Question

(3,5,7,9).(3,5,7,9).

[[ Set B ]]

1. Find ALL possible samples of size 2 which can be drawn with replacement from the population.
2. Solve for the mean μ
3. x
4. μx of the sampling distribution of means.
5. Solve for the variance σ
6. x
7. 2
8. σx2 of the sampling distribution of means.
9. Solve for the standard deviation σ
10. x
11. σx of the sampling distribution of means.

Round all answers to the nearest hundredths.

Answer 1

We have population values 3,5,7,9, population size N=4 and sample size n=2.

Mean of population $(\mu)$ = 

$$\dfrac{3+5+7+9}{4}=6$$

Variance of population 

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

$$=\dfrac{9+1+1+9}{4}=5$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{5}\approx2.24$$

The number of possible samples which can be drawn without replacement is $2^4=16.$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 3 & 1/16 & 3/16 & 9/16\\ \hdashline 4 & 2/16 & 8/16 & 32/16\\ \hdashline 5 & 3/16 & 15/16 & 75/16\\ \hdashline 6 & 4/16 & 24/16 & 144/16\\ \hdashline 7 & 3/16 & 21/16 & 147/16\\ \hdashline 8 & 2/16 & 16/16 & 128/16\\ \hdashline 9 & 1/16 & 9/16 & 81/16\\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{616}{16}-(6)^2=\dfrac{5}{2}=2.5= \dfrac{\sigma^2}{n}$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{5}{2}}\approx1.58$$