Question

Using the number 1,2,3,4,5 and 6 as the element of the population

a. find the mean of the sample of size 3 without replacement

b.construct the sampling distribution of the sample mean

I. The mean

II.The variance

III. The standard deviation of the sampling of the sample means

Answer 1

The number of possible samples is $\binom{N}{n}=\binom{6}{3}=20$.

The possible samples with their means are given below.

$a)$

The sample means are derived from the formula,

$\bar x_i={\sum(x_i)\over3}$

Sample mean

(1,2,3) 2

(1,2,4) 2.33

(1,2,5) 2.67

(1,2,6) 3

(1,3,4) 2.67

(1,3,5) 3

(1,3,6) 3.33

(1,4,5) 3.33

(1,4,6) 3.67

(1,5,6) 4

(2,3,4) 3

(2,3,5) 3.33

(2,3,6) 3.67

(2,4,5) 3.67

(2,4,6) 4

(2,5,6) 4.33

(3,4,5) 4

(3,4,6) 4.33

(3,5,6) 4.67

(4,5,6) 5

$b)$

The sampling distribution is,

$\bar x_i$ 2 2.33 2.67 3 3.33 3.67 4 4.33 4.67 5

$p(\bar x _i)$ ${1\over20}$ ${1\over20}$ ${2\over20}$ ${3\over20}$ ${3\over20}$ ${3\over20}$ ${3\over20}$ ${2\over10}$ ${1\over20}$ ${1\over20}$

$c)$

The mean of the sampling distribution of the means is the population mean given as,

$\mu_{\bar x_i}=\mu={1+2+3+4+5+6\over6}={21\over6}=3.5$

$d)$

To find the variance of the sampling distribution of the means we first determine the the population variance $\sigma^2$, given as,

$\sigma^2={\sum (x^2)-{({\sum (x)})^2\over N }\over N}={91-73.5\over6}=2.91667$

Now, the variance of the sampling distribution of the means is,

$\sigma^2_{\bar x_i}={\sigma^2\over n}={2.91667\over 3}=0.9722$

$e)$

The standard deviation of the sampling of the sample means is,

$\sigma_{\bar x_i}=\sqrt{\sigma^2_{\bar x_i}}=\sqrt{0.9722}=0.9860$