Answer to Question #346616 in Statistics and Probability for Trisha

Question

A group of students got the following scores in a test: 6,9,13,16,18,21. Consider sample size of size 3 that can be drawn from this population.

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

B. Construct the samplong distribution of the sample means.

C. Compute for the mean and standard deviation of the sampling distribution of the sample means.

Answer 1

We have population values 6,9,13,16,18,21, population size N=6 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{6+9+13+16+18+21}{6}=\dfrac{83}{6}$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{216}(2209+841+25$$ $$+169+625+1849)=\dfrac{953}{36}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{953}{36}}\approx5.1451$$

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

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 6,9,13 & 28/3 \\ \hdashline 2 & 6,9,16 & 31/3 \\ \hdashline 3 & 6,9,18 & 33/3 \\ \hdashline 4 & 6,9,21 & 36/3 \\ \hdashline 5 & 6,13,16 & 35/3 \\ \hdashline 6 & 6,13,18 & 37/3 \\ \hdashline 7 & 6,13, 21 & 40/3 \\ \hdashline 8 & 6,16,18 & 40/3 \\ \hdashline 9 & 6,16,21 & 43/3 \\ \hdashline 10 & 6, 18,21 & 45/3 \\ \hdashline 11 & 9,13,16 & 38/3 \\ \hdashline 12 & 9, 13,18 & 40/3 \\ \hdashline 13 & 9, 13, 21 & 43/3 \\ \hdashline 14 & 9,16,18 & 43/3 \\ \hdashline 15 & 9,16,21 & 46/3 \\ \hdashline 16 & 9, 18,21 & 48/3 \\ \hdashline 17 & 13, 16,18 & 47/3 \\ \hdashline 18 & 13, 16, 21 & 50/3 \\ \hdashline 19 & 13, 18,21 & 52/3 \\ \hdashline 20 & 16, 18,21 & 55/3 \\ \hdashline \end{array}$$

B.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 28/3 & 1/20 & 28/60 & 784/180 \\ \hdashline 31/3 & 1/20 & 31/60 & 961/180 \\ \hdashline 33/3 & 1/20 & 33/60 & 1089/180 \\ \hdashline 35/3 & 1/20 & 35/60 & 1225/180 \\ \hdashline 36/3 & 1/20 & 36/60 & 1296/180 \\ \hdashline 37/3 & 1/20 & 37/60 & 1369/180 \\ \hdashline 38/3 & 1/20 & 38/60 & 1444/180 \\ \hdashline 40/3 & 3/20 & 120/60 & 4800/180 \\ \hdashline 43/3 & 3/20 & 129/60 & 5547/180 \\ \hdashline 45/3 & 1/20 & 45/60 & 2025/180 \\ \hdashline 46/3 & 1/20 & 46/60 & 2116/180 \\ \hdashline 47/3 & 1/20 & 47/60 & 2209/180 \\ \hdashline 48/3 & 1/20 & 48/60 & 2304/180 \\ \hdashline 50/3 & 1/20 & 50/60 & 2500/180 \\ \hdashline 52/3 & 1/20 & 52/60 & 2704/180 \\ \hdashline 55/3 & 1/20 & 55/60 & 3025/180 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{830}{60}=\dfrac{83}{6}=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{35398}{180}-(\dfrac{83}{6})^2=\dfrac{953}{180}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{953}{180}}\approx2.3010$$