Is there a method to understand if two lines are (more or less) parallel? I have two lines generated from linear regressions and I would like to understand if they are parallel. In other words, I would like to get the different of the slopes of those two lines.
Is there an R function to calculate this?
EDIT: ... and how can I get the slope (in degrees) of a linear regression line?
I wonder if I am missing something obvious, but couldn't you do this statistically using ANCOVA? An important issue is that the slopes in the two regressions are estimated with error. They are estimates of the slopes in the populations at large. If the concern is whether the two regression lines are parallel or not in the population then it doesn't make sense to compare $a_1$ with $a_2$ directly for exact equivalence; they are both subject to error/uncertainty that needs to be taken into account.
If we think about this from a statistical point of view, and we can combine the data on $x$ and $y$ for both data sets in some meaningful way (i.e. $x$ and $y$ in both sets are drawn from the two populations with similar ranges for the two variables it is just the relationship between them that are different in the two populations), then we can fit the following two models:
$$\hat{y} = b_0 + b_1x + b_2g$$
and
$$\hat{y} = b_0 + b_1x + b_2g + b_3xg$$
Where $b_i$ are the model coefficients, and $g$ is a grouping variable/factor, indicating which data set each observation belongs to.
We can use an ANOVA table or F-ratio to test if the second, more complex model fits the data better than the simpler model. The simpler model states that the slopes of the two lines are the same ($b_1$) but the lines are offset from one another by an amount $b_2$.
The more complex model includes an interaction between the slope of the line and the grouping variable. If the coefficient for this interaction term is significantly different from zero or the ANOVA/F-ratio indicates the more complex model fits the data better then we must reject the Null hypothesis that that two lines are parallel.
Here is an example in R using dummy data. First, data with equal slopes:
set.seed(2)
samp <- factor(sample(rep(c("A","B"), each = 50)))
d1 <- data.frame(y = c(2,5)[as.numeric(samp)] + (0.5 * (1:100)) + rnorm(100),
x = 1:100,
g = samp)
m1 <- lm(y ~ x * g, data = d1)
m1.null <- lm(y ~ x + g, data = d1)
anova(m1.null, m1)
Which gives
> anova(m1.null, m1)
Analysis of Variance Table
Model 1: y ~ x + g
Model 2: y ~ x * g
Res.Df RSS Df Sum of Sq F Pr(>F)
1 97 122.29
2 96 122.13 1 0.15918 0.1251 0.7243
Indicating that we fail to reject the null hypothesis of equal slopes in this sample of data. Of course, we'd want to assure ourselves that we had sufficient power to detect a difference if there really was one so that we were not lead to erroneously fail to reject the null because our sample size was too small for the expected effect.
Now with different slopes.
set.seed(42)
x <- seq(1, 100, by = 2)
d2 <- data.frame(y = c(2 + (0.5 * x) + rnorm(50),
5 + (1.5 * x) + rnorm(50)),
x = x,
g = rep(c("A","B"), each = 50))
m2 <- lm(y ~ x * g, data = d2)
m2.null <- lm(y ~ x + g, data = d2)
anova(m2.null, m2)
Which gives:
> anova(m2.null, m2)
Analysis of Variance Table
Model 1: y ~ x + g
Model 2: y ~ x * g
Res.Df RSS Df Sum of Sq F Pr(>F)
1 97 21132.0
2 96 103.8 1 21028 19439 < 2.2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Here we have substantial evidence against the null hypothesis and thus we can reject it in favour of the alternative (in other words, we reject the hypothesis that the slopes of the two lines are equal).
The interaction terms in the two models I fitted ($b_3xg$) give the estimated difference in slopes for the two groups. For the first model, the estimate of the difference in slopes is small (~0.003)
> coef(m1)
(Intercept) x gB x:gB
2.100068977 0.500596394 2.659509181 0.002846393
and a $t$-test on this would fail to reject the null hypothesis that this difference in slopes is 0:
> summary(m1)
Call:
lm(formula = y ~ x * g, data = d1)
Residuals:
Min 1Q Median 3Q Max
-2.32886 -0.81224 -0.01569 0.93010 2.29984
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.100069 0.334669 6.275 1.01e-08 ***
x 0.500596 0.005256 95.249 < 2e-16 ***
gB 2.659509 0.461191 5.767 9.82e-08 ***
x:gB 0.002846 0.008047 0.354 0.724
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.128 on 96 degrees of freedom
Multiple R-squared: 0.9941, Adjusted R-squared: 0.9939
F-statistic: 5347 on 3 and 96 DF, p-value: < 2.2e-16
If we turn to the model fitted to the second data set, where we made the slopes for the two groups differ, we see that the estimated difference in slopes of the two lines is ~1 unit.
> coef(m2)
(Intercept) x gB x:gB
2.3627432 0.4920317 2.8931074 1.0048653
The slope for group "A" is ~0.49 (x in the above output), whilst to get the slope for group "B" we need to add the difference slopes (give by the interaction term remember) to the slope of group "A"; ~0.49 + ~1 = ~1.49. This is pretty close to the stated slope for group "B" of 1.5. A $t$-test on this difference of slopes also indicates that the estimate for the difference is bounded away from 0:
> summary(m2)
Call:
lm(formula = y ~ x * g, data = d2)
Residuals:
Min 1Q Median 3Q Max
-3.1962 -0.5389 0.0373 0.6952 2.1072
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.362743 0.294220 8.031 2.45e-12 ***
x 0.492032 0.005096 96.547 < 2e-16 ***
gB 2.893107 0.416090 6.953 4.33e-10 ***
x:gB 1.004865 0.007207 139.424 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.04 on 96 degrees of freedom
Multiple R-squared: 0.9994, Adjusted R-squared: 0.9994
F-statistic: 5.362e+04 on 3 and 96 DF, p-value: < 2.2e-16
The first question is actually from geometry. If you have two lines of the form:
$$y=a_1x+b_1$$ $$y=a_2x+b_2$$
then they are parallel if $a_1=a_2$. So if the slopes are equal then then the lines are parallel.
For the second question, use the fact that $\tan \alpha=a_1$, where $\alpha$ is the angle the line makes with $x$-axis, and $a_1$ is the slope of the line. So
$$\alpha=\arctan a_1$$
and to convert to degrees, recall that $2\pi=360$. So the answer in the degrees will be
$$\alpha=\arctan a_1\cdot \frac{360}{2\pi}.$$
The R function for $\arctan$ is called atan.
Sample R code:
> x<-rnorm(100)
> y<-x+1+rnorm(100)/2
> mod<-lm(y~x)
> mod$coef
(Intercept) x
0.9416175 0.9850303
> mod$coef[2]
x
0.9850303
> atan(mod$coef[2])*360/2/pi
x
44.56792
The last line is the degrees.
Update. For the negative slope values conversion to degrees should follow different rule. Note that the angle with the x-axis can get values from 0 to 180, since we assume that the angle is above the x-axis. So for negative values of $a_1$, the formula is:
$$\alpha=180-\arctan a_1\cdot \frac{360}{2\pi}.$$
Note. While it was fun for me to recall high-school trigonometry, the really useful answer is the one given by Gavin Simpson. Since the slopes of regression lines are random variables, to compare them statistical hypothesis framework should be used.
... following up on @mpiktas' answer, here's how you would extract slope from a lm object and apply the above formula.
# prepare some data, see ?lm
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group <- gl(2,10,20, labels=c("Ctl","Trt"))
weight <- c(ctl, trt)
lm.D9 <- lm(weight ~ group)
# extract the slope (this is also used to draw a regression line if you wrote abline(lm.D9)
coefficients(lm.D9)["groupTrt"]
groupTrt
-0.371
# use the arctan*a1 / (360 / (2*pi)) formula provided by mpiktas
atan(coefficients(lm.D9)["groupTrt"]) * (360/(2 * pi))
groupTrt
-20.35485
180-atan(coefficients(lm.D9)["groupTrt"]) * (360/(2 * pi))
groupTrt
200.3549
