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Among patients still alive at the same time after surgery, chemotherapy has an estimated lung-cancer death rate 3.71 times that of radiotherapy, assuming proportional cause-specific hazards. The estimated 95% confidence interval for this hazard ratio is $(1.583,8.697)$. Thus the fit suggests a higher instantaneous lung-cancer death rate on chemotherapy, with appreciable uncertainty in its size.

The null hypothesis of no treatment effect on that hazard is $H_0:{\gamma }_1=0$ against ${\gamma }_1\ne 0$. Its Wald test uses $Z=1.3113/0.4346\simeq 3.018$, approximately standard normal under the null hypothesis, with two-sided $p=0.00255$. This is evidence against equal lung-cancer death hazards in the fitted model. It is not a statement that five-year lung-cancer death probability is multiplied by 3.71 or that survival duration is divided by 3.71. The cumulative incidence function depends on both cause-specific hazards, and the model needs proportional-hazards and independent observation-censoring assumptions. In a randomised comparison the assigned treatment contrast is protected against baseline confounding, but the instantaneous comparison concerns the groups who have survived to the time in question; it does not identify a treatment effect within a common latent survivor subgroup.

For a one-sided Wald test use the signed normal Wald statistic, rather than its square. The maximum-likelihood estimate is the sample mean ${\bar{Y}}_n$, with exact variance ${\sigma }^2/n$, soA sum of independent normal random variables is normal, giving ${\bar{Y}}_n\sim N(\delta ,{\sigma }^2/n)$ and hence $W_n\sim N(\sqrt{n}\delta /\sigma ,1)$. Under the null hypothesis its mean is zero:Thus its null distribution is exactly the standard normal distribution, without an asymptotic approximation. If the squared Wald statistic convention is used, $W_n^2$ has a chi-squared distribution with one degree of freedom; the signed form is needed to distinguish the two directions.

Let $Y_i$ be the number of maintenance jobs, $t_i$ average temperature and $p_i$ average precipitation. The fitted Poisson regression assumes independent conditional responses withThe logarithm is the Poisson canonical link, so fitted means are always positive. The estimated regression coefficients areAt fixed precipitation, one unit of temperature multiplies the fitted mean by $e^{-0.006162}\simeq 0.99386$, a decrease of about $0.614\%$. The fitted precipitation multiplier is $e^{-0.002922}\simeq 0.99708$ per unit, but its large p-value gives little evidence for that effect. These are conditional associations, not established causal effects.

The intercept-only model has $n-1=23$ residual degrees of freedom, and the full model has $n-3=21$. Both imply $n=24$ months.

For an approximate deviance goodness-of-fit test of the full Poisson regression, compare residual deviance $23.527$ with ${\chi }_{21}^2$. Its upper-tail p-value $0.3165361$ supplies no evidence of lack of fit. The ratio $23.527/21\simeq 1.120$ also gives no striking indication of overdispersion, though deviance per degree of freedom is only a rough dispersion check. Fitted count means around fifty make the usual chi-squared approximation plausible. Independence between months, the conditional variance-equals-mean assumption and the absence of residual structure still need diagnostics. Failure to reject is not proof that the model is correct.

The null deviance $37.969$ on $23$ degrees of freedom has $p=0.02566753$, suggesting the constant-mean model is inadequate. In the sequential analysis of deviance for nested generalized linear models, adding temperature to that model reduces deviance by $14.204$ on one degree of freedom, giving $p=0.000164$. Adding precipitation after temperature reduces it by only $0.238$, with $p=0.625677$. Therefore retain temperature; these data give no reason to retain precipitation after temperature. The temperature-only residual deviance is $23.765$ on $22$ degrees of freedom. This prediction-oriented reduced model should be refitted before reporting its coefficients: the printed temperature coefficient belongs to the full model. The Wald test results, $p=0.000833$ and $0.625582$, broadly support the same conclusion, while the sequential deviance tests answer the stated nested-model questions.

Let $a_i$ be age in years, $g_i$ the male indicator and $m_i$ the indicator for the Messiah group. The fitted Cox proportional-hazards model iswith an unspecified common baseline hazard. Its coefficients were estimated by Cox partial likelihood, using the Breslow approximation for tied event times. There are $100$ participants, $96$ recorded completions and four censored observations.

A higher hazard function means a greater instantaneous chance of completion among those not yet completing, so it describes faster completion rather than greater mortality. Holding gender and group fixed, an extra year of age multiplies the completion hazard by $1.0066$, about a $0.66\%$ increase. The 95% confidence interval is $(0.9865,1.0270)$ and the Wald test has $p=0.5241$, so there is little evidence of an age association.

Holding age and group fixed, male participants have a hazard ratio $0.5469$ relative to female participants, about a $45.3\%$ lower completion hazard. Its 95% confidence interval is $(0.3560,0.8401)$ and $p=0.00586$. Under the fitted model this corresponds to slower completion for males. It is not a ratio of mean or median completion times.

Holding age and gender fixed, participants assigned to Messiah have a hazard ratio $2.1283$ relative to The Kingdom, with 95% confidence interval $(1.3653,3.3176)$ and $p=0.000854$. This is strong evidence of a group difference, with faster completion in the Messiah group under the fitted model. Random assignment supports interpreting a group contrast as an effect of the assigned condition, subject to the assumptions about follow-up and censoring; age and gender contrasts remain observational associations.

Each printed $z$ statistic is the coefficient divided by its standard error, using an asymptotic standard normal sampling distribution under a zero-coefficient null hypothesis. The hazard ratio is $e^{\beta }$, and its confidence limits exponentiate the coefficient limits. Proportional hazards assumes these covariate-specific hazard multipliers stay constant over follow-up, together with the specified linear age effect and independent censoring conditional on covariates.

For an independent subgroup contrast in meta-analysis, the difference of disjoint independent subgroup estimates has estimated variance equal to the sum of the two variances. The subgroup contrast and standard error areThe Wald test of equal subgroup mean log odds ratios therefore givesUnder the null, $Z$ is approximately standard normal and $Z^2$ has a chi-squared distribution with one degree of freedom. A quantile-2 confidence interval for the difference is $(0.29-0.8,0.29+0.8)=(-0.51,1.09)$. There is no statistically persuasive evidence of a subgroup difference. The result does not establish equal effects. Nor does a between-trial subgroup comparison, by itself, identify a causal advantage of one technique: other study characteristics may differ.

The Poisson regression uses independent counts with mean $m_i$ and logarithmic link function. Writing $c$ for concentration, the fitted predictors for strains $0,1,2$ are respectivelyThe interaction formula includes concentration, strain indicators and their products. The intercept is the baseline log mean at zero toxin; exponentiating it gives about $87.5$ fleas. Strain coefficients compare log means at zero concentration. Interaction coefficients compare concentration slopes with the common strain. A unit concentration increase multiplies the three means by about $0.751$, $1.057$ and $0.769$ respectively.

For design matrix $X$, the fitted expected Fisher information is $X^{T}WX$, $W=\mathrm{diag}({\hat{m}}_i)$, since a Poisson count has variance equal to its mean. Standard errors are square roots of diagonal entries of $(X^{T}WX{)}^{-1}$. Each reported Wald test divides its coefficient estimate by its standard error and compares with an approximate standard normal distribution under a zero-coefficient null, conditional on the other terms. Concentration has strong evidence of a negative baseline effect. At zero concentration strain 1 is not significant at 5% ($p=0.087$), while strain 2 is ($p=0.030$). The strain-1 slope differs strongly from the baseline ($p=0.000193$); the strain-2 slope has no detected difference ($p=0.808$). A nonsignificant difference is not proof of equality. The very significant intercept is a test of mean one at zero toxin, with little scientific relevance.

Relabelling factor levels merges strains 0 and 2 into one group and leaves strain 1 separate. Model $M_2$ retains three intercepts but allows only two slopes: strain 1 versus the common slope of strains 0 and 2. Thus it has five parameters, compared with six in $M_1$. Model $M_3$ merges both intercepts and slopes for strains 0 and 2, retaining only four parameters. The natural nesting is $M_3\subset M_2\subset M_1$.

For these nested generalized linear models, the difference in Poisson deviance is an asymptotic likelihood-ratio test statistic with degrees of freedom equal to the parameter difference. Comparing $M_2$ with $M_1$ gives $56.93-56.87=0.06$ on one degree of freedom, far below $3.84$: the additional strain-2 slope is unnecessary. Comparing $M_3$ with $M_2$ gives $76.98-56.93=20.05$ on one degree of freedom, far above $3.84$: merging the strain-0 and strain-2 intercepts is rejected. Comparing $M_3$ with $M_1$ uses $20.11$ on two degrees of freedom and requires a two-degree-of-freedom threshold, about $5.99$, rather than the supplied $3.84$.

$M_2$ is the most appropriate of the three models: retain different strain baselines and the special strain-1 slope, but pool the unsupported strain-2 slope difference. Residual degrees of freedom are $74,75,76$; deviances of this magnitude give no obvious evidence of overdispersion from these summaries alone.

The initial Poisson regression contains altitude, temperature, and their interaction term, with a logarithmic link function. Its interaction Wald test has $p=0.521927$, suggesting removal of that term while initially preserving both main effects. The resulting additive model has a negligible temperature effect ($p=0.874$), suggesting a second simplification to altitude alone. The nested analysis of deviance for nested generalized linear models gives the same provisional decisions: deviance increases are $83.326-82.916=0.410$ and $83.351-83.326=0.025$, each for one removed coefficient, with approximate ${\chi }_1^2$ upper-tail probabilities $0.522$ and $0.874$. The Akaike information criterion also decreases at each step.

For independent island counts the final proposed model iswith fitted relationThis is an exponential mean relation on the count scale, not a linear relation between count and altitude. One additional metre multiplies the fitted expected count by $e^{0.01898}\approx 1.01916$. The null model's insignificance of individual main effects in the interaction fit does not require discarding every predictor: correlated design columns can make those conditional tests imprecise. These reductions concern relative mean complexity under the working Poisson regression; the large residual deviance in all three models warns that their variance or mean assumptions still need investigation.

This is a sequential analysis of deviance for nested generalized linear models. Adding age to the intercept-only logistic regression reduces deviance by $6.5187$, with approximate ${\chi }_1^2$ p-value $0.01067$. Adding frailty to the age model reduces it by another $16.6474$, with p-value $4.501\times 10^{-5}$. Thus both variables have evidence of association with surgical success, under the model assumptions, and neither should be dropped solely on these results.

The second test assesses frailty adjusted for age, whereas the first assesses age before frailty is included. Sequential tests depend on term order. Evidence for age adjusted for frailty is additionally provided by its full-model Wald test, $p=0.01390$. Both fitted coefficients are negative: age and frailty are associated with lower fitted success probabilities. On the odds scale, one extra year at fixed frailty multiplies the odds by $e^{-0.08942}\approx 0.9145$, and frailty at fixed age multiplies them by $e^{-3.15336}\approx 0.04271$. These are odds ratios, not risk ratios, and an observational statistical association does not by itself establish a causal effect.

Use the usual clinical trial sampling assumption that all patients' responses are independent, with fixed positive arm sizes and positive known variances. Marginal normal distributions alone would not specify the variance of a treatment contrast without this assumption. Put ${\overline{Y}}_i=n_i^{-1}{\sum }_j{Y}_{ij}$, $v_i={\sigma }_i^2/n_i$, and $s_i=\sqrt{v_i+v_0}$. The sample mean contrast estimates ${\delta }_i$ and has standard error $s_i$. Thus the one-sided Wald statistics areLarge positive values of these Wald test statistics provide evidence against the respective null hypotheses.

Couple the Wald test statistics as $W_i=Z_i+{\delta }_i/s_i$, where the joint centered bivariate normal distribution of $(Z_1,Z_2)$ is fixed. Increasing either ${\delta }_i$ enlarges, pointwise in this coupling, the event $\{W_1>c\text{ or }{W}_2>c\}$. Its probability is consequently nondecreasing in each treatment effect.

If both null hypotheses hold, then ${\delta }_1,{\delta }_2\le 0$, so this probability is maximal at the joint boundary $(0,0)$. To check the familywise error rate across all configurations, suppose only hypothesis $i$ is true. A Type I error then occurs only if $W_i>c$, with probability at most $1-\Phi (c)$, where $\Phi$ is the standard normal distribution function. This is no larger than the any-rejection probability at $(0,0)$, since that joint event contains $\{Z_i>c\}$. If neither null hypothesis is true, the familywise error rate is zero. Hence the least-favourable null configuration for Type I error iswhere ${\Phi }_2$ is the centered bivariate normal distribution function with correlation coefficient $\rho$.

Choose a group sequential design. Randomize patients among the three arms, schedule interim analyses after prespecified amounts of information, and calculate the two Wald test statistics at each look. Prespecify efficacy boundaries and a futility rule; allow early stopping of an ineffective experimental arm or of the trial when sufficiently convincing efficacy evidence is obtained. Calibrate the boundaries for both repeated looks and the two shared-control comparisons so that the familywise error rate remains controlled. An alpha-spending function is one way to allocate the error probability across looks.

Two potential benefits are lower expected recruitment and shorter duration when a treatment is clearly effective or futile, and earlier identification of benefit or lack of benefit, reducing continued exposure to an ineffective experimental treatment. These benefits concern expected performance; the maximum sample size may still be required.

Two drawbacks are the extra operational work needed for rapid, reliable outcomes and confidential interim analyses, and more complicated inference: repeated testing requires adjusted boundaries, and ordinary fixed-sample effect estimates and confidence intervals can be misleading after data-dependent stopping. Continuing two comparisons against a shared control also requires care if an arm is stopped.

Under the null hypothesis of consistency, ${\hat{\delta }}_D-{\hat{\delta }}_I$ has approximately zero expectation and variance $v_D+v_I$, since the displayed trials are independent. Thus a two-sided Wald test usesThe observed difference is $-0.07$, and $Z=-0.07/\sqrt{0.29}\simeq -0.130$. Its two-sided p-value is approximately $0.897$, so there is no detectable discrepancy here. Failure to reject is not proof of consistency: this test has limited statistical power, and transitivity in network meta-analysis also needs substantive assessment. If evidence sources overlap, replace $v_D+v_I$ by $v_D+v_I-2\mathrm{Cov}({\hat{\delta }}_D,{\hat{\delta }}_I)$.

Use one joint hypothesis test for $H_0:{\beta }_{\lambda }={\beta }_{\mu }=0$. Fit the unrestricted model with statistical parameters $({\lambda }_0,{\mu }_0,{\beta }_{\lambda },{\beta }_{\mu })$, where ${\lambda }_i={\lambda }_0{e}^{{\beta }_{\lambda }{z}_i}$ and ${\mu }_i={\mu }_0{e}^{{\beta }_{\mu }{z}_i}$ for treatment indicator $z_i$. Refit under $H_0$, estimating the two common baseline intensities; do not fix them at their unrestricted estimates. The likelihood-ratio test statistic isUnder regular identifiable positive-rate models, sufficient independent patient trajectories, and the null hypothesis, Wilks theorem gives two degrees of freedom because two coefficients are constrained. Reject if $D$ exceeds the chosen upper chi-squared distribution quantile. Alternatively a two-dimensional Wald test uses ${\hat{\beta }}^T\widehat{\mathrm{Cov}}(\hat{\beta }{)}^{-1}\hat{\beta }$, including the off-diagonal covariance. Two separate tests do not provide this single calibrated assessment. The point estimates alone cannot produce a numerical p-value without fitted likelihoods or a coefficient covariance matrix.

Put $v_k=p_k(1-p_k)$ and assume $0<p_k<1$. For fixed $n$ and a fixed nonzero difference $\Delta =p_0-p_1$, the approximate statistical power of the Wald test increases as its noncentrality $|\Delta |/\sqrt{V}$ increases, whereDifferentiating with respect to the continuous allocation givesHence the unique minimum and its associated minimum variance areThis is Neyman allocation. Integer allocations use a feasible adjacent integer to the continuous optimum, comparing the resulting variances; $n_0,n_1\ge 1$ must still hold. It is a large-sample normal approximation argument, not an exact finite-sample power theorem for arbitrary response-adaptive randomization. Under the null $p_0=p_1$, the approximate power is the significance level for any allocation, although the same allocation still minimizes the stated variance.

Under a correctly specified Poisson regression with adequate expected counts, the residual Poisson deviance is approximately chi-squared distribution with $n-p=107-5=102$ residual degrees of freedom. The observed $D=319.86$ givesThus the ordinary Poisson model gives a very poor fit under this goodness-of-fit approximation. This is consistent with overdispersion, although a wrong mean model or dependence can also cause lack of fit; quasi-Poisson variance inflation alone does not establish that the mean model is adequate.

For Quasi-Poisson regression, $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson dispersion estimator isIt uses squared Pearson residuals, not the residual deviance divided by 102. The fitted coefficients are unchanged, and their estimated covariance matrix is multiplied by $\hat{\varphi }$. With $\hat{\varphi }=3.24$, standard errors increase by $1.8$, so the main treatment coefficient hasThe main Progabide coefficient is not significant at 5% after this dispersion adjustment. R uses a Student's t-distribution when dispersion is estimated, as described in its GLM summary documentation; the large-sample normal approximation also gives the same decision.

Because an interaction term is present, this tests the treatment contrast at baseline zero. An overall absence of treatment influence requires the joint null ${\beta }_T={\beta }_{TB}=0$. The printed marginal standard errors do not give their covariance, so they are insufficient to calculate that joint Wald test, or the treatment contrast test at an arbitrary nonzero baseline.

PutAt a fixed alternative $\theta$, the approximate power of the Wald test increases as its variancedecreases. For fixed total sample size $n=n_0+n_1$, the allocation problem is thereforewith integer rounding applied after solving the continuous problem. The second-order condition is positivity of the second derivative at the stationary point.

None of the displayed one-coefficient Wald tests for team1_form rejects at the $5\%$ significance level. That does not establish that formation has no effect: the relevant null hypothesis sets all nine displayed team1_form coefficients to zero simultaneously, and several individually imprecise estimates may still be jointly significant. One should compare the fitted model with the nested model omitting team1_form by an analysis of deviance for nested generalized linear models, equivalently a likelihood-ratio test of those joint restrictions.

For equal arm size $n$, write $v^{\ast }=p_1(1-p_1)+p_0(1-p_0)$ at the clinically relevant alternative. A one-sided level-$\alpha$ Wald test rejects when $Z>z_{1-\alpha }$, and its approximate statistical power isEquating this to $1-\beta$ gives the per-arm sample sizerounded up. If the design uses a null-based critical standard error $v_0$ but an alternative standard error $v^{\ast }$, the corresponding more general formula is

Choose a parametric baseline hazard $h_0(t;\eta )$ and fitUnder independent right censoring, the full survival likelihood isEstimate $(\eta ,\beta )$ by maximum likelihood estimation and test $H_0:\beta =0$ with a likelihood-ratio test, Wald test, or score test. Equality of the two event-time distributions is exactly $\beta =0$ within this model.