A second example of Chi Square
• Imagine that the managers of a particular
factory are interested in whether each line
in their assembly process is equally
accurate in making parts.
• The plant has four assembly lines.
What is the population in this
situation?
• Since the question centers on product that
comes off of the assembly line the
population here is…
– All parts from the four assembly lines.
Why is this a Chi Square problem?
• Since the data will represent the number
of scrap parts in each line it is
representing the number falling into each
one of four categories – the categories
being each production line.
What would be the frequency
expected?
• The plant managers are asking whether each
of the four production lines has equal number
of errors in part production.
– Since there are four production lines, if the errors
occur equally across all lines each line should
produce ¼ of all the errors.
– This means that the f e should be 25% for each
line.
What should be the frequency
observed?
• This needs to represent the data that the
plant managers collect from each line.
– Assume that over a set period of time they
record the number of scrapped parts that
come from each of the four lines.
– These data are:
Line A
108
Line B
140
Line C
133
Line D
119
What is the total n for this
sampling?
Line A
Line B
Line C
108
140
133
Line D
119
108 140 133 119  500
n for the sample
The actual frequency expected
Given this sample of 500 scrap parts…
f e  .25500
f e  125
Computing Chi Square
 
2
 fe  fo 
2
fe
Line A has 108 scrap parts:
125  108
2
125
 2.312
Computing Chi Square cont.
 
2
 fe  fo 
2
fe
Line B has 140 scrap parts:
125  140
2
125
 1.800
Computing Chi Square
 
2
 fe  fo 
2
fe
Line C has 133 scrap parts:
125  133
2
125
 .512
Computing Chi Square
 
2
 fe  fo 
2
fe
Line D has 119 scrap parts:
125  119
2
125
 .288
Computing Chi Square cont.
 
2
 fe  fo 
2
fe
 2  2.312  1.800  .512  .288
 2  4.912
Evaluating Chi Square
H o : fe  fo
H1 : f e  f o
Critical value
df = k – 1
In this situation there were 4 groups (k=4) so
df = 3
Probability of exceeding the critical value
df
1
2
3
4
5
6
7
8
9
10
0.10
2.706
4.605
6.251
7.779
9.236
10.645
12.017
13.362
14.684
15.987
0.05
3.841
5.991
7.815
9.488
11.070
12.592
14.067
15.507
16.919
18.307
0.025
0.01
0.001
5.024 6.635 10.828
7.378
9.210 13.816
9.348 11.345 16.266
11.143 13.277 18.467
12.833 15.086 20.515
14.449 16.812 22.458
16.013 18.475 24.322
17.535 20.090 26.125
19.023 21.666 27.877
20.483 23.209 29.588
Critical value
Evaluating Chi Square
H o : fe  fo
H1 : f e  f o
Critical value: 7.815
Computed value: 4.912
Conclusion: There are no significant differences
between the production lines in terms of # of
scrap parts produced.