= Solution
The two independent <binomial distributions> give fitted event probabilities $24/99$ and $12/100$. Their <odds ratio> and <log odds ratio> are
$$
\widehat{\mathrm{OR}}=\frac{24\cdot88}{75\cdot12}=2.34667,
\qquad \widehat\delta=\log\widehat{\mathrm{OR}}=0.852996.
$$
Using the <log odds ratio variance from a two-by-two table>,
$$
\widehat{\operatorname{Var}}(\widehat\delta)
=\frac1{24}+\frac1{75}+\frac1{12}+\frac1{88}
=0.149697,\qquad \operatorname{SE}(\widehat\delta)=0.386907.
$$
The stated <normal approximation> and quantile approximately two give the <confidence interval>
$$
\boxed{\delta\in0.852996\pm2(0.386907)=(0.07918,1.62681).}
$$
Exponentiation gives the corresponding <odds ratio> interval $(1.0824,5.0876)$. Its lower endpoint exceeds one.
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