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Check your answer after each question. Use a calculator where needed.
Inheritance · ecology · comparing treatments
Statistical tests and conclusions
Choose chi-squared, Spearman’s rank or a t-test for suitable data. Interpret supplied critical values, null hypotheses and statistical significance.
Method and worked examples ↓Learn the method
How to approach these questions
- Chi-squared can compare observed category counts with counts expected under a hypothesis.
- Spearman’s rank tests for a monotonic association between two paired variables.
- An unpaired t-test compares the means of two independent samples when its assumptions are met.
- If the test meets the stated rejection rule, reject the null hypothesis. Otherwise, do not reject it. A non-significant result does not prove the null hypothesis.
Worked examples
A chi-squared statistic is 6.2 and the supplied 5% critical value is 5.99. The rejection rule is statistic ≥ critical value. What follows?
6.2 ≥ 5.99, so reject the null hypothesis at the 5% level. There is evidence that the observed category frequencies differ from the expected frequencies.
Spearman’s rₛ = −0.72. The supplied critical magnitude is 0.65. What follows?
|−0.72| = 0.72, which exceeds 0.65. There is a statistically significant negative association. This does not establish that one variable causes the other.
Keep practising
Topic map and study list · Practical data guides · Active recall decks
Course resources
AQA A-level Biology · OCR A A-level Biology · Textbook answers
Skill coverage follows the common A-level science subject content. Check your board’s current requirements: AQA 7402 and OCR A H420.