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Let $z_p={\Phi }^{-1}(p)$ denote a quantile of the standard normal distribution. Reject when $W_n>z_{1-\alpha }$. The statistical power at the specified positive effect isEquating this to $1-\beta$ and using $z_{\beta }=-z_{1-\beta }$ gives $\sqrt{n}{\delta }^{\ast }/\sigma =z_{1-\alpha }+z_{1-\beta }$. Thus, for the usual target $1-\beta >\alpha$,Rounding up ensures at least the target statistical power. This normal-mean sample size calculation assumes a positive integer sample size; if a requested power is at most $\alpha$, every positive sample size already exceeds that target for ${\delta }^{\ast }>0$, and one should not square a negative quantile sum to impose an unnecessary lower bound.

Three distinct benefits of a meta-analysis are greater precision, an explicit synthesis of the effect, and assessment of variation across studies.

Combining compatible independent estimates can reduce the standard error, improve statistical power and provide more informative confidence intervals than individual small trials. A prespecified pooled estimate summarises the direction and magnitude of the evidence quantitatively, instead of informally counting statistically significant results. Finally, between-study heterogeneity and subgroup comparisons can reveal whether effects vary with populations or interventions, helping assess generalisability. These benefits require comparable estimands and an appraisal of study validity; pooling cannot repair systematic within-study bias.

Let $P_0,P_1$ be simple hypotheses with densities $p_0,p_1$ (their Radon-Nikodym derivatives) relative to a common dominating measure. A possibly randomised statistical test is a measurable function $\phi \in [0,1]$; its size of a statistical test is $E_0\phi$ and its statistical power is $E_1\phi$. The Neyman-Pearson lemma says that a likelihood ratio threshold testchosen to have size $\alpha$, is a most powerful test among tests of size at most $\alpha$. A constant randomisation on the equality set suffices to reach the required size; sets where $p_0=0<p_1$ are always rejected. For $0<\alpha <1$, a quantile of the likelihood ratio under $P_0$ gives such a threshold. If the threshold is zero, randomisation where $p_1=0$ may be needed to use the remaining size. At $\alpha =0$, one can reject only a $P_0$-null set and optimally reject the part supporting $P_1$; at $\alpha =1$, always reject.

For the proof, every competing $\phi$ satisfiespointwise: the signs agree off the equality set, where the product is zero. Integrating givesThus the likelihood-ratio threshold test has maximum power at the stated level. If $c>0$, a competing most-powerful test must have size $\alpha$ and agree with ${\phi }_{\ast }$ off the equality set, apart from sets null for $P_0+P_1$; both inequalities must then be equalities. For $c=0$, equal size is not necessary for optimality, because changing decisions where $p_1=0$ does not affect power. These qualifications account for ties and singular supports rather than assuming all densities are strictly positive.

Under the null hypothesis $\theta =0$, $X$ has the uniform distribution on $[0,1]$. Against $\theta =1$, the likelihood ratio is $2x$, increasing on that interval. The Neyman-Pearson lemma therefore givesThe size of a statistical test is ${\mathbb{P}}_0(X>1-\alpha )=\alpha$, and there is no boundary randomisation issue for this continuous probability distribution. Its statistical power against $\theta =1$ is ${\int }_{1-\alpha }^{1}2xdx=2\alpha -{\alpha }^2$.

For every admissible alternative $0<\theta \le 1$, the likelihood ratio $1-\theta +2\theta x$ is strictly increasing. The same upper-tail rejection set has size $\alpha$ and is a most powerful test against each fixed such alternative, again by the Neyman-Pearson lemma. Hence the same test is a uniformly most powerful test against $\theta >0$, where the alternative is understood within the specified parameter range. Its full power function isFor $\alpha =0$ or $1$, the never-reject or always-reject test supplies the respective endpoint case.

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 the usual normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design $X\in {\mathbb{R}}^{n\times p}$, $n>p$, and $\epsilon \sim N(0,{\sigma }^2{I}_n)$. These Gaussian distribution and constant-variance hypotheses are necessary for an exact finite-sample Student's t-test. Put $V=(X^{T}X{)}^{-1}$ and $\nu =n-p$. The vector of ordinary least squares estimators satisfiesWith $s^2=\Vert Y-X\hat{\beta }{\Vert }^2/\nu$, the test statistic isUnder the null hypothesis, $T$ has Student's t-distribution with $\nu$ degrees of freedom. A two-sided level-$5\%$ Student's t-test rejects when $|T|>t_{\nu ,0.975}$. If the intended test is one-sided against ${\beta }_0>0$, it rejects when $T>t_{\nu ,0.95}$; the statistical power conclusion below holds for either convention.

The orthogonal projections of the Gaussian noise onto the column space of $X$ and its orthogonal complement are independent. Consequently ${\hat{\beta }}_0$ and the residual sum of squares are independent, andThus under the alternative,This is the noncentral t-distribution. The rejection region is fixed by the design, $\nu$ and the chosen level, while this probability distribution depends on $\beta$ only through ${\beta }_0$. Therefore the statistical power does not depend on the other regression coefficients, with the design and $\sigma$ fixed.

A simple hypothesis specifies a single probability distribution. For a test with rejection region $R$, its size of a statistical test is the rejection probability under the simple null hypothesis, while its statistical power against the simple alternative is the rejection probability under that alternative. The Neyman-Pearson lemma states that among tests of size at most $\alpha$, a test rejecting for the largest values of the likelihood ratio $f_1/f_0$ has greatest power, with boundary randomization if needed.

Herewhich decreases with $|x|$. The best test therefore rejects for $|X|\le c$. Under $H_0$, ${\mathbb{P}}_0(|X|\le c)=c$, so size $0.05$ requires $c=0.05=1/20$. Its power isThus reject $H_0$ exactly when $|X|\le 0.05$.

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.

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

Suppose each collaborating centre contributes $n_0$ independently randomized participants per arm with specified event probabilities. At 50% statistical power, the usual normal-approximation alternative quantile is zero. The required size per arm is $2\bar{p}(1-\bar{p})z_{\ast }^2/(p_A-p_B{)}^2$, with $z_{\ast }=z_{1-\alpha /2}$ for a two-sided test or $z_{1-\alpha }$ for a one-sided test. Divide by the per-centre arm size and round up. Endpoints must use the same time origin and follow-up window, and cluster dependence requires an additional adjustment.

In a fixed full-rank normal linear model, the studentized estimator of one coefficient has a noncentral t-distribution with noncentrality equal to that coefficient divided by its standard deviation. The fitted and residual orthogonal projections of the Gaussian noise are independent. Therefore the statistical power of its exact Student's t-test does not depend on the other regression coefficients, with design and variance fixed.

A statistical test is a measurable rule for deciding whether to reject a null hypothesis. Allowing randomisation, it is represented by a function $\phi (x)\in [0,1]$ giving the conditional rejection probability after observing $x$. A deterministic test takes only values zero and one. Under a simple null, $E_0\phi$ is its size of a statistical test; under an alternative, $E_1\phi$ is its statistical power. Randomisation on a likelihood ratio equality set allows a Neyman-Pearson lemma threshold test to attain a prescribed size even for discrete data.