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A likelihood-ratio test rejects for small $\lambda$. The derivative of the natural logarithm of its variable factor isThus $t^n(1-t{)}^m$ increases up to $n/(n+m)$ and decreases thereafter, tending to zero at both endpoints. Every nontrivial lower level set of $\lambda$ therefore consists of two tails. For a prescribed test size $0<\alpha <1$, the exact critical region isWriting $F_{n,m}$ for the cumulative distribution function of the beta distribution with parameters $(n,m)$, the required two equations areThe first makes both cutoffs have the same likelihood ratio; the second calibrates the null hypothesis rejection probability. Equivalently, the second left-hand side isAs the common likelihood ratio threshold increases from zero to one, this probability increases continuously and strictly from zero to one; hence the equations determine unique cutoffs. The two tail probabilities need not each equal $\alpha /2$ when $n\ne m$. Exact calibration holds for every common rate because the given null distribution of $T$ is independent of that rate. Boundary equalities have zero probability.

Put $A=-{\sum }_i\log x_i>0$, $B=-{\sum }_j\log y_j>0$, and $N=n+m$. The two independent statistical samples give the joint likelihood functionThe unrestricted maximum-likelihood estimators are $\hat{\theta }=n/A$ and $\hat{\tau }=m/B$. Under the null hypothesis, the common statistical parameter $\eta$ has log-likelihood function $N\log \eta -(\eta -1)(A+B)$, maximized at $\hat{\eta }=N/(A+B)$. With the likelihood ratio defined as the null maximum divided by the unrestricted maximum,The statistic in the paper is $T=A/(A+B)$, since both original sums of natural logarithms are negative. HenceIt has maximum one at $T=n/N$, corresponding to equal fitted rates. This is the likelihood-ratio test for equality of beta shape parameters.

False. For the test associated with a p-value, reject at significance level $\alpha$ when $p\le \alpha$. Here $0.1>0.05$, so do not reject at level 0.05. The p-value is neither the posterior probability of the null hypothesis nor the probability of a Type I error conditional on these observed data.

False. A test size constrains the probability of rejection conditional on the null hypothesis being true:for an exact size-$\alpha$ test of a simple hypothesis. It does not assign a posterior probability to that statistical hypothesis after observing nonrejection. In the frequentist interpretation, the statistical parameter is fixed and the data are random. Nonrejection can occur even when the alternative hypothesis is true; for example, it is particularly likely for an alternative against which the test has low statistical power. A posterior probability would require a prior distribution and a Bayesian model.

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 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 $\theta \to \infty$.

Put $S={\sum }_i(x_i-\bar{x}{)}^2$ and $q=S/{\sigma }_0^2$. Under the null hypothesis, maximizing over the nuisance mean gives ${\hat{\mu }}_0=\bar{x}$. The unrestricted normal mean and variance maximum-likelihood estimators are $\hat{\mu }=\bar{x}$, $\widehat{{\sigma }^2}=S/n$. Thus, for $S>0$,The likelihood-ratio test statistic is consequentlyThe usual model assumes $n\ge 2$. The event $S=0$ has probability zero under a nondegenerate normal distribution; at that observation the unrestricted likelihood is unbounded as the variance tends to zero, and the ratio has limiting value zero. This is the likelihood-ratio test for a normal variance with unknown mean.

For a critical region $C$, the size of a statistical test of the simple null hypothesis $P_0$ is $\alpha =P_0(\mathbf{X}\in C)$, the Type I error probability. Its statistical power against the simple alternative hypothesis $P_1$ is $P_1(\mathbf{X}\in C)$, the probability of correctly rejecting that null. For a randomized test function $\phi (\mathbf{x})\in [0,1]$, replace these by ${\mathbb{E}}_0\phi$ and ${\mathbb{E}}_1\phi$.

Test homogeneity of the four category probabilities among the three samples. The null hypothesis is $p_{iA}=p_{iB}=p_{iC}$ for every category $i$; the alternative allows at least one difference. This is a Pearson chi-squared test of homogeneity, assuming independently sampled grains with consistent classification.

The row totals are $(195,52,64,16)$, the sample totals $(131,135,61)$ and the grand total $N=327$. Under the null hypothesis, $E_{ij}=O_{i+}{O}_{+j}/N$, giving the expected value tableFor example, $E_{\mathrm{undulatory},C}=52(61)/327=9.7003$, so that cell contributes $(4-9.7003{)}^2/9.7003=3.34974$. The complete cell contributions areso $X^2=6.87754$. There are $(4-1)(3-1)=6$ statistical degrees of freedom. The supplied $5\%$ critical value is $12.592$, and $6.87754<12.592$, so do not reject homogeneity at $\alpha =0.05$. The approximate upper-tail p-value is $0.3323$.

One of twelve expected cells is below five, with none below one; the asymptotic approximation is broadly defensible by a common expected-count guideline but is not exact. An exact/conditional simulation could check it, especially if grain counts are clustered. Do not pool categories solely to obtain a desired result. The conclusion is that these counts do not demonstrate different category proportions; they do not prove that the rocks share one geological formation. Different formations can share such proportions, and real grains within one rock can fail the independent-sampling assumption.

For a known background rate $b$ observed for exposure $T$, independent events give a Poisson distribution with mean ${\mu }_b=bT$. An excess-count test of the background-only null hypothesis uses the upper-tail P-value in the title. The tail, rather than the probability of exactly $n$, counts all outcomes at least as discrepant in the chosen direction.

This probability is conditional on the null model; it is not the posterior probability that the null model is true. A stipulated discovery threshold is met only when the tail is below that threshold. Uncertainty in the background rate or selecting among many searches needs additional modelling rather than being absorbed into the same known-mean formula.

A rejection region is the set of sample outcomes on which a hypothesis test rejects its null hypothesis. Its probability under the null determines the type I error rate.

A scan statistic searches a family of candidate locations or subsets and reports the largest local statistic. Under a null hypothesis, a union bound converts a tail estimate for each candidate into a bound for the maximum; no independence between local statistics is required. Structured alternatives often permit a chi-squared testing lower bound using the overlap geometry of two candidates.

For an independent normal sample with known standard deviation $\sigma$, the signed Wald statistic for testing mean zero is $W=\sqrt{n}\bar{Y}/\sigma$. It has the standard normal distribution under the null hypothesis and distribution $N(\sqrt{n}\delta /\sigma ,1)$ under mean $\delta$. Its square has a chi-squared distribution; retaining the sign permits a one-sided test.

The size of a test is the supremum, over distributions in its null hypothesis, of its rejection probability. For a simple null it is just the null rejection probability; a level-$\alpha$ test has size at most $\alpha$.

A statistical hypothesis is a specified set of possible data-generating probability distributions, usually expressed as a restriction on model parameters. A simple statistical hypothesis specifies one distribution; a composite statistical hypothesis permits several. A hypothesis test compares a null hypothesis with an alternative hypothesis using a statistic calibrated under the null.

Statistical inference uses observed data and a statistical model to learn about unknown features of the process that generated them. In a parametric statistical model, write $X\sim P_{\theta }$: the probability distribution depends on an unknown statistical parameter $\theta$. An estimand specifies the target, such as a mean or a difference of means. A statistic is a function of the observations; an estimator is a statistic used to estimate the target. Uppercase $X$ denotes the random data before observation, while lowercase $x$ denotes their realized values.

Three questions lead to different outputs. Statistical estimation asks which target value the observations suggest. A confidence interval asks for a range produced by a procedure with calibrated coverage probability. A hypothesis test asks whether the observations depart sufficiently from a specified null hypothesis to enter its rejection region. These outputs complement one another: an estimated effect size describes magnitude, while a standard error describes the sampling distribution of an estimator. A p-value describes a tail probability under the null hypothesis.

The appeal of statistical inference is that it turns variable observations into reproducible uncertainty statements. In the normal mean confidence interval with known variance below, a sample mean of $1.2$ has standard error $0.5$: both the estimated shift and its precision matter. The calibration depends on the statistical model, including independence and the specified probability distribution. Increasing sample size improves precision under those assumptions; it does not repair a systematically misleading sampling procedure. Bayesian inference provides another framework, using a prior distribution and a likelihood function to obtain a posterior distribution.

A result is statistically significant at a specified significance level $\alpha$ when its valid p-value is at most $\alpha$ and the corresponding null hypothesis is rejected. Significance does not measure effect size, establish practical importance, or give the probability that the null is true. Software stars encode thresholds for $p$-values; they do not supply an additional statistic.

A statistical test is a measurable rule for deciding whether to reject a null hypothesis. Allowing randomisation, it is represented by a function $\phi (x)\in [0,1]$ giving the conditional rejection probability after observing $x$. A deterministic test takes only values zero and one. Under a simple null, $E_0\phi$ is its size of a statistical test; under an alternative, $E_1\phi$ is its statistical power. Randomisation on a likelihood ratio equality set allows a Neyman-Pearson lemma threshold test to attain a prescribed size even for discrete data.

The two-sample Kolmogorov-Smirnov test compares two empirical distribution functions from independent samples, each internally independent and identically distributed, under the null hypothesis that their underlying probability distributions coincide. Both curves are estimated from data, so calibration depends on both sample sizes; the usual asymptotic scaling is $\sqrt{nm/(n+m)}{D}_{n,m}$. Tied data can be handled by an appropriate permutation test under the required exchangeability, rather than by assuming the untied continuous-null law.^[ref]