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Lesson M04.L03: Confidence Intervals for OLS Coefficients

Field Detail
Module Module 04 — Statistical Inference in OLS
Level Graduate (ECMT5001, S1 2027)
Duration 35 min
Learning Objective Construct 95% and 99% confidence intervals for regression coefficients; correctly interpret them; connect CIs to hypothesis tests; understand what makes intervals wider or narrower
Provenance Wooldridge, Introductory Econometrics, 7e, Chapter 4.3 (pp. 140–146)

Explanation

The Confidence Interval Formula

From Lesson M04.L01, we know that under MLR.1–MLR.6:

\[\frac{\hat{\beta}_j - \beta_j}{se(\hat{\beta}_j)} \sim t_{n-k-1}\]

A \(100(1-\alpha)\%\) confidence interval for \(\beta_j\) is constructed by inverting this pivotal quantity: find the values of \(\beta_j\) that are not rejected by a two-sided \(\alpha\)-level t-test:

\[\boxed{\hat{\beta}_j \pm t_{n-k-1,\, \alpha/2} \cdot se(\hat{\beta}_j)}\]

where \(t_{n-k-1,\, \alpha/2}\) is the upper \(\alpha/2\) critical value of the \(t_{n-k-1}\) distribution.

Common confidence levels and their critical values (large \(n\)):

Confidence level \(\alpha\) \(\alpha/2\) \(t^*\) (large \(n\))
90% 0.10 0.05 1.645
95% 0.05 0.025 1.960
99% 0.01 0.005 2.576

For finite samples, always use the exact \(t_{n-k-1, \alpha/2}\) value from tables.


Correct Interpretation

The correct frequentist interpretation of a 95% CI is:

If we were to repeat the sampling and estimation procedure many times, 95% of the intervals constructed this way would contain the true value \(\beta_j\).

This is a statement about the procedure, not about any single realised interval.

The parameter \(\beta_j\) is a fixed (non-random) unknown constant. It does not have a probability distribution. Once a specific interval \([L, U]\) has been computed from data, it either contains \(\beta_j\) or it does not — there is no probability in between. The randomness lives entirely in \(\hat{\beta}_j\) (and hence in \(L\) and \(U\)), not in \(\beta_j\).


Relationship Between CIs and Hypothesis Tests

There is a direct duality between a two-sided t-test and a confidence interval:

\[\text{Reject } H_0: \beta_j = a_j \text{ at level } \alpha \iff a_j \notin \text{ } 100(1-\alpha)\% \text{ CI for } \beta_j\]

This means: - If \(0\) lies outside the 95% CI, we reject \(H_0: \beta_j = 0\) at the 5% level. - If \(0\) lies inside the 95% CI, we fail to reject \(H_0: \beta_j = 0\) at the 5% level.

The CI provides richer information than the hypothesis test — it tells you not just whether to reject, but also the range of values consistent with the data at the specified confidence level.


What Makes a CI Wider or Narrower?

The width of a CI is \(2 \cdot t_{n-k-1,\, \alpha/2} \cdot se(\hat{\beta}_j)\).

Since \(se(\hat{\beta}_j) = \dfrac{\hat{\sigma}}{\sqrt{SST_j(1-R_j^2)}}\), the CI is wider when:

Factor Effect on CI width Why
Smaller \(n\) Wider Less data → less precision; also larger \(t^*\) with fewer df
Larger \(\hat{\sigma}^2\) (noisier data) Wider Regression errors are large
Less variation in \(x_j\) (small \(SST_j\)) Wider Can't pin down slope without variation in regressor
High multicollinearity (large \(R_j^2\)) Wider \(x_j\) is nearly explained by other regressors; little unique information
Higher confidence level (99% vs 95%) Wider Larger \(t^*\) to capture more of the distribution

Worked Example

Setup. A graduate student estimates the following model of log earnings using \(n = 935\) workers with \(k = 4\) regressors (\(df = 935 - 4 - 1 = 930\)):

\[\log(\text{earnings}_i) = \beta_0 + \beta_1 \text{educ}_i + \beta_2 \text{exper}_i + \beta_3 \text{exper}^2_i + \beta_4 \text{female}_i + u_i\]

The OLS output is:

Variable \(\hat{\beta}_j\) \(se(\hat{\beta}_j)\)
Intercept 0.316 0.085
educ 0.078 0.006
exper 0.033 0.006
exper² −0.0005 0.0001
female −0.252 0.028

With \(df = 930\), we use \(t^* \approx 1.963\) for 95% CIs and \(t^* \approx 2.581\) for 99% CIs (approximately equal to \(z\) critical values given large \(n\)).


95% CI for \(\beta_1\) (return to education):

\[\hat{\beta}_1 \pm t^* \cdot se(\hat{\beta}_1) = 0.078 \pm 1.963 \times 0.006$$ $$= 0.078 \pm 0.012$$ $$= (0.066,\; 0.090)\]

Interpretation: We are 95% confident that the true return to an additional year of education lies between 6.6% and 9.0% (log-point change in earnings), holding other variables constant.

Since 0 is far outside this interval, we reject \(H_0: \beta_1 = 0\) at the 5% level (consistent with \(t = 0.078/0.006 = 13.0\), which is very large).


99% CI for \(\beta_4\) (gender gap):

\[\hat{\beta}_4 \pm t^* \cdot se(\hat{\beta}_4) = -0.252 \pm 2.581 \times 0.028$$ $$= -0.252 \pm 0.072$$ $$= (-0.324,\; -0.180)\]

Interpretation: We are 99% confident that the female earnings penalty lies between \(-32.4\%\) and \(-18.0\%\) log earnings, controlling for education and experience. The entire interval lies below zero, confirming the gender gap is significant at the 1% level.

Comparing 95% vs 99%: The 99% CI for \(\beta_4\) would be:

\[-0.252 \pm 2.581 \times 0.028 = (-0.324, -0.180)\]

vs the 95% CI:

\[-0.252 \pm 1.963 \times 0.028 = (-0.307, -0.197)\]

The 99% interval is wider by \(2 \times (2.581 - 1.963) \times 0.028 = 2 \times 0.017 = 0.035\) — a modest widening given the large sample.


CI for \(\beta_2\) (experience, where \(0\) might plausibly be inside):

\[\hat{\beta}_2 \pm 1.963 \times 0.006 = 0.033 \pm 0.012 = (0.021,\; 0.045)\]

Zero is outside the interval, so we reject \(H_0: \beta_2 = 0\) at 5%. But also note: zero is very far from 0.033 relative to \(se = 0.006\). The experience effect is precisely estimated.


Common Misconception

"There is a 95% probability that \(\beta_j\) lies in the interval \((L, U)\)."

This is a common but incorrect statement. In the frequentist framework:

  • \(\beta_j\) is a fixed unknown constant, not a random variable.
  • The interval endpoints \(L = \hat{\beta}_j - t^* \cdot se(\hat{\beta}_j)\) and \(U = \hat{\beta}_j + t^* \cdot se(\hat{\beta}_j)\) are random — they depend on the random sample.
  • Once we observe the data and compute, say, \(L = 0.066\) and \(U = 0.090\), the probability that \(\beta_1 \in (0.066, 0.090)\) is either 0 or 1 (it either does or doesn't). We just don't know which.

The correct statement: "The interval was constructed by a procedure that, in repeated samples, would contain the true \(\beta_j\) in 95% of cases."

Bayesian note: A statement like "95% probability that \(\beta_j \in (L, U)\)" has a precise meaning in Bayesian statistics (it would be a 95% credible interval). But under frequentist OLS, this language is incorrect.


Practice Prompts

Q1. A model of city crime rates has \(n = 46\) cities, \(k = 3\) regressors, \(\hat{\beta}_{\text{unemployment}} = 0.82\), \(se = 0.41\). Construct a 95% CI and a 90% CI. Does the 95% CI include zero? What does this imply for a two-sided test at 5%?

Answer $df = 46 - 3 - 1 = 42$. From t-tables: $t_{42, 0.025} \approx 2.018$ (95%), $t_{42, 0.05} \approx 1.682$ (90%). **95% CI:** $0.82 \pm 2.018 \times 0.41 = 0.82 \pm 0.828 = (-0.008,\; 1.648)$. **90% CI:** $0.82 \pm 1.682 \times 0.41 = 0.82 \pm 0.690 = (0.130,\; 1.510)$. The 95% CI barely includes zero (lower bound $\approx -0.008$), so we **fail to reject** $H_0: \beta_{\text{unemployment}} = 0$ at the 5% level. However, the 90% CI excludes zero, so we **would reject** at the 10% level. The result is marginally insignificant at 5%: $t = 0.82/0.41 = 2.00 < t_{42,0.025} = 2.018$.

Q2. For the same regression, a colleague claims: "Because the 95% CI is \((-0.008, 1.648)\), there is only a 5% chance that the true effect is negative." Correct this statement.

Answer The statement is incorrect. The parameter $\beta_{\text{unemployment}}$ is a fixed unknown — it is either positive or negative; there is no probability attached to its sign under frequentist inference. The correct interpretation: the interval $(-0.008, 1.648)$ was produced by a procedure that yields intervals containing the true $\beta$ in 95% of repeated samples. We do not know whether this particular realised interval is one of the 95% that contain the truth, or one of the 5% that do not. The fact that 95% of the interval lies in the positive region reflects the distribution of $\hat{\beta}$, not a probability over $\beta$.

Q3. A researcher reports: \(\hat{\beta} = 0.120\), \(se = 0.048\), \(n = 80\), \(k = 2\). Construct the 95% CI and verify that the result is consistent with the t-test outcome at the 5% level.

Answer $df = 80 - 2 - 1 = 77$. $t_{77, 0.025} \approx 1.991$. **95% CI:** $0.120 \pm 1.991 \times 0.048 = 0.120 \pm 0.096 = (0.024,\; 0.216)$. Zero is **outside** the interval, so we **reject** $H_0: \beta = 0$ at 5%. **Verification via t-test:** $t = 0.120 / 0.048 = 2.50 > 1.991$. Reject $H_0$. ✓ Both approaches yield the same decision, confirming the duality between CIs and hypothesis tests.

Further Resources

  • Wooldridge (2019), §4.3: "Confidence Intervals" (pp. 140–146)
  • Wooldridge (2019), Appendix B.4: "Confidence Intervals and Hypothesis Testing"
  • Stock & Watson (2020), Introduction to Econometrics, §7.3: "Confidence Intervals for a Single Coefficient"
  • Freedman, Pisani & Purves (2007), Statistics, Ch. 26: for intuition on what confidence intervals do and don't mean