Past exam of the statistics course of the University of Oxford / 2021 / 2

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.

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. 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.

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.

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.

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.

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 $\theta \to \infty$.