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The Neyman-Pearson lemma concerns a simple hypothesis with density $f_0$ against a simple alternative with density $f_1$. Allow a randomized test $0\le \phi \le 1$. Choose a threshold $c\ge 0$ and, if necessary, randomize on the equality set so thatThen ${\phi }_{\ast }$ is a most powerful test among all tests of size at most $\alpha$. To prove it, let $E_0\phi \le \alpha$. Pointwise, $({\phi }_{\ast }-\phi )(f_1-c{f}_0)\ge 0$: above threshold both factors are nonnegative; below threshold both are nonpositive; on the threshold their product vanishes. ThereforeThis proves the lemma, including regions with $f_0=0$, where a positive alternative density must be rejected. It also shows that ties in the likelihood ratio are the only discretionary region when the inequality is strict off the threshold.

For the stated family, $R=|X|$ has the Gamma distribution with shape $\theta$ and unit rate. At $\theta =1$ it has the exponential distribution, and the likelihood ratio for $\theta =2$ against $\theta =1$ isbecause $\Gamma (1)=\Gamma (2)=1$. The most powerful test therefore rejects for large $|X|$. Since ${\mathrm{Pr}}_1(|X|>c)=e^{-c}$, its exact size-$\alpha$ critical value and rejection rule areThe distribution is continuous, so boundary randomization is unnecessary. Under $\theta =2$, integration by parts gives the power function of a statistical testFor every fixed $\theta >1$, the likelihood ratio against the same null is $|x{|}^{\theta -1}/\Gamma (\theta )$, strictly increasing in $|x|$. The size constraint still selects exactly the same critical value $c=-\log \alpha$, independent of the alternative. The Neyman-Pearson lemma therefore makes this single test most powerful against each $\theta >1$ separately. Hence it is uniformly most powerful for the composite alternative $\theta >1$.

The power function of a statistical test for Test 1 is

Put $a=1/\sqrt{10}$. Test 2 rejects whenUsing the two branches of the triangular distribution givesBoth functions equal $0.05$ at the least favourable null value $\theta =0$.

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.

Euler's criterion states that for $p\nmid a$,For a multiple of $p$, both the Legendre symbol and that positive power vanish modulo $p$, so the congruence extends to that case as well.

Write $q=p^r$, where $p$ is the characteristic. In a splitting field of $X^{q^m}-X$ over $F$, let $K$ be its set of roots. The derivative of this polynomial is $-1$, so all $q^m$ roots are distinct. The Frobenius power is additive and multiplicative: $(x+y{)}^{q^m}=x^{q^m}+y^{q^m}$ and $(xy{)}^{q^m}=x^{q^m}{y}^{q^m}$. Thus roots are closed under addition, negatives and multiplication; for a nonzero root, $(x^{-1}{)}^{q^m}=x^{-1}$. They include $0,1$ and form a field.

Every $a\in F$ satisfies $a^q=a$, by Lagrange's theorem on $F^{\times }$ and separately at zero. Iterating gives $a^{q^m}=a$, so $F\subseteq K$. The constructed field therefore contains $F$ and has exactly $\boxed{q^m\text{ elements}}$.

By the allowed cyclicity theorem, $K^{\times }$ has a generator $z$. Every nonzero element of $K$ is a power of $z$, and zero also belongs to $F(z)$. Thus $K\subseteq F(z)\subseteq K$, giving $\boxed{K=F(z)}$. This is a primitive element argument for finite fields using the multiplicative group, not a claim that the additive group is cyclic.

The circle theorem says that if a complex potential $f$ has no singularities on or inside $|z|=a$ (and has the required analytic continuation there), thenadds image singularities inside the circle and makes it a streamline. On the circle $a^2/\bar{z}=z$, so $w=f(z)+\overline{f(z)}$ is real and its imaginary part is constant. This gives zero normal velocity. The reflected term is holomorphic where defined, as follows from its power series in $1/z$, and its singularities are precisely the reflected ones; the external original singularities are retained. This proves the construction, with an additive constant and any independently prescribed circulation handled separately.

For the fluid vortex use $f(z)=-i\Gamma \log (z-b)/(2\pi )$. Reflecting gives an opposite image at $d=a^2/b$ and also a term $-i\Gamma \log z/(2\pi )$, up to constants. The zero-circulation circle-theorem construction has both these terms. To impose the specified circulation $-\Gamma$ about the obstacle, add $+i\Gamma \log z/(2\pi )$, canceling the central term. ThusIt has the stated circulation and the same impermeable boundary, since the added central-fluid vortex streamfunction is constant on the circle.

For the given Blasius integral take the obstacle contour counterclockwise. Inside it the sole singularity ofis at $d$. Its residue is ${\Gamma }^2/[2{\pi }^2(d-b)]$, from the cross term. The residue theorem then givesTherefore $\boxed{F_x=\rho {\Gamma }^{2}b/[2\pi (b^2-a^2)],F_y=0}$. The force on the obstacle points toward the external fluid vortex.

For a rejection function $\phi (x)\in [0,1]$, its power is $\beta (\theta )={\mathbb{E}}_{\theta }\phi (X)$. A test has level at most $\alpha$ if ${\sup }_{\theta \in H_0}\beta (\theta )\le \alpha$; its size is that supremum. In a simple null the significance probability is just the rejection probability under that null.

For two simple hypotheses with joint sample densities $f_0,f_1$, the Neyman–Pearson lemma states that a test rejecting where $f_1>k{f}_0$, accepting where $f_1<k{f}_0$, and randomizing if necessary on equality to have size $\alpha$, is most powerful among level-$\alpha$ tests. Let ${\phi }_{\ast }$ be that test and $\phi$ any competitor. Pointwise,Integration givessince $k\ge 0$, the first test has size $\alpha$, and the other has size at most $\alpha$. This proves the claimed power bound, allowing null-density-zero regions as well. Equality implies agreement away from the likelihood-ratio boundary, up to null sets, when the inequalities are strict.

For a fixed $\lambda <{\lambda }_0$, the likelihood ratio against the null isstrictly increasing in $S={\sum }_i{x}_i$. The Neyman–Pearson critical region is therefore an upper tail in $S$, and its null threshold depends only on the desired size, not on which $\lambda <{\lambda }_0$ was chosen. Thus the same test is uniformly most powerful for the entire one-sided alternative.

Under the null $S\sim \mathrm{Gamma}(n,{\lambda }_0)$ with the second parameter interpreted as rate. Its continuous distribution permits the exact thresholdUnder any fixed alternative its rejection probability is $\boxed{Q(\lambda )=1-F(c_{\alpha };n,\lambda )}$. Applying the lemma separately to each such $\lambda$ shows that every level-$\alpha$ test has power at most this value there. The constructed test attains the bound for all $\lambda <{\lambda }_0$ simultaneously.