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A possibly randomized statistical test has rejection probability $\phi (x)\in [0,1]$ after observing $x$. Its statistical power at parameter $\theta$ is ${\mathbb{E}}_{\theta }\phi$, and its test size is the supremum of that statistical power over the null parameter set.

For simple hypotheses with densities $f_0,f_1$, the Neyman-Pearson lemma states that a level-$\alpha$ test is most powerful if it rejects where $f_1>k{f}_0$, accepts where $f_1<k{f}_0$, and randomizes on equality to obtain null rejection probability $\alpha$. Here $k\ge 0$ is chosen to reach that test size. To prove optimality, let ${\phi }_{\ast }$ be this test and $\phi$ any test of level at most $\alpha$. Pointwise,Integrating and using $\int ({\phi }_{\ast }-\phi )f_0\ge 0$ gives ${\mathbb{E}}_1{\phi }_{\ast }-{\mathbb{E}}_1\phi \ge 0$. This proves maximal statistical power. For $k>0$, equality also forces any equally powerful test to use all its allowed test size and agree with the indicated decisions away from the likelihood-ratio tie set, up to null sets.

For the Poisson sample, $S={\sum }_i{X}_i\sim \mathrm{Poisson}(n\mu )$ and, for every fixed ${\mu }_1>{\mu }_0>0$, the likelihood ratio isIt increases strictly with $S$. Choose an integer $k$ so that ${\mathbb{P}}_{{\mu }_0}(S>k)\le \alpha \le {\mathbb{P}}_{{\mu }_0}(S\ge k)$, and reject for $S>k$, with boundary rejection probabilityThis has exact test size $\alpha$. Its rejection ordering and calibration do not depend on which ${\mu }_1>{\mu }_0$ is selected, so the lemma proves it uniformly most powerful for the whole one-sided alternative. Randomization is generally needed for an exact arbitrary test size in a discrete model.

For the approximate design, the central limit theorem under ${\mu }_0=1$ gives critical sum $n+1.645\sqrt{n}$. Under ${\mu }_1$, the sum is approximately normal with mean $n{\mu }_1$ and variance $n{\mu }_1$. Target statistical power $0.9$ therefore givesor $\sqrt{n}({\mu }_1-1)\approx 1.645+1.282\sqrt{{\mu }_1}$. Thus the requested design expression isRound upward to obtain an integer design. This is a normal approximation to the requested statistical power, rather than an exact discrete sample-size guarantee; the exact Poisson test above provides its exact calibration.

The central limit theorem states that for independent identically distributed variables of finite mean $\mu$ and positive finite variance ${\sigma }^2$,For a continuous strictly increasing cumulative distribution function, its inverse is defined on $(0,1)$. Thus $\mathbb{P}(F(X)\le y)=\mathbb{P}(X\le F^{-1}(y))=y$ for $0<y<1$, proving the uniform law.

If $T=-\log (1-Y)$, then for $t\ge 0$, $\mathbb{P}(T\le t)=\mathbb{P}(Y\le 1-e^{-t})=1-e^{-t}$. Hence $f_T(t)=e^{-t}{1}_{(0,\infty )}(t)$, an exponential density of rate one. Its first two moments are one and two, by integration by parts, so $\mathbb{E}T=1$ and $\mathrm{Var}T=1$.

Although the original variables may have different distributions, their separate transforms $Y_k=F_k(X_k)$ are independent and all uniform. Consequently $T_k=-\log (1-Y_k)$ are independent identically distributed exponentials of rate one, and the statistic is exactlyThe stated central limit theorem therefore yields $\boxed{Z_n\stackrel{d}{\to }N(0,1)}$.

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

For a normal sample of size $m$,Then $\bar{X}\sim N({\mu }_X,{\sigma }^2/m)$ and $c{S}_X^2\sim {\chi }_{m-1}^2$ with $c=(m-1)/{\sigma }^2$. These two statistics are independent.

Combining the difference distribution and pooled variance calculation gives, with $\nu =m+n-2$,The printed statistic has the sign of the centering term reversed. Its numerator subtracts ${\mu }_Y-{\mu }_X$, so it is not generally a central Student statistic; the corrected centering above is necessary. Under the null hypothesis ${\mu }_X={\mu }_Y$, the two expressions coincide, so the requested test is unaffected. Reject the two-sided null at significance level $\alpha$ whenHere $\nu =8$ and $|T|=2.19<2.31$, so do not reject at the 5% level. This does not establish equality of the population means.

Age matching means that the two observations in a pair share an age effect, while ages vary across pairs. The independent, identically distributed two-sample model no longer represents this design: random pair effects induce within-pair dependence, and fixed differing ages can instead produce different individual means. Pairing by itself need not cause stochastic dependence when ages are treated as fixed. The appropriate alternative is a paired t-test on the within-pair differences. Removing shared age variation can reduce the noise and improve power when age strongly affects volume and the members of a pair are positively associated. Improvement is not automatic, since the paired test also uses fewer degrees of freedom.

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

For a normal sample of size $m$,Then $\bar{X}\sim N({\mu }_X,{\sigma }^2/m)$ and $c{S}_X^2\sim {\chi }_{m-1}^2$ with $c=(m-1)/{\sigma }^2$. These two statistics are independent.

Combining the difference distribution and pooled variance calculation gives, with $\nu =m+n-2$,The printed statistic has the sign of the centering term reversed. Its numerator subtracts ${\mu }_Y-{\mu }_X$, so it is not generally a central Student statistic; the corrected centering above is necessary. Under the null hypothesis ${\mu }_X={\mu }_Y$, the two expressions coincide, so the requested test is unaffected. Reject the two-sided null at significance level $\alpha$ whenHere $\nu =8$ and $|T|=2.19<2.31$, so do not reject at the 5% level. This does not establish equality of the population means.

Age matching means that the two observations in a pair share an age effect, while ages vary across pairs. The independent, identically distributed two-sample model no longer represents this design: random pair effects induce within-pair dependence, and fixed differing ages can instead produce different individual means. Pairing by itself need not cause stochastic dependence when ages are treated as fixed. The appropriate alternative is a paired t-test on the within-pair differences. Removing shared age variation can reduce the noise and improve power when age strongly affects volume and the members of a pair are positively associated. Improvement is not automatic, since the paired test also uses fewer degrees of freedom.

Write $Z_t={\sum }_{j=1}^{N_t}{Y}_j$, with an empty sum equal to zero. Conditioning on the independent Poisson random variable $N_t$, the moment-generating function isThis is the compound Poisson distribution. Differentiating its log moment-generating function at zero, when the first two moments are finite, givesThe variance includes variability in the number of coaches, and so uses the second moment, not merely $\mathrm{Var}(Y)$.

Put $D^{\ast }=D-A$, with $A=(X^{T}X{)}^{-1}{X}^T$. Then $D^{\ast }X=0$, and therefore $D^{\ast }{A}^T=D^{\ast }X(X^{T}X{)}^{-1}=0$; its transpose also vanishes. Expanding givesHenceThe difference from the least-squares covariance is positive semidefinite, since $v^T{D}^{\ast }{D}^{\ast T}v=\Vert D^{\ast T}v{\Vert }^2\ge 0$. In particular its $i$th diagonal entry is ${\sigma }^2{\sum }_j(D_{ij}^{\ast }{)}^2$, proving every stated variance inequality. For estimating ${\theta }_1$, prefer least squares: among linear unbiased estimators it has minimum variance, and therefore minimum mean squared error in that class. A competitor can tie if the first row of $D^{\ast }$ is zero. This is the Gauss-Markov theorem; it does not assert optimality against all biased or nonlinear estimators.

Under the null, replace both parameters by a common $r$. The likelihood becomesThe pooled maximum-likelihood estimate isUnder an interior null $p=q=r$, part (ii) gives a centered asymptotic normal difference with variance $r(1-r)(1/(2M)+1/(2N))$. The law of large numbers makes $\hat{r}\to r$ in probability. By Slutsky's theorem, replacing $r$ by this consistent estimate givesFor the two-sided alternative, reject when $|Z|>z_{1-\alpha /2}$, the corresponding standard-normal quantile. This has asymptotic significance level $\alpha$.

Differentiate the probability generating function:ThereforeUsing the relation between the second factorial moment and variance,

The Gaussian linear model has log likelihood function, up to an additive constant,For each fixed positive ${\sigma }^2$, maximize this by minimizing the residual sum of squares. Differentiation gives the normal equationsThe full column rank of $X$ makes $X^{T}X$ positive definite, hence invertible, soThe quadratic objective is strictly convex, which proves uniqueness. The joint maximum-likelihood estimate has the same coefficient estimate, except on the probability-zero event of an exactly zero residual when the positive-variance likelihood has no finite joint maximizer.

The event $S\ge 500$ is contained in $|S-400|\ge 100$. Thus Chebyshev's inequality givesThe variance bound is rigorous and uses no normal approximation.

The central limit theorem suggests approximating $S$ by a normal variable with mean $400$ and variance $560$. ThusSince $S$ is integer-valued, the usual continuity correction replaces $500$ by $499.5$:These are approximation formulas rather than consequences of Chebyshev's bound; the standard deviation is $\sqrt{560}$, not $560$.

The sample mean has expectation $\theta$ and variance ${\theta }^2/n$. For $g(z)=\log z$, $g^{\prime }(\theta )=1/\theta$. The first-order delta-method expansion isTaking its mean and variance givesThese are large-sample approximations. Retaining the quadratic Taylor term would add the leading mean correction $-1/(2n)$, but it does not change the requested first-order expression.

Chebyshev's inequality states $\mathbb{P}(|Y-\mu |\ge r)\le {\sigma }^2/r^2$ for $r>0$ when $Y$ has mean $\mu$ and finite variance ${\sigma }^2$. Here each summand has mean zero and variance one, so independence gives $\mathbb{E}{S}_n=0$ and $\mathrm{Var}{S}_n=n$. Taking $r=an$ gives

The standardized observations $X_i/\sigma$ are independent standard normal random variables. By the definition of the chi-squared distribution,Its distribution is continuous, so $F_n(c_2)-F_n(c_1)=1-\alpha$ gives the required probability. Put $S={\sum }_i{X}_i^2$. Solving $c_1<S/{\sigma }^2<c_2$ for the positive variance reverses the reciprocal order. The exact confidence interval isThe mean is known to be zero, so the degrees of freedom are $n$, not $n-1$. Coverage is exactly $1-\alpha$ under the stated normal model.

For $S={\sum }_i{x}_i^2$, the likelihood ratio in favor of the larger variance isBecause ${\sigma }_1>{\sigma }_0$, this is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper-tail rejection region. Under the null, $S/{\sigma }_0^2\sim {\chi }_n^2$, so size $\alpha$ is obtained withContinuity of the chi-squared distribution makes the boundary have probability zero. This is a fixed-mean normal variance test.

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.