Library Biology 6 (IAL) WBI16 Chi-squared: Observed vs Expected
A2 Level · Biology 6 (IAL) WBI16

Chi-squared: Observed vs Expected

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Edexcel IAL Biology  •  Unit 6: Practical Skills in Biology II

Chi-squared: Observed vs Expected

🎲 Big idea: Chi-squared asks whether the counts you got are far enough from the counts you predicted to rule out chance. A 9:3:3:1 cross never comes out exactly 9:3:3:1 — the question is whether the deviation is the kind you would expect from sampling, or something the hypothesis cannot explain.

Summary — What This Topic Covers

  • When chi-squared is the right test — and why it needs raw counts
  • Working out expected values from a stated hypothesis
  • The formula, and setting the calculation out in a table
  • Degrees of freedom as categories − 1
  • Writing the conclusion, and what rejecting H₀ actually tells you

1. When to Use It

Use chi-squared when
you have counts (frequencies) in discrete categories, and a hypothesis that predicts what those counts should be.
✓ offspring phenotypes from a genetic cross vs a 9:3:3:1 prediction ✓ woodlice in damp vs dry halves of a choice chamber vs 50:50 ✓ numbers of a species in three habitat types vs an even distribution ✗ mean shell height on two shores → t-test ✗ is height related to mass? → Spearman's rank ✗ percentages, means or rates → chi-squared cannot be used
The absolute requirement
Chi-squared works on raw counts. 60 % could be 6 out of 10 or 600 out of 1000 — utterly different χ² values. If you recorded percentages, the data cannot be tested and cannot be recovered.
The other condition
No expected value should be less than 5. If one is, categories must be combined or a larger sample taken.

2. Working Out the Expected Values

Expected values come from the hypothesis, applied to your actual total.

A dihybrid cross predicts 9 : 3 : 3 : 1. Total offspring = 160. total ratio parts = 9 + 3 + 3 + 1 = 16 one part = 160 ÷ 16 = 10 expected: 9 × 10 = 90 3 × 10 = 30 3 × 10 = 30 1 × 10 = 10 the expected values must always ADD UP to the observed total
Easy marks
Check that your expected values sum to the same total as your observed values. If they do not, you have made an arithmetic error, and the check takes five seconds.

3. The Calculation

The formula
χ² = Σ [ (O − E)² ÷ E ]

where O is the observed count and E the expected count, summed across all categories.
category O E O − E (O − E)² (O − E)² ÷ E ────────────────────────────────────────────────────────────── round yellow 95 90 +5 25 0.278 round green 28 30 −2 4 0.133 wrinkl yellow 30 30 0 0 0.000 wrinkl green 7 10 −3 9 0.900 ────────────────────────────────────────────────────────────── totals 160 160 χ² = 1.311
What mark schemes look for
Set the working out as a table with these columns. Marks are given for the table structure and the intermediate values, not only for the final χ².

4. Degrees of Freedom and the Decision

Degrees of freedom
df = number of categories − 1

Four phenotypes gives df = 3. Two categories in a choice chamber gives df = 1.
χ² calculated = 1.311 df = 4 − 1 = 3 critical value = 7.815 at p = 0.05 1.311 < 7.815 → ACCEPT the null hypothesis NOTE the direction is OPPOSITE to the t-test: χ² BELOW critical → accept H₀ (no significant difference) χ² ABOVE critical → reject H₀ (significant difference)
The confusion examiners rely on
For the t-test, a large value means significant. For chi-squared, a large value means your observed counts are far from what the hypothesis predicted — so a large value means the hypothesis fails. The comparison direction is the same; what it means is not.

5. Writing the Conclusion

"χ² = 1.311, which is less than the critical value of 7.815 at 3 degrees of freedom and p = 0.05. The null hypothesis is accepted: there is no significant difference between the observed and expected numbers, so the results are consistent with a 9 : 3 : 3 : 1 ratio and the genes are inherited independently."
  • Quote χ², df and the critical value
  • State the decision about H₀ in words
  • Say what it means biologically — that is the final mark
  • Accepting H₀ means the data is consistent with the hypothesis, not that the hypothesis is proved

Practice Questions

Practice Question 1

A student counts 45 woodlice in the damp half of a choice chamber and 15 in the dry half. Calculate chi-squared and state the conclusion. The critical value at df = 1, p = 0.05 is 3.841.

Practice Question 2

Explain why a chi-squared test cannot be carried out on data recorded as percentages.

Practice Question 3

A cross gives 152 offspring. The hypothesis predicts a 3:1 ratio. Calculate the expected values and explain how you would check them.

Practice Question 4

A chi-squared test gives a value above the critical value. Explain what this means and what it does not mean.

What to Memorise

χ² = counts in categories RAW COUNTS ONLY — never percentages No expected value below 5 Expected must sum to the observed total χ² = Σ (O − E)² ÷ E df = categories − 1 χ² BELOW critical → accept H₀ χ² ABOVE critical → reject H₀ Set the working out as a table

Concepts Checklist

Exam Tips

What mark schemes look for
The full working table. Marks are given for the intermediate columns, so a bare final value of χ² throws most of them away.
The trap
Reversing the decision. For chi-squared, a value above the critical value means reject H₀ — the opposite of what students often assume from the t-test.
Easy marks
Check that expected values sum to the observed total. It catches arithmetic errors before they propagate through the whole table.
Worth remembering
df is categories minus one — not n minus one. Four phenotypes gives 3, not 159.
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