Answer to Question #349504 in Statistics and Probability for Aashna

Question

If a finite population has four elements: 6, 1, 3, 2.

(a) How many different samples of size n = 2 can be selected from this population if you sample without replacement?

(b) List all possible samples of size n = 2.

(c) Compute the sample mean for each of the samples given in part b.

(d) Find the sampling distribution of x.

(e) Compute standard error.

(f) If all four population values are equally likely, calculate the value of the population mean μ . Do any of the samples listed in part (b) produce a value of x exactly equal to μ ?

Answer 1

We have population values 1,2,3,6, population size N=4 and sample size n=2.

(a) 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 $^{N}C_n=^{4}C_2=6.$

(b)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} no & Sample \\ \hline 1 & 1,2 \\ \hdashline 2 & 1,3 \\ \hdashline 3 & 1,6 \\ \hdashline 4 & 2,3 \\ \hdashline 5 & 2,6 \\ \hdashline 6 & 3,6 \\ \hdashline \end{array}$$

(c)

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

(d)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 3/2 & 1/6 & 3/12 & 9/24 \\ \hdashline 4/2 & 1/6 & 4/12 & 16/24 \\ \hdashline 5/2 & 1/6 & 5/12 & 25/24 \\ \hdashline 7/2 & 1/6 & 7/12 & 49/24 \\ \hdashline 8/2 & 1/6 & 8/12 & 64/24 \\ \hdashline 9/2 & 1/6 & 9/12 & 81/24 \\ \hdashline \end{array}$$

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

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{4+1+0+9}{4}=3.5$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{3.5}\approx1.8708$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{36}{12}=3=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{244}{24}-(3)^2=\dfrac{7}{6}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

(e)

$$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{\dfrac{7}{6}}\approx1.0801$$

(f)

$$\mu=E(X)=1(\dfrac{1}{4})+2(\dfrac{1}{4})+3(\dfrac{1}{4})+6(\dfrac{1}{4})=3$$

 No sample produces a value of $\bar{x}$ exactly equal to $\mu.$