Download Sampling Distribution handout

QMM 384 - Sampling Distribution Handout
This handout is an example problem to show the difference between the population of interest and the sampling
distribution of X (the sample mean). The population of interest is the salary (in thousands of dollars) for 5
employees of a company. Each Xi represents a salary for an employee. To compute the mean and standard
deviation use: Nx = 5 = number of observations in Population 1
Employee
1. Al
2. Ben
3. Chuck
4. Dan
5. Ed
Population 1
Xi
(Xi- µx)2
39
1
41
1
25
225
55
225
40
0
Xi =200 (Xi- µx)2 = 452
µx = 200/5 = 40
X 
 X
i
  x 2
Nx

452

5
90 .4  9.5079
This is the Sampling Distribution of X :
Population 2
Sample Group
1. ABC
2. ABD
3. ABE
4. ACD
5. ACE
6. ADE
X
35
45
40
39.6667
34.6667
44.6667
7. BCD
8. BCE
9. BDE
10. CDE
40.3333
35.3333
45.3333
40
 X i =400
𝝈𝒙̅ = √
̅ 𝐢 − µ𝐱̅ )𝟐
(𝐗
25
25
0
0.1111
28.4444
21.7778
0.1111
21.7778
28.4444
0
150.6667
n = 3 = number of observation in each X
𝐍𝐗̅ = 10 = size of the population of X s
 X = 400/10 = 40
̅ 𝒊 − 𝝁𝑿̅ )𝟐
∑(𝑿
𝟏𝟓𝟎. 𝟔𝟔𝟔𝟕
=√
= √𝟏𝟓. 𝟎𝟔𝟔𝟕 = 𝟑. 𝟖𝟖𝟏𝟔
𝑵𝑿̅
𝟏𝟎
Note the relationship  X   X
or 40 = 40
Note the relationship of  X and  X (standard error of X ):
X 
x
Factor of
n
in general, however when we have a finite Population It is important to note we must include the Finite Po
Nx  n
whenever n ≥ .05N so
N x 1
X 
5  3 9.5079

 .7071 5.48945  3.88158
5 1
3