Chapter 06.03
Linear Regression-More Examples
Industrial Engineering
Example 1
As machines are used over long periods of time, the output product can get off target. Below
is the average value of how much off target a product is getting manufactured as a function
of machine use.
Table 1 Off target value as a function of machine use.
30
33
34
35
39
44
45
Hours of Machine Use, t
1.21
1.25 1.23 1.30
1.40
1.42
Millimeters Off Target, h 1.10
Regress the data to h  a0  a1t . Find when the product will be 2 mm off target.
Solution
Table 2 shows the summations needed for the calculation of the constants of the regression
model.
Table 2 Tabulation of data for calculation of needed summations.
I
−
1
2
3
4
5
6
7
t
Hours
30
33
34
35
39
44
45
h
Millimeters
1.10
1.21
1.25
1.23
1.30
1.40
1.42
t2
Hours 2
900
1089
1156
1225
1521
1936
2025
th
Millimeter-Hour
33
39.93
42.50
43.05
50.70
61.6
63.9
260
8.91
9852
334.68
7

i 1
06.03.1
06.03.2
Chapter 06.03
n7
7
7
7
n t i hi  t i  hi
a1 
i 1
i 1
i 1


n t    t i 
i 1
 i 1 
7
7
2
2
i

7334.68  2608.91
79852  260
 0.019179 mm-h
2
7
_
h
h
i
i 1
n
8.91

7
 1.2729 mm
7
_
t
t
i 1
i
n
260

7
 37.143 h
_
_
a 0  h  a1 t
 1.2729  0.01917937.143
 0.56050 mm-h
Linear Regression-More Examples: Industrial Engineering
h  0.56050  0.019179t
Figure 1 Linear regression of hours of use vs. millimeters off target.
Solving for h  2 mm , the regression model is h  0.56050  0.019179t
2  056050  0.019179t
2  0.56050
t
0.019179
t  75.056 hours
06.03.3