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A Poisson model assumes independent counts between rats conditional on time, a common mean at a given time, and conditional variance equal to that mean. Comparable observation units and consistent counting are needed; unexplained rat-to-rat heterogeneity is excluded by this simple specification. The quadratic Poisson regression isThe printed nested analysis of deviance tests $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$. Its likelihood-ratio statistic is $G^2=3.8548$, approximately ${\chi }_1^2$ under the null hypothesis and regular Poisson asymptotics. The reported $p=0.0496$ gives borderline rejection of the log-linear time effect at 5%, conditional on the Poisson assumptions. This is curvature of the log mean, rather than an ordinary quadratic model for the count itself.

There are only three distinct time values, so the intercept, linear term, and quadratic term span every possible set of three log means. This is Poisson mean saturation at three distinct times: the quadratic fit is saturated for the time-group means, which are $14/7=2$, $17/8=2.125$, and $47/7\simeq 6.7143$. A grouped deviance comparison against unrestricted time-group means has zero statistical degrees of freedom: the quadratic functional form cannot be tested at these three times. The individual-observation deviance 24.515 still describes within-group departure from the Poisson distribution, but small means and zeros make the naive ${\chi }_{19}^2$ goodness-of-fit calibration doubtful. Poisson variation itself can be checked using the replicates, preferably by a parametric bootstrap; it is not rendered untestable by the saturated mean specification.

Overdispersion means conditional variance exceeding the variance assumed by the Poisson mean model. For example, rats may have different latent susceptibility, so if $Y\mid \Lambda \sim \mathrm{Pois}(\Lambda )$, the law of total variance gives $\mathrm{Var}(Y)=\mathbb{E}\Lambda +\mathrm{Var}(\Lambda )>\mathbb{E}Y$. A quasi-Poisson model instead uses $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson-residual dispersion estimate is $\hat{\varphi }=1.254068$, using $n-p=22-3=19$.

For the conventional quasi-likelihood comparison with estimated dispersion, useIts upper-tail probability is about 0.096, so the evidence for curvature no longer reaches 5%. The F-distribution is an approximate calibration, not an exact small-sample result for arbitrary overdispersed counts; quasi-likelihood does not supply a full Poisson likelihood. With known dispersion one would instead compare $G^2/\varphi$ approximately with ${\chi }_1^2$. The displayed dispersion code uses the undefined name acf.glm2; its fitted values must be those of aom.glm2 for the stated calculation.

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.

The likelihood-ratio statistic isThe null hypothesis $\pi =0$ is at the boundary of the permitted mixture weights. Assuming the remaining parameters are identifiable and interior, including $\tau >0$, an efficient normalized score for $\pi$ is asymptotically $Z\sim N(0,1)$ under the null hypothesis. The locally unconstrained quadratic log-likelihood maximizer is proportional to $Z$, while the constrained maximizer truncates it at zero. Thus $G^2\Rightarrow (\max (0,Z){)}^2$, giving the one-sided boundary likelihood-ratio limitHere ${\delta }_0$ is a point mass at zero. For a positive observed statistic the upper-tail probability is half the ordinary ${\chi }_1^2$ upper tail. The 5% critical value is the 90th percentile of ${\chi }_1^2$, about 2.7055, displayed as 2.71. Hence the reported likelihoods give a very marginal rejection of $\pi =0$ at 5%, with $p\simeq 0.04955$.

Using the ordinary 95th-percentile cutoff 3.84 would be inappropriate for this one-sided boundary test. The conclusion is close enough to the threshold that likelihood rounding matters: since each log-likelihood is printed to two decimals, the unrounded statistic can differ from 2.72 by nearly 0.02. The intended calculation rejects using the reported values, but an exact decision from the fitted data requires unrounded likelihoods. If the nuisance dispersion is also on its boundary or the mixture is not identifiable, the stated half-and-half null distribution needs reconsideration.

A sufficient statistic is one for which the conditional distribution of the full sample, given its value, is independent of the parameter. For independent observations the likelihood is ${\prod }_{i=1}^6{p}_i^{n_i}$, depending on the data only through the counts. The factorization criterion for sufficiency therefore makes $(n_1,\dots ,n_6)$ sufficient. Directly, all ordered samples with given counts have the same probability, so their conditional distribution is uniform over the $n!/\prod n_i!$ possible arrangements, independently of the $p_i$.

The unrestricted maximum-likelihood estimator is ${\hat{p}}_i=n_i/n$. Thus the generalized likelihood-ratio test useswith zero-count terms interpreted by continuity. Reject for small $\Lambda$, equivalently large $D$.

Write $n_i=n/6+{\delta }_i$, so $\sum {\delta }_i=0$. A Taylor series expansion of $(n/6+{\delta }_i)\log (1+6{\delta }_i/n)$ gives ${\delta }_i+3{\delta }_i^2/n+O(|{\delta }_i{|}^3/n^2)$. Summing cancels the linear terms, and yieldsUnder the null hypothesis, the multinomial central limit theorem places the standardized count deviations in the five-dimensional subspace with coordinate sum zero, with identity covariance on that subspace. Their squared norm consequently tends to the chi-squared distribution with five degrees of freedom. This gives the Pearson chi-squared goodness-of-fit test.

For $T=8.12$, the asymptotic p-value is about $0.150$. The 5% critical value of ${\chi }_5^2$ is approximately $11.07$. Do not reject at the usual 5% level, or at 10%. A significance level is not specified in the paper, so an unconditional yes-or-no decision is not determined: using this approximation, rejection would require a chosen level at least about $15\%$.

Let $I_0$ be the indices of the true null hypotheses, let $m_0=|I_0|$, and let $V$ count false rejections. The familywise error rate isThe Bonferroni correction rejects a null exactly when its p-value is at most $\alpha /m$. The union bound givesThis controls the familywise error rate for every configuration of true and false nulls, with arbitrary dependence between their p-values. Marginal super-uniformity, $\mathbb{P}(P_i\le t)\le t$, would also suffice.

For the Holm step-down procedure, inspect the ordered p-values in ascending order and stop at the first failed comparison. If the first comparison fails, reject none; this is the convention $k=0$ when the defining set is empty. Ties can be ordered by any fixed rule.

If $m_0=0$ there can be no false rejection. Otherwise let $j$ be the rank of the first true null. Since at most $m-m_0$ false nulls precede it, $j\le m-m_0+1$ and thus $m-j+1\ge m_0$. If any true null is rejected, the step-down rule must have rejected this first true null, which requiresUsing the union bound on the true-null p-values provesThis first true null argument for Holm control also requires no independence. Holm controls the familywise error rate at level $\alpha$ under arbitrary dependence.

Now let $R$ count all rejections. The false discovery rate is the expected proportion of rejections that are false, with zero assigned when nothing is rejected:It differs from the familywise error rate: several false rejections can still represent a small proportion of a large collection of discoveries.

The BH procedure is a step-up rule. SetIf $K=0$, reject nothing; otherwise reject all hypotheses with $P_i\le \alpha K/m$. Exactly $K$ hypotheses are rejected: if more than $K$ p-values were below that threshold, the next ordered value would also satisfy its own larger threshold, contradicting maximality. Thus $R=K$. Unlike the step-down rule, a failed early comparison does not make this procedure stop.

For the proof, assume that each true-null p-value is uniform on $[0,1]$ and independent of the entire vector of the other p-values. Joint independence of all p-values is a sufficient condition; the false-null marginal distributions can be arbitrary. Mere uniformity of the true-null marginals without a dependence condition is insufficient for this argument or for general unmodified BH control.

Fix a true null $i$. Replace its p-value by zero and let $R^{(i)}$ be the number of rejections made by the Benjamini-Hochberg procedure on the modified vector. This variable depends only on the other p-values, and $R^{(i)}\ge 1$. The Benjamini-Hochberg leave-one-out identity isTo prove it, suppose $i$ is rejected with $R=r$. Decreasing its p-value to zero leaves every ordered value above rank $r$ unchanged, since it was already among the first $r$ values. Those higher ranks still fail their thresholds, while rank $r$ still succeeds; hence $R^{(i)}=r$. Conversely, suppose $R^{(i)}=r$ and $P_i\le \alpha r/m$. Restoring $P_i$ still leaves at least $r$ values at most $\alpha r/m$, so $R\ge r$. Increasing one p-value cannot increase the maximal successful rank, so $R\le R^{(i)}=r$. This gives equality and rejection of $i$.

Independence and uniformity now giveSumming over the $m_0$ true nulls proves the exact false discovery rate under independent null p-values:If the independent true-null p-values are only super-uniform, the same calculation gives the inequality instead of equality. The distinction between the two types of error control and the dependence conditions is essential: Bonferroni and Holm have the preceding guarantees without independence, while the BH proof here explicitly uses it.

A predictive discrepancy statistic is a specified measurable function $T$ of a possible dataset, chosen to detect a feature relevant to the null hypothesis. Under the fully specified null density $p_0$, compare the observed $T(\mathbf{x})$ with its reference distribution, derived analytically or from replicated datasets ${\mathbf{X}}^{\mathrm{rep}}\sim p_0$.

For discrepancies whose large values indicate disagreement, useA lower-tail or two-sided discrepancy requires the corresponding comparison. The checking function is a statistic, not a density; no unknown parameter is fitted when the null is fully specified.

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.

Write $s=\sigma /\sqrt{n}$, $\mu =\delta /s$, and let $Z_1,Z_2$ be independent standard normal random variables obtained by centering and scaling the separate stage means. ThenThe second statistic uses all patients, so the statistics are correlated even though the stages' new observations are independent. Their covariance is $1/\sqrt{2}$ and each variance is one. This gives the exact bivariate normal distributionUnder the null hypothesis both means are zero. At $\delta ={\delta }^{\ast }$ replace $\mu$ by $\sqrt{n}{\delta }^{\ast }/\sigma$; the mean vector is $(\sqrt{n}{\delta }^{\ast }/\sigma ,\sqrt{2n}{\delta }^{\ast }/\sigma {)}^T$. In this group sequential design the full-sample statistic may be viewed as a potential statistic from the underlying sequence of outcomes, even on paths where recruitment stops.

Set $z=z_{1-\alpha }$. Under the null hypothesis rejection requires both $W_1\ge f$ and $W_2>z$. Since $W_2$ has the standard normal distribution,The bivariate normal distribution in the preceding calculation has a nonsingular covariance matrix and strictly positive statistical probability density everywhere. For every finite futility boundary $f$ and $0<\alpha <1$, the open rectangle $\{W_1<f,W_2>z\}$ has positive probability. ThereforeFor an explicit expression, conditional on $W_1=w$ the full-sample statistic is $N(w/\sqrt{2},1/2)$ under the null hypothesis, yieldingThe Type I error is reduced because some otherwise rejecting paths stop for futility. Equality is approached as $f\to -\infty$, but does not hold at a finite futility boundary.

The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

Let $W_{aj\ell }$ be recall for age level $a$, processing level $j$, and replicate $\ell =1,\dots ,10$. Write $u_a=1$ for Younger and $0$ for Older. With Older and A as the reference levels in a regression factor, the additive two-factor normal linear model under treatment coding iswhere the errors are independent $N(0,{\sigma }^2)$, with common positive variance. The design matrix is fixed; subjects supply independent observations; the specified conditional mean is correct; and the absence of an interaction term means the age difference is constant across processing levels. These are model assumptions, not consequences of random assignment. The treatment allocation supports independence between processing assignment and background characteristics, but age itself was not randomized.

An older subject in A has $u_a=0$ and ${\gamma }_A=0$, so the estimated mean is $11.35$ words. Its standard error is the intercept's $0.7632$. There are $100-6=94$ residual degrees of freedom, giving the 95% confidence interval for this mean:This is a mean-response confidence interval, not a prediction interval for a new individual's recall; the latter also includes the new observation's error variance.

The second two-factor normal linear model adds age-by-processing interaction terms:Its ten free mean parameters are equivalent to one mean for each of the ten cells. To compare the fits, the null hypothesis is ${\delta }_B={\delta }_C={\delta }_D={\delta }_E=0$. The full residual sum of squares is $722.30$ on $90$ degrees of freedom. Adding interaction reduces the additive model's residual sum of squares by $190.30$, so the reduced value is $912.60$. The nested-model F-test statistic isIts p-value is $0.0002793$. Reject additivity and retain the interaction model. The significant interaction means that a single age effect averaged over all processing methods does not adequately summarize the data.

The regression intercept $11.0$ is the older-A mean. The age coefficient $3.8$ is the younger-minus-older contrast specifically in A. The processing coefficients $(-4.0,2.4,1.0,-4.1)$ compare B, C, D and E with A specifically among older subjects. The interaction coefficients $(-4.3,0.4,3.5,-3.1)$ are differences between the age contrast in each of those methods and its value in A. Adding the relevant coefficients givesThus A, C and especially D favour younger subjects in their estimated means, whereas B and E have little estimated age difference. In the younger group, D has the largest estimated recall, followed by C and A; B and E are much lower. Among older subjects C is largest, D and A are intermediate, and B and E are lowest. Comparing these orders is a description of estimates, not a claim that every pairwise difference is significant.

Each coefficient's printed standard error estimates its uncertainty; its $t$ statistic divides the estimate by that error, and its two-sided p-value uses $t_{90}$ under the relevant zero-contrast null hypothesis. For example, the older B-versus-A and E-versus-A contrasts have $p=0.00217$ and $0.00170$. The C-versus-A and D-versus-A older contrasts have $p=0.06139$ and $0.43201$. The B interaction has $p=0.01846$, while the D interaction is borderline at $0.05387$ and the other two individual interaction tests are not significant at $5\%$. These individual tests do not override the joint interaction F-test, and multiple comparisons require care. A simple younger-versus-older contrast outside A combines coefficients and must use their estimated covariance, rather than adding their marginal errors.

The residual standard error $2.833$ estimates $\sigma$. The coefficient of determination $0.7293$ is the fraction of centered sample recall variation explained by the ten-cell model. The overall $F=26.93$ with null law $F_{9,90}$ tests all nine non-intercept coefficients jointly against zero, giving extremely strong evidence that the cell means are not all equal. The sequential analysis of variance separates age, process, and their interaction; the balanced design makes the main-effect sums orthogonal, whereas the coefficient table uses the specified reference-level contrasts.

To assess pooling into three processing types, retain age-by-type interaction because the previous test rejects additivity. The reduced model $W\sim \mathrm{Age}\ast \mathrm{Type}$ has six cell means. Its null hypothesis says A and C have equal means within each age, and B and E have equal means within each age: four restrictions. Compare it with the ten-cell model usingIn fact the supplied fitted cell means and equal cell sizes determine the increase without raw observations. Pooling two ten-person cells with means $m_1,m_2$ adds $5(m_1-m_2{)}^2$ to the residual sum of squares. Here the four differences are $2.4,2.8,-0.1,1.1$, soTherefore $F_{\mathrm{pool}}=2.3083$ and $p\simeq 0.0640$. At $5\%$, the data do not reject the three-type reduction, though the result is borderline and is not proof that the pooled means are identical. This factor-level pooling test keeps the previously supported interaction structure; reducing the categories and removing all interaction simultaneously would test a different set of restrictions.

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.

The restricted maximum likelihood procedure estimates covariance parameters from error contrasts that remove the unknown fixed effects. Let $X$ have rank $p$ and choose an $n\times (n-p)$ full-rank matrix $A$ with $A^{T}X=0$. Then $A^{T}Y\sim N(0,A^{T}V(\vartheta )A)$ has a likelihood involving the variance parameters $\vartheta$ but not $\beta$. Maximizing that likelihood is restricted maximum likelihood. An equivalent restricted likelihood from orthogonal error contrasts, up to terms independent of $\vartheta$, iswhereAfter estimating $\vartheta$, use generalized least squares for $\hat{\beta }=(X^T{\hat{V}}^{-1}X{)}^{-1}{X}^T{\hat{V}}^{-1}Y$ and the conditional modes from part (b). In an independent-error normal linear model, this covariance procedure reduces to estimating ${\sigma }^2$ by $\mathrm{RSS}/(n-p)$ instead of $\mathrm{RSS}/n$.

The two models being compared have different fixed-effect spaces, so their error contrasts, dimensions, and restricted likelihoods differ. Their REML values are not likelihoods of the same transformed data and must not be used directly for this fixed-effect likelihood-ratio test. Refit both with ordinary maximum likelihood, integrating out the random effects but retaining the original response vector, and maximize each model's marginal likelihood.

The null hypothesis is ${\beta }_1=0$, leaving the random slope variance free; the alternative permits a nonzero population stirring effect. The supplied test has $2({\ell }_1-{\ell }_0)=7.9588$, one extra fixed coefficient, and approximate ${\chi }_1^2$ p-value $0.004786$. Retain the fixed stirring-rate effect at 5%. A mean-zero random slope cannot substitute for an estimated nonzero population slope. The asymptotic calibration assumes regular nuisance covariance parameters and should be interpreted cautiously with only three furnaces; simulation can provide a better finite-sample calibration.

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.

Let ${\varphi }_{\rho }$ denote the centered unit-variance bivariate normal distribution density,The complement of rejecting either null hypothesis is $W_1\le c,W_2\le c$. Therefore the probability of at least one rejection isWhen both null hypotheses are true, this is the familywise error rate. If one null hypothesis is false, a rejection of that hypothesis is not a Type I error, so the displayed probability is then an any-rejection probability rather than the familywise error rate.