Answer to Question #342345 in Statistics and Probability for max

Question

Construct the sampling distribution of the sample mean for the population that consists of the scores 15, 18, 11, 14, 17, with a sample size of 3. Compute as well the mean of the sampling distribution using the expected value of the distribution and verify the result by computing for the population mean.

Answer 1

We have population values 15, 18, 11, 14, 17, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{15+18+11+14+17}{5}=15$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{0+9+16+1+4}{5}=6$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{6}\approx2.4495$$

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 $^{N}C_n=^{5}C_3=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 11,14,15 & 40/3 \\ \hdashline 2 & 11,14,17 & 42/3 \\ \hdashline 3 & 11,14,18 & 43/3 \\ \hdashline 4 & 11,15,17 & 43/3 \\ \hdashline 5 & 11,15,18 & 44/3 \\ \hdashline 6 & 11,17,18 & 46/3 \\ \hdashline 7 & 14,15,17 & 46/3 \\ \hdashline 8 & 14, 15,18 & 47/3 \\ \hdashline 9 & 14,17,18 & 49/3 \\ \hdashline 10 & 15, 17,18 & 50/3 \\ \hdashline \end{array}$$

B.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 40/3 & 1/10 & 40/30 & 1600/90 \\ \hdashline 42/3 & 1/10 & 42/30 & 1764/90 \\ \hdashline 43/3 & 2/10 & 86/30 & 3698/90 \\ \hdashline 44/3 & 1/10 & 44/30 & 1936/90 \\ \hdashline 46/3 & 2/10 & 92/30 & 4232/90 \\ \hdashline 47/3 & 1/10 & 47/30 & 2209/90 \\ \hdashline 49/3 & 1/10 & 49/30 & 2401/90 \\ \hdashline 50/3 & 1/10 & 50/30 & 2500/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{450}{30}=15=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{20340}{90}-(15)^2=1= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{1}=1$$