For a sample size $m\ge 2$ define $\overline{X}=m^{-1}{\sum }_i{X}_i$ and $S_X^2=(m-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The normal-sample theorem givesIt also states that the mean and sample variance are independent. Combining (i) and (ii), and the independence across samples, the correctly centred pivot isThe source reverses the mean difference inside the subtraction. Its literal statistic is generally noncentral when ${\mu }_X\ne {\mu }_Y$, rather than having a central Student distribution. Under the null hypothesis both differences are zero, so this misprint does not change the requested null test.

For a two-sided level-$\alpha$ test reject when $|T|>t_{m+n-2,1-\alpha /2}$. In the numerical example the pooled variance is $0.038$, the standard error is $\sqrt{0.038(2/5)}\approx 0.12329$, and $T\approx -2.190$. There are eight degrees of freedom and the supplied critical value is $2.31$. Thus do not reject equal mean cell volume at the 5% level. This does not establish equality of the means; it states that the observed difference does not cross this test's two-sided threshold.

Age matching introduces a shared block factor: the two observations in a pair can be associated, and their age-dependent means also undermine the simple unblocked independent-samples model. A paired t-test on the within-pair drug-minus-placebo differences accounts for that design, assuming independent pairs and the corresponding normal-difference model. Shared age variation cancels in those differences; positive within-pair association reduces their variance and can increase power, although the reduction in degrees of freedom means improvement is not automatic. The paired test cannot be computed from the two marginal means and variances alone: the within-pair differences or covariance are needed.

Assume independent survival times and positive rate $\lambda$. With $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$ on the observed nonnegative data. The score and negative expected second derivative are $n/\lambda -S$ and $n/{\lambda }^2$. For $S>0$, which holds almost surely under the model,The usual independent-sample likelihood asymptotics give $\sqrt{n}(\hat{\lambda }-\lambda )\Rightarrow N(0,{\lambda }^2)$. Thus an approximate Wald interval isFor a positive interval with the same first-order justification, apply the delta method to $\log \hat{\lambda }$, obtaining endpoints $\hat{\lambda }\exp (\pm z_{1-\alpha /2}/\sqrt{n})$.

A single individual survives until $T$ with probability $s=e^{-\lambda T}$. Thus $N_T\sim \mathrm{Bin}(n,s)$ and the new likelihood is $\left(\genfrac{}{}{0ex}{}{n}{n_T}\right)e^{-\lambda T{n}_T}(1-e^{-\lambda T}{)}^{n-n_T}$. For $0<n_T<n$, its maximum occurs at $s=n_T/n$, givingIf $n_T=0$, the supremum occurs as $\lambda \to \infty$ and there is no finite maximum. If $n_T=n$, it occurs at the boundary $\lambda =0$ of the extended parameter space, or is only a supremum when the rate is required to be strictly positive.

A binomial experiment has information $n/[s(1-s)]$ for $s$. Since $ds/d\lambda =-Ts$, parameter transformation givesAs quantified by Fisher information in an exponential survival snapshot, the ratio of this information to the complete-data information is $z^2/(e^z-1)$ with $z=\lambda T>0$, strictly less than one by the supplied inequality. Hence every fixed survival snapshot loses information. Under regular large-sample conditions its efficient rate estimator has larger asymptotic variance, and correspondingly wider uncertainty intervals, than the estimator based on all individual survival times. This comparison concerns information and asymptotic uncertainty, not a claim that every realized interval in one experiment is wider.

The moment generating function is $M_X(t)=\mathbb{E}(e^{tX})$ wherever that expectation is finite. A useful continuity theorem says that if the moment generating functions $M_n$ are finite on a common open interval containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual continuity at zero, and obtain the uniquely determined limiting law.

For the gamma density, integrating the exponential against the density and changing variable to $(\lambda -t)x$ givesFor the waiting time to one head, independence of successive tosses gives $\mathrm{Pr}(Y=j)=p(1-p{)}^{j-1}$ for $j\ge 1$. Summing the geometric series givesThe waiting times between consecutive heads are independent copies of $Y$, so their sum $N$ has moment generating function $M_Y(t{)}^k$. ConsequentlyFor fixed $t<1/2$, numerator and denominator have leading terms $p$ and $p(1-2t)$ as $p\downarrow 0$. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small $p$, so the continuity theorem applies. The limiting distribution isequivalently a chi-squared distribution with $2k$ degrees of freedom.

Take source strength $Q$ to mean total two-dimensional flux, so an isolated source has potential $Q\log (z-a)/(2\pi )$. Let $a=1+i$, $b=1/\bar{a}=(1+i)/2$, and put $c_Q=Q/(2\pi )$. The circle theorem gives, up to an additive constant,The derivative is single valued, which is what the force integrals require. With a positively oriented contour around the cylinder, the force form of Blasius' theorem isOnly the poles at zero and $b$ are enclosed; the physical source at $a$ lies outside. The residues of $q^2/c_Q^2$ are $2(1/a+1/b)$ at zero and $2[1/(b-a)-1/b]$ at $b$. Their sum is $2[1/a+1/(b-a)]=-2/a=-(1-i)$. ThereforeFor the second integral, the residue of $z{q}^2/c_Q^2$ at zero is one. At $b$ it is $1+2b[1/(b-a)-1/b]=-3$. Their sum is $-2$. Using precisely the supplied complex-moment convention givesThe vanishing physical moment $M$ is also consistent with pressure acting normally to a circle, hence radially about its centre. If strength is instead defined as the coefficient of $\log (z-a)$, replace $c_Q$ by that strength throughout; stating the flux convention fixes all factors of $2\pi$.

For wall-induced motion of a planar potential dipole, the complex velocity convention is $F^{\prime }(z)=v_x-i{v}_y$. A singularity transported by the surrounding flow therefore satisfies $\dot{\bar{a}}=F_{\mathrm{reg}}^{\prime }(a)$, where its own term $m/(z-a)$ is removed before differentiating. The images at a given instant depend on that instantaneous position, but the spatial derivative treats their parameters as fixed.

For the circular wall, the derivative of the regular image is $\bar{m}/(1-\bar{a}z{)}^2$. For the quadrant, differentiate its three image terms. For the sloping wall, differentiate its single image term. This gives respectivelyThese are equations for $d{\bar{z}}_1/dt$, as printed. Conjugating gives equations for $d{z}_1/dt$ if desired; no self-induced singular velocity is included.