Chi-Square Test
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Chi-square analysis methods are among the most commonly used
of all statistical techniques
The method introduced will be used to examine “frequency” or
“count” data
The methods are conceptually simple although manual computation
can be tedious
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Goodness of Fit Test
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Goodness of Fit
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Suppose we don’t know the underlying distribution of the population
We wish to test the hypothesis that a particular distribution will be
satisfactory as a population model
The purpose of goodness-of-fit tests is to evaluate whether a particular
probability distribution is adequate for modeling the behavior of the
process under consideration
Hypotheses
•
•
H0 : p1 = p*1, p2 = p*2, ... , pk = p*k
H1 : at least one pi ≠ p*i
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Test Statistics
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Suppose we take a random sample of size n from the population whose
probability distribution is unknown
These n observations are arranged in a histogram, having k cells or class
intervals
Let Oi be the observed frequency in the ith cell
Let Ei be the expected frequency in the ith cell computed from the
hypothesized probability distribution
Each Ei should not be less than 5. If a cell has an Ei <5, group it with an
another cell.
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Let m be the number of parameters of the hypothesized distribution
estimated by sample statistic
Then
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has approximately a chi-square distribution with k-m-1 degrees of
freedom
Reject H0 if X02 > c2a, k-m-1
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5
Example
Day
Mon
Tue
Wed
Thr
Fri
sum
No. of
accidents
65
43
48
41
73
270
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Example
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Suppose that historic failure rates are:
Due to A: .20
Due to B: .35
Due to C: .30
Due to D: .15
The manufacturer has worked on A, B, and C and believes that failures due
to these causes has been reduced, so that, while fewer failure will occur, it
is more likely that when one occurs, it will be due to D
To examine this claim the manufacturer will sample 200 failed disk drives
manufactured since process changes were made
If the changes had no impact then the number of these failed drives that
were due to causes A, B, C, and D that would be EXPECTED would be:
EA = npA0 = 200(.20) = 40
EC = npC0 = 200(.30) = 60
EB = npB0 = 200(.35) = 70
ED = npD0 = 200(.15) = 30
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H0: pA = .20, pB = .35, pC = .30, pD = .15
H1: at least one pi ≠ pi0 for i = A, B, C, D
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We can conclude that the historic failure mode distribution no
longer applies (reject H0 in favor of H1)
So how has the distribution changed?
The answer is embedded in the individual category
contributions to X02 ... larger contributions indicate where the
changes have occurred: reductions in A and C, no obvious
change in B, the various failures that make-up D now
comprise a proportionally larger amount of the failures
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Example
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A sample of 120 minutes selected during rush periods at a store gave the
following number of customers arriving during each of those 120 minutes
No. of
customers
0
1
2
3
4
5
6
Observed
Frequency
25
44
35
8
5
2
1
Question 1: Is this data consistent with a Poisson distribution with a mean
of 1.7 customers per minute? Test the appropriate hypothesis at the a
= .10 level of significance
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Recall that Poisson Distribution with mean λ is
Customers/
minute
probability
Observed
Expected
(O-E)2/E
0
P(X=0)=0.1827
25
21.922
0.4322
1
P(X=1)=3106
44
37.267
1.2163
2
P(X=2)=0.2640
35
31.677
0.3485
3
P(X=3)=0.1496
8
17.950
5.5158
4 or more
P(X≥4)=0.0913
8
10.958
1.1467
sum
8.6595
DF=k-m-1=5-0-1=4
This data is not consistent with a Poisson distribution with a mean of 1.7
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Question 2: Is this data consistent with a Poisson distribution ? Test the
appropriate hypothesis at the a = .10 level of significance.
In this case, we need to estimate the mean of the number of customers per
minute.
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In this case the degree of freedom for the chi-square distribution is
DF=k-m-1=5-1-1=3
Customers/
minute
probability
Observed
Expected
(O-E)2/E
0
P(X=0)=0.2346
25
28.148
0.3522
1
P(X=1)=0.3401
44
40.815
0.2485
2
P(X=2)=0.2466
35
29.591
0.9887
3
P(X=3)=0.1192
8
14.302
2.7771
4 or more
P(X≥4)=0.0588
8
7.051
0.1735
sum
4.5400
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Example
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Oil & Gas Exploration is both expensive and risky
The average cost of a “dry hole” is in excess of $20 million
New technologies are always under development in an effort to
reduce the likelihood of drilling a “dry hole” with the result being
increased profitability
Suppose an experimental technology has been developed that claims
to have an 80% success rate
This technology was tested by drilling four holes and counting the
number of productive wells
This was done 100 times, each time counting the number of
productive wells
The data is recorded below:
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Number of productive wells
Observed Frequency
0
1
2
3
4
3
6
22
41
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Test the appropriate hypothesis at the a = .01 level of significance
 H0: the new technology delivers success according to a binomial
distribution with p = .8
or equivalently
H0: p(0 or 1) = .0272 p (2) = .1536
p (3) = .4096
p (4) = .4096
 H1: the new technology does not deliver success according to a binomial
distribution with p =.8
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No. of productive
wells
Observed (Oi)
Expected (Ei)
(O-E)2/E
0 or 1
9
2.72
14.50
2
22
15.36
2.87
3
41
40.96
0.00
4
28
40.96
4.10
21.47
That is, we do not have enough evidence to say that the new
technology does not deliver success according to a binomial
distribution with p = .8
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If p were unknown, then it would have to be estimated from the data as
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df = k – m - 1 = 4-1-1=2 (a lost a DF)
The modified calculations follow
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Example
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The following data are tensile strength of concrete
320 380 340 410 380 340 360 350 320 370 350 340 350
360 370 350 380 370 300 420 370 390 390 440 330 390
330 360 400 370 320 350 360 340 340 350 350 390 380
340 400 360 350 390 400 350 360 340 370 420 420 400
350 370 330 320 390 380 400 370 390 330 360 380 350
330 360 300 360 360 360 390 370 370 370 350 390 370
370 340 370 400 360 350 380 380 360 340 330 370 340
360 390 400 370 410 360 400 340 360
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Test if normal distribution can safely be assumed for the data
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From the data,
=> use these for standardization
We make 10 cells and compute chi-square statistic as the following:
DF=10-2-1=7
Thus, we can assume a normal distribution for this data.
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Independence Test
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Two-way contingency table is a set of frequencies that summarize how
a set of objects simultaneously classified under two different
categorizations.
Two factors A and B are independent if P(A and B)=P (A) P(B)
Suppose A has categories A1, A2, ..., Aa and B has categories B1, B2, ...,
Bb
we must have P(AiBj) = P(Ai)P (Bj) for i = 1, 2, ..., a and j = 1, 2, ..., b
i.e. micro- independence to establish macro-independence
The Chi-Square Test of Independence is used to determine whether two
factors are related to one another and, if so, to identify the nature of the
relationship
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Hypotheses
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Let A = a “row” factor with a categories
Let B = a “column” factor with b categories
Let the probability that an observation is classified into the ith row be pi.
Let the probability that an observation is classified into the jth column be
p.j
Let pij be the probability that an observation is classified into the ith row
and jth column
Given the preceding development, the concept of independence is
formally expressed as:
 H0 : pij = pi. p.j for all i,j combinations
 H1: pij  pi. p.j for at least one i,j combination
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Test Statistics
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Let ni. be the number of items in the ith row
Let n.j be the number of items in the jth column
Let n be the total number of sample items
Let Oij be the number of items at the intersection of the ith row & jth column
pi. is estimated by ^pi. = ni. / n
p.j is estimated by ^p.j = n.j / n
pij is estimated by ^
pij = Oij / n if the two factors are dependent
pij is estimated by ^
pij = ^
pi. ^p.j if the two factors are independent
To test H0 vs H1 we compare the two estimates of pij , after each estimate
has been weighted by the amount of evidence that we have, n
Doing this, squaring the comparisons and standardizing the result yields
the c2 statistic for independence
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The chi-square statistic for independence is:
X02 = (Oij - Eij)2/Eij
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^ j = (ni.n.j)/ n is the expected number of values at the
where Eij = n p^i. p.
intersection of the ith row & jth column under independence
Examination of X02 indicates that it will be “small” in value if Oij and Eij are
close - which would support independence
If X02 is “large”, it is due to discrepancy between Oij and Eij in at least one
cell, and perhaps numerous cells. This supports dependence
The cells which contribute most greatly to X02 likely have the most to say
about the nature of any dependence structure
This test has (a-1)(b-1) degrees of freedom
Critical values of X02 are found from the chi-square distribution with
degrees of freedom (a-1)(b-1)
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Example
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Suppose that we wish to determine whether of the opinions of the voting
residents of the state of Illinois concerning a new tax reform are
independent of their levels of income
A random sample of 1000 registered voters from the state of Illinois are
classified as to whether they are in a low, medium or high income bracket
and whether or not they favor a new tax reform
The observed frequencies are presented below
Income Level
Tax Reform
Low
Medium
High
Total
For
182
598*336/1000
=200.93
213
598*351/1000
=209.90
203
598*313/1000
=187.17
598
Against
154
402*336/1000
=135.07
138
402*351/1000
=141.10
110
402*313/1000
=125.83
402
Total
336
351
313
1000
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DF=(3-1)(2-1)=2
That is, there’s no evidence that the opinions of the voting residents
of the state of Illinois concerning a new tax reform are not
independent
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Example
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The organization responsible for administration of the Customer Satisfaction Index
in Sweden examined customer satisfaction and employee empowerment indices for
a sample of 500 Swedish companies
Categories for each index were “very low”, “low”, “moderate”, “high” and “very
high”
Employ
Customer Empower
Satisfaction
Very low
Low
Moderate
High
Very high
Total
Very Low
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E=54*40/500
=4.32
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8
5
3
40
Low
10
18
19
12
6
65
Moderate
18
32
42
44
34
170
High
12
16
34
57
61
180
Very High
1
3
8
14
17
45
Total
54
80
111
132
123
500
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Is there a discernable relationship between customer and employee
satisfaction?
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