Introduction

Contrast coefficients for Trend Analysis

(Valid when X levels are equally spaced & N's are equal)

                 Coefficients, c(i)

 

  r    Trend   X=1   2   3   4   5   6   7       sum c(i)**2

  ---------------------------------------------------------

  3    Linear   -1   0   1                              2

       Quad      1  -2   1                              6

       --------------------------------------------------

  4    Linear   -3  -1   1   3                         20

       Quad      1  -1  -1   1                          4

       Cubic    -1   3  -3   1                         20

       --------------------------------------------------

  5    Linear   -2  -1   0   1   2                     10

       Quad      2  -1  -2  -1   2                     14

       Cubic    -1   2   0  -2   1                     10

       Quartic   1  -4   6  -4   1                     70

       --------------------------------------------------

  6    Linear   -5  -3  -1   1   3   5                 70

       Quad      5  -1  -4  -4  -1   5                 84

       Cubic    -5   7   4  -4  -7   5                180

       Quartic   1  -3   2   2  -3   1                 28

       --------------------------------------------------

  7    Linear   -3  -2  -1   0   1   2   3             28

       Quad      5   0  -3  -4  -3   0   5             84

       Cubic    -1   1   1   0  -1  -1   1              6

       Quartic   3  -7   1   6   1  -7   3            154

       --------------------------------------------------

  8    Linear   -7  -5  -3  -1   1   3   5  7         168

       Quad      7   1  -3  -5  -5  -3   1  7         168

       Cubic    -7   5   7   3  -3  -7  -5  7         264

       Quartic   7 -13  -3   9   9  -3 -13  7         616

       --------------------------------------------------

  9    Linear   -4  -3  -2  -1   0   1   2   3   4     60

       Quad     28   7  -8 -17 -20 -17  -8   7  28   2772

       Cubic   -14   7  13   9   0  -9 -13  -7  14    990

       Quartic  14 -21 -11   9  18   9 -11 -21  14   2002

       --------------------------------------------------

 http://www.ats.ucla.edu/stat/sas/library/SASAnova_mf.htm

Program

data have;

  infile datalines missover;

  input group @;

  do i =1 to 10;

    input y @;

    output;

    end;

datalines;

1  260 200 240 170 270 205 190 200 250 200

2  310 310 190 225 170 210 280 210 280 240

3  320 260 360 310 270 380 240 295 260 250

;

proc glm data =have;

  class group;

  model y =group;

  contrast 'linear trend' group -1 0 1;

  run; quit;

 

#R code

 

#input and formulate data

BLP.Grp1 <- c(260,200,240,170,270,205,190,200,250,200)

BLP.Grp2 <- c(310,310,190,225,170,210,280,210,280,240)

BLP.Grp3 <- c(320,260,360,310,270,380,240,295,260,250)

#Response and Factor(group)

BLP.Grps <- c(BLP.Grp1, BLP.Grp2, BLP.Grp3)

Group    <- rep(1:3, times =1, each =10)

BLP      <- transform(cbind(Group,BLP.Grps), Group =as.factor(Group))

#Construct Contrasts

contrasts(BLP$Group)=contr.poly(levels(BLP$Group)) 

#Fit and printout model

Results

SAS 

Contrast	DF	Contrast SS	Mean Square	F Value	Pr > F
linear trend	1	28880.00000	28880.00000	14.72	0.0007

 

Figure 1: Boxplot for group variable.

http://s2/mw690/002ZOYCigy6FCdOJv5T01&690Trend Test in R/SAS" TITLE="262.Linear Trend Test in R/SAS" />

R

Call:

lm(formula = BLP.Grps ~ Group, data = BLP)

 

Residuals:

   Min     1Q Median     3Q    Max

 -72.5  -32.5  -15.5   36.0   85.5

 

Coefficients:

            Estimate Std. Error t value Pr(>|t|)   

(Intercept)  251.833      8.088  31.136  < 2e-16 ***

 

Group.L       53.740     14.009   3.836 0.000682 *** <<Linear test

Group.Q       11.431     14.009   0.816 0.421654   

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

 

Residual standard error: 44.3 on 27 degrees of freedom

Multiple R-squared:  0.3629,    Adjusted R-squared:  0.3157

F-statistic: 7.691 on 2 and 27 DF,  p-value: 0.002272 #############################################################

 Notice that the results from R and SAS are identical, i.e., t^2 (R) = F, 3.836^2 =14.72(SAS).