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Use the Pearson chi-squared test of homogeneity for two multinomial samples. Under equal preference distributions, estimate the common probabilities from pooled counts, giving $(1/4,1/4,1/2)$. With row totals forty and sixty, the expected table isThe Pearson chi-squared statistic for contingency tables isThere are $(2-1)(3-1)=2$ degrees of freedom, after fitting the common category probabilities. All expected counts exceed five. ThusAt five percent, do not reject the common preference distribution. These data are consistent with that hypothesis; failure to reject does not prove equality. This conclusion differs from part a because a common nonuniform distribution, especially an overall preference for chocolate, is allowed here.

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.

Under the null hypothesis the three outcome probabilities are the same in the two independent samples. The pooled row totals are $50,20,30$, and both column totals are $50$, out of a total of $100$. Thus the expected counts in each column are $25,10,15$. These exceed the usual small-count threshold for the asymptotic Pearson chi-squared test of homogeneity.

The Pearson chi-squared statistic for contingency tables isUnder the null, its asymptotic chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. Equivalently, the two independent three-category samples have four probability parameters unrestricted and two under the common-probability null. The $1\%$ upper-tail critical value is $9.21$.

Since $9.12<9.21$, do not reject the equal-effect hypothesis at the 1% level. For two degrees of freedom, the approximate p-value is $e^{-9.12/2}\simeq 0.01046$, just above $0.01$. This is a decision at the specified significance level, not evidence that the two outcome distributions have been proved identical.

The Pearson chi-squared test of homogeneity tests whether several independent categorical samples have the same category probabilities. Under the null, expected cell counts equal the product of the corresponding row and column totals divided by the total sample size. The Pearson chi-squared statistic for contingency tables $\sum (O-E{)}^2/E$ has an asymptotic chi-squared distribution with $(r-1)(c-1)$ degrees of freedom, when expected counts are sufficiently large. The sampling design differs from a Pearson chi-squared test of independence, though the statistic has the same form.