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The I-squared statistic of $6.6\%$ is a small estimate of excess variation on the heterogeneity scale. It does not mean that $6.6\%$ of studies are different, nor that this percentage of patients responds differently. The Cochran's Q statistic compares the observed variation with that expected from the studies' sampling errors; its approximate null distribution has five degrees of freedom here. A heterogeneity-test p-value of $0.374$ provides no rejection of the common-effect hypothesis.

The individual confidence intervals in the forest plot overlap substantially, although some point estimates are larger than others. Taken together, there is little evidence of between-study heterogeneity in these six studies. Neither the nonsignificant test nor the small I-squared statistic proves equal effects: with only six studies, detecting moderate heterogeneity can be difficult. A study being individually significant while another is not is also not itself a test of their difference.

Only sets 4, 5, 7 and 8 have an earlier failure with both members at risk. Sets 4 and 8 have B fail first and favor A, giving $6+3=9$ informative pairs. Sets 5 and 7 have A fail first and favor B, giving $18+9=27$. Both-censored pairs contribute zero; in sets 3 and 6 the earlier observation is censoring, so the later failure occurs with no remaining comparator and also contributes zero.

There are thus 36 informative independent strata. The stratified log-rank statistic, its null variance, and its standardized value areEquivalently the squared statistic is nine. The normal approximation gives a two-sided p-value $2[1-\Phi (3)]<0.0027$ using the stated bound. The exact conditional sign test instead gives $2{\sum }_{k=0}^9\left(\genfrac{}{}{0ex}{}{36}{k}\right)2^{-36}\approx 0.00393$; the slight difference is discreteness rather than a different direction of effect. There is strong evidence favoring Method B, which is associated with the later failure in three quarters of the informative pairs, under the test assumptions.

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 $15\%$.

Test $H_0:{\gamma }_d=0$ for both nonreference days, against at least one nonzero day fixed effect, conditional on chocolate. There are two added regression coefficients and $24-6=18$ residual statistical degrees of freedom. The nested-model F-test statistic isUnder $H_0$ and the normal linear model assumptions, $F\sim F_{2,18}$. Its 5% upper critical value is $3.555$, so do not reject the absence of day effects; the p-value is approximately $0.456$. Thus the missing row has two statistical degrees of freedom, mean square $4.875$, and the F-test value above. These observations do not provide evidence that including day improves the chocolate-adjusted mean model. They do not prove that every possible day effect or chocolate–day interaction term is absent.

The analysis of variance provides evidence of a chocolate effect both with day included ($F_{3,18}=3.9665$, p-value $0.02473$) and with day omitted ($F_{3,20}=4.039$, p-value $0.02133$). Because every chocolate–day cell has the same replication, balanced factorial orthogonality separates the two main effects; the chocolate row is meaningful despite being entered first.

The chocolate-only fitted group means are approximately $14.17,11.67,9.33,11.33$ boards for A, B, C, D respectively. Thus A has the largest fitted lecturing speed and C the smallest. Relative to A, the fitted differences are $-2.50,-4.83,-2.83$ boards. The printed individual Student t-tests give strong evidence for the A–C contrast (p-value $0.00245$); B and D versus A have p-values $0.08835$ and $0.05582$, respectively. Those latter contrasts are not significant at 5%, and the output does not test all other pairwise comparisons. Simultaneous claims would require accounting for multiple hypothesis testing.

There is little evidence of a day effect after adjusting for chocolate. A chocolate-only normal linear model is therefore a reasonable simpler summary, with residual standard deviation about $2.42$ boards and explained variation $R^2\simeq 0.377$. This is an association under the additive statistical model; a causal claim would additionally require an appropriate assignment of chocolate and checks of the regression residuals and possible interaction terms.

Under the more plausible Quasi-Poisson regression, the year Wald statistic is $-1.686$, with reported p-value $0.0922$. An approximate 95% confidence interval isThus the estimated fall is about 9.35%, but the data do not establish a reduction at the 5% level after allowing for overdispersion. The interval allows both a sizeable reduction and a small increase. The before–after comparison also lacks a contemporaneous randomized control, so causal inference about the campaign would require addressing other changes between years. The small Poisson regression p-value is not sufficient evidence of a campaign effect when its variance assumption is unsuitable.

The second logistic regression adds ${\gamma }_2{Y}_{i,t-2}$ while retaining the first lag and baseline covariates. The models are nested under $H_0:{\gamma }_2=0$. Their likelihood-ratio test statistic is the reduction in binomial deviance,Under the null and regular large-sample conditions for the correctly specified conditional likelihood, this has an approximate chi-squared distribution with one statistical degree of freedom. Since $9.4>3.841$, reject at 5%; the approximate p-value is $0.0022$. Prefer the two-lag model to the one-lag model. Its additional lag captures statistically useful information in the history.

The preferred cumulative-response logistic model estimatesIts logit link describes conditional smoking-free probability given baseline predictors and prior successful weeks. Holding the other predictors fixed, males have $e^{-0.364092}=0.695$ times the female success odds; the reported p-value is $0.0152$, and the approximate 95% odds ratio confidence interval is $(0.518,0.932)$. Age has estimated odds ratio $e^{0.001224}=1.0012$ per additional year, with p-value $0.918$, giving little evidence for an age association in this fit.

The combined treatment has conditional success odds ratio $e^{0.327118}=1.387$ versus the reference treatment, with approximate 95% confidence interval $(1.035,1.859)$ and p-value $0.0285$. The fitted conditional odds are about 39% higher for the combined treatment. The trial randomization supports treatment comparisons, but conditioning on accumulated post-treatment outcomes means this coefficient is not directly the marginal total treatment effect.

Each previous successful week multiplies current success odds by $e^{0.399296}=1.491$, with approximate 95% confidence interval $(1.329,1.673)$ and very small p-value $1.06\times 10^{-11}$. This is strong fitted persistence. It can reflect true state dependence, unobserved heterogeneity, or an omitted calendar-time trend; this fit alone cannot distinguish them. The baseline intercept implies success probability ${\mathrm{logit}}^{-1}(-1.417630)\simeq 0.195$ for a reference-treatment female aged zero with no previous success. That age is outside the study's useful interpretation range, so the intercept chiefly anchors the regression. The Bernoulli distribution dispersion parameter is fixed at one, and the residual binomial deviance is 1122 on 995 statistical degrees of freedom. With individual binary outcomes, comparing that binomial deviance mechanically to a chi-squared distribution is not a reliable general goodness-of-fit test.

Under $H_0$, the seven free covariate coefficients all equal zero. The likelihood-ratio test statistic isThe larger fit has ten free statistical parameters, compared with three in the constant-rate fit, so the reference chi-squared distribution has seven statistical degrees of freedom. Its 95th percentile is $14.06714$. Therefore reject $H_0$ at 5%; the p-value is approximately $0.0043$, and the covariate model is preferred. The two constrained sex coefficients add no free statistical parameters.

For the dementia-to-death arrow, the hazard ratio for higher versus lower education is $e^{-1.490}=0.2254$, with approximate 95% confidence interval $(e^{-2.547},e^{-0.4327})=(0.0783,0.6488)$. Thus higher education is associated with a roughly 77.5% lower fitted death transition intensity, conditional on diagnosis age; the interval excludes one. This is an observational association and is not a direct multiplicative statement about death probability.

Each additional year of age at diagnosis multiplies the dementia-to-death transition intensity by $e^{0.07984}=1.0831$, with confidence interval $(e^{-0.007765},e^{0.1674})=(0.9923,1.1822)$. The estimate suggests an increase, but the interval includes one, so this individual age coefficient is not significant at 5%. Sex has hazard ratio one for this transition by the model's imposed constraint. The output supplies no estimated sex effect or test of that constraint for this arrow. The nonzero sex coefficient for $1\to 2$ must not be mistaken for a sex effect on $2\to 3$.

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.