4. A book on robust statistical methods published in June 2006 considers regres-
ID: 3376455 • Letter: 4
Question
4. A book on robust statistical methods published in June 2006 considers regres- sion models for a data set taken from Jalali-Heravi and Knouz (2002). The aim of this modeling is to predict a physical property of chemical compounds called the Krafft point based on four potential predictor variables using a data set of size n = 32. According to Maronna, Martin and Yohai (2006, p. 380)
8 11.5 11.5 11.5 10.5 10.5 10.5 2 9.5 (2 9.5 2 9.5 1.1 1.2 1.3 1.4 1.5 2 1.2 1.6 2.0 2.4 tCylinders 4.5 5.0 5.5 6.0 tHorsepower 8 3 tEngineSize 11.5 11.5 11.5 ^ 10.5 O 10.5 10.5 c2 9.5 2 9.5 0.02 0.03 0.04 0.05 2 2000 3000 4000 2 4.55 4.65 4.75 tHighwayMPG Weight tWheelBase 11.5 10.5 c2 9.5 9.5 10.5 1 11.5 Fitted Values Figure 6.57 Marginal model plots from model (6.37)Explanation / Answer
SolutionA:
Regression model is
KPOINT=70.31+10.47RA+9038VTINV-1826DIPINV+0.3550 HEAT
F=115
p=0.0000
model is significant.
Can be used to predict KPOINT with the above regression eq.
Solutionb:
Upper left: a “residuals plot” to check if the pattern is linear.
Upper right: a probability plot to check the normality of residuals.
Lower left: a plot of standard error versus fitted values, to check whether variance is changing.
Constant variance assumption is violated.
noncompliance with assumptions was dealt with by transforming the KPOINT variable, often using logarithms. Logs frequently produce a straight-line pattern, near-normal residuals and constant variance.
solutionc:
rs q=0.9446
=94.46% varaition in KPOINT is explained by model.Good model
Ad j R sq is also close to R sq.
r sq=0.9446
r=correlation coeffciient=sqrt(0.9446)=0.9719
there exists a strng relationship between KPOINT and other independent variables.
F=115
degrres of Freedom numerator=4
degrees of freedom denominator=27
p=0.0000
p<0.05
model is significant.
means we can use this regression equation to predict KPOINT.
standard error of estimate=3.919
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