Answer to Question #345295 in Statistics and Probability for Jcanne

Question

A population consists of five numbers 1, 2, 3, 4, 5 and 6. Suppose samples of size 2 are drawn from this population.

(a) Find the mean and variance of the population

(b) Describe the sampling distribution of the sample means.

(c) Find the mean and variance of the sampling distribution of the sample means.

Answer 1

1.

 We have population values $1,2,3,4,5,6$ population size $N=6$

$$\mu=\dfrac{1+2+3+4+5+6}{6}=3.5$$

$$\sigma^2=\dfrac{1}{6}((1-3.5)^2+(2-3.5)^2+(3-3.5)^2$$ $$+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2)=2.92$$

The number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{6}{2}=15$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,2 & 1.5 \\ \hdashline 2 & 1,3 & 2 \\ \hdashline 3 & 1,4 & 2.5\\ \hdashline 4 & 1,5 & 3 \\ \hdashline 5 & 1,6 & 3.5 \\ \hdashline 6 & 2,3 & 2.5 \\ \hdashline 7 & 2,4 & 3 \\ \hdashline 8 & 2,5& 3.5 \\ \hdashline 9 & 2,6 & 4 \\ \hdashline 10 & 3,4 &3.5\\ \hdashline 11 & 3,5 & 4 \\ \hdashline 12 & 3,6 & 4.5 \\ \hdashline 13&4,5 & 4.5 \\ \hdashline 14 & 4,6 & 5 \\ \hdashline 15 & 5,6 &5.5 \\ \hline \end{array}$$

2.

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 1.5 & 1 & 1/15 &0.1& 0.15 \\ \hdashline & 2& 1 & 1/15 & 0.13 & 0.266 \\ \hdashline & 2.5& 2 & 2/15 & 0.333& 0.8333 \\ \hdashline & 3 & 2 & 2/15 & 0.4& 1.2 \\ \hdashline & 3.5 & 3 & 3/15 & 0.7 & 2.45 \\ \hdashline & 4 & 2 & 2/15& 0.533 & 2.133 \\ \hdashline & 4.5 & 2 & 2/15 & 0.6 & 2.7 \\ \hdashline & 5 & 1 & 1/15 & 0.333 & 1.666 \\ \hdashline & 5.5 & 1 & 1/15 & 0.366 & 2.0166\\ \hdashline Total & & 15 & 1 & 3.5 & 13.415\\ \hline \end{array}$$

4.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=3.5$$

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

$$E(\bar{X})=\mu_{\bar{X}}=3.5=\mu$$

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

$$={13.415}-(3.5)^2=1.165$$