Nonlinear Relationships using Polynomial Coding

Blindsided by nonlinearity

We submitted a paper on the relationship between conscientiousness and test performance. In contrast to much prior research, we did not find a significant positive relationship. We suggested the lack of significance might be due to low power. Here’s what we said in the original submission . . .

“One explanation for the finding of insignificant relationships in the above analyses is lack of power. Assuming the population correlation of conscientiousness to performance is .2, the power of the test of the conscientiousness-performance relationship was .65, somewhat smaller than the .80 that is the typical recommendation. So lack of power is one potential explanation for above results.”

A reviewer had the following comments . . .

“Regarding the zero correlation of conscientiousness and performance the authors offer an explanation using power. This is not very convincing since they base their conclusion on the correlation between the factor score and performance which is troubled with the issue described above. However, there are also other possibilities. The authors should check the scatter plot for outliers which distort the relationship. A theoretical explanation could be given if more were known about the intelligence distribution in the sample. There is evidence for a curvilinear relationship between conscientiousness and intelligence (LaHuis, Martin, & Avis, 2005). The reasoning is that intelligent people do not need to be conscientious in order to achieve. Thus, if the participants were very intelligent, this might also explain the small correlation.

One of the things you learn when attempting to publish articles in peer reviewed journals is the ins and outs of reviewers. I whole-heartedly support the process. But it is sometimes a pain.

We thought that the logic of the reviewer’s argument was a stretch, but had to address the issue for the sake of getting published. So, to address the reviewer’s comments we examined the nature of the relationship of conscientiousness to test score.

We tested the hypothesis that the relationship was quadratic, which is the type of relationship mentioned in the LaHuis, Martin, & Avis (2005) reference given by the reviewer. More on quadraticness later. The point here is that we were forced to address the issue of nonlinearity.

Here is what we said in the revision that was resubmitted . . (It worked – the paper was accepted)

“FourThree potential explanations for the nonsignficant relationships involving conscientiousness were considered. First, the conscientiousness-test scatterplot revealed no outliers. Second, following LaHuis, Martin, & Avis (2005), adding a quadratic component to the conscientiousness-performance relationship did not result in a significant increase in r2(t=1.745, p > .05). Third, power of the test of the conscientiousness-test correlation was computed. Assuming the population value to be .2, a conservative estimate of the conscientiousness-performance relationship from previous studies, power was found to be .65. None of these considerations provided a definitive explanation for the small correlations shown in Table 2.” (I’m not sure what the fourth potential explanation was.)

By the way . . . LaHuis et al. (2005) did NOT find a curvilinear relationship between INTELLIGENCE and Conscientiousness. They found a curvilinear relationship between Job Performance and Conscientiousness. Here’s a figure from their paper. The vertical axis is Job Performance, not Intelligence. So, the reviewer had his/her facts mixed up.

This is an example of a situation in which we needed to be able to investigate nonlinear relationships.

The issue of nonlinearity has been studied fairly frequently in psychology. For example . . .

Le, H., Oh, I., Robbins, S. B., Ilies, R., Holland, E., & Westrick, P. (2011). Too much of a good thing: Curvilinear relationships between personality traits and job performance. Journal of Applied Psychology, 96, 113-133.

Arneson, J. J., Sackett, P. R., & Beatty, A. S. (2011). Ability-performance relationships in education and employment settings: Critical tests of the more-is-better and the good-enough hypotheses. Psychological Science, ??, ???-???. DOI: 10.1177/0956797611417004 (Don’t have page numbers in my version of paper.)

There are other references at the end of this lecture.

A Couple of the 1000s of possible nonlinear relations

LogarithmicY = k*log10(X)

data list free /x.

begin data.

1 10 20 30 40 50 100 200 500 1000

end data.

compute ylog = lg10(x).

graph /scatterplot=x with ylog.

ExponentialY = k*ea+bx

data list free /x.

begin data.

0 .5 1 1.5 2 2.5 3 3.5 4

end data.

compute yexp = exp(x).

graph /scatterplot=x with yexp.

If
Problem – there are 1000s of possibilities

Do we have to try each different type in order to determine whether or not a relationship is nonlinear?

We need a way to subsume them into one general technique.

Solution – Approximate with Polynomial relations

General form: Y = a + b1X1 + b2X2 + b3X3 + . . .

Special cases of the general form . . .

Linear:Y = a + b1X1

Quadratic:Y = a + b1X1 + b2X2

A quadratic function has the general form of a huge valley or a huge hill depending on the value of b2.

b1 neg; b2 posb1 neg; b2 neg

If b2 is positive, the opening of the “cup” is pointing upwards, as on the left – a huge valley.

If b2 is negative, the opening of the cup is pointing downwards, as on the right – a huge hill.

Key fact: Different values of b1 and b2 yield different shapes that look almost like the logarithmic or exponential above.

The point of this is that depending on the values of b1 and b2, a quadratric function may look almost identical to many different nonlinear relationships. So we can use quadratic functions to mimic those nonlinear relationships. All we have to estimate is a few parameters – b1, b2, perhaps b3 and b4.

Cubic:Y = a + b1X + b2X2 + b3X3

Cubic functions have both an upward pointing “cup” and a downward pointing “cup”, as in the following . . .

Neither quadratic, cubic, quartic, quintic functions are of much interest in their own right. They’re used primarily to approximate specific naturally occurring shapes that might have more complex expressions.

Most of the time, our data cover only part of the range of a function, so we rarely see the full curve of a quadratic or cubic function.

The bottom line is that polynomial functions are like shape chameleons – they can change their shape to match their surroundings (the data).

We can perform statistical tests of the nonlinear aspects of those shapes.
Making our statistical package think it’s working with a linear equation

The standard formulas for solving for regression parameters assume that all of the predictors are raised to only the first power – linear relationships. So that raises the question of how to estimate coefficients in

That is – our regression programs work only with variables raised to power of 1.

But the equations we’re interested in have X raised to the 2nd, 3rd, etc power . . .

Y = a + B1X + B2X2 + B3X3

Solution: Put the powered values of X into the statistical package. Don’t tell it they’re powered.

Let V11=X

V21=X2

V31=X3

Note the difference – V’s are not “powered”.

What the statistical package sees: Y = a + B1V1 + B2V2 + B3V3

So, the Regression procedure thinks it’s analyzing a linear relationship.

(Like putting pills for your dog in peanut butter.)

Data matrix

YV1 (X in disguise)V2 (X2 in disguise)V3 (X3 in disguise)

1864512

3152253375

62984124389

Etc.

We are doing multiple LINEAR regression with the Vs. The program doesn’t know that the Vs are actually Xs raised to powers. (Shh – don’t tell it.)

The Vs are like aliens who have invaded the world and disguised themselves as regular humans. We (the regression procedure) cannot see the real shape of these creatures.

When we get a solution in terms of the Vs, if we need to, we then can covert that solution into an equivalent solution in terms of the Xs (like unmaking the aliens).

Another Technical Issue

Vs are often highly correlated with each other. They are multicollinear.

This means that it won’t work to simply enter all of them into the equation at the same time – in what is called a simultaneous analysis of the predictors.

3 ways of doing polynomial regression

1. Hierarchical analysis with the V variables – for quick and dirty analysis

2.Hierarchical analysis with centered V variables – for publication

3.Substituting Vs with OrthogonalCoefficients – for publication

Examples of each way of doing polynomial regression . . .

1. Performing a hierarchical analysis with the V variables.

a. Enter V1 only and test it. This is a test for linear trend.

b. Enter V1 and V2 and test only V2. This is a test for the existence of quadratic trend.

c. Enter V1, V2, and V3 and test only V3. This tests for the existence of cubic trend.

d. Enter V1, V2, V3 and V4 and test only V4. This tests for the existence of a quartic trend.

And so forth.

e. Stop when last V tested is not significant and the next-to-last V WAS significant. Conclude that the power of the polynomial equation is that of the next-to-last V. I sometimes test for two powers above the last significant coefficient.

If V1 is significant, test V2. If V2 is not, conclude that there is only a linear trend.

If V2 is significant, test V3. If V3 is not, conclude that there is a quadratic trend (V2) and interpret the coefficients of the quadratic equation.

If V3 is significant, test V4. If V4 is not, conclude that there is a cubic relationship (V3) and interpret it. Otherwise (if you’re young) continue.

Example Hierarchical Analysis of Vs

Pedhazur. P.522. The effect of practice time (PRACTIME) on performance (NUMRIGHT).

Scatterplot of relationship of NUMRIGHT to PRACTIME.

It appears that the relationship of NUMRIGHT to PRACTIME is curved downward.

Issues . . .

1. Is the curvilinearity significant?

2. What is the prediction formula?

NUMRIGHT PRACTIME X2 X3

Y V1 V2 V3

4 2 4 8

6 2 4 8

5 2 4 8

7 4 16 64

10 4 16 64

10 4 16 64

13 6 36 216

14 6 36 216

15 6 36 216

16 8 64 512

17 8 64 512

21 8 64 512

18 10 100 1000

19 10 100 1000

20 10 100 1000

19 12 144 1728

20 12 144 1728

21 12 144 1728

Note that the predictors do not have to be called “Vs”. Calling them Vs is just a pedagogical device introduced by me to represent the idea that SPSS does not know that the variables are actually powers of the Xs.

Syntax of the regression

regression variables = numright practime timesqr timecub

/descriptives = default / statistics = default cha

/dep=numright

/enter practime

/enter timesqr

/enter timecub.

Regression

The high correlations among X, X2, X3, etc is the reason we have to enter the predictors hierarchically.


Note that the coefficients and the ts and ps for each lower level variable change as the higher level variables were added to the equation. This means that it’s very important to interpret only the appropriate set of coefficients – that set just before the model in which the highest power is not significant – the second model in this case.

The final prediction equation

YHAT = a + B1 * V1 + B2*V2

YHAT = a + B1 *X1 + B2*X2

So, YHAT = -1.9000 + 3.495*X - .138*X2

Graphing predicted and observed points.

(Note: You could rerun regression, specifying the quadratic term and have SPSS compute and save Y-hats.)

First, create predicted points using the Compute Variable... command.

Computing YHAT in SPSS . . .In Data Editor, Transform -> Compute…

Creating the scatterplot of predicted and observed points.

Create an overlay plot of

a. Y’s vs. X’s, and

b. Yhats vs. X’s.

Graphs -> Scatterplot -> Overlay

Difficulty with the hierarchical entry method: Vs may be so highly correlated that computer won’t perform the analysis.

Warning: One of the assignment problems on polynomials is like this. This leads to the 2nd solution . . .

2. Center the predictors about the mean of the Xs and perform a hierarchical analysis.

Center the Xsby subtracting the mean from each X. Call the result, Centered V1 or CV1,

Square CV1 to form CV2

Cube CV1 to form CV3,

and so forth.

So we begin with

XX2X3X4etc. in the data editor

We create

X-M(X-M)2(X-M)3(X-M)4etc.

CV1CV2CV3CV4etc.

The same hierarchical analysis described above is performed.

Hierarchical Analysis of Centered Variables

(The mean of the Xs is 7. It just happened to be a whole number – coincidence.)

Y X (X-7)1 (X-7)2 (X-7)3

NUMRIGHTPRACTIMECENTTIM1CENTTIM2CENTTIM3

4 2 -5 25 -125

6 2 -5 25 -125

5 2 -5 25 -125

7 4 -3 9 -27

10 4 -3 9 -27

10 4 -3 9 -27

13 6 -1 1 -1

14 6 -1 1 -1

15 6 -1 1 -1

16 8 1 1 1

17 8 1 1 1

21 8 1 1 1

18 10 3 9 27

19 10 3 9 27

20 10 3 9 27

19 12 5 25 125

20 12 5 25 125

21 12 5 25 125

126

regression variables = numright centtim1 centtim2 centtim3

/descriptives = default /sta = default cha

/dep=numright /enter centtim1 /enter centtim2 /enter centtim3.

Regression

Model Summary
Model / R / R Square / Adjusted R Square / Std. Error of the Estimate / Change Statistics
R Square Change / F Change / df1 / df2 / Sig. F Change
1 / .940a / .883 / .876 / 2.051 / .883 / 121.029 / 1 / 16 / .000
2 / .971b / .943 / .935 / 1.483 / .060 / 15.604 / 1 / 15 / .001
3 / .973c / .946 / .935 / 1.487 / .003 / .911 / 1 / 14 / .356
a. Predictors: (Constant), CENTTIM1
b. Predictors: (Constant), CENTTIM1, CENTTIM2
c. Predictors: (Constant), CENTTIM1, CENTTIM2, CENTTIM3
ANOVAa
Model / Sum of Squares / df / Mean Square / F / Sig.
1 / Regression / 509.186 / 1 / 509.186 / 121.029 / .000b
Residual / 67.314 / 16 / 4.207
Total / 576.500 / 17
2 / Regression / 543.507 / 2 / 271.754 / 123.551 / .000c
Residual / 32.993 / 15 / 2.200
Total / 576.500 / 17
3 / Regression / 545.524 / 3 / 181.841 / 82.185 / .000d
Residual / 30.976 / 14 / 2.213
Total / 576.500 / 17
a. Dependent Variable: NUMRIGHT
b. Predictors: (Constant), CENTTIM1
c. Predictors: (Constant), CENTTIM1, CENTTIM2
d. Predictors: (Constant), CENTTIM1, CENTTIM2, CENTTIM3


Coefficientsa
Model / Unstandardized Coefficients / Standardized Coefficients / t / Sig.
B / Std. Error / Beta
1 / (Constant) / 14.167 / .483 / 29.303 / .000
CENTTIM1 / 1.557 / .142 / .940 / 11.001 / .000
2 / (Constant) / 15.781 / .538 / 29.342 / .000
CENTTIM1 / 1.557 / .102 / .940 / 15.215 / .000
CENTTIM2 / -.138 / .035 / -.244 / -3.950 / .001
3 / (Constant) / 15.781 / .539 / 29.256 / .000
CENTTIM1 / 1.814 / .288 / 1.095 / 6.294 / .000
CENTTIM2 / -.138 / .035 / -.244 / -3.939 / .001
CENTTIM3 / -.013 / .013 / -.166 / -.955 / .356
Nota. Dependent Variable: NUMRIGHT
Excluded Variablesa
Model / Beta In / t / Sig. / Partial Correlation / Collinearity Statistics
Tolerance
1 / CENTTIM2 / -.244b / -3.950 / .001 / -.714 / 1.000
CENTTIM3 / -.166b / -.681 / .506 / -.173 / .127
2 / CENTTIM3 / -.166c / -.955 / .356 / -.247 / .127
a. Dependent Variable: NUMRIGHT
b. Predictors in the Model: (Constant), CENTTIM1
c. Predictors in the Model: (Constant), CENTTIM1, CENTTIM2

Transforming the Centered results into Raw results . . .

YHAT = 15.781 + 1.557*(practime-7) - .138*(practime-7)2

(Recall that 7 = mean of the X’s.)

We want to show the relationship of YHATS to Xs along with the relationship of Ys to Xs.

To create the appropriate graph, you have to translate the above equation expressing YHAT as a function of (PRACTIME-7) into an equation expressing YHAT as a function of X: PRACTIME.

The algebra using X instead of PRACTIME to simplify the symbols is as follows

YHAT = 15.781 + 1.557(X-7) - .138(X-7)2.

YHAT = 15.781 + 1.557X – 10.889 - .138(X2 – 14X + 49)

(When’s the last time you squared a binomial?)

YHAT = 15.781 + 1.557X – 10.889 - .138X2 + 1.932X – 6.762

YHAT = 15.781 – 10.889 – 6.762 + 1.557X + 1.932X - .138X2.

YHAT = -1.870 + 3.489X - .138X2.

This is theoretically the same equation we got with uncentered X’s.

What we got above was: YHAT = -1.9000 + 3.495*X - .138*X2

It’s slightly different due to rounding error.

Use TRANSFORM -> COMPUTE…to create the YHAT and do the same overlay plot as done above.

3. Using orthogonal polynomial contrasts

The orthogonal polynomial proceduredescribed here works best if there are just a few X values – say no more than 10 or so.

Procedure . . .

1. Treat each set of cases with the same value of X as a group. The X values must be equally spaced on the number line – 1, 2, 3, 4, 5, 6, . . . or 2, 4, 6, 8, 10, . .. or 0, 3, 6, 9, 12, . . . etc.

2. Create (i.e., get from a table prepared by some really smart person) K-1 orthogonal contrast codes between the groupsordered on X. Differences between X values of adjacent groups must be equal.

3. Perform a simultaneous(not hierarchical) regression of the dependent variable onto the K-1 contrast coefficients.

You may perform the regression on fewer than K-1 contrasts if you wish. You’ll lose power to detect relations if you do.

Coefficients for each contrast can be obtained from tables.

Below is a table of orthogonal polynomial contrast codes from (which no longer is available.)

# X values

GroupLinearQuadraticCubicQuarticQuintic

1 -1 1

2 0 -2

3 1 1

1 -3 1 -1

2 -1 -1 3

3 1 -1 -3

4 3 1 1

1 -2 2 -1 1

2 -1 -1 2 -4

3 0 -2 0 6

4 1 -1 -2 -4

5 2 2 1 1

1 -5 5 -5 1 -1

2 -3 -1 7 -3 5

3 -1 -4 4 2 -10

4 1 -4 -4 2 10

5 3 -1 -7 -3 -5

6 5 5 5 1 1

The Analysis using Just 3 Orthogonal Polynomials

Even though the independent variable PRACTIME varies from 2 to 12, it has only 6 discrete values, so orthogonal polynomials for a 6-level IV were used. Only the first 3 of the 5 possible contrasts are analyzed here. The contrast values from the first column above are in the column named ORTHTIM1. The contrast values from the 2nd column above are in ORTHTIM2, etc.

NUMRIGHT PRACTIME ORTHTIM1 ORTHTIM2 ORTHTIM3

Y X LIN QUAD CUB

4 (G1) 2 -5 5 -5

6 2 -5 5 -5

5 2 -5 5 -5

7 (G2) 4 -3 -1 7

10 4 -3 -1 7

10 4 -3 -1 7

13 (G3) 6 -1 -4 4

14 6 -1 -4 4

15 6 -1 -4 4

16 (G4) 8 1 -4 -4

17 8 1 -4 -4

21 8 1 -4 -4

18 (G5)10 3 -1 -7

19 10 3 -1 -7

20 10 3 -1 -7

19 (G6)12 5 5 5

20 12 5 5 5

21 12 5 5 5

Number of cases read: 18 Number of cases listed: 18

Part of the table on the previous page for reference . . .

GroupLinearQuadraticCubicQuarticQuintic

1 -5 5 -5 -5 -1

2 -3 -1 7 -3 5

3 -1 -4 4 2 -10

4 1 -4 -4 2 10

5 3 -1 -7 -3 -5

6 5 5 5 1 1

The analysis using just 3 orthogonal polynomials

Since the coefficients are orthogonal, we do NOT have to use hierarchical analysis.

We can enter all of the predictors at the same time.

regression variables = numright orthtim1 orthtim2 orthtim3

/descriptives = default /statistics = default cha

/dep = numright /enter orthtim1 orthtim2 orthtim3.

Regression

Descriptive Statistics
Mean / Std. Deviation / N
NUMRIGHT / 14.17 / 5.823 / 18
ORTHTIM1 / .00 / 3.515 / 18
ORTHTIM2 / .00 / 3.850 / 18
ORTHTIM3 / .00 / 5.636 / 18
Correlations
NUMRIGHT / ORTHTIM1 / ORTHTIM2 / ORTHTIM3
Pearson Correlation / NUMRIGHT / 1.000 / .940 / -.244 / -.059
ORTHTIM1 / .940 / 1.000 / .000 / .000
ORTHTIM2 / -.244 / .000 / 1.000 / .000
ORTHTIM3 / -.059 / .000 / .000 / 1.000
Variables Entered/Removeda
Model / Variables Entered / Variables Removed / Method
1 / ORTHTIM3, ORTHTIM2, ORTHTIM1b / . / Enter
a. Dependent Variable: NUMRIGHT
b. All requested variables entered.
Model Summary
Model / R / R Square / Adjusted R Square / Std. Error of the Estimate / Change Statistics
R Square Change / F Change / df1 / df2 / Sig. F Change
1 / .973a / .946 / .935 / 1.487 / .946 / 82.185 / 3 / 14 / .000
a. Predictors: (Constant), ORTHTIM3, ORTHTIM2, ORTHTIM1
ANOVAa
Model / Sum of Squares / df / Mean Square / F / Sig.
1 / Regression / 545.524 / 3 / 181.841 / 82.185 / .000b
Residual / 30.976 / 14 / 2.213
Total / 576.500 / 17
a. Dependent Variable: NUMRIGHT
b. Predictors: (Constant), ORTHTIM3, ORTHTIM2, ORTHTIM1
Coefficientsa
Model / Unstandardized Coefficients / Standardized Coefficients / t / Sig.
B / Std. Error / Beta
1 / (Constant) / 14.167 / .351 / 40.407 / .000
ORTHTIM1 / 1.557 / .103 / .940 / 15.170 / .000
ORTHTIM2 / -.369 / .094 / -.244 / -3.939 / .001
ORTHTIM3 / -.061 / .064 / -.059 / -.955 / .356
a. Dependent Variable: NUMRIGHT
Since the predictors (the polynomial coefficients) are orthogonal, hierarchical analysis is not required.
Just one single analysis, in which all predictors are entered, is sufficient.

The same analysis using all possible orthogonal polynomials – all 5

NUMRIGHT LIN QUAD CUBE QUAR QUIN

4 -5 5 -5 1 -1

6 -5 5 -5 1 -1

5 -5 5 -5 1 -1

7 -3 -1 7 -3 5

10 -3 -1 7 -3 5

10 -3 -1 7 -3 5

13 -1 -4 4 2 -10

14 -1 -4 4 2 -10

15 -1 -4 4 2 -10

16 1 -4 -4 2 10

17 1 -4 -4 2 10

21 1 -4 -4 2 10

18 3 -1 -7 -3 -5

19 3 -1 -7 -3 -5

20 3 -1 -7 -3 -5

19 5 5 5 1 1

20 5 5 5 1 1

21 5 5 5 1 1

regression variables = numrightlin quad cube quar quin

/descriptives = default /statistics = default cha

/dep = numright/enter lin quad cube quar quin.

Regression

Descriptive Statistics
Mean / Std. Deviation / N
NUMRIGHT / 14.17 / 5.823 / 18
LIN / .00 / 3.515 / 18
QUAD / .00 / 3.850 / 18
CUBE / .00 / 5.636 / 18
QUAR / .00 / 2.223 / 18
QUIN / .00 / 6.669 / 18
Correlations
NUMRIGHT / LIN / QUAD / CUBE / QUAR / QUIN
Pearson Correlation / NUMRIGHT / 1.000 / .940 / -.244 / -.059 / .068 / .023
LIN / .940 / 1.000 / .000 / .000 / .000 / .000
QUAD / -.244 / .000 / 1.000 / .000 / .000 / .000
CUBE / -.059 / .000 / .000 / 1.000 / .000 / .000
QUAR / .068 / .000 / .000 / .000 / 1.000 / .000
QUIN / .023 / .000 / .000 / .000 / .000 / 1.000
Variables Entered/Removeda
Model / Variables Entered / Variables Removed / Method
1 / QUIN, QUAR, CUBE, QUAD, LINb / . / Enter
a. Dependent Variable: NUMRIGHT
b. All requested variables entered.
Model Summary
Model / R / R Square / Adjusted R Square / Std. Error of the Estimate / Change Statistics
R Square Change / F Change / df1 / df2 / Sig. F Change
1 / .975a / .951 / .931 / 1.528 / .951 / 47.014 / 5 / 12 / .000
a. Predictors: (Constant), QUIN, QUAR, CUBE, QUAD, LIN
ANOVAa
Model / Sum of Squares / df / Mean Square / F / Sig.
1 / Regression / 548.500 / 5 / 109.700 / 47.014 / .000b
Residual / 28.000 / 12 / 2.333
Total / 576.500 / 17
a. Dependent Variable: NUMRIGHT
b. Predictors: (Constant), QUIN, QUAR, CUBE, QUAD, LIN
Coefficientsa
Model / Unstandardized Coefficients / Standardized Coefficients / t / Sig.
B / Std. Error / Beta
1 / (Constant) / 14.167 / .360 / 39.347 / .000
LIN / 1.557 / .105 / .940 / 14.772 / .000
QUAD / -.369 / .096 / -.244 / -3.835 / .002
CUBE / -.061 / .066 / -.059 / -.930 / .371
QUAR / .179 / .167 / .068 / 1.071 / .305
QUIN / .020 / .056 / .023 / .357 / .727
a. Dependent Variable: NUMRIGHT

Note that the p-values are not identical to those for the raw and centered predictors. That’s because we used 5 predictors for this equation instead of 3 in the above. This is a more nearly thorough analysis than those above. The bonus is that it allowed us to absolutely rule out 4th and 5th order predictors, something which we simply assumed in the above analyses.
The above Centered Variables Analysis using Rcmdr

With Rcmdr, you have to perform each regression in the hierarchy separately.

R -> Load Rcmdr

Data --> Import Data --> from SPSS data set --> InclassDatasets/Nonlinear Relationships Pedhazur Data.sav

Statistics --> Fit models --> Linear Model . . .

Output Window . . .

> LinearModel.2 <- lm(numright ~ centtim1, data=Nonlinear1)

> summary(LinearModel.2)

Call:

lm(formula = numright ~ centtim1, data = Nonlinear1)

Residuals:

Min 1Q Median 3Q Max

-2.9524 -1.2738 0.2190 0.9976 5.2762

Coefficients:

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

(Intercept) 14.1667 0.4835 29.3 2.48e-15 ***

centtim1 1.5571 0.1415 11.0 7.17e-09 ***

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.051 on 16 degrees of freedom

Multiple R-squared: 0.8832,Adjusted R-squared: 0.8759