 Solution

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.