Past exam of the options I course of the University of Oxford / 2006 / O (Statistics)

For independent daily counts $X_1,\dots ,X_N$ modeled by a Poisson distribution, the log-likelihood isDifferentiating gives the maximum-likelihood estimator $\hat{\mu }=\overline{X}$. Here the observed frequencies sum to $N=1096$, and the expected frequencies are consistent with $\hat{\mu }\approx 2.157$ (their ratio $E_1/E_0$ estimates the same parameter). The final class is open-ended: its single observation is only reported as at least 9, so its exact contribution to the sample total cannot be recovered from this grouped table alone. The underlying ungrouped daily observations determine the exact sample mean.

Fit the Poisson distribution using the estimated daily mean. For $k=0,\dots ,8$, its expected frequency isand $E_{\ge 9}=N\left(1-{\sum }_{k=0}^8{e}^{-\hat{\mu }}{\hat{\mu }}^k/k!\right)$. Each row's contribution to the Pearson chi-squared statistic is $(O_k-E_k{)}^2/E_k$. For example the zero-count class contributes $(162-126.784{)}^2/126.784\approx 9.781$. Summing the class contributions gives the test statistic; expected frequencies depend on the fitted model, while observed frequencies come directly from counting days.

The expected counts in the final two separate classes are too small for a reliable chi-squared asymptotic approximation. Pooling the last three classes gives the tail class $\ge 7$, with observed frequency $8+3+1=12$ and expected frequency $5.464+1.473+0.447=7.384$. This exceeds the usual expected-count guideline of 5. Pooling only the last two would leave an expected count of $1.920$, still too small. Pooling requires recomputing the contribution from the pooled totals, rather than adding the old separate contributions.

After pooling, there are eight categories: $0,1,\dots ,6,\ge 7$. Their frequencies have one constraint because they sum to $N$, and one model parameter was fitted. Thus the reference chi-squared distribution hasdegrees of freedom. The reported statistic satisfies $25.914>22.46$, the stated upper $0.001$ cutoff for six degrees of freedom. Therefore reject the Poisson model at the 0.1% significance level under the independence and common-mean assumptions. The discrepancy is especially pronounced in the zero-count class, but the test assesses the whole fitted distribution. The conclusion uses the approximate Pearson chi-squared goodness-of-fit test, not an exact finite-sample tail probability.

Assume each sampled response is truthful, card selection is independent of the student's status, and responses follow the usual independent sampling model. In randomized response, a yes answer occurs either for a user receiving the first question or for a nonuser receiving the complementary question. Its probability isConsequently $Y\sim \mathrm{Bin}(n,q)$, and the likelihood isFor $x\ne 1/2$, first maximize the binomial likelihood in the unrestricted probability $q$: its maximum occurs at $Y/n$. Inverting the affine relation givesFor an interior value $0<Y<n$, the same result follows from the scoreThe endpoint cases follow by monotonicity. This estimator has expected value $p$ and varianceThese identities follow immediately from $\mathbb{E}Y=nq$ and $\mathrm{Var}(Y)=nq(1-q)$.

The printed claim needs a parameter-space qualification. The admissible yes probabilities lie in $[\min (x,1-x),\max (x,1-x)]$, which may exclude $Y/n$. The constrained maximum-likelihood estimator isThe displayed untrimmed estimator is the maximum-likelihood estimator only when it already lies in $[0,1]$. Its unbiased estimator property does not carry over to the clipped estimator in general. For example, if $n=1$, $x=3/4$ and $p=0$, the clipped estimate equals 1 exactly when the answer is yes, so its expectation is $1/4$, not zero. At $x=1/2$, $q=1/2$ for every $p$: the parameter is unidentifiable and the printed quotient is undefined. Thus the unbiased formula requires $x\ne 1/2$, and the constrained likelihood assertion requires the stated clipping.