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For independent $N(0,{\sigma }^2)$ observations, testing a fixed variance against larger variances uses the sum of squares $S={\sum }_i{X}_i^2$. The likelihood ratio for any simple larger variance is strictly increasing in $S$. The Neyman-Pearson lemma therefore gives the displayed rejection rule, where $F_n$ is the chi-squared distribution function with $n$ degrees of freedom. Its threshold is independent of the selected larger variance, so it is a uniformly most powerful test for the one-sided composite alternative. Its power function is $1-F_n(({\sigma }_0^2/{\sigma }^2)F_n^{-1}(1-\alpha ))$, increasing continuously from zero to one and equaling $\alpha$ at the null. A known nonzero mean is handled by replacing each observation by its difference from that mean.

For a family with a monotone likelihood ratio in $T$, a level-$\alpha$ upper-tail test in $T$ is a uniformly most powerful test for the corresponding one-sided hypothesis.

At level $\alpha$, a most powerful test against a specified simple alternative maximises statistical power among statistical tests whose size of a statistical test is at most $\alpha$. The Neyman-Pearson lemma constructs it by a likelihood ratio threshold. A uniformly most powerful test is most powerful simultaneously against every member of the specified alternative family; maximising power at one alternative alone need not give uniform optimality.

Represent a possibly randomized statistical test by a function $\phi (Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its power function is $\pi (\rho )={\mathbb{E}}_{\rho }\phi (Y)$, and its size of a statistical test is the supremum of this probability over the null hypothesis. At a prescribed significance level $\alpha$, a uniformly most powerful test has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.

Put $S={\sum }_{j=1}^n{Y}_j$. For any fixed alternative $0<{\rho }_1<{\rho }_0$, the likelihood ratio iswhich is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects the upper tail $S>c_{\alpha }$. Its cutoff is determined solely by the null distribution, not by ${\rho }_1$, so the same statistical test is most powerful against every allowed alternative and hence uniformly most powerful.

Under rate $\rho$, $S$ has a gamma distribution with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is ${\rho }^n{s}^{n-1}{e}^{-\rho s}/(n-1)!$, convolving once more gives ${\rho }^{n+1}{s}^n{e}^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently reject precisely when $S>c_{\alpha }$, whereand the power function isFor $0<\alpha <1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.

Put $S={\sum }_i{X}_i$, and let $z_{.95}$ be the exact standard-normal 95th percentile. Under parameter $\theta$, $S=n\theta +\sqrt{n}Z$ with $Z$ standard normal. This distribution can be justified directly: the joint density of the centered observations is proportional to $\exp [-{\sum }_i(X_i-\theta {)}^2/2]$. An orthogonal change of coordinates with first row $(1,\dots ,1)/\sqrt{n}$ has unit absolute Jacobian and preserves the sum of squares. Its density factors into $n$ one-dimensional standard-normal densities, making the first coordinate $Z=(S-n\theta )/\sqrt{n}$ standard normal.

Consider the test ${\varphi }_{\ast }=1_{\{S>\sqrt{n}{z}_{.95}\}}$. Its rejection probability isincreasing in $\theta$. Therefore its maximum rejection probability over the null $\theta \le 0$ is achieved at zero and equals $\alpha =1/20$. This proves it has the required test size.

To prove the uniformly most powerful test property, rather than quote the Neyman-Pearson lemma, fix any ${\theta }_1>0$. The likelihood ratio of the full sample against $\theta =0$ isstrictly increasing in $S$. Set $K=\exp ({\theta }_1\sqrt{n}{z}_{.95}-n{\theta }_1^2/2)$. For any competing possibly randomized test $0\le \varphi \le 1$ with composite-null size at most $\alpha$, we have ${\mathbb{E}}_0\varphi \le \alpha ={\mathbb{E}}_0{\varphi }_{\ast }$. Pointwise,because both factors are nonnegative above the threshold and both are nonpositive below it. Integrating givesThis comparison proves the likelihood-ratio optimality result needed here in full. Since the same critical region works for every ${\theta }_1>0$, it proves the uniformly most powerful test property over the entire alternative.

For $n=9$, the exact critical region and its quoted-table approximation areUsing the supplied rounded percentile $1.65$ gives $\sum X_i>4.95$. The exact cutoff uses $z_{.95}\approx 1.644854$ and hence $\overline{X}>0.548285$; the rounded cutoff is the intended numerical approximation, not a literally exact size-$0.05$ threshold. The Gaussian distribution is continuous, so equality at the cutoff has probability zero and no randomization there is needed.

Interpret $\varphi (x)$ as the conditional rejection probability. The Type I error is $\alpha =E_0\varphi (X)$, the Type II error is $\beta =E_1[1-\varphi (X)]$, the size is $\alpha$, and the power is $1-\beta$. Their weighted sum isPointwise minimization over $0\le \varphi \le 1$ forces rejection where $p_1>c{p}_0$ and acceptance where $p_1<c{p}_0$, with arbitrary randomization on equality. Conversely those choices minimize every integrand. The equivalence holds almost everywhere, since changing a test on a common null set cannot change its errors. With both weights strictly positive, any simultaneous improvement of the two errors, strict in at least one, would lower this weighted sum. Every such positive-threshold likelihood ratio test is admissible for the two simple hypotheses.

For the specified densities the likelihood ratio isIt increases strictly on $(\theta ,1)$ from zero to $1/(1-\theta )$. For $0<c<1/(1-\theta )$, the likelihood-ratio test rejects when $x>\theta /[1-c(1-\theta {)}^2]$. Larger thresholds accept almost surely. Threshold zero rejects everywhere above $\theta$ and permits arbitrary rejection below $\theta$, where both weighted choices tie; values at individual endpoints do not affect errors. These describe all nonnegative-threshold likelihood ratio tests, including their limiting accept-all and reject-all cases.

The upper-tail test ${\varphi }_k$ has size $1-k^2$. If $\theta <k$, it is a positive-threshold likelihood ratio test and is admissible. If $\theta =k$, it rejects exactly the alternative's support, with power one; no test can retain power one with smaller size, so it remains admissible. If $\theta >k$, it still has power one but wastes rejection probability on $(k,\theta )$, where the alternative density vanishes. The test ${\varphi }_{\theta }$ has the same power and smaller size, soNevertheless ${\varphi }_k$ is a uniformly most powerful test at size $1-k^2$ against all $0<\theta <1$: for $\theta <k$, apply the Neyman-Pearson lemma; for $\theta \ge k$, its power is already the maximum one. It is also admissible against the composite alternative. Indeed choose any $0<\theta <k$. Strict monotonicity of its likelihood ratio makes ${\varphi }_k$ the unique positive-threshold minimizer up to null sets. A test dominating it for the composite problem would either improve the weighted simple risk at this $\theta$, impossible, or tie that risk and coincide almost everywhere. Since $p_0>0$ throughout $(0,1)$, such coincidence also precludes improvement at any other $\theta$.

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.

A level-$\alpha$ test is uniformly most powerful for a composite alternative when its power is at least that of every other level-$\alpha$ test at every parameter value in that alternative.

For every fixed ${\lambda }_1<{\lambda }_0$, the likelihood ratio is increasing in the same statistic $S$, and the size-$\alpha$ threshold is always the null quantile $n{c}_{\alpha }$, independent of ${\lambda }_1$. The Neyman-Pearson lemma therefore makes the test from part (a) most powerful against every individual ${\lambda }_1<{\lambda }_0$. Consequently it is

The normalized density isFor $\theta =1$ against $\theta =0$, the sample likelihood ratio isIt is strictly decreasing in $S$. Under $H_0$, parts (a) and (b) giveIf $q_{\alpha }$ is the lower $\alpha$-quantile of this gamma distribution, the Neyman-Pearson lemma gives the most powerful size-$\alpha$ critical region

For any fixed $\theta >0$,is again strictly decreasing in the same statistic $S$. Thus the same critical region is most powerful against every $\theta >0$ and is consequently a uniformly most powerful test of $H_0:\theta =0$ against $H_1:\theta >0$.

A possibly randomized test is a function $\phi (X)\in [0,1]$, interpreted as the conditional probability of rejecting $H_0$. Its power function isIt has size at most $\alpha$ whenSuch a test is a uniformly most powerful test of size $\alpha$ if, for every other test $\psi$ of size at most $\alpha$,Thus one test maximizes power simultaneously at every parameter value in the alternative.

For $S={\sum }_i{X}_i$, the ratio of the two sample likelihood functions isFor $0<{\mu }_0<{\mu }_1$ it is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper tail:Under the null, $S\sim \mathrm{Poisson}(n{\mu }_0)$, so choose $k$ with this upper-tail probability at most $\alpha$. If exact size $\alpha$ is required, randomize at the adjacent boundary value as needed. The rejection rule is the same for every ${\mu }_1>{\mu }_0$, giving a uniformly most powerful test for the one-sided alternative.

For any fixed $p>1/4$ below 1, the ratio to the null isIt increases with $y$. Its ordering is therefore the same for every such alternative, while the size is determined only by the null. The very same rule, reject when $Y\ge 4$, is most powerful against each alternative and hence is a uniformly most powerful test. At $p=1$ its power is 1, also optimal.

More generally, for $0<p_0<1$, the binomial likelihood ratio against any $p>p_0$ increases with $y$. Choose an integer cutoff $c$ so thatand reject for $Y>c$, randomize with probabilitywhen $Y=c$, and accept below. This gives exact size $\alpha$ and is a uniformly most powerful test for every positive integer $n$. Exact prescribed size may require this boundary randomization; a deterministic cutoff alone need not attain it. For $p_0=0$, reject all $Y>0$ and reject at $Y=0$ with probability $\alpha$, which is likewise optimal. At $p_0=1$ the stated one-sided alternative is empty. The cutoff generally depends on $p_0,n,\alpha$; there is no claim that one fixed cutoff serves every null parameter.

For a fixed alternative $0<\lambda <{\lambda }_0$, the joint likelihood ratio isIt is strictly increasing in $S={\sum }_i{x}_i$. By the Neyman-Pearson lemma, the most powerful level-$\alpha$ rejection region is therefore $S\ge c_{\alpha }$. Under the null, $S\sim \mathrm{Gamma}(n,{\lambda }_0)$ with the second parameter interpreted as a rate. Its continuous distribution givesThis threshold depends on the null and $\alpha$, but not on the particular smaller alternative $\lambda$. Thus the same test is most powerful for every $0<\lambda <{\lambda }_0$, making it uniformly most powerful.

Under a true alternative $\lambda$, $S\sim \mathrm{Gamma}(n,\lambda )$, so its attained power isFor each $\lambda <{\lambda }_0$, the Neyman-Pearson lemma bounds the power of every other level-$\alpha$ test by this value. Hence the displayed function is not only an upper bound but the attainable power envelope on the specified alternative.

Yes, it is a uniformly most powerful test for the one-sided alternative. For every fixed ${\sigma }_1^2>{\sigma }_0^2$, the likelihood ratio is increasing in the same statistic $S$. The size-$\alpha$ threshold found in part (c) depends only on the null variance and $\alpha$, not on the selected alternative variance. Thus one and the same test is most powerful against each simple member of the composite alternative, which is exactly uniform most powerfulness.

The statistical power of a test $\phi$ is $\beta (\theta )={\mathbb{E}}_{\theta }\phi (X)$, its rejection probability at parameter $\theta$. Its size is ${\sup }_{\theta \in H_0}\beta (\theta )$; for a simple null this is just $\beta ({\theta }_0)$.

The Neyman-Pearson lemma states that for testing a simple null probability density function $f_0$ against a simple alternative probability density function $f_1$, a test rejecting where $f_1/f_0>k$, accepting where it is below $k$, and if needed randomizing on equality to attain the desired size is most powerful among all tests of that size. It follows by integrating $(\phi -{\phi }_{\ast })(f_1-k{f}_0)\le 0$ and using the size constraint.

The proposed probability density function is nonnegative, andFor an independent sample, cancellation of the factors $(x_i+1)$ in the likelihood ratio givesFor ${\theta }_1<{\theta }_0$ this is increasing in $S={\sum }_i{X}_i$. The most powerful test therefore has rejection regionThe sum has a continuous positive probability density function on $(0,\infty )$, so no boundary randomization is required. The threshold depends on the null and chosen size, but not on which ${\theta }_1<{\theta }_0$ was specified. The same test is therefore a uniformly most powerful test for the composite alternative $0<\theta <{\theta }_0$.

For $S={\sum }_i{X}_i$, the ratio of the two sample likelihood functions isFor $0<{\mu }_0<{\mu }_1$ it is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper tail:Under the null, $S\sim \mathrm{Poisson}(n{\mu }_0)$, so choose $k$ with this upper-tail probability at most $\alpha$. If exact size $\alpha$ is required, randomize at the adjacent boundary value as needed. The rejection rule is the same for every ${\mu }_1>{\mu }_0$, giving a uniformly most powerful test for the one-sided alternative.

Let $S={\sum }_i{X}_i$. For the exponential distribution with mean $\theta$, the likelihood function is $L(\theta )={\theta }^{-n}{e}^{-S/\theta }$ on positive observations. Fix any $0<{\theta }_1<{\theta }_0$. The alternative-to-null likelihood ratio isSince $1/{\theta }_1-1/{\theta }_0>0$, this is strictly decreasing in $S$. The Neyman-Pearson lemma says to reject for small $S$. Under the null hypothesis, the given sum rule yields $S\sim \mathrm{Gamma}(n,1/{\theta }_0)$, where the second argument is a rate, and henceLet $q_{\alpha }$ be the lower $\alpha$ quantile of this chi-squared distribution. The required critical region isIts test size is exactly $\alpha$ because the null distribution is continuous. Its cutoff is the same for every ${\theta }_1<{\theta }_0$, so it is simultaneously most powerful against every such value: it is a uniformly most powerful test. Equivalently, reject if $S$ is below the lower $\alpha$ quantile of the specified gamma distribution.

For any fixed $p>1/4$, the alternative-to-null likelihood ratio iswhich is strictly increasing in $S$ for $1/4<p<1$. Consequently the Neyman-Pearson lemma selects the same upper-tail ordering for every such alternative hypothesis. At the null hypothesis value,Thereforeis a uniformly most powerful test of the required exact test size. Its power function is $5p^4(1-p)+p^5=5p^4-4p^5$. At $p=1$ its statistical power is one, so including that boundary alternative preserves the conclusion.

Here the gamma distribution uses a rate statistical parameter $\theta$, with known shape $\lambda$. Write $S={\sum }_i{X}_i$ and $a=n\lambda$. For each fixed ${\theta }_1>{\theta }_0$, the alternative-to-null likelihood ratio isstrictly decreasing in $S$. The Neyman-Pearson lemma therefore gives the same lower-tail critical region $S<c_{\alpha }$ for every larger rate. Since $S$ has a continuous gamma distribution under the null hypothesis, there is a unique positive $c_{\alpha }$ satisfyingThus reject when $S<c_{\alpha }$ is a uniformly most powerful test of exact test size $\alpha$. If $G_a$ denotes the unit-rate gamma distribution cumulative distribution function with shape $a$, thenThis is the uniformly most powerful test for a gamma rate.

When $\lambda =1/n$, the sum has shape one and hence an exponential distribution with rate $\theta$. ThereforeIn particular $\pi ({\theta }_0)=\alpha$, and the power function is strictly increasing, exceeding $\alpha$ for $\theta >{\theta }_0$ and tending to one as $\theta \to \infty$.

For an independent statistical sample from a gamma distribution with known shape $\lambda >0$ and unknown rate $\theta >0$, test $\theta ={\theta }_0$ against $\theta >{\theta }_0$. The alternative-to-null likelihood ratio for any larger rate ${\theta }_1$ is $({\theta }_1/{\theta }_0{)}^{n\lambda }\exp \{-({\theta }_1-{\theta }_0)S\}$, where $S={\sum }_i{X}_i$. This decreases in $S$, so the Neyman-Pearson lemma gives one common lower-tail ordering for all alternatives.

Let $G_a$ be the unit-rate gamma distribution cumulative distribution function with shape $a$. The uniformly most powerful test of exact test size $\alpha \in (0,1)$ rejects when $S<c_{\alpha }$, where $c_{\alpha }=G_{n\lambda }^{-1}(\alpha )/{\theta }_0$. Its power function is displayed above. If $n\lambda =1$, the sum has an exponential distribution, so $c_{\alpha }=-\log (1-\alpha )/{\theta }_0$ and $\pi (\theta )=1-(1-\alpha {)}^{\theta /{\theta }_0}$. Monotonicity of $G_a(\theta c_{\alpha })$ also makes this test valid and uniformly most powerful for the larger null hypothesis $\theta \le {\theta }_0$.