Answer to Question #337981 in Statistics and Probability for San

Question

Q.3A population consists of five numbers 1, 3, 4, 5, and 7. (i) Enumerate all possible samples of size two which can be drawn from the population without replacement. (ii) Calculate the mean and variance of the population. (iii) Show that the mean of the sampling distribution of the sample means is equal to the population mean. (iv) Calculate the standard error of the mean.

Answer 1

We have population values 1,3,4,5 and 7 population size N=5 and sample size n=2.

(i) The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_2=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,3 & 4/2 \\ \hdashline 2 & 1,4 & 5/2 \\ \hdashline 3 & 1,5 & 6/2 \\ \hdashline 4 & 1,7 & 8/2 \\ \hdashline 5 & 3,4 & 7/2 \\ \hdashline 6 & 3,5 & 8/2 \\ \hdashline 7 & 3,7 & 10/2 \\ \hdashline 8 & 4,5 & 9/2 \\ \hdashline 9 & 4,7 & 11/2 \\ \hdashline 10 & 5,7 & 12/2 \\ \hdashline \end{array}$$

(ii) Mean of population $(\mu)$ = 

$$\dfrac{1+3+4+5+7}{5}=4$$

Variance of population 

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

$$\sigma=\sqrt{\sigma^2}=\sqrt{4}=2$$

(iii)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 4/2 & 1/10 & 4/20 & 16/40\\ \hdashline 5/2 & 1/10 & 5/20 & 25/40\\ \hdashline 6/2 & 1/10 & 6/20 & 36/40\\ \hdashline 7/2 & 1/10 & 7/20 & 49/40\\ \hdashline 8/2 & 2/10 & 16/20 & 128/40\\ \hdashline 9/2 & 1/10 & 9/20 & 81/40\\ \hdashline 10/2 & 1/10 & 10/20 & 100/40\\ \hdashline 11/2 & 1/10 & 11/20 & 121/40\\ \hdashline 12/2 & 1/10 & 12/20 & 144/40\\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=4=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{700}{40}-(4)^2=\dfrac{3}{2}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

(iv) The standard error of the mean

$$\sigma_{\bar{X}}=\sqrt{\dfrac{3}{2}}\approx1.2247$$