S6 - Scatter graphs and correlation

S6 - Scatter graphs and correlation

Plot paired numerical data, describe the direction and strength of correlation, and draw an estimated line of best fit. Use it for predictions while distinguishing interpolation from extrapolation and association from causation.

Plotting paired numerical data

Can a shallow scatter-graph trend be stronger than a steep one? Yes: steepness describes a rate, while strength describes how closely points follow a pattern. We will separate those ideas before trusting any prediction.

You already know

Coordinates are ordered pairs: horizontal first, vertical second. Read each axis scale and label both the quantity and its units.

A teacher records practice time and quiz score for each pupil. The two measurements from one pupil must stay paired. Sorting each column separately would invent relationships that were never observed.

Practice time, xx (minutes)Quiz score, yy (out of 2020)
101033
202077
303088
40401212
50501313
60601717

Bivariate data and scatter graph

Bivariate data record two variables for each individual or object. A scatter graph plots each paired observation as one point on two numerical axes. For example, each pupil contributes one practice-time-and-score point. A two-way table of favourite colours and school houses is also bivariate, but those categorical labels do not make a numerical scatter graph.

Plot small crosses or clear points. Do not join successive points: these are different pupils, not one quantity changing through time. Two pupils may share the same practice time but have different scores, so two points can lie above the same horizontal position.

Worked example 1: plotting one pupil

Question. Plot the pupil who practised for 2020 minutes and scored 77. Non-calculator.

Plan. Use practice time horizontally and score vertically.

x=20horizontal value in minutesy=7vertical scorepoint=(20,7)keep this pupil’s pair\begin{aligned} x&=20 &&\text{horizontal value in minutes}\\ y&=7 &&\text{vertical score}\\ \text{point}&=(20,7) &&\text{keep this pupil's pair} \end{aligned}

Why this step? Reversing the order would put a score on the time axis and a time on the score axis.

Answer. Mark a cross at (20,7)(20,7).

Check. Its horizontal projection reaches 2020 minutes and its vertical projection reaches 77 marks.

Change only the pupil selected from the table. The axis roles stay fixed.

Worked example 2: another observation

Question. Plot the pupil who practised for 4040 minutes and scored 1212. Non-calculator.

Plan. Read both values from the same row and locate their intersection.

x=40same horizontal variabley=12matching scorepoint=(40,12)one cross for the pair\begin{aligned} x&=40 &&\text{same horizontal variable}\\ y&=12 &&\text{matching score}\\ \text{point}&=(40,12) &&\text{one cross for the pair} \end{aligned}

Why this step? Each row supplies one point. Combining the time from one row with the score from another creates false data.

Answer. Mark a cross at (40,12)(40,12).

Check. The point is to the right of and above (20,7)(20,7), agreeing with both larger values.

[DIAGRAM: asset_name: Practice time and quiz scores; asset_slug: aqa8300_s05_scatter; caption: Each cross represents one pupil's paired measurements, without joining the crosses; description: A scatter graph with practice time from 0 to 70 minutes horizontally and score from 0 to 20 vertically plots 10,3 then 20,7 then 30,8 then 40,12 then 50,13 then 60,17. The points 20,7 and 40,12 are labelled.]
Each cross represents one pupil's paired measurements, without joining the crosses

These plotting conventions are not on the formulae sheet: learn them.

Watch out

Wrong: sort the scores and times independently before plotting. Right: keep each original pair together. A scatter graph is about the link between the two measurements on the same individual.

The next two pairs belong to two different cyclists.

Direction and strength of correlation

Read a scatter graph from left to right. If larger xx-values tend to go with larger yy-values, the association is positive. If they tend to go with smaller yy-values, it is negative. A tendency allows exceptions; it does not require every point to rise or fall in order.

Correlation

Positive correlation means that as one variable increases, the other tends to increase. Negative correlation means it tends to decrease. No correlation means no clear increasing or decreasing association is evident. Strong correlation has points close to a clear trend; weak correlation has a more widely scattered trend.

Use both direction and strength when requested: “strong positive correlation”, for example. “Going up” describes an impression but does not give the statistical term. A single unusual point may be an outlier; it does not automatically cancel the pattern among the others.

[DIAGRAM: asset_name: Types and strengths of correlation; asset_slug: aqa8300_s05_correlation; caption: Direction and closeness to a trend are separate features; description: Four labelled scatter panels show strong positive correlation, weak positive correlation, strong negative correlation and no correlation. Identical axis ranges make direction easy to compare, but the captions emphasise spread around the trend rather than steepness.]
Direction and closeness to a trend are separate features

Here is a useful surprise: converting practice time from minutes to seconds makes every horizontal number 6060 times larger and the numerical gradient 6060 times smaller. It does not weaken the relationship between the pupils' measurements. Points exactly on a rising straight line show perfect positive correlation whether that line is shallow or steep.

Key idea

Strength is closeness to a trend, not the size of the gradient. Give direction and strength as separate pieces of information.

These terms are not on the formulae sheet: learn them. With no useful trend, a straight-line prediction has little support; do not draw a line simply because a question contains a scatter graph.

Drawing an estimated line of best fit

An estimated line of best fit summarises a roughly straight trend. Draw one straight line through the middle of the point cloud, following its direction, with points reasonably balanced above and below across the range.

It need not pass through a particular data point or the origin. Two extreme observations do not automatically fix it. Do not join the crosses with a zigzag. Use the overall pattern and a ruler; small differences between sensible estimated lines are expected.

Estimated line of best fit

A line representing the overall linear trend in paired numerical data. It supports estimates, not exact outcomes for every individual. It is appropriate when a roughly straight association is visible; a pronounced curve needs a different model.

The line drawn below for the practice data has equation y=0.25x+1.5y=0.25x+1.5. An equation is useful here because earlier work on straight lines lets us check readings; drawing an estimated line does not require calculating an exact regression formula.

Worked example 3: following the middle of the cloud

Question. Describe a reasonable estimated line of best fit for the practice-time data. Non-calculator.

Plan. Follow the rising central trend, then check the balance of the points.

line at x=10: y=0.25(10)+1.5=4left-end checkline at x=60: y=0.25(60)+1.5=16.5right-end checkpoints above/below=3 above, 3 belowcheck the whole range\begin{aligned} \text{line at }x=10&:\ y=0.25(10)+1.5=4 &&\text{left-end check}\\ \text{line at }x=60&:\ y=0.25(60)+1.5=16.5 &&\text{right-end check}\\ \text{points above/below}&=3\text{ above},\ 3\text{ below} &&\text{check the whole range} \end{aligned}

Why this step? A line through the cloud should balance the deviations, rather than merely connect the first and last crosses.

Answer. A straight line near (10,4)(10,4) and (60,16.5)(60,16.5) is a reasonable estimate. Extend it across the observed range.

Check. Early, middle and late observations appear on both sides. The line is not forced through (0,0)(0,0).

Change only one observation by adding a pupil at (35,19)(35,19). That point lies well away from the existing pattern.

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