Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 4 5J Solution Created 2026-09-24 Updated 2026-10-03
With Control as the reference group, the fitted normal linear model has meanThus the control-group rate is , while the treatment-group rate is . The estimates areandweight units per day. The control slope has and , so there is no statistically significant evidence of weight loss over time in the control group at conventional significance levels.
The plotted regression residuals for each mouse form smooth time patterns rather than an unstructured cloud: nearby observations on the same mouse have similar residuals, and the mice also show persistent individual offsets. This is strong evidence of autocorrelation and omitted mouse-level variation. The ordinary linear model treats all 400 measurements as independent after accounting for cage, so it commits pseudoreplication and will generally underestimate the coefficient standard errors. The reported -tests and -values should therefore not be trusted. A random-intercept linear mixed model with a within-mouse correlation structure would respect the repeated-measures design.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 13J c Solution Created 2026-09-24 Updated 2026-09-29
The null hypothesis is that price and score are independent, equivalently that the independence log-linear model for a two-way contingency table is correct. Against the saturated alternative, the reported Poisson deviance is the likelihood-ratio test statisticUnder the null and the usual large-sample regularity assumptions, Wilks theorem givesThe observations must be independent, the cell probabilities must not lie on the boundary of the parameter space, and the expected counts must be large enough for the chi-squared asymptotic approximation. The fitted counts are either or , so the customary expected-count check is comfortably satisfied. Since(equivalently, the p-value is about ), we do not reject the null at the significance level. The data provide no significant lack of fit for independence.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 205 2 Solution Created 2026-09-28 Updated 2026-10-03
The familywise error rate isThe Bonferroni correction rejects when . Since a valid p-value is super-uniform under its null, the union bound givesNo independence assumption is needed.
For the closed testing procedure, form the intersection hypothesis for every nonempty and choose a level- local test for each . Reject exactly when every with is rejected by its local test. If any true is rejected, then the intersection of all true nulls is rejected. Since its local test has significance level ,This is the closed-testing control of the familywise error rate.
Procedure (B) is the Weighted Holm step-down procedure. Let be the first rank in the ordering whose hypothesis is a true null, and put . If any true null is rejected, the procedure reaches step andbecause every true null remains among ranks . Henceand another union bound gives FWER at most .
At step , procedure (B) divides by the total weight still under consideration, which is no larger than . Its critical values therefore increase as hypotheses are rejected. Moreover, if procedure (A) would reject a hypothesis, every earlier ordered also passes the initial threshold, so procedure (B) reaches and rejects it. Thus (B) contains every rejection of (A) and can make strictly more rejections while retaining the same strong FWER control.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 13J ii Solution Created 2026-09-24 Updated 2026-09-25
None of the displayed one-coefficient Wald tests for
team1_form rejects at the significance level. That does not establish that formation has no effect: the relevant null hypothesis sets all nine displayed team1_form coefficients to zero simultaneously, and several individually imprecise estimates may still be jointly significant. One should compare the fitted model with the nested model omitting team1_form by an analysis of deviance for nested generalized linear models, equivalently a likelihood-ratio test of those joint restrictions. Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 13J iv Solution Created 2026-09-24 Updated 2026-09-25
Under , the slope in this through-origin linear regression is . Its observed Wald statistic isThe corresponding two-sided p-value is about , so the fit does not give evidence against at the significance level.
The usual least-squares standard error assumes constant conditional variance. Goal counts are counts, and under the natural Poisson distribution model their variance equals their mean , so observations with different expected goals are heteroscedastic. The reported homoscedastic standard error is therefore doubtful; a heteroscedasticity-robust calculation or a suitable count model would be more credible.