How do I calculate a Confidence Interval for a Regression Intercept?

Simple linear regression is used to quantify the relationship between a predictor variable and a response variable.

This method finds a line that best “fits” a dataset and takes on the following form:

ŷ = b₀ + b₁x

where:

• ŷ: The estimated response value
• b₀: The intercept of the regression line
• b₁: The slope of the regression line
• x: The value of the predictor variable

Often we’re interested in the value for b₁, which tells us the average change in the associated with a one unit increase in the predictor variable.

However, in rare circumstances we’re also interested in the value for b₀, which tells us the average value of the response variable when the predictor variable is equal to zero.

We can use the following formula to calculate a confidence interval for the value of β₀, the true population intercept:

Confidence Interval for β₀: b₀ ± t_{α/2, n-2} * se(b₀)

The following example shows how to calculate a confidence interval for an intercept in practice.

Example: Confidence Interval for Regression Intercept

Suppose we’d like to fit a simple linear regression model using hours studied as a predictor variable and exam score as a response variable for 15 students in a particular class:

The following code shows how to fit this simple linear regression model in R:

#create data frame
df <- data.frame(hours=c(1, 2, 4, 5, 5, 6, 6, 7, 8, 10, 11, 11, 12, 12, 14),
                 score=c(64, 66, 76, 73, 74, 81, 83, 82, 80, 88, 84, 82, 91, 93, 89))

#fit simple linear regression model
fit <- lm(score ~ hours, data=df)

#view summary of model
summary(fit)

Call:
lm(formula = score ~ hours, data = df)

Residuals:
   Min     1Q Median     3Q    Max 
-5.140 -3.219 -1.193  2.816  5.772 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   65.334      2.106  31.023 1.41e-13 ***
hours          1.982      0.248   7.995 2.25e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3.641 on 13 degrees of freedom
Multiple R-squared:  0.831,	Adjusted R-squared:  0.818 
F-statistic: 63.91 on 1 and 13 DF,  p-value: 2.253e-06

Using the coefficient estimates in the output, we can write the fitted simple linear regression model as:

Score = 65.334 + 1.982*(Hours Studied)

We can use the following formula to calculate a 95% confidence interval for the intercept:

• 95% C.I. for β₀: b₀ ± t_{α/2, n-2} * se(b₀)
• 95% C.I. for β₀: 65.334 ± t_{.05/2, 15-2} * 2.106
• 95% C.I. for β₀: 65.334 ± 2.1604 * 2.106
• 95% C.I. for β₀: [60.78, 69.88]

We interpret this to mean that we’re 95% confident that the true population mean exam score for students who study for zero hours is between 60.78 and 69.88.

Note: We used the to find the t critical value that corresponds to a 95% confidence level with 13 degrees of freedom.

Cautions on Calculating a Confidence Interval for a Regression Intercept

We often don’t calculate a confidence interval for a regression intercept in practice because it usually doesn’t make sense to interpret the value of the intercept in a regression model.

For example, suppose we fit a regression model that uses height of a basketball player as a predictor variable and average points per game as a response variable.

It’s not possible for a player to be zero feet tall, so it wouldn’t make sense to interpret the intercept literally in this model.

There are countless scenarios like this where a predictor variable can’t take on a value of zero so it doesn’t make sense to interpret the intercept value of the model or create a confidence interval for the intercept.

For example, consider the following potential predictor variables in a model:

• Square footage of a house
• Length of a car
• Weight of a person

Each of these predictor variables can’t take on a value of zero, so it wouldn’t make sense to calculate a confidence interval for the intercept of a regression model in any of these circumstances.

The following tutorials provide additional information about linear regression: