Past exam of the options I course of the University of Oxford / 2007 / O (Statistics)

A frequentist confidence interval is a pair of data-dependent endpoints whose random interval covers the fixed parameter with probability at least $1-\alpha$ under every allowed parameter value. In repeated uses of the same sampling procedure, at least the advertised fraction of intervals cover the true value. It is not a posterior probability assigned to the parameter after observing the data.

For a sample of size $n$ from a Poisson law with mean $\mu$, the sum $S$ is Poisson with mean $n\mu$. At observed $s>0$, define the lower and upper limits byAt $s=0$ set $L=0$ and use the same upper equation. This inverts the two exact one-sided tail tests. Discreteness means the resulting central interval is generally conservative: its coverage is at least the nominal value, not identically equal to it.

Here one observes a four-year total of 39, so the lower endpoint satisfies ${\varphi }_L(38)=0.975$ and the upper satisfies ${\varphi }_U(39)=0.025$. Selecting the corresponding entries supplied in the source givesThis interval concerns the expected four-year total, not an annual rate. The other two supplied numbers correspond to the wrong inclusive tail for one of these endpoints.

The power of a test at a parameter value is its probability of rejecting the null there. Its size is the supremum of this probability over the null parameter set. The Neyman-Pearson lemma states that for simple null and alternative densities $f_0,f_1$, a test rejecting where $f_1>c{f}_0$, accepting where $f_1<c{f}_0$, and randomizing as needed on equality to attain size $\alpha$, has greatest power among tests of size at most $\alpha$.

The specified function is nonnegative. Its integral isFor independent observations write $S={\sum }_i{X}_i$. The likelihood ratio of the smaller alternative parameter to the null isIt is strictly increasing in $S$. Thus the most powerful level-$\alpha$ test rejects forThe continuous distribution permits this size without randomization for $0<\alpha <1$. Its threshold depends only on the null and the desired size, not on which ${\theta }_1<{\theta }_0$ was chosen. The same test is therefore uniformly most powerful against the entire one-sided alternative $\theta <{\theta }_0$.