= Solution
Let $p_{\rm old}$ and $p_{\rm new}$ be the underlying full-recovery probabilities. The two-sided <statistical hypothesis test> is
$$
H_0:p_{\rm old}=p_{\rm new},\qquad
H_1:p_{\rm old}\ne p_{\rm new}.
$$
Use the <Pearson chi-squared test of homogeneity> for the two independent groups. Under $H_0$, the pooled recovery estimate is $\widehat p=56/100=0.56$, so each row has expected counts 28 recovered and 22 not recovered. All expected counts comfortably exceed five. The <Pearson chi-squared statistic for contingency tables> is
$$
X^2=2\left(\frac{3^2}{28}+\frac{3^2}{22}\right)
=\frac{225}{154}\approx1.461.
$$
The asymptotic reference distribution is a <chi-squared distribution> with $(2-1)(2-1)=1$ degree of freedom. Since $1.461<3.84$, \b[do not reject equal recovery probabilities at the 5% level]. The observed recovery proportions, 0.50 and 0.62, do not provide statistically significant evidence of a difference in this trial. This conclusion does not prove that the procedures have identical recovery probabilities.
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