Past exam of the options I course of the University of Oxford / 2005 / O / 1

Under independence, the cell probability is the product of its row and column probabilities. Estimate these by the sample proportions. If $R_i,C_j$ are row and column totals and $N$ is the grand total, the fitted expected count isHere the row totals are $123,76,163,90$, the column totals $304,65,83$, and $N=452$. For example $E_{11}=123\cdot 304/452=82.72566\dots$, rounding to 82.73 as printed. The other entries follow identically.

Pearson's statistic isFor independent observations under the independence null hypothesis, its large-sample null distribution is approximately ${\chi }^2$ with $(4-1)(3-1)=6$ statistical degrees of freedom. The fitted row and column probabilities account for the reduction in statistical degrees of freedom. All expected counts exceed five, so the usual approximation has adequate expected cell sizes here. Reject independence when the statistic exceeds the corresponding upper-tail critical value.

There are six statistical degrees of freedom, so the supplied 1% critical value is $q_6=16.81$. Since $42.25>16.81$, reject independence at the 1% significance level. The data give strong evidence of an association between the two consumption classifications in this group of participants.