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For each observed event $i$ with $v_i=1$, let $R_i=\{j:x_j\ge x_i\}$ be its risk set. The Cox partial likelihood isMaximize it to obtain $\hat{\beta }$, estimate its variance from the observed partial information, and test $H_0:\beta =0$ using a partial likelihood-ratio test, Wald test, or score test. A positive fitted coefficient means the group with $z=1$ has the larger hazard.

Under a country-specific constant hazard survival model, let $D_k$ be the number of deaths and $Y_k$ the total person-time at risk in country $k$. The likelihood is proportional toso the maximum-likelihood estimator is ${\hat{\lambda }}_k=D_k/Y_k$. Test $H_0:{\lambda }_0={\lambda }_1$ by a likelihood-ratio test, score test, or Wald test; equivalently, give $D_k$ a Poisson distribution with mean ${\lambda }_k{Y}_k$ and test the country coefficient in a Poisson regression with offset $\log Y_k$.

The pooled Wald test statistic isIn asymptotic calculations one replaces $\bar{p}$ by its probability limit $\bar{\pi }=(n_1{\pi }_1+n_0{\pi }_0)/(n_1+n_0)$.

The estimates maximize the grouped-binomial likelihood, numerically obtained by iteratively reweighted least squares. At the fitted probabilities, the estimated covariance matrix is the inverse observed information, approximatelyThe Std. Error column is the square root of its diagonal. Each z value is Estimate/Std. Error, and the two-sided Wald test p-value isfor the null hypothesis that the corresponding coefficient is zero.

Thus ${\hat{\beta }}_0=1.6855$ is the fitted log odds for a nonmalignant patient at site A. The site coefficients compare B and C with A, while ${\hat{\beta }}_M=-0.9048$ compares malignant with nonmalignant tumours after adjustment for site. In odds-ratio form,At level $0.05$, the B-versus-A contrast is not significant ($p=0.0624$), the C-versus-A contrast is not significant ($p=0.2610$), and malignancy has a significant negative association with survival ($p=0.0175$). The intercept test merely compares the baseline survival probability with $1/2$.

The null deviance $11.693$ compares the intercept-only fit with the saturated model and has $6-1=5$ residual degrees of freedom. The residual deviance $0.85048$ compares the four-parameter fitted model with the saturated model and has $6-4=2$ degrees of freedom. Finally, the Akaike information criterion is

For independent effect estimates $1.01$ with standard error $0.50$, and $0.99$ with standard error $0.51$, one two-sided Wald test is significant at five percent and the other is not. But their difference is $0.02$ with standard error $\sqrt{0.50^2+0.51^2}\approx 0.714$, giving a negligible test statistic. Hence different significance labels do not establish an interaction or between-study heterogeneity; compare the effects directly.

In a Poisson regression with $\log \mu ={\beta }_0+{\beta }_{T}T+{\beta }_{B}b+{\beta }_{TB}Tb$, the treatment logarithm of the mean ratio at baseline $b$ is ${\beta }_T+b{\beta }_{TB}$. The main treatment coefficient tests the contrast at $b=0$. An overall no-treatment test requires both ${\beta }_T=0$ and ${\beta }_{TB}=0$, with their joint covariance matrix for a Wald test.

Independent normal arm means with variances $v_0,v_1,v_2$ give treatment-contrast Wald test statistics whose correlation coefficient is $v_0/\sqrt{(v_0+v_1)(v_0+v_2)}$. The common control mean creates positive dependence between comparisons.