# Data analysis

> CIE A-Level Physics · Practical skills (AS)
> Source: https://www.owlsprep.com/study/cie-9702-u15-data-analysis/

This module covers core data analysis skills for CIE AS Level Physics practical exams, including plotting graphs, calculating gradients and intercepts, and linearising non-linear relationships, a heavily weighted skill in Paper 2.

**Prerequisites:** Algebraic rearrangement of linear equations; Basic understanding of experimental uncertainty

## Learning objectives

- Follow CIE marking conventions for graph plotting to secure full method marks
- Calculate gradients and intercepts of linear graphs correctly
- Rearrange non-linear experimental relationships into linear form
- Calculate absolute uncertainty in gradient and intercept from worst acceptable lines

## Graph Plotting Conventions for CIE Exams

CIE examiners follow strict marking criteria for graph plots, so following standard conventions guarantees you all method marks even if your final calculation is slightly off.

**Best-fit line** — A straight line (or curve) that passes as close as possible to all plotted points, with an approximately equal number of points scattered on either side of the line

*Example:* For 6 plotted points, 3 lie above and 3 lie below your best fit line

- Label both axes with the quantity and unit, e.g. $I \text{ (A)}$ not just $I$
- Choose a scale that uses at least half the grid in both $x$ and $y$ directions; uneven scales are allowed
- Plot each point with a sharp cross or circled dot; avoid thick blurry points
- Draw error bars that span the full range of the absolute uncertainty for each point

**Worked example:** A student collects current-potential difference data: (0.2 A, 1.1 V), (0.4 A, 2.1 V), (0.6 A, 3.2 V), (0.8 A, 4.0 V). The graph grid available is 10 × 10 squares. Propose an appropriate scale for the y-axis (V) that meets CIE requirements.

1. Calculate the full range of y-values from the data:
2. $$V_{max} - V_{min} = 4.0 - 1.1 = 2.9 \text{ V}$$
3. Divide the range by the number of squares available to get the value per square:
4. $$2.9 \text{ V} / 10 = 0.29 \text{ V per square}$$
5. Round to a simple, easy-to-use value: 0.3 V per square, which uses the full grid and meets CIE requirements. A 0.5 V per square scale would only use 6 of 10 squares, which is rejected.

> **tip**
>
> You do not need to start your axis at (0,0) if your data is all far from zero. Just clearly mark the axis break to avoid losing marks.

## Gradient and Intercept Calculation

Gradient and intercept are the most commonly required calculations in CIE data analysis, and there is a specific method you must follow to get full marks.

**Gradient of a line** — The rate of change of the y-variable with respect to the x-variable, equal to the change in y divided by the change in x between two points on the best fit line

*Notation:* $m = \frac{\Delta y}{\Delta x}$

*Example:* For $y = mx + c$, $m$ is the gradient

> **tip**
>
> Always use two points on your best fit line that are as far apart as possible. Never use raw data points unless they lie exactly on the line. This reduces percentage uncertainty in your gradient.

**Worked example:** A linear graph of $I$ (y-axis, units A) against $V$ (x-axis, units V) has a best fit line passing through two widely separated points on the line: $(1.2, 0.30)$ and $(4.8, 1.26)$. Calculate the gradient with correct units.

1. Calculate the change in y ($\Delta I$) and change in x ($\Delta V$):
2. $$\Delta I = 1.26 - 0.30 = 0.96 \text{ A}$$
3. $$\Delta V = 4.8 - 1.2 = 3.6 \text{ V}$$
4. Compute gradient as $\frac{\Delta y}{\Delta x}$:
5. $$m = \frac{\Delta I}{\Delta V} = \frac{0.96}{3.6} = 0.27 \text{ A V}^{-1}$$
6. Round the answer to 2-3 significant figures, matching the precision of the raw data.

## Linearising Non-Linear Relationships

Most experimental relationships in CIE practical exams are non-linear, so you need to rearrange them into the standard linear form $y = mx + c$ to find unknown constants from gradient and intercept.

**Linear form** — Rearrangement of a non-linear equation into the form $Y = mX + c$, where $Y$ and $X$ are combinations of measured variables, and $m$ and $c$ give the unknown constants

*Example:* For $PV = k$, we write $P = k (1/V)$, so plotting $P$ against $1/V$ gives gradient $k$

**Worked example:** The relationship between pendulum period $T$ and length $l$ is given by $T = 2\pi \sqrt{\frac{l}{g}}$, where $g$ is an unknown constant. Show how you would rearrange this into linear form, and state what quantities you would plot to find $g$.

1. Square both sides of the equation to eliminate the square root:
2. $$T^2 = 4\pi^2 \frac{l}{g}$$
3. Rearrange to match $y = mx + c$, where $y$ and $x$ are the plotted variables:
4. $$T^2 = \left( \frac{4\pi^2}{g} \right) l + 0$$
5. Identify plotted variables and the relation of gradient to $g$:
6. Plot $T^2$ (y-axis) against $l$ (x-axis). The gradient $m = \frac{4\pi^2}{g}$, so rearranged $g = \frac{4\pi^2}{m}$.

> **tip**
>
> Always add units to your gradient that match your plotted variables, not the original variables from the question.

## Uncertainty in Gradient and Intercept

CIE frequently asks you to calculate the absolute uncertainty in the gradient or intercept from your graph. The standard method uses worst acceptable lines to find the maximum possible uncertainty.

**Worst acceptable line** — The steepest or shallowest possible line that passes through all the error bars of your data points, used to find the range of possible gradients/intercepts

**Worked example:** The best fit line gradient is calculated as $0.27 \text{ A V}^{-1}$. The steepest worst acceptable line has gradient $0.30 \text{ A V}^{-1}$ and the shallowest has gradient $0.24 \text{ A V}^{-1}$. Calculate the absolute uncertainty in the gradient.

1. Calculate the difference between the best fit gradient and each worst gradient:
2. $$\Delta m_{\text{steep}} = 0.30 - 0.27 = 0.03$$
3. $$\Delta m_{\text{shallow}} = 0.27 - 0.24 = 0.03$$
4. The absolute uncertainty is equal to the largest difference calculated:
5. $\Delta m = 0.03 \text{ A V}^{-1}$, so the final gradient is written as $0.27 \pm 0.03 \text{ A V}^{-1}$.

## Common pitfalls

- **Wrong:** Using raw data points to calculate gradient instead of points on the best fit line
  - Why it fails: Raw data points have random error and often do not lie on the best fit line, leading to an incorrect gradient
  - Correct: Always use two points that lie exactly on your best fit line, positioned as far apart as possible
- **Wrong:** Choosing a scale that uses less than half the graph grid
  - Why it fails: CIE explicitly awards a mark for using most of the grid, and small scales increase uncertainty in gradient calculations
  - Correct: Calculate the range of your data and choose a scale that fills at least half the grid in both axes
- **Wrong:** Using original variables instead of plotted variables to calculate unknown constants
  - Why it fails: After linearisation, the gradient relates to the plotted (transformed) variables, so the calculation will be wrong
  - Correct: Always base your calculation of the unknown constant on the gradient of the transformed plotted variables
- **Wrong:** Forcing axes to start at 0 when all data is far from 0
  - Why it fails: This wastes grid space, forces you to use a smaller scale, and makes intercept readings less accurate
  - Correct: Use a broken axis to start near your minimum data value, clearly marking the break on the axis
- **Wrong:** Calculating gradient uncertainty as half the difference between the two worst lines
  - Why it fails: This underestimates uncertainty, as each worst line is already at the extreme of acceptable error
  - Correct: The absolute uncertainty equals the full difference between the best fit gradient and either worst line gradient

## Cheatsheet

| Step | Required Action | Key Rule |
| --- | --- | --- |
| Graph Plotting | Label axes with quantity + unit, fill ≥ half grid | Mark all error bars fully |
| Gradient Calculation | Use two far points on best fit line, calculate Δy/Δx | Never use raw data points |
| Linearisation | Rearrange to Y = mX + c, identify Y/X | Unknown = constant × gradient or constant / gradient |
| Uncertainty | Draw two worst lines through all error bars | Δm = \|m_worst - m_best\| |
| Intercept | Read directly if x=0 is plotted, else calculate from m | Solve c = y - mx for any point on the line |

## What's next

Data analysis is the core of all CIE AS and A Level practical assessment, and the skills you learn here are carried forward to A Level practical work and Paper 5 planning and analysis. Mastering these marking conventions ensures you get all the easy method marks that many students lose through careless mistakes, which can make a huge difference to your overall grade. The same linearisation and gradient calculation skills are used in all practical papers, so practicing these now will pay off when you move to more advanced practical assessment at A Level.

- [Uncertainty Analysis](https://www.owlsprep.com/study/cie-9702-u15-uncertainty-analysis/)
- [Evaluation of Results](https://www.owlsprep.com/study/cie-9702-u15-evaluation-of-results/)
- [Motion in a circle](https://www.owlsprep.com/study/cie-9702-u16-overview/)

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