AQA · A-Level · Biology
Course version: 7402 · Version from Sept 2016
Use of the chi-squared ( Χ ) test to compare the goodness of fit of…
Full specification wording
Inheritance (A-level only): Use of the chi-squared ( Χ ) test to compare the goodness of fit of observed phenotypic ratios with expected ratios.
The chi-squared test evaluates whether the difference between observed offspring numbers and those predicted by a genetic diagram is significant or due to chance.
Checked against the AQA 7402 specification (opens in a new tab) on .
- Point 7 of 7
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
The chi-squared test evaluates whether the difference between observed offspring numbers and those predicted by a genetic diagram is significant or due to chance. Begin with the null hypothesis: there is no significant difference between observed and expected results. Calculate expected numbers by applying the predicted ratio to the total offspring count. Calculate chi-squared as the sum of (O - E) squared divided by E, using actual numbers, not percentages. Degrees of freedom equal the number of phenotype classes minus one. Compare the calculated value to the critical value at p = 0.05 to determine significance.
What earns the marks
- Expected values calculated by applying the predicted ratio to the total number of offspring counted.
- Correct substitution into the sum of (O - E) squared divided by E, with the working shown.
- The degrees of freedom given as the number of phenotype classes minus one.
- Comparing the calculated value with the critical value at p = 0.05 and stating whether the difference is significant.
- A conclusion written in terms of the null hypothesis, saying whether the difference is due to chance.
Common mistakes and how to put them right
Using percentages or proportions in the calculation instead of the actual numbers of individuals.
Dividing by the observed value rather than by the expected value.
Using the total number of individuals as the degrees of freedom instead of the number of classes minus one.
Writing that the results are significant, when the mark is for the difference between observed and expected results being significant.
Concluding that a non-significant result proves the genetic diagram is correct, rather than that there is no evidence of a real difference.
Exam tips
- Lay the calculation out as a table with columns for O, E, O - E, (O - E) squared and (O - E) squared divided by E; the method marks come from that table.
- Write down the null hypothesis before you start, even if the question does not ask for it, because your conclusion has to refer back to it.
- Read the critical value from the row for your degrees of freedom and the p = 0.05 column, and quote both numbers in the conclusion.
Try the lesson
The start of the Inheritance (A-level only): Use of the chi-squared ( Χ ) test to compare the goodness of fit of observed phenotypic ratios with expected ratios. lesson, made from checked teaching notes for this specification point.
Have a go first
A chi-squared test is used to compare observed phenotypic ratios with expected ratios. What is the null hypothesis in this type of test?
Listen
The chi-squared test evaluates whether the difference between observed offspring numbers and those predicted by a genetic diagram is significant or due to chance. Begin with the null hypothesis: there is no significant difference between observed and expected results. Calculate expected numbers by applying the predicted ratio to the total offspring count. Calculate chi-squared as the sum of (O - E) squared divided by E, using actual numbers, not percentages. Degrees of freedom equal the number of phenotype classes minus one. Compare the calculated value to the critical value at p = 0.05 to determine significance. A strong answer earns marks for these points. expected values calculated by applying the predicted ratio to the total number of offspring counted. correct substitution into the sum of (O - E) squared divided by E, with the working shown. the degrees of freedom given as the number of phenotype classes minus one. comparing the calculated value with the critical value at p = 0.05 and stating whether the difference is significant. Exam tip: Lay the calculation out as a table with columns for O, E, O - E, (O - E) squared and (O - E) squared divided by E; the method marks come from that table.
Key words
Say what each word means before you turn the card.
null hypothesis
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degrees of freedom
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critical value
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statistical significance
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See every definition
- null hypothesis
- There is no significant difference between observed and expected results.
- degrees of freedom
- The number of phenotype classes minus one.
- critical value
- The value read from the row for your degrees of freedom and the p = 0.05 column, compared with the calculated value to determine significance.
- statistical significance
- Whether the difference between observed and expected results is significant, determined by comparing the calculated value with the critical value at p = 0.05.
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