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Model the two samples as independent multinomial counts with fixed row totals $N_A=N_B=500$ and three category probabilities. The null hypothesis is $p_{Aj}=p_{Bj}=p_j$ for every category $j$. Up to combinatorial factors the likelihood is ${\prod }_{ij}{p}_{ij}^{O_{ij}}$. Without the null restriction its maximizers are $O_{ij}/N_i$; under the null, maximizing ${\sum }_j(O_{Aj}+O_{Bj})\log p_j$ subject to ${\sum }_j{p}_j=1$ givesThe observed column totals are $243,281,476$, giving fitted expected counts $(121.5,140.5,238)$ in each row.

To derive the null distribution, put $D=O_A-O_B$. Under the null its mean is zero and covariance is $1000(\mathrm{diag}p-p{p}^T)$. The multivariate central limit theorem gives, asymptotically, the standardized contrast $Z_j=D_j/\sqrt{1000p_j}$ a centered normal distribution with covarianceThis is the orthogonal projection onto a two-dimensional plane. In an orthonormal basis of that plane, the two nonzero coordinates are independent standard normal variables, so ${\sum }_j{Z}_j^2$ tends to ${\chi }_2^2$. The pooled estimates are consistent; substituting them for $p_j$ leaves this limit unchanged by Slutsky's theorem. Since $O_{Aj}-E_{Aj}=D_j/2$ and $O_{Bj}-E_{Bj}=-D_j/2$, the resulting statistic is precisely the Pearson chi-squared test of homogeneity statisticThis supplies the two degrees of freedom rather than treating all six cells as independent. Every fitted expected count is large, supporting the approximation.

For these data,It exceeds the two-degree-of-freedom 95th percentile $5.99$, but not the 99th percentile $9.21$. Hence reject equal score distributions at the 5% level, but not at the 1% level. The asymptotic p-value is $\mathrm{Pr}({\chi }_2^2\ge 7.56906)=e^{-7.56906/2}\simeq 0.02272$.

For the first sampling design, treat the fixed groups of 60 and 90 people as two independent, representative samples from the respective populations. Within each sample, assume independent individuals with common eye-colour probabilities and mutually exclusive, exhaustive categories. This gives two independent multinomial distributions. The null hypothesis is that the two populations have the same eye-colour probabilities; its estimated common probabilities are $40/150$, $70/150$ and $40/150$.

The expected counts under the null hypothesis areThe Pearson chi-squared test of homogeneity usesThere are $2(3-1)=4$ unrestricted category parameters and $3-1=2$ under the null hypothesis, so the asymptotic chi-squared distribution has two degrees of freedom. All expected counts exceed five, making the usual large-sample approximation reasonable. Reject equality at the 5% level, but not at the 1% level: $5.99<T<9.21$. The approximate p-value is $\mathbb{P}({\chi }_2^2\ge T)=e^{-25/7}\simeq 0.0281$. Independence and representative sampling, rather than merely the count sizes, are essential to population inference; the result does not establish a causal explanation.

In the second design, the total sample size is fixed but both row and column classifications are observed for each sampled person. Assume an independent, representative sample with a common joint distribution of handedness and eye colour. The six cell counts then have one multinomial distribution. The null hypothesis is independence of the two classifications, so joint probabilities factor into row and column probabilities. Their maximum-likelihood estimates are the observed marginal proportions, again giving exactly $E$ above.

The unrestricted joint distribution has five free parameters; independence leaves $(2-1)+(3-1)=3$. The Pearson chi-squared test of independence again has two degrees of freedom, the same statistic, and the same significance conclusion. The numerical analysis is unchanged, but the sampling model changes from two fixed-row samples to one joint sample. If handedness groups had instead been deliberately sampled with fixed row sizes, the first, homogeneity formulation would apply to those groups too.

Treat the counts as independent binomial distributions, $R_i\sim \mathrm{Bin}(n_i,{\pi }_i)$, conditional on the known numbers of operations. The null hypothesis is that the probabilities ${\pi }_i$ are all equal, with no assumed numerical value. Centre 4 has no operations in this period and hence contributes likelihood one, supplying no information about its mortality probability. The empty groups in a binomial homogeneity test principle leaves 11 contributing centres and 10 test degrees of freedom.

Under the common-probability model the maximum-likelihood estimator isUse the two-column contingency table of deaths and survivors, deleting its all-zero row. The Pearson chi-squared test of homogeneity has statisticUnder the null, the large-sample reference distribution is ${\chi }_{10}^2$. The expected counts are adequate for this approximation here, including at the smallest nonempty centre. Calculating givesAn alternative is the likelihood-ratio test against separate centre probabilities, with binomial devianceThe unrestricted probabilities are $r_i/n_i$; subtracting one fitted common probability leaves the same 10 degrees of freedom. Here $D=39.5033$ and $p\approx 2.07\times 10^{-5}$. Thus there is strong evidence against a common mortality probability. This omnibus test does not identify Centre 1 as the sole source of the heterogeneity. If its comparison with the remaining centres is the primary scientific question, specify a separate planned one-versus-rest comparison; inspection of many pairwise comparisons calls for consideration of multiple testing.

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.

Treat the four fixed row totals as four independent samples from binomial distributions. Under independence, each has the same unknown probability $p$ of a third-child boy. The pooled maximum-likelihood estimator is $\hat{p}=100/200=1/2$, so each expected cell count is $25$. The Pearson chi-squared test of homogeneity statistic isThere are four independently varying row probabilities under the alternative, and one estimated common probability under the null; the degrees of freedom are therefore $4-1=3$. Equivalently this is the $(4-1)(2-1)$ contingency-table count. The fact that the pooled estimate happens to equal one half does not make it a specified null probability and does not change the degrees of freedom to four.

Since $10.08>7.8147$, reject independence at the 5% level using the stated chi-squared calibration. All expected counts are sufficiently large for the usual asymptotic test.

Let $S_i$ be singleton deliveries among $C_i$ treatment cycles. Under a common-probability binomial distribution model, $S_i\sim \mathrm{Bin}(C_i,p)$ independently across years. The pooled maximum-likelihood estimator isThe Pearson chi-squared test of homogeneity compares singleton and non-singleton counts across all fourteen years, usingAll expected cells are large. One probability has been estimated, so the reference chi-squared distribution has 13 degrees of freedom. Numerically,The constant singleton-per-cycle probability is decisively rejected under the independent-binomial model. The observed proportion rises from $1712/18201=0.0941$ to $3626/23794=0.1524$. Dependence between repeated treatments or patient heterogeneity can invalidate literal binomial precision, but they should be addressed by an explicit variance model rather than ignored.

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.

Test homogeneity of the four category probabilities among the three samples. The null hypothesis is $p_{iA}=p_{iB}=p_{iC}$ for every category $i$; the alternative allows at least one difference. This is a Pearson chi-squared test of homogeneity, assuming independently sampled grains with consistent classification.

The row totals are $(195,52,64,16)$, the sample totals $(131,135,61)$ and the grand total $N=327$. Under the null hypothesis, $E_{ij}=O_{i+}{O}_{+j}/N$, giving the expected value tableFor example, $E_{\mathrm{undulatory},C}=52(61)/327=9.7003$, so that cell contributes $(4-9.7003{)}^2/9.7003=3.34974$. The complete cell contributions areso $X^2=6.87754$. There are $(4-1)(3-1)=6$ statistical degrees of freedom. The supplied $5\%$ critical value is $12.592$, and $6.87754<12.592$, so do not reject homogeneity at $\alpha =0.05$. The approximate upper-tail p-value is $0.3323$.

One of twelve expected cells is below five, with none below one; the asymptotic approximation is broadly defensible by a common expected-count guideline but is not exact. An exact/conditional simulation could check it, especially if grain counts are clustered. Do not pool categories solely to obtain a desired result. The conclusion is that these counts do not demonstrate different category proportions; they do not prove that the rocks share one geological formation. Different formations can share such proportions, and real grains within one rock can fail the independent-sampling assumption.

For observed counts $O_{ij}$ and positive expected counts $E_{ij}$ in a contingency table, the statistic is $X^2={\sum }_{ij}(O_{ij}-E_{ij}{)}^2/E_{ij}$. Under common category probabilities or statistical independence, estimate $E_{ij}$ by the product of the row and column totals divided by the grand total. With large expected counts, this gives the test statistic for the Pearson chi-squared test of homogeneity or the Pearson chi-squared test of independence, with asymptotic chi-squared distribution on $(r-1)(c-1)$ degrees of freedom.