Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 2 / 12D (Statistics)

For a generalized likelihood-ratio test, maximize the likelihood separately over the null parameter space and the unrestricted space. Define $\Lambda ={\sup }_{H_0}L/\sup L\le 1$, and reject for small $\Lambda$, equivalently large $T=-2\log \Lambda$. Choose the rejection threshold to achieve the desired null significance level, using an exact distribution when available or a justified asymptotic calibration; nuisance parameters are fitted under the respective restrictions.

For independent Poisson distributions, the likelihood is $L={\prod }_i{e}^{-{\lambda }_i}{\lambda }_i^{X_i}/X_i!$. The unrestricted maximizers are ${\hat{\lambda }}_i=X_i$, allowing zero as a boundary value. Under the common-mean null, maximizing $e^{-n\lambda }{\lambda }^{{\sum }_i{X}_i}$ gives $\hat{\lambda }=\bar{X}$. The exponential and factorial factors cancel in the ratio, yieldingIf all counts are zero, set $\Lambda =1$ and $T=0$; no division by $\bar{X}$ is then needed.

For positive $\bar{X}$, put $X_i=\bar{X}+{\delta }_i$, so ${\sum }_i{\delta }_i=0$. Expanding the logarithm givesConsequentlyIt is the quadratic Pearson approximation to the Poisson deviance. The approximated statistic is the logarithmic ratio $T$, not the bounded raw ratio $\Lambda$.

Under the null in the large-common-count regime, this statistic has an approximate chi-squared distribution with $n-1$ degrees of freedom: fitting the common mean removes one independent residual direction. For $n=7$, use six degrees of freedom. At $T=27.3$, the upper-tail probability in that approximation isThus reject the equal-mean null at conventional 5% or 1% significance levels, provided the expected counts justify the chi-square approximation. No particular significance level or common count was supplied, so the numerical decision is stated with those assumptions.