 Solution

Let $p_{\mathrm{old}}$ and $p_{\mathrm{new}}$ be the underlying full-recovery probabilities. The two-sided statistical hypothesis test isUse the Pearson chi-squared test of homogeneity for the two independent groups. Under $H_0$, the pooled recovery estimate is $\hat{p}=56/100=0.56$, 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 isThe asymptotic reference distribution is a chi-squared distribution with $(2-1)(2-1)=1$ degree of freedom. Since $1.461<3.84$, 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.