For a normal sample of size $m$,Then $\bar{X}\sim N({\mu }_X,{\sigma }^2/m)$ and $c{S}_X^2\sim {\chi }_{m-1}^2$ with $c=(m-1)/{\sigma }^2$. These two statistics are independent.

Combining the difference distribution and pooled variance calculation gives, with $\nu =m+n-2$,The printed statistic has the sign of the centering term reversed. Its numerator subtracts ${\mu }_Y-{\mu }_X$, so it is not generally a central Student statistic; the corrected centering above is necessary. Under the null hypothesis ${\mu }_X={\mu }_Y$, the two expressions coincide, so the requested test is unaffected. Reject the two-sided null at significance level $\alpha$ whenHere $\nu =8$ and $|T|=2.19<2.31$, so do not reject at the 5% level. This does not establish equality of the population means.

Age matching means that the two observations in a pair share an age effect, while ages vary across pairs. The independent, identically distributed two-sample model no longer represents this design: random pair effects induce within-pair dependence, and fixed differing ages can instead produce different individual means. Pairing by itself need not cause stochastic dependence when ages are treated as fixed. The appropriate alternative is a paired t-test on the within-pair differences. Removing shared age variation can reduce the noise and improve power when age strongly affects volume and the members of a pair are positively associated. Improvement is not automatic, since the paired test also uses fewer degrees of freedom.

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.

The moment generating function is $M_X(\theta )=\mathbb{E}(e^{\theta X})$ where finite. The continuity theorem states that, if moment generating functions are finite on a common open interval containing zero and converge there to the moment generating function of a probability law, the corresponding distributions converge weakly to that law.

Integrating the gamma density and using the gamma integral givesFor a geometric waiting time, on $1,2,\dots$, summing the geometric series giveswhen $0<p<1$. For $p=1$, $Y=1$ deterministically. The waiting times between successive heads are independent and identically distributed, so $N=Y_1+\cdots +Y_k$ and $M_N=M_Y^k$. ThereforeIndeed the denominator is $p(1-2\theta )+O(p^2)$ and the numerator is $p+O(p^2)$. On any sufficiently small fixed neighbourhood of zero these moment generating functions exist for all sufficiently small $p$. The limit is the moment generating function of a gamma law with shape $k$ and rate $1/2$, so

Put $\kappa =Q/(2\pi )$, $a=1+i$ and $b=1/\bar{a}=a/2$. The circle theorem gives, up to an additive constant,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.

For each case omit the dipole's own singular field when evaluating the surrounding complex velocity. With the convention $W^{\prime }=u-iv$ this gives $d\bar{a}/dt=W_{\mathrm{image}}^{\prime }(a)$, which is the conjugated equation requested in the source. In case (a), differentiating $\bar{m}z/(1-\bar{a}z)$ givesFor case (b), differentiate the three image terms:For case (c), differentiation givesThe physical equation for $da/dt$ is the complex conjugate of the displayed equation. These expressions apply away from the walls, where the point-dipole model is nonsingular.

The source-free Maxwell equations in vacuum areTheir real and imaginary parts are exactlyPut $\theta =\omega t-K\cdot x$. Differentiating gives $\nabla \cdot W=(\Omega \cdot K)\sin \theta$ and $\nabla \times W=-(\Omega \times K)\sin \theta$, while $(i/c){\partial }_{t}W=-i\omega \Omega \sin \theta /c$. Thus the necessary and sufficient conditions areHere the complex dot product is bilinear, without complex conjugation. Taking the dot product of the second equation with $\Omega$ gives $\Omega \cdot \Omega =0$. Crossing it with $K$ givesso, since $\Omega \ne 0$, $|K{|}^2={\omega }^2/c^2$. Writing $\Omega =e+icb$, the real conditions areBoth physical fields are linearly polarized: $E=e\cos \theta$ and $B=b\cos \theta$ each oscillate along a fixed real direction. Combining two fields as $E+icB$ does not turn the electric field into a circularly polarized field.