Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2023 / Paper 1 / 9 / e

For a one-factor analysis of variance, let $n_j,{\bar{x}}_j,s_j$ be the four seasonal sizes, means and standard deviations. The weighted grand arithmetic mean isThis agrees with the rounded annual mean $5.07$. The between-season and within-season sums of squares areThere are $4-1=3$ between-season statistical degrees of freedom, because the four fitted seasonal means are compared subject to their weighted grand-mean constraint. There are $N-4=6882$ within-season statistical degrees of freedom, because one mean was estimated in each group. The total has $N-1=6885$, equal to their sum. HenceUnder $H_0:{\mu }_W={\mu }_{Sp}={\mu }_{Su}={\mu }_F$, compare with the F-distribution on $(3,6882)$ statistical degrees of freedom. The $5\%$ critical value is about $2.60620$, with p-value approximately $8.9\times 10^{-63}$ from the rounded summaries. Reject equality of all four means very strongly under the stated model. The omnibus hypothesis test establishes that at least one mean differs; it does not identify every pairwise difference.

The summer variance is substantially lower than the other three, so the exact common-variance ANOVA assumptions deserve checking, as does temporal independence. Raw observations would permit diagnostics and a heteroscedastic or time-dependent analysis. As a cross-check on arithmetic, $S{S}_B+S{S}_W=64499.774183$ implies annual sample standard deviation $3.06074$, matching the table's $3.06$. The numerical significance must not be mistaken for verification of the unobserved Attachment C or of the sampling model.