Question

standard data system will be based on the use of a mathema Welding ume depends on three variables in the job: (1) the length of the weld path, (2) the TICKness, 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 into bon 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. (a) Use regres- sion analysis to derive an equation that relates welding time to these work variables. Let 7 the welding time, min; L-length of weld path, mit = plate thickness, mm; and nenum- ber of parts in the welded assembly. (b) Test your equation with the following values: 5.0 m, 8 mm, and n = 6 parts. Length, L (m) Thickness, (mm) Number of Parts.n, Welding Time, T (min) 2.47 10.85 0.68 3.26 1.45 2.10 2.66 4.75 0.81 1.82 1.49 2.05 3.79 2.92 4.22 3.66 1.78 8.69 8.71 16.35 3.30 701 5.61 699 10.56 9.65 14.65 12.45 6,61

Answer 1

I have used Excel for finding the regression equation.

Enter Time , Length, Thickness, Number of parts in Columns A, B,C and D.

Select Data -> Data Analysis -> Regression ->Ok -> a dialogue box appears -> In 'Input Y Range' select the range containing the data of Time -> In 'Input X Range' select the range containing the data of the other 3 variables ->Enable 'output range' -> select any blank column in the sheet -> Ok -> The output will appear as follows:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.997582
R Square	0.995171
Adjusted R Square	0.993854
Standard Error	0.308441
Observations	15
ANOVA
	df	SS	MS	F	Significance F
Regression	3	215.6464	71.88213	755.5757	5.14E-13
Residual	11	1.046491	0.095136
Total	14	216.6929
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-0.8193	0.421674	-1.94298	0.078045	-1.7474	0.108795	-1.7474	0.108795
X Variable 1	2.777736	0.082069	33.8463	1.79E-12	2.597103	2.958369	2.597103	2.958369
X Variable 2	0.083002	0.054656	1.518623	0.157062	-0.0373	0.203298	-0.0373	0.203298
X Variable 3	0.420802	0.060562	6.948289	2.43E-05	0.287506	0.554098	0.287506	0.554098

From the above output, the regression line can be constructed as:

T=-0.8193+2.777736 L+0.83002 t+0.420802 n_p

The estimated time for L=5.0 m, t=8 mm, n_p= 6 parts is

T= -0.8193+2.777736*5+0.83002*8+0.420802*6 = 22.234 min