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## 5.2 Confidence Intervals for Regression Coefficients

As we already know, estimates of the regression coefficients \(\beta_0\) and \(\beta_1\) are subject to sampling uncertainty, see Chapter 4. Therefore, we will *never* exactly estimate the true value of these parameters from sample data in an empirical application. However, we may construct confidence intervals for the intercept and the slope parameter.

A \(95\%\) confidence interval for \(\beta_i\) has two equivalent definitions:

- The interval is the set of values for which a hypothesis test to the level of \(5\%\) cannot be rejected.
- The interval has a probability of \(95\%\) to contain the true value of \(\beta_i\). So in \(95\%\) of all samples that could be drawn, the confidence interval will cover the true value of \(\beta_i\).

We also say that the interval has a confidence level of \(95\%\). The idea of the confidence interval is summarized in Key Concept 5.3.

### Key Concept 5.3

### A Confidence Interval for \(\beta_i\)

Imagine you could draw all possible random samples of given size. The interval that contains the true value \(\beta_i\) in \(95\%\) of all samples is given by the expression

\[ \text{CI}_{0.95}^{\beta_i} = \left[ \hat{\beta}_i - 1.96 \times SE(\hat{\beta}_i) \, , \, \hat{\beta}_i + 1.96 \times SE(\hat{\beta}_i) \right]. \]

Equivalently, this interval can be seen as the set of null hypotheses for which a \(5\%\) two-sided hypothesis test does not reject.

### Simulation Study: Confidence Intervals

To get a better understanding of confidence intervals we conduct another simulation study. For now, assume that we have the following sample of \(n=100\) observations on a single variable \(Y\) where

\[ Y_i \overset{i.i.d}{\sim} \mathcal{N}(5,25), \ i = 1, \dots, 100.\]

```
# set seed for reproducibility
set.seed(4)
# generate and plot the sample data
rnorm(n = 100,
Y <-mean = 5,
sd = 5)
plot(Y,
pch = 19,
col = "steelblue")
```

We assume that the data is generated by the model

\[ Y_i = \mu + \epsilon_i \]

where \(\mu\) is an unknown constant and we know that \(\epsilon_i \overset{i.i.d.}{\sim} \mathcal{N}(0,25)\). In this model, the OLS estimator for \(\mu\) is given by \[ \hat\mu = \overline{Y} = \frac{1}{n} \sum_{i=1}^n Y_i, \] i.e., the sample average of the \(Y_i\). It further holds that

\[ SE(\hat\mu) = \frac{\sigma_{\epsilon}}{\sqrt{n}} = \frac{5}{\sqrt{100}} \]

(see Chapter 2) A large-sample \(95\%\) confidence interval for \(\mu\) is then given by

\[\begin{equation} CI^{\mu}_{0.95} = \left[\hat\mu - 1.96 \times \frac{5}{\sqrt{100}} \ , \ \hat\mu + 1.96 \times \frac{5}{\sqrt{100}} \right]. \tag{5.1} \end{equation}\]

It is fairly easy to compute this interval in `R` by hand. The following code chunk generates a named vector containing the interval bounds:

```
cbind(CIlower = mean(Y) - 1.96 * 5 / 10, CIupper = mean(Y) + 1.96 * 5 / 10)
#> CIlower CIupper
#> [1,] 4.502625 6.462625
```

Knowing that \(\mu = 5\) we see that, for our example data, the confidence interval covers true value.

As opposed to real world examples, we can use `R` to get a better understanding of confidence intervals by repeatedly sampling data, estimating \(\mu\) and computing the confidence interval for \(\mu\) as in (5.1).

The procedure is as follows:

- We initialize the vectors
`lower`and`upper`in which the simulated interval limits are to be saved. We want to simulate \(10000\) intervals so both vectors are set to have this length. - We use a
`for()`loop to sample \(100\) observations from the \(\mathcal{N}(5,25)\) distribution and compute \(\hat\mu\) as well as the boundaries of the confidence interval in every iteration of the loop. - At last we join
`lower`and`upper`in a matrix.

```
# set seed
set.seed(1)
# initialize vectors of lower and upper interval boundaries
numeric(10000)
lower <- numeric(10000)
upper <-
# loop sampling / estimation / CI
for(i in 1:10000) {
rnorm(100, mean = 5, sd = 5)
Y <- mean(Y) - 1.96 * 5 / 10
lower[i] <- mean(Y) + 1.96 * 5 / 10
upper[i] <-
}
# join vectors of interval bounds in a matrix
cbind(lower, upper) CIs <-
```

According to Key Concept 5.3 we expect that the fraction of the \(10000\) simulated intervals saved in the matrix `CIs` that contain the true value \(\mu=5\) should be roughly \(95\%\). We can easily check this using logical operators.

```
mean(CIs[, 1] <= 5 & 5 <= CIs[, 2])
#> [1] 0.9487
```

The simulation shows that the fraction of intervals covering \(\mu=5\), i.e., those intervals for which \(H_0: \mu = 5\) cannot be rejected is close to the theoretical value of \(95\%\).

Let us draw a plot of the first \(100\) simulated confidence intervals and indicate those which *do not* cover the true value of \(\mu\). We do this via horizontal lines representing the confidence intervals on top of each other.

```
# identify intervals not covering mu
# (4 intervals out of 100)
which(!(CIs[1:100, 1] <= 5 & 5 <= CIs[1:100, 2]))
ID <-
# initialize the plot
plot(0,
xlim = c(3, 7),
ylim = c(1, 100),
ylab = "Sample",
xlab = expression(mu),
main = "Confidence Intervals")
# set up color vector
rep(gray(0.6), 100)
colors <- "red"
colors[ID] <-
# draw reference line at mu=5
abline(v = 5, lty = 2)
# add horizontal bars representing the CIs
for(j in 1:100) {
lines(c(CIs[j, 1], CIs[j, 2]),
c(j, j),
col = colors[j],
lwd = 2)
}
```

For the first \(100\) samples, the true null hypothesis is rejected in four cases so these intervals do not cover \(\mu=5\). We have indicated the intervals which lead to a rejection of the null red.

Let us now come back to the example of test scores and class sizes. The regression model from Chapter 4 is stored in `linear_model`. An easy way to get \(95\%\) confidence intervals for \(\beta_0\) and \(\beta_1\), the coefficients on `(intercept)` and `STR`, is to use the function `confint()`. We only have to provide a fitted model object as an input to this function. The confidence level is set to \(95\%\) by default but can be modified by setting the argument `level`, see `?confint`.

```
# compute 95% confidence interval for coefficients in 'linear_model'
confint(linear_model)
#> 2.5 % 97.5 %
#> (Intercept) 680.32312 717.542775
#> STR -3.22298 -1.336636
```

Let us check if the calculation is done as we expect it to be for \(\beta_1\), the coefficient on `STR`.

```
# compute 95% confidence interval for coefficients in 'linear_model' by hand
summary(linear_model)
lm_summ <-
c("lower" = lm_summ$coef[2,1] - qt(0.975, df = lm_summ$df[2]) * lm_summ$coef[2, 2],
"upper" = lm_summ$coef[2,1] + qt(0.975, df = lm_summ$df[2]) * lm_summ$coef[2, 2])
#> lower upper
#> -3.222980 -1.336636
```

The upper and the lower bounds coincide. We have used the \(0.975\)-quantile of the \(t_{418}\) distribution to get the exact result reported by `confint`. Obviously, this interval *does not* contain the value zero which, as we have already seen in the previous section, leads to the rejection of the null hypothesis \(\beta_{1,0} = 0\).