Answer to Question #145891 in Statistics and Probability for Adnil

Question

Given the probability distribution of infinite population below
X = x 1 2 3 4 5 Total
P(X = x) 1/5 1/5 1/5 1/5 1/5 1
a. Assuming that you sample without replacement, select all possible samples of
size 2 and set up the sampling distribution of the mean. Compute the mean of all
the sample means and also compute the population mean. Are they equal? What
is this property called?
b. Do (a) for a possible sample of size 3.
c. Compare the shape of the sampling distribution of the mean obtained in (a) and
(b). Which sampling distribution seems to have less variability, those in (a) or
(b)? Why?

Answer 1

$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 \\ p(x) & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \end{matrix}$

(a)

$S_2=\{(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2,5),$

$(3, 4), (3, 5), (4,5)\}$

$$1.5(1/10)+2(1/10)+2.5(1/10)+3(1/10)+2.5(1/10)+$$

$$+3(1/10)+3.5(1/10)+3.5(1/10)+4(1/10)+4.5(1/10)=$$

$$=3$$

$$1(1/5)+2(1/5)+3(1/5)+4(1/5)+5(1/5)=3$$

The mean of all the sample means for n = 2 without replacement and the population mean are equal. This property is called the unbiased property of the population mean.

b)

$S_3=\{(1, 2,3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5),$

$(1,4, 5), (2,3, 4), (2,3,5), (2, 4, 5), (3, 4, 5)\}$

$$2(\dfrac{1}{10})+\dfrac{7}{3}(\dfrac{1}{10})+\dfrac{8}{3}(\dfrac{1}{10})+\dfrac{8}{3}(\dfrac{1}{10})+$$

$$+3(\dfrac{1}{10})+\dfrac{10}{3}(\dfrac{1}{10})+3(\dfrac{1}{10})+\dfrac{10}{3}(\dfrac{1}{10})+$$

$$+\dfrac{11}{3}(\dfrac{1}{10})+4(\dfrac{1}{10})=3$$

$$1(1/5)+2(1/5)+3(1/5)+4(1/5)+5(1/5)=3$$

The mean of all the sample means for n = 3 without replacement and the population mean are equal. This property is called the unbiased property of the population mean.

(c)

The distribution for n=3 without replacement has less variability because the sample means cluster closer to the population mean $\mu.$