Solution Found!
Yield strength of steel alloy. Industrial engineers at
Chapter 12, Problem 155SE(choose chapter or problem)
Problem 155SE
Yield strength of steel alloy. Industrial engineers at theUniversity of Florida used regression modeling as atool to reduce the time and cost associated with developing new metallic alloys (Modelling and Simulationin Materials Science and Engineering, Vol. 13, 2005).To illustrate, the engineers built a regression modelfor the tensile yield strength (y) of a new steel alloy.
The potentially important predictors of yield strength arelisted in the accompanying table.
|
x 1 = Carbon amount (% weight) x 2 = Manganese amount (% weight) x 3 = Chromium amount (% weight) x 4 = Nickel amount (% weight) x 5 = Molybdenum amount (% weight) x 6 = Copper amount (% weight) x 7 = Nitrogen amount (% weight) x 8 = Vanadium amount (% weight) x 9 = Plate thickness (millimeters) x 10 = Solution treating (milliliters) x 11 = Aging temperature (degrees, Celsius) |
a. The engineers discovered that the variable Nickel (x4)was highly correlated with the other potential independent variables. Consequently, Nickel was droppedfrom the model. Do you agree with this decision?Explain.
b. The engineers used stepwise regression on theremaining 10 potential independent variables in orderto search for a parsimonious set of predictor variables.Do you agree with this decision? Explain.
c. The stepwise regression selected the following independent variables: x1 = Carbon, x2 = Manganese,x3 = Chromium, x5 = Molybdenum, x6 = Copper,x8 = Vanadium, x9 = Plate thickness, x10 = Solutiontreating, and x11 = Aging temperature. All thesevariables were statistically significant in the stepwise model, with R2 = .94. Consequently, theengineers used the estimated stepwise model to predict yield strength. Do you agree with this decision?Explain
Questions & Answers
QUESTION:
Problem 155SE
Yield strength of steel alloy. Industrial engineers at theUniversity of Florida used regression modeling as atool to reduce the time and cost associated with developing new metallic alloys (Modelling and Simulationin Materials Science and Engineering, Vol. 13, 2005).To illustrate, the engineers built a regression modelfor the tensile yield strength (y) of a new steel alloy.
The potentially important predictors of yield strength arelisted in the accompanying table.
|
x 1 = Carbon amount (% weight) x 2 = Manganese amount (% weight) x 3 = Chromium amount (% weight) x 4 = Nickel amount (% weight) x 5 = Molybdenum amount (% weight) x 6 = Copper amount (% weight) x 7 = Nitrogen amount (% weight) x 8 = Vanadium amount (% weight) x 9 = Plate thickness (millimeters) x 10 = Solution treating (milliliters) x 11 = Aging temperature (degrees, Celsius) |
a. The engineers discovered that the variable Nickel (x4)was highly correlated with the other potential independent variables. Consequently, Nickel was droppedfrom the model. Do you agree with this decision?Explain.
b. The engineers used stepwise regression on theremaining 10 potential independent variables in orderto search for a parsimonious set of predictor variables.Do you agree with this decision? Explain.
c. The stepwise regression selected the following independent variables: x1 = Carbon, x2 = Manganese,x3 = Chromium, x5 = Molybdenum, x6 = Copper,x8 = Vanadium, x9 = Plate thickness, x10 = Solutiontreating, and x11 = Aging temperature. All thesevariables were statistically significant in the stepwise model, with R2 = .94. Consequently, theengineers used the estimated stepwise model to predict yield strength. Do you agree with this decision?Explain
ANSWER:Step 1 of 4
Let y = tensile yield strength of a new steel alloy. 
