The derivative of the log ratio is $\frac{1}{2}(n/q-1)$, so $\lambda (q)$ increases from zero to one on $(0,n)$ and decreases to zero on $(n,\infty )$. Rejecting for a sufficiently small ratio therefore gives two tails with $0<q_1<n<q_2$.

Under $H_0$, Cochran's theorem gives the exact sampling distribution $Q=S/{\sigma }_0^2\sim {\chi }_{n-1}^2$, independently of the unknown mean. Writing $F_{n-1}$ for its cumulative distribution function, the two cutoffs are determined byThe first equation calibrates the exact size of a statistical test. The second is $\lambda (q_1)=\lambda (q_2)$, so both tails belong to one likelihood ratio cutoff. Hence the exact critical region is $\boxed{\{q\le q_1\}\cup \{q\ge q_2\}}$. Its two tail probabilities are not generally equal: equal-tailed chi-squared cutoffs give a different test.

The full parameter space has dimension two and the null space dimension one. The regular Wilks theorem therefore gives $W\stackrel{d}{\to }{\chi }_1^2$ under $H_0$ as $n\to \infty$. If $F_1$ denotes the cumulative distribution function of the chi-squared distribution with one degree of freedom, the approximate size-$\alpha$ rejection region isThe corresponding approximate p-value isThe one degree of freedom here describes the large-sample likelihood-ratio test statistic; it is different from the exact residual degrees of freedom $n-1$ in the next part.

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.

The estimated standard error is ${\hat{\theta }}^{3/2}/\sqrt{n}$. Using the central 95% standard normal quantiles $\pm 1.96$, a Wald confidence interval isConsistency of $\hat{\theta }$ and Slutsky's theorem justify replacing the unknown scale ${\theta }^{3/2}$ by its estimate. The coverage probability is asymptotic, not exact at finite $n$. Intersecting with the positive parameter space removes impossible values without changing whether the true positive parameter is covered.

Expanding the exponent, the log-likelihood function iswhere $C$ is independent of $\theta$. Its score function and second derivative areThe first is positive below $\bar{y}$ and negative above it, so the unique maximum-likelihood estimator isUsing $\mathbb{E}{\sum }_i{Y}_i=n\theta$, the one-observation Fisher information is $I(\theta )=-n^{-1}\mathbb{E}{\ell }^{\prime \prime }(\theta )={\theta }^{-3}$. The support does not depend on the interior parameter $\theta >0$, and the regular asymptotic normality of a maximum likelihood estimator givesThis is Inverse Gaussian mean estimation with known shape one. Equivalently, the Inverse Gaussian distribution has variance ${\theta }^3$, so the same result follows directly from the central limit theorem applied to its sample mean.

Put $M_0=\alpha +x_0\beta$. As a linear combination of independent normal random variables under the posterior distribution,The shortest 95% credible interval isHere $\Phi$ is the standard normal distribution cumulative distribution function. To justify shortest length, consider all intervals of a fixed length $d$. If $f$ is this normal distribution posterior density, the interval probability ${\int }_a^{a+d}f(t)dt$ has derivative $f(a+d)-f(a)$ with respect to $a$. The normal distribution is symmetric and its density strictly decreases with distance from its mean, so this probability is uniquely maximized when the midpoint is $m_0$. The centered interval can therefore attain 95% posterior probability with the least length. It is both an equal-tailed credible interval and a highest posterior density interval.

The target is the experiment's mean at $x_0$. A posterior predictive distribution for a new noisy observation would instead have variance $v_0+{\sigma }^2$; that extra observation variance does not belong in the requested interval.

Let $S_{xx}={\sum }_i{x}_i^2$, $S_{xy}={\sum }_i{x}_i{y}_i$, and $\bar{y}=n^{-1}{\sum }_i{y}_i$. The independent observations in the normal linear model giveThe centered design condition ${\sum }_i{x}_i=0$ removes the $\alpha \beta$ term:The stated prior density factors into independent normal distributions for the regression intercept $\alpha$ and regression coefficient $\beta$. Multiplication by the likelihood function therefore leaves separate quadratic expressions in the two coefficients. Define the positive posterior variancesand the posterior meansBy completing the square, the joint posterior density isEquivalently,This is Gaussian conjugacy for a normal linear model. Each posterior precision parameter is the sum of its data and prior precision parameters. The formulas remain valid when $S_{xx}=0$: all $x_i$ are then zero, so the posterior distribution of $\beta$ equals its prior distribution.

The prior mean $1/(\beta +1)$ decreases as $\beta$ increases. More strongly, its cumulative distribution function and upper-tail probability areFor every fixed $t$, that upper-tail probability decreases with $\beta$. The statistician increasingly favors smaller values of $\theta$; as $\beta \to \infty$, the prior distribution concentrates near zero. For $\beta >1$, its variance $\beta /[(\beta +1{)}^2(\beta +2)]$ also decreases. This is concentration of a beta prior near zero.

With the data fixed, the derivatives areThus the prior weight increases and the posterior mean decreases. Their limits are $\lambda \to 1$ and ${\tilde{\theta }}_1\to 0$. Both the changing prior mean and the increasing weight contribute to the change in the posterior mean.

The prior mean, posterior mean, and maximum-likelihood estimate areFor $0<S<N$, the last follows by maximizing $S\log \theta +(N-S)\log (1-\theta )$. At $S=0$ or $S=N$, use the usual binomial likelihood convention that admits the endpoints $0$ and $1$ as maximum-likelihood estimates; if the parameter domain is strictly $(0,1)$, these are boundary suprema rather than attained maxima. The posterior mean remains interior in all cases.

TakinggivesSince $N=nm>0$ and $\beta +1>0$, $0<\lambda <1$. Thus the posterior mean is a convex combination of the prior mean and observed binomial proportion, with larger prior weight when $\beta +1$ is large relative to $N$.

Set $S={\sum }_i{x}_i$ and $N=nm$. Conditional independence of the binomial distribution observations gives the likelihood functionThe prior distribution is the beta distribution with parameters $(1,\beta )$. By Bayes' theorem, multiplying the prior density by the likelihood function gives a posterior density proportional toRecognizing and normalizing this beta distribution givesfor $0<\theta <1$. This is Beta-binomial conjugacy; the original binomial coefficients cancel in the normalization because they are independent of $\theta$.

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.

Let $S={\sum }_i{X}_i$, which has a binomial distribution with parameters $(n,p)$. For $0<p_1<1/2$, the alternative-to-null likelihood ratio of the independent Bernoulli trials isSince $p_1/(1-p_1)<1$, this is strictly decreasing in $S$. A threshold on large $R$ is therefore a lower-tail likelihood-ratio test, and a threshold between $R(2)$ and $R(1)$ gives the stated critical region $S<2$. The Neyman-Pearson lemma also proves it is a most powerful test at its attained test size.

Using the binomial probability mass, its test size and statistical power are respectivelyThe associated power function is $\pi (p)=(1-p{)}^n+np(1-p{)}^{n-1}$. These formulas hold for positive integer $n$; when $n=1$ the test always rejects and has size one. If the boundary value $p_1=0$ is admitted, all alternative probability is at $S=0$; the same test still has statistical power one and can be represented using a threshold with ties in the likelihood ratio.

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.

True, with the test's rejection rule fixed. If its rejection region in the space of the test statistic is $C$, its statistical power against the specified simple hypothesis $H_1$ isThus the probability distribution of $T$ under the alternative hypothesis determines the statistical power of this test. The null distribution is used to choose a rule with the desired test size, but statistical power is evaluated under the alternative hypothesis. Different rejection rules can have different statistical power even for the same probability distribution of $T$.

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.

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.

The standard error suggested by part (iii) is $\theta /\sqrt{n}$. Substituting the consistent estimator $T_2$ gives $T_2/\sqrt{n}$; Slutsky's theorem givesUsing the central standard normal quantiles $\pm 1.96$, an approximate 95% Wald confidence interval isIts coverage probability approaches 0.95 for each fixed $\theta >0$ as $n\to \infty$. The intersection enforces positivity and does not change whether the true positive statistical parameter is covered. The value 1.64 would give the wrong central two-sided coverage probability.

The specified asymptotic relative efficiency isFor every fixed $\theta >0$ this lies strictly between zero and one. Prefer $T_2$ on grounds of asymptotic precision: it has the smaller leading asymptotic variance and attains the inverse Fisher information. Both are consistent estimators, by the law of large numbers and the continuous mapping theorem applied respectively to $\bar{X}$ and $A/n$. The efficiency approaches zero as $\theta \downarrow 0$ and one as $\theta \to \infty$.

This preference is an asymptotic comparison. In particular, it does not assert that $T_2$ is an unbiased estimator: since $A$ has a gamma distribution of shape $n$ and rate $\theta$, $\mathbb{E}{T}_2=n\theta /(n-1)$ for $n>1$.