Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 6 / d / Solution

Use the prespecified excess-sharing alternative: affected full siblings linked to a disease genetic locus are expected to share more marker copies identical by descent. The total sharing count is $0(42)+1(98)+2(60)=218$, so $\bar{J}=218/200=1.09$. Under the no-linkage null, independent sibling pairs have $E[J]=1$ and $\mathrm{Var}(J)=1/2$, giving the mean allele-sharing test for affected siblingsThe upper 5% standard normal critical value is approximately $1.64$, so there is evidence of excess sharing, and hence linkage, at the one-sided 5% level. The normal approximation gives upper-tail p-value $1-\Phi (1.8)\simeq 0.0359$.

The conclusion depends on the intended directional test. A two-sided test of the sharing mean uses $1.96$, so it would not reject at 5%. An omnibus Pearson chi-squared goodness-of-fit test of the three counts against $(50,100,50)$ gives $64/50+4/100+100/50=3.32$ with two degrees of freedom and also would not reject. The directional mean-sharing test targets the scientifically specified excess-sharing alternative; these different tests should not be selected after seeing which rejects. Independence of the 200 pairs is assumed: if pairs overlap within families, the standard error needs adjustment.