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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.

An upper-tail monotone test in a real statistic $T$ has rejection probability $\varphi =1_{T>c}+\eta 1_{T=c}$, with $0\le \eta \le 1$; the constant tests correspond to infinite cutoffs. For a family with positive densities on a common support and a monotone likelihood ratio in $T$, its power function can cross that of an arbitrary competing test only from below to above as the parameter increases. If $r=f_{{\theta }_1}/f_{{\theta }_0}$ and ${\theta }_1>{\theta }_0$, choose a positive number $k$ between the ratios on opposite sides of the cutoff. The signs of $\varphi -{\varphi }^{\prime }$ and $r-k$ agree away from $T=c$, and $r=k$ at a boundary with positive mass. Therefore $E_{{\theta }_1}(\varphi -{\varphi }^{\prime })\ge k{E}_{{\theta }_0}(\varphi -{\varphi }^{\prime })$. In particular strict positivity at ${\theta }_0$ propagates to ${\theta }_1$. Positivity of the densities matters for this strict conclusion; with changing supports only a weak conclusion may remain.

Write $H_0,H_1$ for the two simple hypotheses, with prior probabilities ${\pi }_0,{\pi }_1$ and densities $f_0,f_1$. Under zero-one loss, the posterior expected losses of choosing $H_0$ and $H_1$ are respectively the posterior probabilities of $H_1$ and $H_0$. A Bayes decision rule therefore chooses $H_1$ ifchooses $H_0$ for the reverse inequality, and may randomize at equality. With positive priors this is a likelihood-ratio test with cutoff $k={\pi }_0/{\pi }_1$.

For a test $\varphi$, its prior risk is ${\pi }_0{E}_0\varphi +{\pi }_1(1-E_1\varphi )$. Choose $k$ and boundary randomization so that the likelihood-ratio test ${\varphi }_k$ has size $\alpha$, and choose priors with ratio $k$. Equivalence of normal and extensive analyses makes the pointwise posterior minimizer minimize this prior risk over all randomized tests. Thus for any $E_0{\varphi }^{\prime }\le \alpha$,This proves the Neyman-Pearson lemma: a likelihood-ratio test of size $\alpha$ has greatest power among tests of size at most $\alpha$. Such a cutoff and randomization come from a quantile of the likelihood ratio under $H_0$; always reject where $f_0=0<f_1$. Endpoint cutoffs are handled directly or by their limiting rules.

For the remaining assertions use the usual common-support formulation with positive densities. A family has a monotone likelihood ratio in $T$ if for every ${\theta }_1>{\theta }_0$, $f_{{\theta }_1}(x)/f_{{\theta }_0}(x)$ is a nondecreasing function $r(T(x))$. A monotone test here is an upper-cutoff test $\varphi =1_{T>c}+\eta 1_{T=c}$, $0\le \eta \le 1$, including constant tests. It does not mean an arbitrary smoothly increasing randomized rejection function.

Put $D(\theta )=\gamma (\theta )-{\gamma }^{\prime }(\theta )$. For a finite cutoff, choose $k>0$ between the values of $r$ below and above $c$, with $k=r(c)$ when the boundary has positive probability. On $T>c$, $\varphi -{\varphi }^{\prime }\ge 0$ and $r-k\ge 0$; on $T<c$, both are nonpositive. At a boundary with positive mass $r-k=0$. ThereforeWhen there is probability on both sides, finite positive monotone ratios below the cutoff bound those above it away from zero, so $k$ can be positive and finite. A boundary with positive mass instead supplies the positive finite value $r(c)$. If the cutoff yields a constant test on the support, positivity on the common support proves strict propagation directly: a test identically zero cannot have greater power, and for a test identically one any positive-probability nonrejection by the competitor remains positive-probability at the larger parameter. Hence $D({\theta }_0)>0$ implies $D({\theta }_1)>0$.

Let ${\theta }^{\ast }=\inf \{\theta :D(\theta )>0\}$, taking ${\theta }^{\ast }=+\infty$ when this set is empty. Below ${\theta }^{\ast }$ there is no positive difference. Above a finite ${\theta }^{\ast }$ there is a smaller parameter with positive difference, which propagates upward; the same reasoning applies when ${\theta }^{\ast }=-\infty$. ThusNo assertion at ${\theta }^{\ast }$ itself is needed.

For the final request fix any cutoff $a$ and prior density $\pi$. Set $A(x)={\int }_{\theta >a}\pi (\theta )f_{\theta }(x)d\theta$ and $B(x)={\int }_{\theta \le a}\pi (\theta )f_{\theta }(x)d\theta$. The posterior odds are $A(x)/B(x)$. If $T(x_2)\ge T(x_1)$, thenEvery bracket is nonnegative by monotone likelihood ratio. Equal statistics make the bracket zero, so the ratio depends only on $T$. Dividing where the posterior is defined proves the posterior odds are nondecreasing in $T$, with extended values allowed when one posterior mass is zero. This proves that posterior odds are monotone under a monotone likelihood ratio without restricting the prior's shape.

The common positive-support condition is material to the printed strict inequality. Under the broader definition permitting zero densities, take a two-parameter family on $T\in \{0,1,2\}$ with $f_0=(1/3,1/3,1/3)$ and $f_1=(0,0,1)$. Its likelihood ratio $(0,0,3)$ is nondecreasing. The cutoff test rejecting $T\ge 1$ has greater power than the test rejecting only $T=2$ at parameter zero, but both have power one at parameter one. Thus the strict assertion needs the usual positivity convention; it is false without it.

For ${\theta }_1>{\theta }_0$, the likelihood ratio iswhich is increasing in $x$. Thus the logistic location family has a monotone likelihood ratio in $X$.

The upper-tail rejection probability is increasing in the location parameter, since $X=\theta +Z$ with $Z$ having the standard logistic distribution. Hence for the critical value from part (i),The test has size $\alpha$ for the composite null.

Now fix any alternative ${\theta }_1>0$. The Neyman-Pearson lemma applied to the simple hypotheses $\theta =0$ and $\theta ={\theta }_1$ says that this same upper-tail test is most powerful among all tests whose rejection probability at $0$ is at most $\alpha$. Every test of size at most $\alpha$ for the composite null satisfies that restriction. The upper-tail test is therefore most powerful at every ${\theta }_1>0$, so it is the uniformly most powerful upper-tail test for a logistic location.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

For a family with a monotone likelihood ratio in $T$, a prior density $\pi$ and cutoff $a$, the posterior odds of $\theta >a$ against $\theta \le a$ are nondecreasing in $T$. To prove it, write the unnormalized masses as $A(t)$ and $B(t)$. For $t_2>t_1$, the difference $A(t_2)B(t_1)-A(t_1)B(t_2)$ is the double integral over $\theta >a\ge \eta$ of $\pi (\theta )\pi (\eta )[f_{\theta }(t_2)f_{\eta }(t_1)-f_{\theta }(t_1)f_{\eta }(t_2)]$. Each integrand is nonnegative by the likelihood-ratio ordering. Common factors in the full-data likelihood cancel from the odds, so the same argument works when $T$ is a statistic rather than the complete observation. Interpret zero denominators by extended odds whenever the posterior is defined.