population consisting of 1, 2, and 3. Suppose samples of size 3 are drawn from this population with and without replacement.

1. Determine the number of sets of all possible samples.

2. List all the possible samples, and compute the mean of each sample.

3. Construct the sampling distribution of the sample means

Answer 1

We have population values 1,2,3, population size N=3 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{1+2+3}{3}=2$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1+0+1}{3}=\dfrac{2}{3}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{2}{3}}\approx0.8165$$

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

1. The number of possible samples which can be drawn without replacement is $^{N}C_n=^{3}C_3=1.$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,2,3 & 2 \\ \hdashline \end{array}$$

3.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 2 & 1 & 2 & 4 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=2=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=4-(2)^2=0= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{5.25}\approx2.291288$$

b) Select a random sample of size 3 with replacement. We have a sample distribution of sample mean.

1. The number of possible samples which can be drawn with replacement is $N^n=3^3=27.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,1,1 & 3/3 \\ \hdashline 2 & 1,1,2 & 4/3 \\ \hdashline 3 & 1,1,3 & 5/3 \\ \hdashline 4 & 1,2,1 & 4/3\\ \hdashline 5 & 1,2,2 & 5/3 \\ \hdashline 6 & 1,2,3 & 6/3 \\ \hdashline 7 & 1,3,1 & 5/3 \\ \hdashline 8 & 1,3,2 & 6/3 \\ \hdashline 9 & 1,3,3 & 7/3 \\ \hdashline 10 & 2,1,1 & 4/3 \\ \hdashline 11 & 2,1,2 & 5/3 \\ \hdashline 12 & 2,1,3 & 6/3 \\ \hdashline 13 & 2,2,1 & 5/3 \\ \hdashline 14 & 2,2,2 & 6/3 \\ \hdashline 15 & 2,2,3 & 7/3 \\ \hdashline 16 & 2,3,1 & 6/3 \\ \hdashline 17 & 2,3,2 & 7/3 \\ \hdashline 18 & 2,3,3 & 8/3 \\ \hdashline 19 & 3,1,1 & 5/3 \\ \hdashline 20 & 3,1,2 & 6/3 \\ \hdashline 21 & 3,1,3 & 7/3 \\ \hdashline 22 & 3,2,1 & 6/3 \\ \hdashline 23 & 3,2,2 & 7/3 \\ \hdashline 24 & 3,2,3 & 8/3 \\ \hdashline 25 & 3,3,1 & 7/3 \\ \hdashline 26 & 3,3,2 & 8/3 \\ \hdashline 27 & 3,3,3 & 9/3 \\ \hdashline \end{array}$$

3.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 3/3 & 1/27 & 3/81 & 9/243 \\ \hdashline 4/3 & 3/27 & 12/81 & 48/243 \\ \hdashline 5/3 & 6/27 & 30/81 & 150/243 \\ \hdashline 6/3 & 7/27 & 42/81 & 252/243 \\ \hdashline 7/3 & 6/27 & 42/81 & 294/243 \\ \hdashline 8/3 & 3/27 & 24/81 & 192/243 \\ \hdashline 9/3 & 1/27 & 9/81 & 81/243 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{162}{81}=2=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1026}{243}-(2)^2=\dfrac{2}{9}= \dfrac{\sigma^2}{n}$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{2}{9}}=\dfrac{\sqrt{2}}{3}\approx0.4714$$