linear regression

Proportions as Dependent Variable in Regression–Which Type of Model?

When the dependent variable in a regression model is a proportion or a percentage, it can be tricky to decide on the appropriate way to model it.

The big problem with ordinary linear regression is that the model can predict values that aren’t possible–values below 0 or above 1. But the other problem is that the relationship isn’t linear–it’s sigmoidal. A sigmoidal curve looks like a flattened S–linear in the middle, but flattened on the ends. So now what?

The simplest approach is to do a linear regression anyway. This approach can be justified only in a few situations.

1. All your data fall in the middle, linear section of the curve. This generally translates to all your data being between .2 and .8 (although I’ve heard that between .3-.7 is better). If this holds, you don’t have to worry about the two objections. You do have a linear relationship, and you won’t get predicted values much beyond those values–certainly not beyond 0 or 1.

2. It is a really complicated model that would be much harder to model another way. If you can assume a linear model, it will be much easier to do, say, a complicated mixed model or a structural equation model. If it’s just a single multiple regression, however, you should look into one of the other methods.

A second approach is to treat the proportion as a binary response then run a logistic or probit regression. This will only work if the proportion can be thought of and you have the data for the number of successes and the total number of trials. For example, the proportion of land area covered with a certain species of plant would be hard to think of this way, but the proportion of correct answers on a 20-answer assessment would.

The third approach is to treat it the proportion as a censored continuous variable. The censoring means that you don’t have information below 0 or above 1. For example, perhaps the plant would spread even more if it hadn’t run out of land. If you take this approach, you would run the model as a two-limit tobit model (Long, 1997). This approach works best if there isn’t an excessive amount of censoring (values of 0 and 1).

Reference: Long, J.S. (1997). Regression Models for Categorical and Limited Dependent Variables. Sage Publishing.

Interpreting Interactions in Regression

by Karen Grace-Martin 31 Comments

Adding interaction terms to a regression model has real benefits. It greatly expands your understanding of the relationships among the variables in the model. And you can test more specific hypotheses. But interpreting interactions in regression takes understanding of what each coefficient is telling you.

The example from Interpreting Regression Coefficients was a model of the height of a shrub (Height) based on the amount of bacteria in the soil (Bacteria) and whether the shrub is located in partial or full sun (Sun). Height is measured in cm, Bacteria is measured in thousand per ml of soil, and Sun = 0 if the plant is in partial sun, and Sun = 1 if the plant is in full sun.

SPSS GLM: Choosing Fixed Factors and Covariates

by Karen Grace-Martin 87 Comments

The beauty of the Univariate GLM procedure in SPSS is that it is so flexible. You can use it to analyze regressions, ANOVAs, ANCOVAs with all sorts of interactions, dummy coding, etc.

The down side of this flexibility is it is often confusing what to put where and what it all means.

So here’s a quick breakdown.

The dependent variable I hope is pretty straightforward. Put in your continuous dependent variable.

Fixed Factors are categorical independent variables. It does not matter if the variable is [Read more…] about SPSS GLM: Choosing Fixed Factors and Covariates

Centering for Multicollinearity Between Main effects and Quadratic terms

by Karen Grace-Martin 8 Comments

One of the most common causes of multicollinearity is when predictor variables are multiplied to create an interaction term or a quadratic or higher order terms (X squared, X cubed, etc.).

Why does this happen? When all the X values are positive, higher values produce high products and lower values produce low products. So the product variable is highly correlated with the component variable. I will do a very simple example to clarify. (Actually, if they are all on a negative scale, the same thing would happen, but the correlation would be negative).

In a small sample, say you have the following values of a predictor variable X, sorted in ascending order:

2, 4, 4, 5, 6, 7, 7, 8, 8, 8

It is clear to you that the relationship between X and Y is not linear, but curved, so you add a quadratic term, X squared (X2), to the model. The values of X squared are:

4, 16, 16, 25, 49, 49, 64, 64, 64

The correlation between X and X2 is .987–almost perfect.

To remedy this, you simply center X at its mean. The mean of X is 5.9. So to center X, I simply create a new variable XCen=X-5.9.

These are the values of XCen:

-3.90, -1.90, -1.90, -.90, .10, 1.10, 1.10, 2.10, 2.10, 2.10

Now, the values of XCen squared are:

15.21, 3.61, 3.61, .81, .01, 1.21, 1.21, 4.41, 4.41, 4.41

The correlation between XCen and XCen2 is -.54–still not 0, but much more managable. Definitely low enough to not cause severe multicollinearity. This works because the low end of the scale now has large absolute values, so its square becomes large.

The scatterplot between XCen and XCen2 is:

Plot of Centered X vs. Centered X squared

If the values of X had been less skewed, this would be a perfectly balanced parabola, and the correlation would be 0.

Tonight is my free teletraining on Multicollinearity, where we will talk more about it. Register to join me tonight or to get the recording after the call.

Regression Through the Origin

by Karen Grace-Martin 3 Comments

I just wanted to follow up on my last post about Regression without Intercepts.

Regression through the Origin means that you purposely drop the intercept from the model. When X=0, Y must = 0.

The thing to be careful about in choosing any regression model is that it fit the data well. Pretty much the only time that a regression through the origin will fit better than a model with an intercept is if the point X=0, Y=0 is required by the data.

Yes, leaving out the intercept will increase your df by 1, since you’re not estimating one parameter. But unless your sample size is really, really small, it won’t matter. So it really has no advantages.

Regression models without intercepts

by Karen Grace-Martin 8 Comments

A recent question on the Talkstats forum asked about dropping the intercept in a linear regression model since it makes the predictor’s coefficient stronger and more significant. Dropping the intercept in a regression model forces the regression line to go through the origin–the y intercept must be 0.

The problem with dropping the intercept is if the slope is steeper just because you’re forcing the line through the origin, not because it fits the data better. If the intercept really should be something else, you’re creating that steepness artificially. A more significant model isn’t better if it’s inaccurate.