Simple solution that I overlooked. I know the slope and the intercept so I can just pass them to abline directly:
abline(0.22483, -0.07115)
Question
Guided by the answer to this post:
Linear Regression with a known fixed intercept in R
I have fit an explicit intercept value to some data, but also an explicit slope, thus:
intercept <- 0.22483
fit <- lm(I(Response1 - intercept) ~ 0 + offset(-0.07115*Continuous))
Where Response1 is my dependent variable and Continuous is my explanatory variable, both from this dataset.
I want to plot an abline for my relationship. When only the intercept has been specified the post above recommends:
abline(intercept, coef(fit))
However I have no coefficients in the model "fit" as I specified them all. Is there a way to plot the abline for the relationship I specified?
Solution
Simple solution that I overlooked. I know the slope and the intercept so I can just pass them to abline directly:
abline(0.22483, -0.07115)
OTHER TIPS
Based on your comment, you can do this programmatically, without having to manually put in values. Here's an example with sample data from two similar dataframes:
df1 <- data.frame(Response1=rnorm(100,0,1), Continuous=rnorm(100,0,1))
df2 <- data.frame(Response1=rnorm(100,0,1), Continuous=rnorm(100,0,1))
fit1 <- with(df1, lm(Response1 ~ Continuous))
with(df2, plot(Response1 ~ Continuous)) # plot df2 data
abline(coef(fit1)) # plot df1 model over it