Answer to Question #334800 in Statistics and Probability for Ali

Question

A population consisting of 6 measurements 4 ̧ 5 ̧ 8 ̧ 12 ̧ 9 and 12.

a. Solve for the mean of the population.

b. Solve for the variance of the population.

c. Suppose samples of size 4 are to be taken from this population ̧ what is

the mean of the sampling distribution?

d. What is its variance of the sampling distribution?

e. What is the standard error of the mean?

f. Determine whether the estimate for the population mean is good or bad.

Support your answer.

Answer 1

We have population values 4, 5, 8, 12, 9, 12, population size N=6 and sample size n=4.

a. Mean of population $(\mu)$ = $\dfrac{4+5+8+12+9+12}{6}=25/3$

b. Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{86}{9}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{86}{9}}=\dfrac{\sqrt{86}}{3}$$

c. The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_4=15.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 4,5,8,12 & 29/4 \\ \hdashline 2 & 4,5,8,9 & 26/4 \\ \hdashline 3 & 4,5,8,12 & 29/4 \\ \hdashline 4 & 4,5,12,9 & 30/4 \\ \hdashline 5 & 4,5,12,12 & 33/4 \\ \hdashline 6 & 4,5,9,12 & 30/4 \\ \hdashline 7 & 4,8,12,9 & 33/4 \\ \hdashline 8 & 4,8,12,12 & 36/4 \\ \hdashline 9 & 4,8,9,12 & 33/4 \\ \hdashline 10 & 4,12,9,12 & 37/4 \\ \hdashline 11 & 5,8,12,9 & 34/4 \\ \hdashline 12 & 5,8,12,12 & 37/4 \\ \hdashline 13 & 5,8,9,12 & 34/4 \\ \hdashline 14 & 5,12,9,12 & 38/4 \\ \hdashline 15 & 8,12,9,12 & 41/4 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X}) \\ \hline 26/4 & 1/15 & 26/60 & 676/240 \\ \hdashline 29/4 & 2/15 & 58/60 & 1682/240 \\ \hdashline 30/4 & 2/15 & 60/60 & 1800/240 \\ \hdashline 33/4 & 3/15 & 99/60 & 3267/240 \\ \hdashline 34/4 & 2/15 & 68/60 & 2312/240 \\ \hdashline 36/4 & 1/15 & 36/60 & 1296/240 \\ \hdashline 37/4 & 2/15 & 74/60 & 2738/240 \\ \hdashline 38/4 & 1/15 & 38/60 & 1444/240 \\ \hdashline 41/4 & 1/15 & 41/60 & 1681/240 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=25/3$$

d. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{16896}{240}-(\dfrac{25}{3})^2=\dfrac{43}{45}$$

e.

$$\sigma_{\bar{X}}=\sqrt{\dfrac{43}{45}}\approx0.9775$$

f.

$$\mu_{\bar{X}}=E(\bar{X})=25/3=\mu$$

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\dfrac{43}{45}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$