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Let $y_i$ be each trial's estimated log risk ratio, with estimated sampling variance $s_i^2$. A fixed-effect meta-analysis models the independent estimates as $y_i\simeq N(\mu ,s_i^2)$, with one common true log risk ratio $\mu$. Differentiating the Gaussian log-likelihood, or minimizing ${\sum }_i(y_i-\mu {)}^2/s_i^2$, gives inverse-variance weights $w_i=1/s_i^2$ andThe variance follows directly by adding independent variances of the weighted estimates: ${\sum }_i{w}_i^2{s}_i^2/({\sum }_i{w}_i{)}^2=1/{\sum }_i{w}_i$.

Using the printed rounded estimates and standard errors, ${\sum }_i{w}_i=50.2923$, soThe pooled risk ratio is about $0.902$. An interval using the same quantile two is $(-0.3848,0.1792)$ on the log scale, so it includes no effect. Recomputing all trial estimates from the event counts would change the last digits because the printed log estimates and errors are rounded.

The key assumptions are independent trials, a common true treatment effect on this chosen log scale, approximately unbiased and approximately normal trial estimates, and suitable sampling-variance estimates. The common-effect assumption is stronger than merely studying the same named treatment: systematic population, treatment or design differences may produce genuine heterogeneity. For an unbiased interpretation of the pooled evidence, inclusion of studies must also not depend selectively on their results. Treating the plug-in variances as fixed is the usual approximation in the displayed variance formula.

Publication bias occurs when the availability or publication of a study depends on its results, for example their direction or statistical significance. The published studies then need not represent all studies meeting the review's eligibility criteria. This is different from unbiased studies simply having large sampling errors.

If mortality reductions favouring transfusion are more likely to be published than null or unfavourable findings, the observed log risk ratios will tend to be too negative and the pooled risk ratio too small. The apparent benefit will then be exaggerated. An ordinary meta-analysis interval ignores uncertainty about the missing evidence and may have poor coverage even if its sampling-variance calculation is correct for the studies included. The direction of bias depends on which results are preferentially made available; publication bias does not invariably favour a particular treatment.

A numerical power calculation needs a significance level, desired power, sidedness and allocation; these are not specified in this part. For a concrete planning illustration, assume independent batches, equal samples per supplier, a two-sided 5% test with 80% power, and true rates $p_B=0.02$ and $p_A=0.04$. The observed 2% from B and C is being used as a planning value for B, not as proof that B's population rate is exactly known. There is also a numerical inconsistency in the stated frame: at six batches per working day, B and C together produce only 120 batches in two weeks, and their proposed schemes would test 24. The asserted 3,000 sampled batches cannot literally come from that frame. The calculation treats $60/3000$ as a stipulated planning estimate; its collection would require a larger frame or longer period.

For two independent sample proportions, the approximate null variance of their difference is $2\bar{p}(1-\bar{p})/n$ and its variance at the alternative is $[p_A(1-p_A)+p_B(1-p_B)]/n$, with $\bar{p}=(p_A+p_B)/2$. Separating the null critical value from the alternative mean by the required power quantile gives the sample size for comparing two proportions:With $z_{0.975}=1.960$, $z_{0.8}=0.842$ and $\bar{p}=0.03$, this gives $n\simeq 1140.83$, so the normal-approximation calculation rounds to 1,141 batches per supplier, or approximately 1,150 for a practical planning target. This is per supplier, not the combined total. A different power or a one-sided test changes the answer. Positive clustering of sampled batches requires a cluster-aware calculation or inflation, so the independent-batch calculation should not simply be applied to C's one-day clusters.

At stage 1, the maximum-likelihood estimators of the two normal arm means are their sample means. Their difference estimates the treatment effect:Each arm has $n_1$ patients and variance ${\sigma }^2/n_1$, so independence across arms givesThe quantity $I_1$ is the Fisher information for the mean difference when the common variance is known. The PDF's preliminary expression uses $n_i$ for the variance denominator, although $i$ indexes arms and $n_0$ is not defined. The stage-specific sample size is $n_j$; the calculation uses this intended reading, consistent with the specified $n_1,n_2$ per-arm recruitment and the later weighted estimator. Fresh stage-2 patients are independent of stage-1 patients in the usual trial model; the cumulative stage means themselves are not independent across stages.

Both alternatives can be made explicit. For a conditional bias-corrected normal mean estimate, let $m$ be the observed pooled estimate and $w=n_1/n_2$, with $0<w<1$. Invert its conditional mean:This can be solved by bracketing or by Newton iteration $d_{k+1}=d_k-[g(d_k)-m]/g^{\prime }(d_k)$. Since $r^{\prime }(x)=-r(x)[x+r(x)]$,The variance of a standard normal conditional on exceeding $-x$ is $1-r(x)[x+r(x)]$, strictly between zero and one. Positivity follows from nondegeneracy; the upper bound follows because the conditional mean $r(x)$ exceeds the truncation threshold $-x$. Hence $1-w<g^{\prime }(d)<1$, so the equation has at most one root. As $d\to \infty$, $g(d)\sim d$; as $d\to -\infty$, the truncated stage-1 mean approaches its threshold and $g(d)\sim (1-w)d+w{f}_1/\sqrt{I_1}$, so a root exists for every finite $m$. Equivalently, differentiating the conditional likelihood divides the ordinary likelihood by ${\mathrm{Pr}}_d(C)$ and gives the same score equation. This conditional-likelihood correction is not exactly conditionally unbiased merely because it inverts a mean.

For an illustration, take $I_1=1$, $w=1/2$, $f_1=0$ and $m=0.5$. The equation is $d+\frac{1}{2}\varphi (d)/\Phi (d)=0.5$. Since $g(0)=0.39894$ and $g(0.5)\simeq 0.75458$, the corrected estimate lies between zero and $0.5$; numerical solution gives $\tilde{\delta }\simeq 0.14649$, below the selected ordinary estimate.

For the uniform minimum variance conditionally unbiased estimator, put $I_2=n_2/(2{\sigma }^2)$, $J=I_2-I_1>0$, $A={\hat{\delta }}_1$ and $V={\hat{\delta }}_2^{\ast }$. The fresh estimate $V$ is conditionally unbiased because it is independent of continuation. LetBefore selection, conditional Gaussian calculations give $A\mid T=t\sim N(t,s^2)$. Conditional on $C$ as well, this normal variable is truncated below $c$, soUsing $V=(I_{2}T-I_{1}A)/J$, Rao-Blackwellization therefore givesIts conditional expectation is $\mathbb{E}[V\mid C]=\delta$, and its conditional variance cannot exceed that of $V$. To justify uniform minimum variance, the joint conditional density of $(A,V)$ is a base density on $\{A\ge c\}$ multiplied by $\exp (\delta I_{2}T-{\delta }^2{I}_2/2)/{\mathrm{Pr}}_{\delta }(C)$. Thus $T$ is a complete sufficient statistic in the one-parameter conditional exponential family, whose natural parameter ranges over an open real interval. The Lehmann–Scheffé theorem proves the claim. The orthogonal pooled arm-average statistic is independent of the entire difference process and carries the nuisance common mean. Together with $T$, it gives a complete sufficient statistic in the selected two-parameter normal family, with an open natural-parameter space. Thus allowing that nuisance statistic does not improve the conditional unbiased estimate of the difference. For the same illustration, $I_2=2$, $J=1$ and $s=1/\sqrt{2}$ give ${\hat{\delta }}_{\mathrm{U}}\simeq 0.21102$. It differs from the conditional-likelihood estimate because exact conditional unbiasedness is a different criterion.

Keeping the event-free A individual under observation through day 193 leaves them in the risk set at both later B events. Now the calculations are

ThereforeThe negative score now indicates lower A event hazard, relative to B, in the contribution after day 160. The observed event counts have not changed. The change is in what events A was expected to contribute: its additional event-free exposure makes the two later B events informative comparisons, producing negative A score terms. No conclusion about the complete study's significance follows from this late contribution without the earlier scores and variances.

At day 165 there are two A and two B individuals at risk. The event is in A, with expected A count $2/4=1/2$, so its score contribution is $1-1/2=1/2$ and its variance contribution is $1/4$. The remaining A individual is censored at 173 and leaves the risk set before either later event. The risk-set calculations are

ThusThe positive score indicates higher A event hazard in this late contribution. Although B has more observed events, its two events occur with no A individual at risk, so those times do not compare the groups. This is why raw event totals alone are insufficient for the log-rank test.

The log-rank test compares survival experience through the allocation of events within the successive risk sets. Its null is equal hazard functions between the groups over follow-up, with independent individuals and noninformative censoring within groups. At each distinct event time $t_j$, let $n_{Aj},n_{Bj}$ be the numbers at risk immediately before that time, and let $d_{Aj},d_{Bj}$ be event counts. Under the null, conditional on the total $d_j=d_{Aj}+d_{Bj}$, the expected number of A events is $e_{Aj}=d_j{n}_{Aj}/(n_{Aj}+n_{Bj})$.

The signed log-rank statistic is the observed-minus-expected scoreA positive score indicates relatively more A events than expected, hence a higher event hazard for A; B's score is its negative. Censoring times do not produce score terms, but remove people from all later risk sets.

At each event time the conditional null allocation is hypergeometric, as if the $d_j$ events were sampled without replacement from the risk set. Its variance gives the score-increment variance, with the finite-population correction for ties. Summing these conditional variances gives $V$, since distinct-time martingale score increments have zero cross covariance under the null. For sufficiently many informative events, $U_A/\sqrt{V}$ is approximately standard normal and $U_A^2/V$ approximately chi-squared with one degree of freedom. In the no-tie case, each variance contribution is $n_{Aj}{n}_{Bj}/(n_{Aj}+n_{Bj}{)}^2$. A time at which only one group remains at risk contributes no comparative information.

The numerical parts below report signed-score and variance contributions after day 160. They cannot be added as separate chi-squared statistics to the earlier follow-up: add the scores and variances first, and standardize once for the complete dataset. The log-rank test is particularly effective for proportional hazards alternatives; crossing hazards can produce cancelling score contributions.

A completely randomized design uses $2m$ eligible animals, randomly selecting $m$ to receive $P$ and assigning the remainder to $Q$. Use comparable follow-up and assess the same response variable in both groups. Its advantage is simplicity and freedom from previous-treatment carryover effects; its disadvantage is that between-animal variation enters the residual and can make a treatment contrast imprecise.

A randomized complete block design with matched pairs first forms $m$ pairs using pre-treatment characteristics such as initial disease severity, age or breed. Independently choose which animal in each pair receives $P$, with its partner receiving $Q$. The experimental units are animals; pairs are blocks. The average within-pair difference estimates the treatment contrast, and positive within-pair similarity can reduce its variance. Its advantage is control of known heterogeneity; its disadvantage is the need for useful matching, with fewer residual statistical degrees of freedom and little gain if the matching variables are uninformative. Do not construct pairs using outcomes observed after assignment.

A two-period crossover design randomly assigns half the animals to sequence $PQ$ and half to $QP$. Each animal receives both treatments in separate periods, with a scientifically justified interval between them and the same outcome assessment after each period. Animal blocks remove persistent between-animal differences, while the two sequences balance treatment against period. Its advantage is potentially high precision from within-animal comparisons; its disadvantage is vulnerability to carryover effects, changing disease state and irreversible effects. It is suitable only when comparing the treatments in both periods remains meaningful and residual effects of the first treatment are adequately controlled. A two-period crossover design does not by itself disentangle arbitrary treatment-specific carryover effects.

Under the usual normal linear model for independent orchard means with common error variance, the treatment ratios have null F-distributions with denominator six statistical degrees of freedom. Their upper-tail P-values are approximatelyUse the unrounded ratios when evaluating these P-values. Thus none of spray, pruning or their interaction term is significant at a 5% significance level. Spray gives modest evidence if a 10% significance level was chosen in advance, but this is not strong evidence.

Failure to reject does not establish absence of treatment effects. Only two orchards per combination and six residual statistical degrees of freedom leave considerable uncertainty and potentially low statistical power. The table alone cannot give the direction or magnitude of individual effects: that requires treatment means. These model-based F-tests also rely on appropriate orchard allocation and comparable residual variation; they should not be interpreted as unconditional conclusions from the table alone.