Testing and Confidence Intervals for 2SLS Coefficients

Why Inference for 2SLS Coefficients Needs Care

A 2SLS estimate is often reported as a coefficient together with a standard error, a test, or a confidence interval. The estimate alone does not show how precisely the coefficient has been estimated. Moreover, the standard error used for inference must reflect the sampling structure: the usual second-stage OLS standard error does not generally do so for 2SLS. Here we take the appropriate 2SLS variance estimate as given and focus on how to use it for tests and intervals.

The procedures in this tutorial are conventional Wald procedures. Their familiar large-sample calibration relies on the 2SLS estimator being approximately normally distributed around the parameter under the null, and on the chosen robust covariance estimator consistently estimating its sampling variance. The instruments and model must also satisfy the conditions that give the coefficient its intended structural interpretation. Passing a coefficient test does not establish those conditions.

Definition: For a 2SLS coefficient $\hat{\beta}_j$, let $SE(\hat{\beta}_j)$ be the square root of the corresponding diagonal element of an appropriate estimated 2SLS covariance matrix. The Wald statistic for testing $H_0:\beta_j=\beta_{j,0}$ is $$ z=\frac{\hat{\beta}_j-\beta_{j,0}}{SE(\hat{\beta}_j)}. $$ With a large-sample normal reference distribution, a two-sided p-value is $2P(Z\geq |z|)$, where $Z\sim N(0,1)$. A two-sided confidence interval with confidence level $1-\alpha$ is $$ \hat{\beta}_j\ \mathbin{\pm}\ z_{1-\alpha/2}SE(\hat{\beta}_j), $$ where $z_{1-\alpha/2}$ is the corresponding standard-normal quantile.

Setting Up a Coefficient Test

A test concerns a specified coefficient and a specified null value. The usual two-sided test is $H_0:\beta_j=\beta_{j,0}$ against $H_a:\beta_j\neq\beta_{j,0}$. A one-sided alternative is appropriate only when the scientific question and analysis call for a directional test; it should not be chosen after seeing the estimate.

The numerator of the Wald statistic is the distance between the estimate and the null value, measured in coefficient units. Dividing by its standard error expresses that distance in estimated standard-error units. For a two-sided test, large absolute values of $z$ provide evidence against the null. The p-value is the probability, under the null and the reference approximation, of a statistic at least as far from zero as the observed statistic. It is not the probability that the null is true.

Worked Example: Testing a Coefficient Against a Nonzero Value

Suppose a hypothetical 2SLS analysis estimates an outcome change in points per one-unit increase in treatment. The estimated coefficient is $1.80$ points per treatment unit, and its heteroskedasticity-robust standard error is $\sqrt{0.51}$, approximately $0.7141$ points per treatment unit. Test $H_0:\beta_j=1.00$ against $H_a:\beta_j\neq1.00$ at the 5% level. Assume the large-sample normal approximation is appropriate and the robust covariance estimator is consistent.

The Wald statistic is $$ z=\frac{1.80-1.00}{\sqrt{0.51}} =\frac{0.80}{0.714142\ldots} \approx 1.1202. $$

The two-sided p-value is $2P(Z\geq 1.1202)\approx 0.2626$. Since $0.2626>0.05$, this test does not reject the null at the 5% level. The data do not provide strong evidence that the coefficient differs from $1.00$ points per treatment unit under the stated model and approximation. This result does not establish that the coefficient equals $1.00$.

Confidence Intervals and Their Interpretation

The confidence interval for a single coefficient uses the same estimated standard error as the Wald test. For a 95% large-sample interval, $z_{0.975}$ is approximately $1.96$. The interval is centered at the estimate, with a margin of error equal to $1.96$ times its standard error. It gives a range of coefficient values compatible with the data under the model and approximation; it does not assign a 95% probability to the fixed parameter being inside this particular interval.

For the two-sided test of $H_0:\beta_j=\beta_{j,0}$ at level $\alpha$, the corresponding $1-\alpha$ Wald interval excludes $\beta_{j,0}$ exactly when the test rejects, provided both use the same standard error and reference distribution. This follows by rearranging the test rule, as shown below.

Derivation: The two-sided normal Wald test rejects when $$ \left|\frac{\hat{\beta}_j-\beta_{j,0}}{SE(\hat{\beta}_j)}\right|>z_{1-\alpha/2}. $$ Because the standard error is positive, multiplying by it and rearranging gives $$ \left|\hat{\beta}_j-\beta_{j,0}\right|>z_{1-\alpha/2}SE(\hat{\beta}_j). $$ This is equivalent to $\beta_{j,0}$ lying outside the interval $$ \left[\hat{\beta}_j-z_{1-\alpha/2}SE(\hat{\beta}_j),\ \hat{\beta}_j+z_{1-\alpha/2}SE(\hat{\beta}_j)\right]. $$ Thus the test and interval express the same Wald comparison when their level and standard-error calculation match.

Worked Example: A 95% Interval in Outcome Units

In a hypothetical study, the outcome is systolic blood pressure in millimeters of mercury, and treatment is measured in increments of 10 milligrams. A 2SLS estimate is $-2.40$ millimeters of mercury per 10-milligram increase, with a heteroskedasticity-robust standard error of $0.900$ in the same units. Assume a large-sample normal approximation is appropriate.

The 95% interval is $$ -2.40\mathbin{\pm}1.96(0.900) =-2.40\mathbin{\pm}1.764 =[-4.164,-0.636]. $$

Under the model and inference assumptions, the interval estimates a decrease of between $0.636$ and $4.164$ millimeters of mercury in systolic blood pressure per 10-milligram increase in treatment. Zero is not in the interval, so the corresponding two-sided test of $H_0:\beta_j=0$ at the 5% level rejects. Whether the coefficient has a causal interpretation depends on the IV assumptions, not on the interval itself.

Which Standard Error and Reference Distribution?

The choice of standard error should match the plausible dependence in the data. A heteroskedasticity-robust 2SLS covariance estimator allows the error variance to differ across observations, while relying on observations being independent across units under the sampling design. If observations may be correlated within groups, such as patients treated at the same clinic, a cluster-robust covariance estimator allows arbitrary within-cluster dependence under its assumptions.

Cluster-robust inference typically relies on independence across clusters and a sufficiently large number of independent clusters for its large-sample approximation. With few clusters, the usual normal reference distribution and conventional cluster-robust standard errors may be unreliable; small-sample adjustments or other methods may be needed. The degrees of freedom and reference distribution used by software can depend on the particular procedure, so report and follow the method actually used rather than automatically substituting a $t$ critical value.

Covariance estimate	Dependence it accommodates	Key sampling consideration
Heteroskedasticity-robust	Unequal variances across independent observations	Independence across observations for the usual large-sample argument
Cluster-robust	Unequal variances and dependence within specified clusters	Independent clusters and enough clusters for the chosen approximation

Worked Example: A Test Using Cluster-Robust Inference

Suppose a hypothetical analysis uses 480 patients from 48 clinics. The treatment is the number of additional 10-hour staffing increments, and the outcome is patient waiting time in minutes. The estimated 2SLS coefficient is $0.65$ minutes per staffing increment, and the reported cluster-robust standard error is $0.28$ minutes per increment. Assume clinics are independent sampling units, the number of clusters is adequate for a normal approximation, and the covariance estimator is consistent. Test $H_0:\beta_j=0.50$ against a two-sided alternative at the 5% level.

The statistic is $$ z=\frac{0.65-0.50}{0.28} =\frac{0.15}{0.28} \approx0.5357. $$

The two-sided p-value is $2P(Z\geq0.5357)\approx0.5922$. Since this exceeds $0.05$, the test does not reject $H_0$. The estimate is $0.15$ minutes per increment above the null value, but that difference is only about $0.536$ cluster-robust standard errors. The test does not provide strong evidence that the coefficient differs from $0.50$ minutes per staffing increment.

For comparison, the 95% interval from the same standard error is $$ 0.65\mathbin{\pm}1.96(0.28) =0.65\mathbin{\pm}0.5488 =[0.1012,1.1988] $$ minutes per staffing increment. It contains $0.50$, agreeing with the test result.

What the Coefficient Test Does—and Does Not—Say

A coefficient is expressed in the units of the outcome divided by the units of the corresponding regressor, holding the other terms in the specified model fixed. If the treatment is measured in 10-milligram increments and the outcome in millimeters of mercury, the coefficient is in millimeters of mercury per increment—not per milligram unless it is rescaled. A nonzero test result is evidence against a particular coefficient value, not a test of every assumption behind the IV analysis.

For an IV coefficient to represent the intended structural or causal quantity, the instruments must satisfy the relevant IV conditions, the model must be appropriately specified, and the parameter must be identified. A robust standard error addresses a feature of sampling variability; it does not repair an invalid instrument or establish identification. Conventional Wald inference can also be inaccurate when the instruments provide little information about the endogenous regressor. The ordinary intervals and tests here should therefore be read as conditional on their approximation being suitable, not as a remedy for every IV problem.

Common Mistakes

Using the second-stage OLS standard error. It is not generally the correct 2SLS standard error. Use the appropriate 2SLS covariance estimate, robust to heteroskedasticity or clustering as required by the sampling structure.
Testing the estimate against zero by default. The null can be any meaningful value. Subtract the stated null coefficient in the numerator, as in $(\hat{\beta}_j-\beta_{j,0})/SE(\hat{\beta}_j)$.
Leaving out the coefficient's units. State the outcome and regressor units when interpreting an estimate or interval; a coefficient is not unit-free.
Calling a non-rejection proof of equality. A large p-value means the data do not provide sufficient evidence against the specified null at that test level. It does not prove the null or show that the estimate is precise enough for every practical purpose.
Treating robust inference as instrument validation. Robust covariance estimation concerns estimated sampling uncertainty. It does not show that instruments are valid or that conventional inference is reliable under weak identification.

Key takeaway: For an individual 2SLS coefficient, specify the null value, use the appropriate 2SLS standard error, and apply a reference distribution justified by the sampling design and approximation. Report the result in context and in units, while keeping statistical uncertainty distinct from the assumptions that give the coefficient its IV interpretation.

Check Your Understanding

Use the stated null value and inference method in each answer.

A coefficient estimate is $2.1$ with standard error $0.5$. Write the Wald statistic for testing $H_0:\beta_j=1$, and calculate it.
In a large-sample normal Wald analysis, what does a two-sided p-value of $0.08$ imply at significance level $0.05$? What does it not imply?
A 95% normal Wald interval for a coefficient is $[-1.2,-0.2]$. What is the result of the matching two-sided test of a zero coefficient at the 5% level?
Why might a cluster-robust standard error be more appropriate than a heteroskedasticity-robust one when patients are sampled from a modest number of clinics?
What assumptions are not established merely by reporting a robust 2SLS confidence interval?