Question

Solve the given problem. Dont forget to show your solutions.

1.AA pop fourulstion consist of the four numbers 1,2,4 and 5. List all the possible samples of sizes n=3 wich can be drawn with replacement from the population. Find the following : a. Population mean b. Population variance c. Population standard deviation d. Mean of the sampling distribution of sample means e. Variance of the sampling distribution of sample means f standard deviation of the sampling distribution of sample means

Answer 1

The number of the samples is the number of combinations of size n from a set S of size m with replacement:

$N=\begin{pmatrix} n+m-1\\ n \end{pmatrix}=\cfrac{(n+m-1)!}{n!\cdot (m-1)!}=\\ =\cfrac{(4+3-1)!}{3!\cdot (4-1)!}=\cfrac{6!}{3!\cdot 3!}=\cfrac{4\cdot5\cdot6}{2\cdot3}=20.$

All the possible samples of sizes n=3 wich can be drawn with replacement from the population:

$\{ (1,1,1), (1,1,2),(1,1,4),(1,1,5),(1,2,2),\\ (1,2,4),(1,2,5),(1,4,4),(1,4,5),(1,5,5),\\ (2,2,2),(2,2,4),(2,2,5),(2,4,4),(2,4,5),\\ (2,5,5),(4,4,4),(4,4,5),(4,5,5),(5,5,5)\}.$

a. Population mean:

$\mu=\sum x_i\cdot P(x_i)=\\ =1\cdot\cfrac{1}{4}+2\cdot\cfrac{1}{4}+4\cdot\cfrac{1}{4}+5\cdot\cfrac{1}{4}=3.$

b. Population variance:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\begin{Bmatrix} 1-3, 2-3, 4-3, 5-3 \end{Bmatrix}=$

$=\begin{Bmatrix} -2, -1, 1, 2 \end{Bmatrix},$

$\sigma^2=(-2)^2\cdot \cfrac{1}{4}+(-1)^2\cdot \cfrac{1}{4}+1^2\cdot \cfrac{1}{4}+2^2\cdot \cfrac{1}{4}=2.5.$

c. Population standard deviation:

$\sigma=\sqrt{2.5}=1.58.$

For d., e., f., we'll use the properties of sampling distributions of sample means.

d. Mean of the sampling distribution of sample means:

$\mu_{\bar x} =\mu=3.$

e. Variance of the sampling distribution of sample means:

$\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{2.5}{2}=1.25.$

f. standard deviation of the sampling distribution of sample means:

$\sigma_{\bar x}=\cfrac{\sigma}{\sqrt n}=\cfrac{1.58}{\sqrt 2}=1.12.$