Answer to Question #350344 in Statistics and Probability for cessa

Question

Random samples of size 3 are taken from a population of the numbers 3, 4,5,6, 7,8,and 9.

1. How many samples are possible? List them and compute

the mean of each sample. 2. Construct the sampling distribution of the sample means.

3. Construct the histogram of the sampling distribution of

the sample means. Describe the shape of the histogram.

Answer 1

We have population values 3, 4, 5, 6, 7, 8, 9, population size N=7 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{3+4+5+6+7+8+9}{7}=6$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}$$ $$=\dfrac{1}{7}(9+4+1+0+1+4+6)=4$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{4}=2$$

1. Select a random sample of size 4 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{7}C_3=35.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 3,4,5 & 12/3 \\ \hdashline 2 & 3,4,6 & 13/3 \\ \hdashline 3 & 3,4,7 & 14/3\\ \hdashline 4 & 3,4,8 & 15/3 \\ \hdashline 5 & 3,4,9 & 16/3 \\ \hdashline 6 & 3,5,6 & 14/3 \\ \hdashline 7 & 3,5,7 & 15/3\\ \hdashline 8 & 3,5,8 & 16/3 \\ \hdashline 9 & 3,5,9 & 17/3\\ \hdashline 10 & 3, 6,7 & 16/3 \\ \hdashline 11 & 3,6,8 & 17/3 \\ \hdashline 12 & 3,6,9 & 18/3 \\ \hdashline 13 & 3,7,8 & 18/3 \\ \hdashline 14 & 3,7,9 & 19/3 \\ \hdashline 15 & 3,8,9 & 20/3 \\ \hdashline 16 & 4,5,6 & 15/3 \\ \hdashline 17 & 4,5,7 & 16/3 \\ \hdashline 18 & 4,5,8 & 17/3 \\ \hdashline 19 & 4,5,9 & 18/3 \\ \hdashline 20 & 4,6,7 & 17/3 \\ \hdashline 21 & 4,6,8 & 18/3 \\ \hdashline 22 & 4,6,9 & 19/3 \\ \hdashline 23 & 4,7,8 & 19/3 \\ \hdashline 24 & 4,7,9 & 20/3 \\ \hdashline 25 & 4,8,9 & 21/3 \\ \hdashline 26 & 5,6,7 & 18/3 \\ \hdashline 27 & 5,6,8 & 19/3 \\ \hdashline 28 & 5,6,9 & 20/3 \\ \hdashline 29 & 5,7,8 & 20/3 \\ \hdashline 30 & 5,7,9 & 21/3 \\ \hdashline 31 & 5,8,9 & 22/3 \\ \hdashline 32 & 6,7,8 & 21/3 \\ \hdashline 33 & 6,7,9 & 22/3 \\ \hdashline 34 & 6,8,9 & 23/3 \\ \hdashline 35 & 7,8,9 & 24/3 \\ \hdashline \end{array}$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 12/3 & 1/35 & 12/105 & 144/315 \\ \hdashline 13/3 & 1/35 & 13/105 & 169/315 \\ \hdashline 14/3& 2/35 & 28/105 & 392/315 \\ \hdashline 15/3 & 3/35 & 45/105 & 675/315 \\ \hdashline 16/3 & 4/35 & 64/105 & 1024/315\\ \hdashline 17/3 & 4/35 & 68/105 & 1156/315 \\ \hdashline 18/3 & 4/35 & 90/105 & 1620/315 \\ \hdashline 19/3 & 4/35 & 76/105 & 1444/315 \\ \hdashline 20/3 & 4/35 & 80/105 & 1600/315 \\ \hdashline 21/3 & 3/35 & 63/105 & 1323/315 \\ \hdashline 22/3 & 2/35 & 44/105 & 968/315 \\ \hdashline 23/3 & 1/35 & 23/105 & 529/315 \\ \hdashline 24/3 & 1/35 & 24/105 & 576/315 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{630}{105}=6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{11620}{315}-(6)^2=\dfrac{8}{9}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}\approx0.9428$$

3.

Symmetric distribution.