Past exam of the statistics course of the University of Oxford / 2014 / 2

For two simple statistical hypotheses with joint probability density functions $g_0$ and $g_1$, fix a significance level $\alpha$. A randomized statistical test $\phi (x)$ is the conditional probability of rejection given the data. Choose $k\ge 0$ and, if necessary, randomization on the equality set so thatThe Neyman-Pearson lemma says that this is a most powerful test among all tests with test size at most $\alpha$:Thus the optimal ordering rejects where the alternative-to-null likelihood ratio is largest. On null-impossible observations with $g_1>0$, the rejection probability is one. Randomization is dispensable whenever an ordinary threshold already attains the required test size.

For a random sample, write $g_j(x)={\prod }_{i=1}^{n}f(x_i;{\theta }_j)$, $j=0,1$, for its joint probability density functions. Let $\phi$ be any competing statistical test of level at most $\alpha$. Pointwise,Indeed, where $g_1>k{g}_0$, ${\phi }_{\ast }=1\ge \phi$; where $g_1<k{g}_0$, ${\phi }_{\ast }=0\le \phi$, so both factors have the same sign. On the equality set the product is zero, regardless of randomization. Integrate over the joint sample space using its ordinary volume measure:Since $k\ge 0$ and ${\mathbb{E}}_0{\phi }_{\ast }=\alpha \ge {\mathbb{E}}_0\phi$,This proves maximal power. A continuous sample distribution does not itself guarantee that the equality set of the likelihood ratio has zero probability; allowing randomization there makes the proof valid in that case too.

Put $T={\sum }_i{X}_i$. For interior parameters $0<{\theta }_0<{\theta }_1<1$, the alternative-to-null likelihood ratio isThe second base is greater than one, so this is strictly increasing in $T$. The Neyman-Pearson lemma therefore gives an upper-tail rejection rule. Under the null hypothesis, $T\sim \mathrm{Bin}(n,{\theta }_0)$, and an ordinary conservative threshold can be chosen asThere is an important discrete qualification: an integer cutoff need not have test size exactly $\alpha$. To obtain exact size, choose an integer $b$ with ${\mathrm{Pr}}_0(T>b)\le \alpha \le {\mathrm{Pr}}_0(T\ge b)$, reject certainly when $T>b$, and at $T=b$ reject with probabilityThis randomized statistical test is a most powerful test at exact size $\alpha$; when a binomial upper tail equals $\alpha$, no randomization is needed and the stated region is literal. Otherwise the unrandomized upper-tail region has size at most $\alpha$, and the exact-size result includes the boundary randomization. Endpoint parameter cases retain the same upper-tail ordering by taking limits, with ties resolved on the null-supported outcomes.

The sample mean $\overline{X}=T/n$ has mean $\theta$ and variance $\sigma (\theta {)}^2=\theta (1-\theta )/n$. The central limit theorem gives a normal approximation for interior $\theta$ when both expected binomial counts are sufficiently large. Under the null hypothesis, a level-$\alpha$ upper-tail cutoff is approximatelyTo obtain the power function, evaluate this same cutoff under the true statistical parameter $\theta$, standardizing with $\sigma (\theta )$:Here $\Phi$ is the standard normal distribution's cumulative distribution function. The formula treats the binomial threshold as continuous; exact integer cutoffs and boundary randomization are finite-sample refinements.

Here $z_{\alpha }=1.96$, $\sigma (1/2)=1/(2\sqrt{n})$ and $\sigma (3/4)=\sqrt{3}/(4\sqrt{n})$. The desired approximate power is at least $0.975=\Phi (1.96)$ exactly when the argument of the upper tail in part (i) is at most $-1.96$:Rearranging and squaring the positive quantities yieldsThis is a normal approximation to a sample size calculation. It is not a guarantee of the exact finite-sample power for an arbitrary unrandomized binomial cutoff; the discrete test size and power can instead be checked using binomial probabilities.