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AQA A-level Mathematics (7357) · Data presentation and interpretation

Mini-Lesson

Data presentation and interpretation

2 missions. Guess, learn, then check. Your place is saved.

Mission 1Two classes both average 62%. Why might one have far more fails?
Mission 2Chocolate-loving countries win more Nobels. Should you eat more?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Two classes both average 62%. Why might one have far more fails?

Make a guess: Two classes both average 62%. Why might one have far more fails?

Data presentation · single-variable data

Histograms: area means frequency

In a histogram the area of each bar represents the frequency, so bars can have unequal widths.
Plot frequency density on the vertical axis:

frequency density = \frac{frequency}{class width}

Frequency = frequency density × class width. If a question says area is proportional to frequency, find the scale factor from one bar first.
To estimate a frequency over part of a class, take the matching fraction of that bar's area.
Box plots and cumulative frequency diagrams show the median and quartiles directly.

Calculate

?A histogram of journey times

t

minutes has a bar of frequency density 2.4 for

20 \leq t < 30

and a bar of frequency density 0.9 for

30 \leq t < 50

. Estimate the number of journeys with

25 \leq t < 40

Data presentation · single-variable data

Variance and standard deviation

The standard deviation measures spread: roughly the typical distance of a value from the mean. The variance, its square, is the mean squared deviation.
From summary statistics, with $S_{xx} = \sum (x - x ˉ)^{2} = \sum x^{2} - \frac{( \sum x ) ^{2}}{n}$ S_xx = Σ (x − x̄)² = Σ x² − ((Σ x)²)/n:

x ˉ = \frac{\sum x}{n}, σ^{2} = \frac{\sum ( x - x ˉ ) ^{2}}{n} = \frac{\sum x ^{2}}{n} - x ˉ^{2} = \frac{S _{xx}}{n}

For a frequency table use $\sum f x$ Σ fx and $\sum f x^{2}$ Σ fx² over $\sum f$ Σ f. With grouped data use midpoints: the results are estimates.
To compare distributions, compare a measure of location and a measure of spread, in context.

Worked example Data presentation · single-variable data

Standard deviation from summary statistics

Question

The masses,

x

grams, of 20 apples give

\sum x = 2400

and

\sum x^{2} = 289620

. Find the mean and standard deviation of the masses.

1
Mean: $x ˉ = \frac{2400}{20} = 120$ x̄ = 2400/20 = 120
2
Mean of the squares: $\frac{289 620}{20} = 14481$ (289 620)/20 = 14 481
3
Subtract the square of the mean: $σ^{2} = 14481 - 12 0^{2} = 14481 - 14400 = 81$ σ² = 14 481 − 120² = 14 481 − 14 400 = 81
4
Square-root: $σ = 81 = 9$ σ = √81 = 9

Answer

mean 120 g, standard deviation 9 g. (Equivalently

S_{xx} = 289620 - \frac{240 0 ^{2}}{20} = 1620

and

\frac{1620}{20} = 81

⚠️ Watch out

subtract

x ˉ^{2}

after dividing

\sum x^{2}

n

, and remember the final square root. 81 is the variance, not the standard deviation.

Quick check

?For 10 values,

\sum x = 50

and

\sum x^{2} = 310

. Find the standard deviation.

Data presentation · single-variable data

Quartiles, interpolation and outliers

$Q_{1}$ Q₁, median and $Q_{3}$ Q₃ cut ordered data into quarters. IQR $= Q_{3} - Q_{1}$ = Q₃ − Q₁.
For grouped data, estimate the median (or a percentile) by linear interpolation in its class:

$median \approx L + \frac{\frac{n}{2} - F}{f} \times w$ median ≈ L + (n/2 − F)/f × w ( $L$ L lower boundary, $F$ F cumulative frequency below, $f$ f class frequency, $w$ w width.)

A possible outlier lies more than $1.5 \times$ 1.5 × IQR beyond the nearer quartile, or more than 2 standard deviations from the mean. Use the rule the question gives.
Cleaning data: remove errors, handle missing values, and comment on genuine outliers.

Calculate

?The times of 40 runners are grouped. 14 runners took less than 30 minutes and 16 took from 30 up to 35 minutes. Use linear interpolation to estimate the median time. Give your answer to 2 decimal places.

minutes

Sort it

?A data set has

Q_{1} = 24

and

Q_{3} = 36

. An outlier is a value more than

1.5 \times

IQR beyond the nearer quartile. Classify each value.

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Two classes both average 62%. Why might one have far more fails?

Class P scored 55, 58, 60, 62, 62, 64, 66, 69 and class Q scored 38, 45, 52, 62, 66, 72, 78, 83. Q's standard deviation is about 14.9; P's is about 4.2.

Mission 2

Chocolate-loving countries win more Nobels. Should you eat more?

Make a guess: Chocolate-loving countries win more Nobels. Should you eat more?

Data presentation · bivariate data

Scatter diagrams and correlation

Plot the explanatory (independent) variable across and the response (dependent) variable up.
Describe correlation by direction (positive, negative or none) and strength (strong or weak): how closely the points follow a straight line.
Look for distinct sections: two groups (say, adults and children) can each show no pattern, yet together look correlated.
A point far from the pattern is a possible outlier: check it is not an error.

Sort it

?Which type of correlation would you expect between each pair of variables?

Tap a statement, then tap the group it belongs in. Then press Check.

Data presentation · bivariate data

Using a regression line

A regression line of $y$ y on $x$ x, $y = a + b x$ y = a + bx, is used to predict $y$ y from a given $x$ x.
Gradient $b$ b: the change in $y$ y for each increase of 1 in $x$ x, in context and with units.
Intercept $a$ a: the predicted $y$ y when $x = 0$ x = 0; often meaningless if $x = 0$ x = 0 is outside the data.
Interpolation (inside the range of $x$ x) is fairly reliable. Extrapolation (outside it) is not.

Worked example Data presentation · bivariate data

Interpreting a regression line

Question

For 30 students, revision time

h

hours (from 2 to 12) and test mark

m

(out of 100) give the regression line

m = 28 + 4.5 h

. Interpret 4.5 and 28, and comment on the predictions for

h = 8

and

h = 30

1
Gradient: each extra hour of revision is linked with about 4.5 more marks.
2
Intercept: a student doing no revision is predicted to score 28, but $h = 0$ h = 0 is outside the data, so treat this with caution.
3
Interpolate: $m = 28 + 4.5 \times 8 = 64$ m = 28 + 4.5 × 8 = 64, reliable because 8 is within 2 to 12.
4
Extrapolate: $m = 28 + 4.5 \times 30 = 163$ m = 28 + 4.5 × 30 = 163, which is impossible for a mark out of 100.

Answer

about 4.5 marks per extra hour; 28 marks predicted with no revision (unreliable); 64 is a sensible prediction, 163 is not.

⚠️ Watch out

interpret in context with units ("marks per hour"), not just "the gradient is 4.5".

Calculate

?The regression line of daily ice lolly sales

s

on maximum temperature

t

°C is

s = 2.5 t - 18

, from data with

14 \leq t \leq 29

. Estimate the number of lollies sold on a day with a maximum temperature of 22 °C.

Calculate

?Using the same line

s = 2.5 t - 18

, estimate how many more lollies are sold on a 26 °C day than on a 20 °C day.

Data presentation · bivariate data

Correlation is not causation

Two variables can be correlated without one causing the other.
A third (confounding) variable may drive both. Ice cream sales and sea rescues are positively correlated: hot weather increases both.
The correlation may be coincidence, especially with few data points or many variables compared.
Data about groups (countries, schools) may not describe individuals within them.
To claim cause you need a controlled experiment or a clear mechanism, not just a scatter diagram.

Sort it

?The line

s = 2.5 t - 18

came from data with

14 \leq t \leq 29

. Is each conclusion valid?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Chocolate-loving countries win more Nobels. Should you eat more?

Recap

Big ideas

Histogram bars: area represents frequency, with frequency density = frequency ÷ class width.
$σ^{2} = \frac{\sum x ^{2}}{n} - x ˉ^{2}$ σ² = (Σ x²)/n − x̄²; take the square root for the standard deviation. Compare data sets with a measure of location and a measure of spread, in context.
Interpolate grouped data at position $\frac{n}{2}$ n/2; outliers lie beyond $1.5 \times$ 1.5 × IQR from the nearer quartile (or a stated number of standard deviations from the mean).
Describe correlation by direction and strength; watch for distinct sections and outliers.
Interpret the regression gradient and intercept in context; interpolate, don't extrapolate.
Correlation does not imply causation: look for a third variable.

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Mission 1Two classes both average 62%. Why might one have far more fails?
Mission 2Chocolate-loving countries win more Nobels. Should you eat more?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (3)

The specification points and sources behind this lesson.

1 AQA A-level Mathematics 7357 specification v1.3, 3.13 L1, L3, L4
2 AQA A-level Mathematics 7357 specification v1.3, 3.13 L2, L4
3 Messerli F. H. (2012) "Chocolate consumption, cognitive function, and Nobel laureates", New England Journal of Medicine 367:1562–1564, doi:10.1056/NEJMon1211064, PMID 23050509 (primary source); figures restated in Kayser M. (2012) "Editors' pick: Christmas is coming - time for chocolate to get ready for your Nobel Prize", Investigative Genetics 3:26, https://pmc.ncbi.nlm.nih.gov/articles/PMC3544694