The accompanying data resulted from an experiment in which weld diameter x and s
ID: 2923501 • Letter: T
Question
The accompanying data resulted from an experiment in which weld diameter x and shear strength y (in pounds) were determined for five different spot welds on steel. A scatterplot shows a pronounced linear pattern. The least-squares line is [y hat] = -967.38 + 8.60x. Because 1 lb = 0.4536 kg, strength observations can be re-expressed in kilograms through multiplication by this conversion factor: new y = 0.4536(old y). What is the equation of the least-squares line when y is expressed in kilograms? (Give the answer to two decimal places.)
[y hat] =
x 202.9 212.9 222.9 232.9 242.8
y 812.1 783.7 958.8 1116.4 1074.6
Explanation / Answer
Line of Regression Y on X i.e Y = bo + b1 X
converting Y into Kilo grams.
Line of Regression Y on X i.e Y = bo + b1 X
calculation procedure for regression
mean of X = X / n = 222.88
mean of Y = Y / n = 101.0984
(Xi - Mean)^2 = 996.008
(Yi - Mean)^2 = 204.93
(Xi-Mean)*(Yi-Mean) = 451.78923
b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2
= 451.78923 / 996.008
= 0.4536
bo = Y / n - b1 * X / n
bo = 101.0984 - 0.4536*222.88 = 0.00003
value of regression equation is, Y = bo + b1 X
Y'=0.00003+0.4536* X
~
Y'=0.0+0.45* X
Y y = 0.4536(old y). 812.1 368.36856 783.7 355.48632 958.8 434.91168 1116.4 506.39904 1074 487.1664Related Questions
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