The manager of a plate steel fabricator plant wants to develop a standard data s
ID: 3208764 • Letter: T
Question
The manager of a plate steel fabricator plant wants to develop a standard data system to
estimate the time required to arc weld the plates together to form the welded assembly.
The standard data system will be based on the use of a mathematical formula in which the
arc welding time depends on three variables in the job: (1) the length of the weld path, (2)
the plate thickness, and (3) the number of parts in the welded assembly. For a given
assembly, the plate thickness would be the same for all parts. Because the plant
specializes in low carbon plate steel, the welding voltage, current, and other machine
settings are constant for all jobs to be included in the coverage of this standard data
system. The data that have been collected from previous direct time studies are presented
in the table below.
Develop an equation to predict the welding time based on length, thickness, and
number of parts using linear regression in Excel. You may not need to include
every variable. Compute a regression equation r value (correlation coefficient) for
the final equation.
Do you think the equation fits the data well, based on the r value? (Give a yes/no
and a reason.)
Define the range of values for length, thickness, and number of parts for which
your equation is valid
Length L (m) Thickness t (mm) Number of parts np Welding Time T (min) 0.68 6 2 2.47 3.26 8 5 10.85 1.45 6 3 5.21 2.10 10 7 8.69 2.66 6 4 8.71 4.75 6 8 16.35 0.81 10 3 3.30 1.82 8 4 7.01 1.49 8 5 5.61 2.05 6 4 6.99 3.79 6 2 10.56 2.92 8 3 9.65 4.22 10 7 14.65 3.66 10 5 12.45 1.78 8 4 6.61Explanation / Answer
Result:
Develop an equation to predict the welding time based on length, thickness, and
number of parts using linear regression in Excel. You may not need to include
every variable. Compute a regression equation r value (correlation coefficient) for
the final equation.
Regression Analysis
R²
0.995
Adjusted R²
0.994
n
15
R
0.998
k
3
Std. Error
0.308
Dep. Var.
Welding Time T (min)
ANOVA table
Source
SS
df
MS
F
p-value
Regression
215.6464
3
71.8821
755.58
5.14E-13
Residual
1.0465
11
0.0951
Total
216.6929
14
Regression output
confidence interval
variables
coefficients
std. error
t (df=11)
p-value
95% lower
95% upper
Intercept
-0.8193
0.4217
-1.943
.0780
-1.7474
0.1088
Length L (m)
2.7777
0.0821
33.846
1.79E-12
2.5971
2.9584
Thickness t (mm)
0.0830
0.0547
1.519
.1571
-0.0373
0.2033
Number of parts np
0.4208
0.0606
6.948
2.43E-05
0.2875
0.5541
The regression equation
Welding Time =-0.8193 +2.7777 * Length L (m)+ 0.0830 *Thickness t (mm)+ 0.4208 *Number of parts np
R=0.998
Do you think the equation fits the data well, based on the r value? (Give a yes/no and a reason.)
Yes. R square =0.995.
99.5% of variance in welding time is explained by the model. The equation fits the data well.
Define the range of values for length, thickness, and number of parts for which your equation is valid
The equation is valid for,
length (0.68 to 4.75), thickness ( 6 to 10), and number of parts (2 to 8).
Regression Analysis
R²
0.995
Adjusted R²
0.994
n
15
R
0.998
k
3
Std. Error
0.308
Dep. Var.
Welding Time T (min)
ANOVA table
Source
SS
df
MS
F
p-value
Regression
215.6464
3
71.8821
755.58
5.14E-13
Residual
1.0465
11
0.0951
Total
216.6929
14
Regression output
confidence interval
variables
coefficients
std. error
t (df=11)
p-value
95% lower
95% upper
Intercept
-0.8193
0.4217
-1.943
.0780
-1.7474
0.1088
Length L (m)
2.7777
0.0821
33.846
1.79E-12
2.5971
2.9584
Thickness t (mm)
0.0830
0.0547
1.519
.1571
-0.0373
0.2033
Number of parts np
0.4208
0.0606
6.948
2.43E-05
0.2875
0.5541
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