= 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 $\operatorname{Var}(J)=1/2$, giving the <mean allele-sharing test for affected siblings>
$$
\boxed{Z=\frac{1.09-1}{\sqrt{(1/2)/200}}=1.8.}
$$
The upper 5% <standard normal> critical value is approximately $1.64$, so \b[there is evidence of excess sharing, and hence linkage, at the one-sided 5% level]. The <normal approximation> gives upper-tail <p-value> $1-\Phi(1.8)\simeq0.0359$.
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.
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