Answer to Question #337153 in Statistics and Probability for elle

Question

Solve the given problems.

1. A population consists of the four numbers 1, 2, 4 and 5. List all the possible

samples of size n = 2

which can be drawn with replacement from the population.

Find the following:

a. Population mean

b. Population variance

c. Population standard deviation.

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f. Standard deviation of the sampling distribution of sample means

Answer 1

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

Mean of population $(\mu)$ = 

$$\dfrac{1+2+4+5}{4}=3$$

b. Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}$$

$$=\dfrac{4+1+1+4}{4}=\dfrac{5}{2}$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{5}{2}}\approx1.5811$$

d. The number of possible samples which can be drawn with replacement is $2^4=16.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 &1,1 & 2/2 \\ \hdashline 2 & 1,2 & 3/2 \\ \hdashline 3 & 1,4 & 5/2 \\ \hdashline 4 & 1,5 & 6/2 \\ \hdashline 5 & 2,1 & 3/2 \\ \hdashline 6 & 2,2 & 4/2 \\ \hdashline 7 & 2,4 & 6/2 \\ \hdashline 8 & 2,5 & 7/2 \\ \hdashline 9 & 4,1 & 5/2 \\ \hdashline 10 & 4,2 & 6/2 \\ \hdashline 11 & 4,4 & 8/2 \\ \hdashline 12 & 4,5 & 9/2 \\ \hdashline 13 & 5,1 & 6/2 \\ \hdashline 14 & 5,2 & 7/2 \\ \hdashline 15 & 5,4 & 9/2 \\ \hdashline 16 & 5,5 & 10/2 \\ \hdashline \end{array}$$

b.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 2/2 & 1/16 & 2/32 & 4/64\\ \hdashline 3/2 & 2/16 & 6/32 & 18/64\\ \hdashline 4/2 & 1/16 & 4/32 & 16/64\\ \hdashline 5/2 & 2/16 & 10/32 & 50/64\\ \hdashline 6/2 & 4/16 & 24/32 & 144/64\\ \hdashline 7/2 & 2/16 & 14/32 & 98/64\\ \hdashline 8/2 & 1/16 & 8/32 & 64/64\\ \hdashline 9/2 & 2/16 & 18/32 & 162/64\\ \hdashline 10/2 & 1/16 & 10/32 & 100/64\\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=3=\mu$$

e.The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{656}{64}-(3)^2=\dfrac{162}{40}=\dfrac{5}{4}= \dfrac{\sigma^2}{n}$$

f.

$$\sigma_{\bar{X}}=\sqrt{\dfrac{5}{4}}\approx1.1180$$