I used a linear regression on data I have, using the lm
function. Everything works (no error message), but I'm somehow surprised by the result: I am under the impression R "misses" a group of points, i.e. the intercept and slope are not the best fit. For instance, I am referring to the group of points at coordinates x=15-25,y=0-20.
My questions:
lm
to do
that?Following some answers: additionnal information on x and y
x and y are both visual estimates of disease symptoms. There is the same uncertainty on both of them.
The data and code are here:
x1=c(24.0,23.9,23.6,21.6,21.0,20.8,22.4,22.6,
21.6,21.2,19.0,19.4,21.1,21.5,21.5,20.1,20.1,
20.1,17.2,18.6,21.5,18.2,23.2,20.4,19.2,22.4,
18.8,17.9,19.1,17.9,19.6,18.1,17.6,17.4,17.5,
17.5,25.2,24.4,25.6,24.3,24.6,24.3,29.4,29.4,
29.1,28.5,27.2,27.9,31.5,31.5,31.5,27.8,31.2,
27.4,28.8,27.9,27.6,26.9,28.0,28.0,33.0,32.0,
34.2,34.0,32.6,30.8)
y1=c(100.0,95.5,93.5,100.0,98.5,99.5,34.8,
45.8,47.5,17.4,42.6,63.0,6.9,12.1,30.5,
10.5,14.3,41.1, 2.2,20.0,9.8,3.5,0.5,3.5,5.7,
3.1,19.2,6.4, 1.2, 4.5, 5.7, 3.1,19.2, 6.4,
1.2,4.5,81.5,70.5,91.5,75.0,59.5,73.3,66.5,
47.0,60.5,47.5,33.0,62.5,87.0,86.0,77.0,
86.0,83.0,78.5,83.0,83.5,73.0,69.5,82.5,78.5,
84.0,93.5,83.5,96.5,96.0,97.5)
## x11()
plot(x1,y1,xlim=c(0,35),ylim=c(0,100))
# linear regression
reg_lin=lm(y1 ~ x1)
abline(reg_lin,lty="solid", col="royalblue")
text(12.5,25,labels="R result",col="royalblue", cex=0.85)
text(12.5,20,labels=bquote(y== .(5.26)*x - .(76)),col="royalblue", cex=0.85)
# result I would have imagined
abline(a=-150,b=8,lty="dashed", col="red")
text(27.5,25,labels="What I think is better",col="red", cex=0.85)
text(27.5,20,labels=bquote(y== .(8)*x - .(150)),col="red", cex=0.85)
The lm() function is used to fit linear models to data frames in the R Language. It can be used to carry out regression, single stratum analysis of variance, and analysis of covariance to predict the value corresponding to data that is not in the data frame.
lm uses the QR factorization method (a direct rather than iterative method) to solve linear least squares problems. The documentation for lm shows that it solves linear least squares problems and uses QR factorization to do it.
Summary: R linear regression uses the lm() function to create a regression model given some formula, in the form of Y~X+X2. To look at the model, you use the summary() function.
The lm() function Note that the formula argument follows a specific format. For simple linear regression, this is “YVAR ~ XVAR” where YVAR is the dependent, or predicted, variable and XVAR is the independent, or predictor, variable.
Details. For a linear model fitted by lm or aov , the effects are the uncorrelated single-degree-of-freedom values obtained by projecting the data onto the successive orthogonal subspaces generated by the QR decomposition during the fitting process.
Try this:
reg_lin_int <- reg_lin$coefficients[1]
reg_lin_slp <- reg_lin$coefficients[2]
sum((y1 - (reg_lin_int + reg_lin_slp*x1)) ^ 2)
# [1] 39486.33
sum((y1 - (-150 + 8 * x1)) ^ 2)
# [1] 55583.18
The sum of squared residuals is lower under the lm
fit line. This is to be expected, as reg_lin_int
and reg_lin_slp
are guaranteed to produce the minimal total squared error.
Intuitively, we know estimators under squared loss functions are sensitive to outliers. It's "missing" the group at the bottom because it gets closer to the group at the top left that's much further away--and squared distance gives these points more weight.
In fact, if we use Least Absolute Deviations regression (i.e., specify an absolute loss function instead of a square), the result is much closer to your guess:
library(quantreg)
lad_reg <- rq(y1 ~ x1)
(Pro tip: use lwd
to make your graphs much more readable)
What gets even closer to what you had in mind is Total Least Squares, as mentioned by @nongkrong and @MikeWilliamson. Here is the result of TLS on your sample:
v <- prcomp(cbind(x1, y1))$rotation
bbeta <- v[-ncol(v), ncol(v)] / v[1, 1]
inter <- mean(y1) - bbeta * mean(x1)
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