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

Section E: Mathematical Tools.

TODO: Attachment C is missing from the released inputs. Parts (a–c) and (f) use only the printed table or general definitions. Parts (d–e) include conditional numerical work and retain explicit incomplete status until the attachment can be checked.

The arithmetic mean is $\bar{x}=n^{-1}{\sum }_{i=1}^n{x}_i$, the total measured concentration divided by the number of observations. The median is the central value of the sorted observations, or an interpolation between the two central values under the chosen even-sample convention. It is the 50th percentile.

The table's “75th quartile” means the 75th percentile, or upper quartile $Q_3$, with approximately three quarters of observations at or below it. Its “25th quartile” means the 25th percentile, or lower quartile $Q_1$. There are four quarters of a distribution, rather than 75 different quartiles. Quantile interpolation rules can differ for a finite sample; their intended population interpretation is unchanged.

The population variance is ${\sigma }^2=E[(X-EX{)}^2]$, and the population standard deviation is $\sigma =\sqrt{{\sigma }^2}$. For a sample, the usual unbiased estimate of population variance isThe sample standard deviation measures scatter in the original concentration units; variance has squared units. Both are nonnegative, but the sample standard deviation itself is not generally an unbiased estimator of $\sigma$. The table's $\pm 1\sigma$ entries are interpreted as sample standard deviations for the calculations below, rather than standard errors of the mean.

Squaring each seasonal standard deviation gives the seasonal variance in squared concentration units:

No unit is printed in the held table, so a numerical concentration unit should not be invented. The markedly smaller summer variance is also a warning when assessing the common-variance assumption of ordinary ANOVA.

TODO: Attachment C is absent from the held release. Recover it to check any supplied critical-value table, significance convention or additional instructions. This leaf remains incomplete against that missing input. The printed summary statistics nevertheless allow the following conditional numerical answer, using a two-sided $5\%$ Student's t-test and independent observations with common variance.

Test $H_0:{\mu }_W={\mu }_S$ against $H_1:{\mu }_W\ne {\mu }_S$. The pooled sample variance isThe test statistic and statistical degrees of freedom areUsing the Student t-distribution directly gives a two-sided p-value $p\simeq 0.00250$ and a $5\%$ critical value $1.96076$. Thus reject equal seasonal means at $5\%$ and $1\%$; the winter mean is higher by $0.36$ concentration units. This does not establish a large physical effect or its cause.

The two variances are similar, and Welch's t-test gives essentially the same result: $t=3.02344$, $\nu \simeq 2958.36$, $p\simeq 0.00252$. Both calculations assume the counts represent independent observations. Serial dependence, repeated measurements and strong non-normality can invalidate nominal p-values; the summary table alone cannot diagnose them. The unequal means are a conclusion under these statistical assumptions, pending recovery of Attachment C.

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.

Four groups have $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ pairwise comparisons. Testing each at test size $\alpha =0.05$ without adjustment generally increases the familywise error rate, the chance of at least one type I error. If the six rejection events were independent, that chance would be $1-(1-0.05{)}^6\simeq 0.265$; shared samples make the actual comparisons dependent, so this expression is illustrative rather than exact.

A single omnibus ANOVA tests equality of all four means with one stated test size, avoiding that six-test inflation for the global question. If the aim is to identify differing pairs, follow with suitably adjusted comparisons, such as Bonferroni correction or a simultaneous pairwise method. A significant omnibus test does not automatically license six unadjusted claims. The benefit concerns control of type I error, rather than a guarantee that the model's independence and common-variance assumptions hold.