Answer to Question #264356 in Statistics and Probability for joshua picart

Question

Find the  ∑ (x − μ)2
 on the values of (3, 5, 8,9) with the sample size of n=3.

Answer 1

We have population values $3,5,8,9,$ population size $N=4$

$$\mu=\dfrac{3+5+8+9}{4}=6.25$$ $$\sigma^2=\dfrac{1}{4}((3-6.25)^2+(5-6.25)^2+(8-6.25)^2$$

$$+(9-6.25)^2)=5.6875$$

We have population values $3,5,8,19,$ population size $N=4$ and sample size $n=3.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{4}{3}=4$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 3,5,8 & 16/3 \\ \hdashline 2 & 3,5,9 & 17/3 \\ \hdashline 3 & 3,8,9 & 20/3 \\ \hdashline 4 & 5,8,9 & 22/3 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 16/3 & 1 & 1/4 & 6.25 & 256/36 \\ \hdashline & 17/3 & 1& 1/4 & 17/12 & 289/36 \\ \hdashline & 20/3 & 1 & 1/4 & 20/12 & 400/36 \\ \hdashline & 22/3 & 1& 1/4 & 22/12 & 484/36 \\ \hdashline Total & & 4 & 1 & 6.25 & 1429/36 \\ \hline \end{array}$$

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

$$E(\bar{X})=\mu_{\bar{X}}=6.25=\mu$$

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

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

$$=\dfrac{1429}{36}-(6.25)^2=\dfrac{91}{144}$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5.6875}{3}(\dfrac{4-3}{4-1})$$

$$=\dfrac{91}{144}, True$$

$$\sum(\bar{X}-\mu)^2=\dfrac{91}{144}$$