A.1.9 - Chi-squared and inherited ratios

Choosing the test and hypotheses

A chi-squared goodness-of-fit test compares counts observed in categories with the counts predicted by a specified model.

For example, a cross between two heterozygous plants, Rr × Rr, predicts 3 round : 1 wrinkled seed when R is completely dominant to r. The probabilities are $\frac{3}{4}$ round and $\frac{1}{4}$ wrinkled. Each new offspring is a new event under this model: one wrinkled seed does not force the next three to be round. Random fertilisation can therefore produce counts that differ from the expectations even if the model is suitable.

Chi-squared measures the discrepancy between the observed and expected category counts. This is the appropriate test here because the data are frequencies of phenotypes and a proposed ratio supplies expectations. It is not a comparison of mean seed masses or a test of correlation between two continuous measurements.

State the model before examining whether the sample supports it:

• The null hypothesis, $H_0$, is that offspring phenotype probabilities follow the proposed 3 round : 1 wrinkled ratio. Differences in a finite sample can arise through random sampling under this model.
• The alternative hypothesis, $H_1$, is that the offspring phenotype probabilities do not follow that 3:1 ratio.

The null does not predict identical observed and expected counts in every sample. Nor is “there is no correlation” an appropriate null for this question: it does not name the inheritance prediction being tested. The test asks whether the observed discrepancy is unusually large under the stated null model. Pearson's chi-squared support confirms the category-count application.

Expected counts and valid data

Expected counts come from the proposed probabilities and the total number counted, rather than from a preferred answer.

For an original teaching sample of 160 seeds with a proposed 3 round : 1 wrinkled ratio:

The expected counts must sum to the observed total. They can be fractional because they are model expectations, not a claim that part of an organism was counted. For 50 seeds under the same ratio, retain 37.5 and 12.5. Rounding them separately changes the model and may change the total.

Check the design before calculating:

• Independent observations: each offspring contributes once, and one recorded outcome must not determine another. Counting the same seedlings on several days does not create additional independent offspring. Independence concerns the observations; the category totals themselves are constrained to sum to $N$.
• Exclusive, complete categories: every sampled offspring fits one category, with none counted twice or deliberately omitted. Use the same phenotype definitions for observed and expected categories.
• Adequate expected frequencies: the chi-squared distribution is an approximation. Use the conservative condition that every expected count is greater than 5 for the examples here. A larger sample may be needed when a predicted category is rare. Check $E$, rather than assuming a large total or a large observed count is enough.
• Suitable sampling: collect an unbiased sample of the offspring population addressed by the hypothesis. Choosing only vigorous seedlings can change the observed phenotype proportions before the test begins.

The FSC guide supplied by Pearson identifies independent frequency data and expected counts above 5. If expectations are too small, do not apply the usual table uncritically or merge biologically different categories just to obtain a preferred result; obtain an adequate sample or use a justified alternative analysis.

Use counts in the calculation, not 75 and 25 merely because the model is 75% and 25%. Percentages conceal sample size: 15 round out of 20 and 150 out of 200 have the same proportion but provide different amounts of information. If only percentages are supplied, the sample total is needed to recover counts; rounded percentages may not recover them exactly.

Calculating the statistic

Each phenotype contributes a non-negative amount to the chi-squared statistic.

Continue with the original 160-seed sample. Suppose 126 seeds were round and 34 were wrinkled. The proposed ratio is still 3:1, so the expected counts remain 120 and 40. Both expectations exceed 5. For the significance decision, use $\alpha=0.05$ and the supplied critical value 3.841 for one degree of freedom; the next part explains that choice.

Work along each row: subtract $E$ from $O$, square the difference, divide by that row's $E$, and then add the final column.

Thus,

The signed differences sum to zero because both columns total 160. Adding those differences would hide the discrepancy; squaring prevents positive and negative departures from cancelling. Equal absolute differences need not make equal contributions: a difference of six is larger relative to an expectation of 40 than to one of 120.

Divide each squared difference by its own expected count, then sum. Do not divide by $O$, by 160, or by the sum of the expected counts. These are different calculations. The official 2023 Paper 3 mark scheme, Q6(c)(ii) assesses this sequence; its examiner report identifies division by the sample total as a common error.