Suppose I have a data frame that include a factor with two levels. The response is proportional to the predictor, but the slope depends on the factor:
foo <- data.frame(x = rep(0:10, 2),
y = c(5 * 0:10 + rnorm(11),
10 * 0:10 + rnorm(11)),
z = factor(rep(c("Low", "High"), each = 11)))
> foo
x y z
1 0 1.504 Low
2 1 5.294 Low
3 2 8.433 Low
4 3 16.006 Low
5 4 21.094 Low
6 5 27.099 Low
7 6 30.864 Low
8 7 34.439 Low
9 8 39.623 Low
10 9 45.097 Low
11 10 50.719 Low
12 0 -0.777 High
13 1 9.821 High
14 2 18.902 High
15 3 29.413 High
16 4 40.563 High
17 5 48.560 High
18 6 58.661 High
19 7 69.551 High
20 8 79.504 High
21 9 90.469 High
22 10 100.259 High
When I fit a linear model, I obtain estimates for an intercept and for the difference with the second intercept. Same for the slopes:
> summary(lm(y ~ x * z, foo))
Call:
lm(formula = y ~ x * z, data = foo)
Residuals:
Min 1Q Median 3Q Max
-2.1597 -0.3999 0.0324 0.5420 1.6287
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.8229 0.5101 -1.61 0.124
x 10.0723 0.0862 116.83 <2e-16 ***
zLow 1.4977 0.7213 2.08 0.052 .
x:zLow -5.1133 0.1219 -41.94 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.904 on 18 degrees of freedom
Multiple R-squared: 0.999, Adjusted R-squared: 0.999
F-statistic: 6.95e+03 on 3 and 18 DF, p-value: <2e-16
But if I want the slopes for each group of values (grouped on z), I need to compute x + x:zLow and then also need to compute manually the standard error of that estimate.
Is there a way to obtain directly the two different slopes, with the standard error?
