Single-Coefficient Tests With Estimated Dispersion

Why Estimate Dispersion for a Single Coefficient?

In a quasi-binomial or quasi-Poisson model, the variance of each response is allowed to be a dispersion parameter $\phi$ times its variance-function value. When $\phi$ is estimated rather than fixed at one, the estimated coefficient covariance matrix is scaled accordingly. That changes the standard error used to test one coefficient or form its confidence interval.

The calculation resembles a familiar t procedure: divide the coefficient’s distance from the null value by its estimated standard error, then use a t reference distribution with residual degrees of freedom. This is a useful finite-sample calibration, but it is important not to overstate it. Quasi-likelihood alone does not generally imply an exact t distribution. The t reference is an approximation; ordinary binomial and Poisson Wald procedures instead use the standard-normal reference distribution and fix dispersion at one.

Coefficient Standard Error and the t Statistic

Let $\hat{\beta}_j$ be the fitted estimate of coefficient $\beta_j$, and let $C_{jj}$ be the $j$th diagonal element of the fitted coefficient covariance matrix calculated with dispersion fixed at one. This is the unscaled covariance estimate. With estimated dispersion $\hat{\phi}$, the estimated variance and standard error for this coefficient are

\widehat{\operatorname{Var}}(\hat{\beta}_j)=\hat{\phi}C_{jj}, \qquad \operatorname{SE}(\hat{\beta}_j)=\sqrt{\hat{\phi}C_{jj}}.

For a quasi-GLM with $n$ response observations and a fitted mean model of rank $p$, a common residual degrees-of-freedom choice is $\nu=n-p$. For example, $\hat{\phi}$ may be estimated by a Pearson statistic divided by $n-p$; the particular estimator must be stated because different dispersion estimates can produce different standard errors.

Definition: To test $H_0:\beta_j=\beta_{j,0}$, the single-coefficient t statistic is $$ t_{\mathrm{obs}}= \frac{\hat{\beta}_j-\beta_{j,0}}{\sqrt{\hat{\phi}C_{jj}}}. $$ A conventional two-sided t-calibrated test compares $|t_{\mathrm{obs}}|$ with the $1-\alpha/2$ quantile of a $t_\nu$ distribution, or reports the corresponding two-sided $t_\nu$ tail probability.

The matching two-sided confidence interval is

\hat{\beta}_j\ \pm\ t_{\nu,\,1-\alpha/2}\sqrt{\hat{\phi}C_{jj}},

where $t_{\nu,\,1-\alpha/2}$ is the $1-\alpha/2$ quantile of the $t_\nu$ distribution. The estimate and the unscaled covariance come from fitting the mean model; the estimated dispersion multiplies the covariance, not the coefficient itself.

Worked Example: A t-Calibrated Confidence Interval

Suppose a quasi-GLM gives $\hat{\beta}_j=-0.35$, $C_{jj}=0.0225$, and $\hat{\phi}=1.6$. There are 32 observations and 5 fitted parameters, so use $\nu=32-5=27$. Construct a 95% t-calibrated interval.

First scale the unscaled variance and calculate the standard error:

\widehat{\operatorname{Var}}(\hat{\beta}_j) =1.6(0.0225)=0.036, \qquad \operatorname{SE}(\hat{\beta}_j)=\sqrt{0.036}=0.1897367.

The 97.5th percentile of $t_{27}$ is approximately $2.05183$. Thus the margin of error is $2.05183(0.1897367)\approx0.3893$, and the interval is

-0.35\pm0.3893 \ \approx\ (-0.7393,\ 0.0393).

Because the interval includes zero, it does not reject the two-sided null $H_0:\beta_j=0$ at the 5% level under this calibration. The endpoints are in coefficient units; exponentiating them would be appropriate only if the coefficient’s interpretation called for a multiplicative effect.

Why the Interval and Test Agree

The relationship between the two procedures follows by inverting the test. For any proposed value $\beta_{j,0}$, the two-sided test rejects at level $\alpha$ when $|t_{\mathrm{obs}}|>t_{\nu,1-\alpha/2}$. Since the standard error is positive, this is equivalent to

\left|\frac{\hat{\beta}_j-\beta_{j,0}}{\operatorname{SE}(\hat{\beta}_j)}\right| >t_{\nu,1-\alpha/2} \quad\Longleftrightarrow\quad \beta_{j,0}<\hat{\beta}_j-t_{\nu,1-\alpha/2}\operatorname{SE}(\hat{\beta}_j) \ \text{or}\ \beta_{j,0}>\hat{\beta}_j+t_{\nu,1-\alpha/2}\operatorname{SE}(\hat{\beta}_j).

Therefore the values not rejected by the test are exactly those inside the stated confidence interval (with the usual endpoint convention). This is an algebraic equivalence between the specified t-calibrated test and interval; it does not establish that the t reference distribution is exact.

Result: Under the stated t calibration, a two-sided level-$\alpha$ test of $H_0:\beta_j=\beta_{j,0}$ rejects precisely when $\beta_{j,0}$ lies outside the corresponding $100(1-\alpha)\%$ interval. This follows by rearranging the test inequality as above.

Why the t Reference Is Usually Approximate

A t distribution arises exactly in the classical normal linear model under particular assumptions, including normally distributed errors and an independent chi-square estimate of the error variance. A quasi-GLM specifies a mean and a variance relationship, not a full response distribution. Those specifications alone do not supply the distributional facts needed for an exact t result.

For a regular quasi-GLM, the usual large-sample justification is instead that the coefficient estimator is approximately normal when the mean and variance model are correctly specified and the observations are independent. More specifically, suppose the number of coefficients is fixed, the estimating equations are regular, the information matrix is nonsingular, the relevant moments are finite, and the dispersion estimator is consistent. Then the coefficient’s estimation error, standardized by its true standard deviation, is asymptotically standard normal. If $\hat{\phi}/\phi\xrightarrow{p}1$ and the estimated unscaled variance is consistent, replacing the unknown standard error by the estimated one preserves that limit by Slutsky’s theorem. As residual degrees of freedom grow, $t_\nu$ quantiles approach the corresponding standard-normal quantiles.

This large-sample reasoning supports normal or t-calibrated inference as an approximation; it does not imply that the statistic has an exact $t_\nu$ distribution in a finite sample. Using residual degrees of freedom for the t reference is a common practical convention, not a universal theorem for every quasi-GLM.

Worked Example: Testing One Coefficient

Suppose $\hat{\beta}_j=0.50$, $C_{jj}=0.04$, and $\hat{\phi}=2.25$. Use $\nu=24$ residual degrees of freedom to test $H_0:\beta_j=0$ against a two-sided alternative.

The dispersion-scaled standard error is

\sqrt{2.25(0.04)}=\sqrt{0.09}=0.30.

The test statistic is $t_{\mathrm{obs}}=0.50/0.30=1.6667$. The two-sided $t_{24}$ p-value is approximately $0.1086$. Since this exceeds $0.05$, the test does not reject the null at the 5% level. This conclusion uses the t calibration and the stated degrees of freedom; the p-value is not an exact finite-sample probability guaranteed by the quasi-GLM specification.

Contrast With Ordinary Binomial and Poisson Wald Inference

In an ordinary binomial or Poisson GLM, the variance function uses dispersion fixed at $\phi=1$. The usual single-coefficient Wald statistic is therefore

z_{\mathrm{obs}}= \frac{\hat{\beta}_j-\beta_{j,0}}{\sqrt{C_{jj}}}.

Its conventional reference is standard normal. In a quasi-GLM, the coefficient estimate often agrees with the ordinary fit, while the estimated covariance is multiplied by $\hat{\phi}$. When $\hat{\phi}>1$, the resulting standard error is larger, and using a t rather than a normal reference can further widen the interval for finite $\nu$. Neither standard-normal Wald inference nor t-calibrated quasi-inference is generally an exact finite-sample test for a GLM coefficient.

Worked Example: How Dispersion Can Change a Decision

An ordinary Poisson fit gives $\hat{\beta}_j=0.22$ and unscaled standard error $0.10$. Suppose the quasi-Poisson analysis estimates $\hat{\phi}=2.25$, and use $\nu=40$. Test $H_0:\beta_j=0$ at the two-sided 5% level under each conventional procedure.

For the ordinary Poisson Wald test, the standard error is $0.10$, so $z_{\mathrm{obs}}=0.22/0.10=2.20$. Since $2.20>1.96$, the ordinary normal-reference test rejects at 5%.

For the quasi-Poisson calculation, the standard error is

\sqrt{2.25}(0.10)=1.5(0.10)=0.15.

Thus $t_{\mathrm{obs}}=0.22/0.15\approx1.467$. The 97.5th percentile of $t_{40}$ is approximately $2.021$, and $1.467<2.021$, so the quasi-GLM test does not reject at 5%. The coefficient estimate is unchanged in this illustration; estimated overdispersion increases its uncertainty enough to change the decision.

Common Pitfalls

Using the unscaled standard error. In a quasi-GLM, $\sqrt{C_{jj}}$ omits estimated dispersion. Use $\sqrt{\hat{\phi}C_{jj}}$.
Changing the coefficient instead of its uncertainty. Multiplying the coefficient estimate by $\hat{\phi}$ is incorrect; dispersion scales its estimated variance.
Calling the t result exact. A quasi mean–variance specification does not generally imply an exact t distribution. Describe the reference as a t calibration or approximation.
Using a t reference for an ordinary fixed-dispersion Wald calculation without saying so. The conventional ordinary binomial or Poisson Wald reference is standard normal; the quasi procedure estimates dispersion and commonly uses residual degrees of freedom for t calibration.
Ignoring which dispersion estimate and degrees of freedom were used. These choices affect the reported standard error, interval, and p-value and should be made explicit.

Key takeaway: For a single quasi-GLM coefficient, scale its unscaled variance by $\hat{\phi}$, then use the resulting standard error in a t statistic and a matching t-calibrated interval. Treat the t reference as a practical finite-sample approximation, not an exact consequence of quasi-likelihood.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

If $C_{jj}=0.01$ and $\hat{\phi}=4$, what is the dispersion-scaled standard error?
For a test of $H_0:\beta_j=\beta_{j,0}$, write the single-coefficient t statistic and identify the role of $\nu$.
Why does inverting the two-sided t test produce the stated t-calibrated confidence interval?
What distributional information is missing from a quasi-GLM specification that prevents its t calibration from being generally exact?
In the final worked example, which changes between the ordinary Poisson and quasi-Poisson calculations, and why can that change affect the test decision?