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A-Level Biology – Statistics and Practical Skills

Chi-squared, t-test, standard deviation, planning and evaluation.

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12 questions  ·  33 marks  ·  answers included  ·  A-Level Biology (Edexcel / AQA / OCR)
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This worksheet covers:

  • This worksheet covers standard deviation
  • The t-test
  • The chi-squared test
  • Practical planning and evaluation skills required for A-level biology investigations

A calculator is required.

Questions

Section A – Descriptive Statistics

Skills
1

The following are lengths (mm) of five leaves sampled from a plant: $42, 45, 41, 47, 45$. Calculate the mean length.

Show answer
Mean $=\dfrac{42+45+41+47+45}{5}=\dfrac{220}{5}=44\,\text{mm}$.
[2]
2

Calculate the standard deviation of the leaf lengths in the previous question, giving your answer to 2 decimal places. (Use the sample standard deviation, dividing by $n-1$.)

Show answer
Deviations from the mean (44): $-2, 1, -3, 3, 1$. Squared deviations: $4, 1, 9, 9, 1$, summing to $24$. Variance $=\dfrac{24}{5-1}=6$. Standard deviation $=\sqrt{6}=2.45\,\text{mm}$ (2 d.p.).
[4]
3

State one advantage of using standard deviation, rather than range, to describe the spread of a data set.

Show answer
Standard deviation takes into account every value in the data set, not just the two extremes, so it gives a more representative measure of spread and is less affected by a single outlier.
[1]
4

A student measures the resting heart rate of 8 volunteers before and after a 10-minute exercise session. Explain why this is an example of paired (repeated-measures) data.

Show answer
Each volunteer’s heart rate is measured twice, once under each condition (before and after), so each individual acts as their own control; the two sets of data are linked/matched by subject, rather than coming from two separate independent groups.
[2]

Section B – Hypothesis Testing

Exam-style
5

A student investigates whether there is a significant difference in the mean leaf width between plants grown in shade and in sunlight. State a suitable null hypothesis for this investigation.

Show answer
$H_0$: There is no significant difference in the mean leaf width between plants grown in shade and plants grown in sunlight; any difference observed is due to chance.
[2]
6

Explain when a t-test should be used to analyse data, rather than a chi-squared test.

Show answer
A t-test is used to compare the means of two sets of continuous (quantitative) data, typically when the data are normally distributed, e.g. comparing two sample means such as leaf length or heart rate. A chi-squared test is used instead for categorical (discrete) data, to test whether observed frequencies differ significantly from expected frequencies, e.g. in a genetic cross or a species distribution investigation.
[3]
7

In a genetics investigation, a dihybrid cross was expected to produce offspring in a $9:3:3:1$ ratio. From 160 offspring, the observed numbers were $96, 26, 30, 8$ for the four phenotypes. Calculate the expected number of offspring for each phenotype.

Show answer
Expected numbers: $\frac{9}{16}\times160=90$, $\frac{3}{16}\times160=30$, $\frac{3}{16}\times160=30$, $\frac{1}{16}\times160=10$.
[3]
8

Using your answers to the previous question, calculate the chi-squared ($\chi^2$) value for this data. $\chi^2=\sum\dfrac{(O-E)^2}{E}$

Show answer
$\chi^2=\dfrac{(96-90)^2}{90}+\dfrac{(26-30)^2}{30}+\dfrac{(30-30)^2}{30}+\dfrac{(8-10)^2}{10}=0.40+0.53+0+0.40=1.33$ (3 s.f.).
[4]

Section C – Extended / Challenge: Planning and Evaluation

9

The critical value of $\chi^2$ at the 5% significance level for 3 degrees of freedom is $7.82$. Using your calculated $\chi^2$ value from Section B, state whether the null hypothesis (that the offspring ratio fits $9:3:3:1$) should be accepted or rejected, and explain your reasoning.

Show answer
Since the calculated $\chi^2$ value ($1.33$) is less than the critical value ($7.82$), the null hypothesis is accepted. There is no significant difference between the observed and expected results, so the deviation is likely due to chance and the data support a $9:3:3:1$ ratio.
[3]
10

A t-test comparing the mean stomatal density on the upper and lower leaf surfaces of a plant gave a calculated $t$ value of $3.85$, with a critical value of $2.10$ at the 5% significance level (degrees of freedom $=18$). State the conclusion that should be drawn, with reference to the null hypothesis.

Show answer
Since the calculated $t$ value ($3.85$) is greater than the critical value ($2.10$), the null hypothesis is rejected. There is a significant difference between the mean stomatal density on the upper and lower leaf surfaces at the 5% significance level, so the difference is unlikely to be due to chance.
[3]
11

Explain why it is important to use a large sample size when investigating the effect of an environmental factor (e.g. light intensity) on a biological variable (e.g. leaf size).

Show answer
A larger sample size reduces the effect of individual variation and anomalous results on the calculated mean, making it more representative of the true population and reducing the influence of chance. This increases the reliability of any conclusions drawn and makes a statistical test more likely to detect a real significant difference or relationship if one exists.
[3]
12

A student investigating the effect of temperature on enzyme activity records only two repeats at each temperature. Suggest one improvement to the investigation’s reliability, and explain how this would improve the data obtained.

Show answer
Increase the number of repeats at each temperature (e.g. to 5 or more) and calculate a mean. This would reduce the effect of anomalous results and random error on the mean value at each temperature, allow anomalies to be identified and excluded, and allow a standard deviation to be calculated, making the data more reliable and any pattern easier to identify.
[3]

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