# Formula used in confidence intervalle on R's lm function

Does someone know how confidence interval on factor values are computed on R (lm function), here is a simple example :

data

df <- data.frame(c("Male", "Female", "Male","Female"),c(900,600,1200,800)) names(df) <- c("gender","salary") df$$gender <- as.factor(df$$gender) 

model

model <- lm(salary ~ gender, data = df)

Male confidence intervale

predict(model, list(gender="Male"), interval = "confidence")

fit lwr upr 1050 501.5172 1598.483

Female confidence intervale

predict(model, list(gender="Female"), interval = "confidence")

fit lwr upr 700 151.5172 1248.483

How these lower and upper bounds have been computed ?

nb : this question is linked to Confidence interval in R lm function with factors values

• Since factor variables are dummy encoded automatically, in the same way as with continuous variables. Study the source code of predict.lm. – Roland Jan 8 at 14:02

Note that you are in fact estimating an ANOVA model with "treatment contrasts" (the baseline is set to "Female"). This simple ANOVA is equivalent to a simple linear regression model with some dummy-variable encoding: By running lm() you are in fact running a regression of $$y$$ on $$X$$, where $$y=(900,600,1200,800)^T$$ and $$X$$ is given by: $$X=\begin{pmatrix}1 & 1\\ 1 & 0 \\ 1 & 1 \\ 1 & 0 \end{pmatrix}.$$ (see model.matrix(model) to get $$X$$).
Hence, you are estimating a regression line in a simple linear regression model of the form $$y= \alpha + \beta x + \epsilon_i$$ and are interested in a predicition interval of the a mean response at $$x=1$$ (this is NOT at a new observation!).
It is well known that under the usual assumptions (including the normality assumption on $$\epsilon_i$$), the prediction interval for the mean response at a in-sample $$x$$ is then given by $$\widehat{y}_x \pm t_{1-\frac{\alpha}{2},n-2}SE(\widehat{y}_x),$$ where $$t_{\alpha,n}$$ is the $$\alpha$$ quantile of a $$t$$-distribution with $$n$$-degrees of freedem, $$s= \sqrt{\frac{1}{n-2}\sum_{i=1}^n e_i^2}$$, $$\widehat{y}_x = \widehat{\alpha} + \widehat{\beta} x$$ and $$SE(\widehat{y})$$ is given by:: $$SE(\widehat{y}_x) = s \sqrt{\frac{1}{n}+\frac{(x-\overline{x})^2}{\sum(x_i-\overline{x})^2}}.$$
In your case, for $$x=1$$, you have $$\widehat{\alpha} = 700$$, $$\widehat{\beta}=350$$, $$s\approx 180.2776$$, $$SE(\widehat{y}_{x=1}) \approx 127.4755$$, $$t_{0.975,2}\approx 4.302653$$. Hence your $$0.95$$-prediction interval is given by: $$\widehat{y}_{x=1} \pm t_{1-\frac{\alpha}{2},n-2}SE(\widehat{y}_{x=1}) = 1050 \pm 548.4828.$$
• there are indeed different forms of modelling the influence of the factor variable and different corresponding $X$-matrices. however you want to measure two effects (the effect of eacht of the levels of the gender-factor) and hence you need two columns. Try for example: model <- lm(salary ~ gender, data = df,contrasts = list(gender = "contr.sum")) and take a look at the model.matrix(model). – chRrr Jan 15 at 16:34