Practice: Testing Counts Against a Claim

1. The expected counts. They are $E_i=n{\hat{p}}_i$, where ${\hat{p}}_i$ is the Poisson probability of category $i$ evaluated at the estimated mean, and $n=120$. The final category is "5 or more", so its probability is $1$ minus the sum of the others rather than a single Poisson term. A tail category, not a point mass at 5. Before using the statistic, every expected count must be checked against the conventional minimum of about 5. The tail category is the one at risk here: with a mean near 2, $P(X\ge 5)$ is small, and $120$ times a small probability can fall below the threshold. If it does, the remedy is to merge "4" and "5 or more" into "4 or more", decided on substantive grounds and recorded, not chosen after seeing which grouping gives the preferred answer. 2. The degrees of freedom. Six categories. One constraint because the expected counts must total $120$. One more because the mean was estimated from these same counts, which pulled the expected counts toward the observed ones. So

not 5. Using 5 compares $X^2$ against a distribution that is too spread out, which makes the tail probability larger. The reported $p=0.57$ therefore overstates the agreement; the correct degrees of freedom give a smaller $p$-value on the same statistic. The direction matters: the error raises the $p$-value rather than lowering it, so the output gives no sign of the mistake and the mistake survives review. 3. The conclusion. "Admissions are Poisson" is not supported. A goodness-of-fit test accumulates evidence only against the null, so failing to reject reports compatibility, not truth. Several distributions, a negative binomial with modest overdispersion, for one, would also be compatible with 120 shifts of counts like these. The supported statement is: these counts show no evidence of departure from a Poisson distribution, at the sample size available. Worth adding that 120 observations spread over six categories gives limited power, so a large $p$-value here partly reports the study's resolution rather than the model's quality. A Poisson model also implies variance equal to mean, which can be checked directly and is a sharper diagnostic than the omnibus test. 4. The second table. The chi-square test is a poor choice. Two expected counts, $4.8$ and $5.2$, sit right at the conventional threshold, and that threshold is a working convention rather than a guarantee of accuracy. Use Fisher's exact test. For a $2\times 2$ table it computes the null probability directly from the hypergeometric distribution with no approximation, and it is cheap. It should be chosen in advance, not after comparing both $p$-values. Why it matters at these counts: the approximation and the exact test can differ substantially in this range. In the unit's worked case, expected counts of $5.0$, $5.0$, $6.0$ and $6.0$, satisfying the rule, gave $p=0.0102$ by chi-square and $p=0.029973$ exactly, a factor of about three, falling either side of a 1% level. The two nulls differ: the exact test conditions on the observed margins. 5. A different question. A rank-based two-sample test, such as Wilcoxon-Mann-Whitney, does not have "the medians are equal" as its null. Its null is that a randomly chosen observation from one group is equally likely to exceed or fall below one from the other, that the two distributions are the same in stochastic ordering. It coincides with a statement about medians only under an extra assumption, typically that the two distributions differ by a shift and have the same shape. So it does not directly answer the question as posed. If the medians are the quantity of interest, estimate them with a confidence interval for the difference, and if a rank test is used, report what it tested rather than relabelling it a test of medians. The skew is a legitimate reason to avoid assuming normality; it is not a reason to treat the rank test as the same test with weaker requirements.