Answer to Question #348646 in Statistics and Probability for Mariamae

Question

Consider all samples of size 4 from this population: 3, 6, 10, 13, 15, 20

a. Make a sampling distribution of the sample means

b. Compute the mean of the sample means

c. Compute the variance and standard deviation of the sample means

Answer 1

We have population values 3, 6, 10, 13, 15, 20, population size N=6 and sample size n=4.

Mean of population $(\mu)$ = $\dfrac{3+6+10+13+15+20}{6}=\dfrac{67}{6}$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{216}(2401+961+49$$ $$+121+529+2809)=\dfrac{1145}{36}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{1145}{36}}\approx5.64$$

a. 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 $^{N}C_n=^{6}C_4=15.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 3,6,10,13 & 32/4 \\ \hdashline 2 & 3,6,10,15 & 34/4 \\ \hdashline 3 & 3,6,10,20 & 39/4\\ \hdashline 4 & 3,6,13,15 & 37/4 \\ \hdashline 5 & 3,6,13,20 & 42/4 \\ \hdashline 6 & 3,6,15,20 & 44/4 \\ \hdashline 7 & 3,10,13,15 & 41/4\\ \hdashline 8 & 3,10,13,20 & 46/4 \\ \hdashline 9 & 3,10,15,20, & 48/4 \\ \hdashline 10 & 3, 13,15,20 & 51/4 \\ \hdashline 11 & 6,10,13,15 & 44/4 \\ \hdashline 12 & 6, 10,13,20 & 49/4 \\ \hdashline 13 & 6, 10, 15,20 & 51/4 \\ \hdashline 14 & 6,13,15,20 & 54/4 \\ \hdashline 15 & 10,13,15,20 & 58/4 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 32/4 & 1/15 & 32/60 & 1024/240 \\ \hdashline 34/4 & 1/15 & 34/60 & 1156/240 \\ \hdashline 37/4 & 1/15 & 37/60 & 1369/240 \\ \hdashline 39/4 & 1/15 & 39/60 & 1521/240 \\ \hdashline 41/4 & 1/15 & 41/60 & 1681/240 \\ \hdashline 42/4 & 1/15 & 42/60 & 1764/240 \\ \hdashline 44/4 & 2/15 & 88/60 & 3872/240 \\ \hdashline 46/4 & 1/15 & 46/60 & 2116/240 \\ \hdashline 48/4 & 1/15 & 48/60 & 2304/240 \\ \hdashline 49/4 & 1/20 & 49/60 & 2401/240 \\ \hdashline 51/4 & 2/15 & 102/60 & 5202/240 \\ \hdashline 54/4 & 1/15 & 54/60 & 2916/240 \\ \hdashline 58/4 & 1/15 & 58/60 & 3364/240 \\ \hdashline \end{array}$$

b. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{670}{60}=\dfrac{67}{6}=\mu$$

c. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{30690}{240}-(\dfrac{67}{6})^2=\dfrac{229}{72}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{229}{72}}\approx1.78$$