CCEA A-Level Biology 1010
Biological Molecules
Enzymes
Viruses and Cell Structure
Cell Physiology and Transport
Continuity of Cells
Tissues and Organs
Exchange Surfaces and Gas Exchange
Transport in Plants
The Circulatory System in Mammals
Adaptation and Biodiversity
Human Impact on Biodiversity
Homeostasis and Excretion
Immunity
Co-Ordination and Control
Populations and Communities
Ecological Energetics and Nutrient Cycling
Respiration
Photosynthesis
DNA as the Genetic Code
Gene Technology
Genes and Patterns of Inheritance
Population Genetics and Evolution
Evolutionary Trends in Plants
Adaptation and Biodiversity
Handling data: averages, spread and the t-test
- 3 things to know
- Unit AS 2
Overview
Mean, median and mode, standard deviation, standard error and 95% confidence limits, a worked t-test on stomatal counts, correlation and causation, presenting tables and graphs, and writing a bibliography.
What you need to know
Answer each one before you reveal it.
How are 95% confidence limits interpreted when two means are compared?
What are the median and the mode?
Questions from this subtopic
Each one with the model answer underneath.
- 1
What are the mean, median and mode of a data set?
Model answer
The mean is the sum of the values divided by their number, the median is the middle value when they are ranked, and the mode is the most frequent value; the median is least affected by outliers.
- 2
Calculate the standard deviation of the stomatal counts 12, 15, 14, 10 and 9.
Model answer
The mean is 12 and the squared deviations are 0, 9, 4, 4 and 9, totalling 26. σ̂ = √(26 ÷ (n − 1)) = √(26 ÷ 4) = 2.55.
- 3
What is the standard error of the mean, and how is it calculated?
Model answer
It estimates how closely the sample mean is likely to match the true population mean, and equals the standard deviation divided by the square root of the sample size, σ̂/√n.
- 4
Calculate the standard error and 95% confidence limits for a mean of 12 with σ̂ = 2.55 and n = 5 (t = 2.78 for 4 degrees of freedom).
Model answer
SE = 2.55 ÷ √5 = 1.14. The 95% confidence limits are 12 ± 2.78 × 1.14, that is 12 ± 3.2, from 8.8 to 15.2.
- 5
How are 95% confidence limits interpreted when two means are compared?
Model answer
If the confidence intervals of two means do not overlap, the difference between them is likely to be significant; if they overlap widely, it probably is not.
- 6
When is Student's t-test used, and what are its degrees of freedom?
Model answer
To test whether the means of two independent samples of normally distributed, continuous data differ significantly; the degrees of freedom are n₁ + n₂ − 2.
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