Question

Consider a population consisting of 2.3.4 and 6 illustrate the situation by doing the following:

1.Compute the mean and variance of the population

2.list all possible samples of size 2

3.construct the sampling distribution of the sample

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

5.construct the histogram.

Answer 1

We have population values 2,3,4,6, population size N=4 and sample size n=2.

1. Mean of population $(\mu)$ = $\dfrac{2+3+4+6}{4}=3.75$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{4}(3.0625+0.5625$$

$$+0.0625+5.0625)=2.1875=\dfrac{35}{16}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{2.1875}\approx1.47902$$

2. Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{4}C_2=6.$

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

3.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 5/2 & 1/6 & 5/12 & 25/24 \\ \hdashline 6/2 & 1/6 & 6/12 & 36/24 \\ \hdashline 7/2 & 1/6 & 7/12 & 49/24 \\ \hdashline 8/2 & 1/6 & 8/12 & 64/24 \\ \hdashline 9/2 & 1/6 & 9/12 & 81/24 \\ \hdashline 10/2 & 1/6 & 10/12 & 100/24 \\ \hdashline \end{array}$$

4. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{45}{12}=\dfrac{15}{4}=3.75=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{355}{24}-(\dfrac{15}{4})^2=\dfrac{35}{48}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{35}{48}}\approx0.85391$$

5.