Let's take a toy example in R:
set.seed(135)
fEffect <- sample(c(0,1), size = 15, replace = TRUE)
rEffect <- rnorm(15, mean = 3)
y <- 2 * fEffect + 0.4 * rEffect + rnorm(15)
summary(lm(y ~ fEffect + rEffect))
(I set the seed for replicability.) The output from this linear model is:
Call:
lm(formula = y ~ fEffect + rEffect)
Residuals:
Min 1Q Median 3Q Max
-0.98166 -0.83153 -0.08039 0.75780 1.27464
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.2176 1.1684 -0.186 0.85540
fEffect 2.0093 0.4751 4.229 0.00117 **
rEffect 0.5156 0.3496 1.475 0.16605
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.8564 on 12 degrees of freedom
Multiple R-squared: 0.6533, Adjusted R-squared: 0.5956
F-statistic: 11.31 on 2 and 12 DF, p-value: 0.001736
As you can see from this output, the estimate for the fixed effect is 2.0093, and the standard error of the fixed effect is 0.4751, the Student's $t$ statistic for $\alpha = 0.25$ on 12 degrees of freedom is 2.179. Thus, the confidence interval for this estimate is
$$
\beta_1 \in 2.0093 \pm 2.179 \times 0.4751 = (0.9741,3.0445).
$$
As we can see, this interval does not contain 0, so we reject the null claim that $\beta_1 = 0$. Furthermore, while we know that the response, $y$, is in fact related to the random effect (because we created it to be so), the confidence interval for that estimate does contain 0:
$$
\beta_2 \in 0.5156 \pm 2.179 \times 0.3496 = (-0.2462,1.2774).
$$
We therefore do not have sufficient evidence to reject the claim that $\beta_2 = 0$.
If you are interested in finding out how that standard error value is calculated, that question was answered previously in this Cross Validated question.
