Answer to Question #204269 in Statistics and Probability for Jomer Abadiano

Question

  

  A population consists of the five measurements

 2 , 6 , 8 , 0 , and 1 . How many different samples 

 of size n = 2 can be drawn from the population.

 Construct a table for the possible samples and

 sample mean.        

  

Answer 1

1.

We have population values $2,6,8,0,1$ population size $N=5$ and sample size $n=2.$Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{2}=10$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 0,1 & 0.5 \\ \hdashline 2 & 0,2 & 1 \\ \hdashline 3 & 0,6 & 3 \\ \hdashline 4 & 0,8 & 4\\ \hdashline 5 & 1,2 & 1.5 \\ \hdashline 6 & 1,6 & 3.5 \\ \hline 7 & 1,8 & 4.5 \\ \hline 8 & 2,6 & 4 \\ \hline 9 & 2,8 & 5 \\ \hline 10 & 6,8 & 7 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 0.5 & 1 & 0.1 & 0.05 & 0.025 \\ \hdashline 1.0 & 1 & 0.1 & 0.10 & 0.1 \\ \hdashline 1.5 & 1 & 0.1 & 0.15 & 0.225 \\ \hdashline 3.0 & 1 & 0.1 & 0.3 & 0.9 \\ \hdashline 3.5 & 1 & 0.1 & 0.35 & 1.225 \\ \hdashline 4.0 & 2& 0.2 & 0.80 & 3.2\\ \hdashline 4.5 & 1& 0.1 & 0.45 & 2.025 \\ \hdashline 5.0 & 1& 0.1 & 0.5 & 2.5 \\ \hdashline 7.0 & 1& 0.1 & 0.7 & 4.9 \\ \hdashline Total & 10 & 1 & 3.4 & 15.1 \\ \hline \end{array}$$

3.

Mean

$$\mu=\dfrac{2+6+8+0+1}{5}=3.4$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=3.4$$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$$E(\bar{X})=3.4=\mu$$

4.

Variance

$$\sigma^2=\dfrac{1}{5}\big((2-3.4)^2+(6-3.4)^2+(8-3.4)^2$$

$$+(0-3.4)^2+(1-3.4)^2\big)=9.44$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{9.44}\approx3.072458$$

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$

$$=15.1-(3.4)^2=3.54$$

$$\sqrt{Var(\bar{X})}=\sqrt{3.54}\approx1.881489$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{9.44}{2}(\dfrac{5-2}{5-1})$$

$$=3.54, True$$