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IBM- 09: Six Sigma – Tools and Techniques
Dr. A. Ramesh
Department of Management Studies
Indian Institute of Technology Roorkee
χ2 Goodness of Fit Test
• Understand the χχχχ2 goodness-of-fit test and how to use it.
• Analyze data using the χχχχ2 test of independence.
χχχχ2 Goodness-of-Fit Test
The χ2 goodness-of-fit test compares
expected (theoretical) frequencies
of categories from a population distribution
to the observed (actual) frequencies
from a distribution to determine whether
there is a difference between what was
expected and what was observed.
χχχχ2 Goodness-of-Fit Test
( )
data sample thefrom estimated parameters ofnumber =
categories ofnumber
valuesexpected offrequency
valuesobserved offrequency :
- 1 - = df
2
2
c
k
where
ck
eo
f
f
f
ff
e
o
e
=
=
=
=∑−
χ
Month Litres
January 1,610
February 1,585
March 1,649
April 1,590
May 1,540
June 1,397
July 1,410
August 1,350
September 1,495
October 1,564
November 1,602
December 1,655
18,447
Milk Sales Data
Hypotheses and Decision Rules
ddistributeuniformly not are
salesmilk for figuresmilk monthly The :H
ddistributeuniformly are
salesmilk for figuresmilk monthly The :H
a
o
α
χ
=
= − −
= − −
=
=
.
.. ,
01
1
12 1 0
11
24 72501 11
2
df k cIf reject H .
If do not reject H .
Cal
2
o
Cal
2
o
χ
χ
>
≤
24 725
24 725
. ,
. ,
Calculations
for Demonstration Problem 1Month fo fe (fo - fe)
2/fe
January 1,610 1,537.25 3.44
February 1,585 1,537.25 1.48
March 1,649 1,537.25 8.12
April 1,590 1,537.25 1.81
May 1,540 1,537.25 0.00
June 1,397 1,537.25 12.80
July 1,410 1,537.25 10.53
August 1,350 1,537.25 22.81
September 1,495 1,537.25 1.16
October 1,564 1,537.25 0.47
November 1,602 1,537.25 2.73
December 1,655 1,537.25 9.02
18,447 18,447.00 74.38
ef =
=
18447
12
153725.
Cal
2
74 37χ = .
Conclusion
0.01
df = 11
24.725
Non Rejection
region
Cal
2
74 37 24 725χ = >. . , reject H .o
Bank Customer Arrival Data
- Problem 2
Number of
Arrivals
Observed
Frequencies
0 7
1 18
2 25
3 17
4 12
≥≥≥≥5 5
Hypotheses and Decision Rules
for Problem 2
Ho: The frequency distribution is Poisson
H : The frequency distribution is not Poissona
α
χ
=
= − −
= − −
=
=
.
.. ,
05
1
6 1 1
4
9 48805 4
2
df k cIf reject H .
If do not reject H .
Cal
2
o
Cal
2
o
χ
χ
>
≤
9 488
9 488
. ,
. ,
Calculations
for Demonstration Problem.2:
Estimating the Mean Arrival Rate
Number of
Arrivals
X
Observed
Frequencies
f f·X
0 7 0
1 18 18
2 25 50
3 17 51
4 12 48
≥≥≥≥5 5 25
192
λ =⋅
=
=
∑∑
f X
f
192
84
2 3. customers per minute
Mean
Arrival
Rate
Calculations for Demonstration Problem.2:
Poisson Probabilities for λλλλ = 2.3
Number of
Arrivals X
Expected
Probabilities
P(X)
Expected
Frequencies
n·P(X)
0 0.1003 8.42
1 0.2306 19.37
2 0.2652 22.28
3 0.2033 17.08
4 0.1169 9.82
≥5≥5≥5≥5 0.0838 7.04
n f=
=
∑84
Poisson
Probabilities
for λλλλ = 2.3
χχχχ2 Calculations
for Demonstration Problem 2
Cal
2
1 74χ = .Number of
Arrivals
X
Observed
Frequencies
f
Expected
Frequencies
nP(X)
(fo - fe)2
fe
0
1
2
3
4
≥≥≥≥5
7 8.42
18 19.37
25 22.28
17 17.08
12 9.82
5 7.04
84 84.00
0.24
0.10
0.33
0.00
0.48
0.59
1.74
Demonstration Problem 2: Conclusion
0.05
df = 4
9.488
Non Rejection
region
Cal
2
174 9 488χ = ≤. . , do not reject H .o
(((( ))))
(((( )))) (((( )))) (((( ))))
(((( )))) (((( )))) (((( ))))
(((( )))) (((( )))) (((( ))))
(((( )))) (((( )))) (((( ))))
2
2
88 6615 16 24 46 6 16 40
102 87 78 27 32 46 13 2176
36 4513 22 16 69 15 1119
15 38 95 23 14 40 25 9 65
66 15 24 46 16 40
87 78 32 46 21 76
4513 16 69 11 19
38 95 14 40 9 65
70 78
χχχχ ====
==== ++++ ++++ ++++
++++ ++++ ++++
++++ ++++ ++++
++++ ++++
====
−−−−∑∑∑∑∑∑∑∑
−−−− −−−− −−−−
−−−− −−−− −−−−
−−−− −−−− −−−−
−−−− −−−− −−−−
o ef f
fe
2 2 2
2 2 2
2 2 2
2 2 2
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
.
Gasoline Preference Versus Income
Category: χχχχ2 Calculation
Gasoline Preference Versus Income
Category: Conclusion
0.01
df = 6
16.812
Non rejection
region
Cal
2
70 78 16812χ = >. . , reject H .o
