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6.2.2Inquiry Process

Processing Data in Biology — IB Biology HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 157 of 160

📚 IB Biology HL Hub / Topical Revision
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Part 1 of 5Steps in Processing

6.2.2 Processing Data in Biology
1
Calculate a mean for each set of repeats, and a measure of spread such as the standard deviation or range.
2
Convert to rates or percentages where this allows fair comparison, for example volume per minute or percentage change.
3
State the uncertainties and propagate them into any calculated quantity.
4
Plot a suitable graph with axis labels, units, a trend and error bars.
5
Carry out an appropriate statistical test to decide whether any difference or correlation is significant.

Part 2 of 5Choosing and Using Tests

t-test
  • Compares the means of two data sets; the degrees of freedom are n₁ + n₂ − 2,
    • and a calculated t above the critical value at p = 0.05 shows a significant difference.
Chi-squared
  • Compares observed and expected frequencies of categories and uses degrees of freedom of categories minus one.
Correlation coefficient
  • Measures how strongly two variables are associated, with values from −1 to +1.
Null hypothesis
  • The test is set up to decide whether the null hypothesis should be rejected, using p < 0.05 as the usual threshold.
QuestionTest
Do two means differ?t-test
Do counts fit an expected ratio?Chi-squared
Are two variables related?Correlation

Part 3 of 5Interpreting the Results

Interpreting the Results
Pattern
  • Describe the overall trend in the processed data, quoting values, before explaining it.
Significance
  • A significant result means the difference is unlikely to be due to chance, but does not show the size or importance of the effect.
Error bars
  • Overlapping error bars suggest a difference may not be real, even when the means differ.
Link to theory
  • The pattern is explained with biological knowledge and compared with the hypothesis and the literature.

Part 4 of 5Worked Example

Question

  • Three repeats of gas production in 5 minutes are 12, 15 and 13 cm³.
  • Calculate the mean rate.
  • Then a t-test comparing two groups of 5 gives t = 3.1: is it significant at p = 0.05 (critical value 2.31)?

  1. Mean volume = (12 + 15 + 13) ÷ 3 = 13.33 cm³.
  2. Rate = 13.33 ÷ 5 min = 2.67 cm³ min⁻¹.
  3. Degrees of freedom = 5 + 5 − 2 = 8.
  4. Calculated t = 3.1 is greater than the critical value 2.31.

Answer

  • The mean rate is 2.7 cm³ min⁻¹ (2 s.f.);
    • because 3.1 > 2.31, the difference is significant and the null hypothesis is rejected.

Part 5 of 5Exam Checklist & Tips

6.2.2 – Key Points for the Exam

  • Standard deviation A measure of the spread of data around the mean.
  • t-test A test comparing the means of two data sets.
  • Significant Unlikely to have arisen by chance, usually at p < 0.05.
  • Degrees of freedom A value derived from the sample size used to find the critical value.

Common Mistakes

  • Calculating the mean before checking and flagging anomalies.
  • Assuming a significant difference proves the hypothesis is explained.
  • Using an incorrect number of degrees of freedom to read the critical value.
  • Omitting error bars when repeats were taken.

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