问题
I'm running a multivariate regression with 2 outcome variables and 5 predictors. I would like to obtain the confidence intervals for all regression coefficients. Usually I use the function lm
but it doesn't seem to work for a multivariate regression model (object mlm
).
Here's a reproducible example.
library(car)
mod <- lm(cbind(income, prestige) ~ education + women, data=Prestige)
confint(mod) # doesn't return anything.
Any alternative way to do it? (I could just use the value of the standard error and multiply by the right critical t value, but I was wondering if there was an easier way to do it).
回答1:
confint
won't return you anything, because there is no "mlm" method supported:
methods(confint)
#[1] confint.default confint.glm* confint.lm confint.nls*
As you said, we can just plus / minus some multiple of standard error to get upper / lower bound of confidence interval. You were probably going to do this via coef(summary(mod))
, then use some *apply
method to extract standard errors. But my answer to Obtain standard errors of regression coefficients for an “mlm” object returned by lm() gives you a supper efficient way to get standard errors without going through summary
. Applying std_mlm
to your example model gives:
se <- std_mlm(mod)
# income prestige
#(Intercept) 1162.299027 3.54212524
#education 103.731410 0.31612316
#women 8.921229 0.02718759
Now, we define another small function to compute lower and upper bound:
## add "mlm" method to generic function "confint"
confint.mlm <- function (model, level = 0.95) {
beta <- coef(model)
se <- std_mlm (model)
alpha <- qt((1 - level) / 2, df = model$df.residual)
list(lower = beta + alpha * se, upper = beta - alpha * se)
}
## call "confint"
confint(mod)
#$lower
# income prestige
#(Intercept) -3798.25140 -15.7825086
#education 739.05564 4.8005390
#women -81.75738 -0.1469923
#
#$upper
# income prestige
#(Intercept) 814.25546 -1.72581876
#education 1150.70689 6.05505285
#women -46.35407 -0.03910015
It is easy to interpret this. For example, for response income
, the 95%-confidence interval for all variables are
#(intercept) (-3798.25140, 814.25546)
# education (739.05564, 1150.70689)
# women (-81.75738, -46.35407)
回答2:
This comes from the predict.lm example. You want the interval = 'confidence'
option.
x <- rnorm(15)
y <- x + rnorm(15)
predict(lm(y ~ x))
new <- data.frame(x = seq(-3, 3, 0.5))
predict(lm(y ~ x), new, se.fit = TRUE)
pred.w.clim <- predict(lm(y ~ x), new, interval = "confidence")
matplot(new$x, pred.w.clim,
lty = c(1,2,2,3,3), type = "l", ylab = "predicted y")
回答3:
This seems to have been discussed recently (July 2018) on the R-devel list, so hopefully by the next version of R it will be fixed. A workaround proposed on that list is to use:
confint.mlm <- function (object, level = 0.95, ...) {
cf <- coef(object)
ncfs <- as.numeric(cf)
a <- (1 - level)/2
a <- c(a, 1 - a)
fac <- qt(a, object$df.residual)
pct <- stats:::format.perc(a, 3)
ses <- sqrt(diag(vcov(object)))
ci <- ncfs + ses %o% fac
setNames(data.frame(ci),pct)
}
Test:
fit_mlm <- lm(cbind(mpg, disp) ~ wt, mtcars)
confint(fit_mlm)
Gives:
2.5 % 97.5 %
mpg:(Intercept) 33.450500 41.119753
mpg:wt -6.486308 -4.202635
disp:(Intercept) -204.091436 -58.205395
disp:wt 90.757897 134.198380
Personnally, I like it in a clean tibble way (using broom::tidy
would be even better, but has an issue currently)
library(tidyverse)
confint(fit_mlm) %>%
rownames_to_column() %>%
separate(rowname, c("response", "term"), sep=":")
Gives:
response term 2.5 % 97.5 %
1 mpg (Intercept) 33.450500 41.119753
2 mpg wt -6.486308 -4.202635
3 disp (Intercept) -204.091436 -58.205395
4 disp wt 90.757897 134.198380
来源:https://stackoverflow.com/questions/28442141/get-confidence-intervals-for-regression-coefficients-of-mlm-object-returned-by