Regression model point estimation

5
votes

I'd like to retrieve the values of a second order polynomial regression line based on a list of values for a parameter.

Here is the model:

fit <- lm(y ~ poly(age, 2) + height + age*height)

I would like to use a list of values for age and retrieve the value on the regression line, as well as the standard deviation and standard errors. 'age' is a continuous variable, but I want to create an array of discrete values and return the predicted values from the regression line.

Example: 

age <- c(10, 11, 12, 13, 14)
2
rstatisticsregressionlm
Do you mean that you want to get predicted values using your model? - NM_
Yes. But I want to get all of the values based on the variable 'AGE' at discrete points, even though it is a continuous variable. - statsguyz
Your model has 2 variables age and height. In order to get predictions, you need to provide both an age and a height - NM_

2 Answers

4
votes

Since you have an interaction term, the regression coefficients for either the linear or quadratic age term (or both together) only have meaning when you simultaneously specify what value of height is being considered. So to get predictions when the height is at its mean value you would do this:

predict(fit, data.frame(age=c(10, 11, 12, 13, 14), height=mean(height) ) )

bouncyball brings up a good point. You asked about "standard deviation and standard errors", but coefficients and predictions don't have "standard deviations" as the term is usually used, but ratehr "standard errors of the estimate" usually shortened to just standard errors.

predict(fit, data.frame(age=c(10, 11, 12, 13, 14), height=mean(height) ), se.fit=TRUE  )

I suppose if you did a bootstrap run and looked at the standard deviations of the separate coefficients as an estimate of the std error of the coefficients, that might be argued to be a standard deviation, but it would be in the scale of the parameter space rather than on the scale of the variables.

2
votes

Your data has 2 variables, so you need to provide both an age and a height.

For example, using simulated data:

age = sample(10)
height = sort(rnorm(10, 6, 1))
y = sort(rnorm(10, 150, 30))

fit <- lm(y ~ age + poly(age, 2) + height + age*height)

To get predictions specify age and heights and then predict:

# I'm using my own heights, you should choose the values you're interested in
new.data <- data.frame(age=c(10, 11, 12, 13, 14) , 
                  height=c(5.7, 6.3, 5.8, 5.9, 6.0) )

> predict(fit, new.data)
           1            2            3            4            5 
132.76675715 137.70712251 113.39494557 102.07262016  88.84240532

To get confidence bands for each prediction

> predict(fit, new.data, interval="confidence")
           fit            lwr          upr
1 132.76675715  96.0957812269 169.43773307
2 137.70712251  73.2174486246 202.19679641
3 113.39494557  39.5470153667 187.24287578
4 102.07262016   3.5466926099 200.59854771
5  88.84240532 -37.7404171712 215.42522781