Lesson 14: Data Presentation

Data presentation is where statistics meets honesty: the same numbers can illuminate or deceive depending on how they are shown. A-level data presentation covers histograms with unequal class widths, box plots and outliers, cumulative frequency, correlation and regression — with constant attention to correct interpretation. This lesson builds both the techniques and the critical eye: compute correctly, draw accurately, and read sceptically. Every chart in a newspaper is a claim; this lesson teaches you to check it.

Histograms — area means frequency

The histogram's golden rule: frequency is represented by AREA, not height. For equal class widths, height works; for unequal widths, plot frequency density = frequency/class width on the vertical axis. Example: journey times (min) for Leeds commuters: 0–10: 40 people; 10–20: 90; 20–40: 120; 40–60: 30. Class widths 10, 10, 20, 20 → densities 4, 9, 6, 1.5. The 20–40 bar is tallest in frequency but only density 6 — drawing it at height 120 would quadruple its visual area and lie.

Worked exam-style: estimate the mean from the histogram using midpoints: (5 × 40 + 15 × 90 + 30 × 120 + 50 × 30)/280 = (200 + 1350 + 3600 + 1500)/280 = 6650/280 ≈ 23.75 min. This assumes uniform spread within classes — state the assumption. Modal class: 10–20 (highest density, not highest frequency — with unequal widths, the modal class is the densest). Median from histogram: find the class containing the 140th value: cumulative frequencies 40, 130, 250 → median in 20–40; linear interpolation: 20 + ((140 − 130)/120) × 20 ≈ 21.67 min.

Common errors: labelling the axis 'frequency' when plotting density (it must say frequency density); forgetting that bar areas (not heights) compare; and using midpoints with open-ended classes ('60+' — assume a reasonable width or state you cannot estimate). Examiners check axis labels first — wrong label, lost mark.

Box plots, outliers, and cumulative frequency

A box plot shows min, lower quartile Q1, median Q2, upper quartile Q3, max — five numbers summarising distribution shape. IQR = Q3 − Q1 measures spread; outliers lie beyond Q1 − 1.5 × IQR or Q3 + 1.5 × IQR (the 1.5 × IQR rule — examinable). Example: house prices (£000s) in a Newcastle street: 120, 135, 140, 150, 155, 160, 175, 190, 420. Sorted (n = 9): median (5th) = 155; Q1 (median of lower 4) = (135 + 140)/2 = 137.5; Q3 = (175 + 190)/2 = 182.5; IQR = 45; upper fence = 182.5 + 67.5 = 250 → 420 is an outlier (a mansion among terraces). Box plot: box 137.5–182.5, line at 155, whisker to 190, point at 420.

Skew: median nearer Q1 → positive skew (tail right, e.g. incomes); symmetric → median centred. Comparing box plots: compare medians (averages), IQRs (spread), skew, and outliers — in context. 'Heathrow's temperatures have higher median and smaller IQR than Leuchars' — warmer and more consistent. Always compare like with like and use the numbers, not just 'bigger box'.

Cumulative frequency graphs: plot running totals against upper class boundaries, then read medians (at n/2), quartiles (n/4, 3n/4), and percentiles by drawing across. Interpolation on the graph estimates values; box plots can be drawn from the graph's quartiles. The ogive's S-shape steepens where data concentrates — steep middle, flat tails for bell-shaped data.

Correlation — scatter diagrams

A scatter diagram plots bivariate data; correlation describes the trend: positive (upward), negative (downward), or none. Strength: strong (points hug a line) to weak. Example: hours studied vs exam score for 20 pupils — strong positive; umbrella sales vs temperature — negative; shoe size vs maths grade — none. Correlation is not causation: ice-cream sales correlate with drownings (both driven by hot weather — the confounding variable).

Worked example (interpreting r with a confounder). A study of 200 children finds r = 0.61 between shoe size and reading score. Tempting headline: 'Bigger feet make better readers.' The confounder is age: older children have bigger feet and read better. Within a single year group, the correlation would collapse towards 0. A-level answers must name a plausible confounder and state that correlation alone cannot establish causation — the two-sentence structure (mechanism + caution) is what full marks look like.

Product moment correlation coefficient r (given in the formula booklet): r = Σ(x − x̄)(y − ȳ)/√(Σ(x − x̄)²Σ(y − ȳ)²), between −1 and 1. |r| near 1: strong linear; near 0: weak/none. r measures LINEAR association only: y = x² on [−2, 2] gives r ≈ 0 despite perfect (nonlinear) association — always plot first. Worked: for small data sets, compute via the booklet formula; interpret: r = 0.87 → strong positive linear correlation.

Exam interpretation: 'Describe the correlation and interpret in context.' Strong answer: 'Strong negative correlation (r = −0.82): as daily maximum temperature rises, gas consumption falls — consistent with less heating demand in warm weather. This does not prove temperature causes the change (insulation, prices also matter), but the physical mechanism is plausible.' Context + mechanism + caution = full marks.

Regression — the line of best fit

The regression line of y on x, y = a + bx, minimises squared vertical residuals (least squares). Formulae (booklet): b = Σ(x − x̄)(y − ȳ)/Σ(x − x̄)², a = ȳ − b x̄ — the line passes through (x̄, ȳ). Example: advertising spend x (£000s): 2, 4, 5, 7, 9; revenue y (£000s): 40, 55, 58, 72, 85. x̄ = 5.4, ȳ = 62. Compute b: deviations x: −3.4, −1.4, −0.4, 1.6, 3.6; y: −22, −7, −4, 10, 23. Numerator: 74.8 + 9.8 + 1.6 + 16 + 82.8 = 185; denominator: 11.56 + 1.96 + 0.16 + 2.56 + 12.96 = 29.2. b ≈ 6.336, a = 62 − 6.336(5.4) ≈ 27.79. So y = 27.8 + 6.34x: each £1,000 of advertising predicts £6,340 extra revenue.

Interpolation (within the data range) is reasonably safe; extrapolation (beyond it) is dangerous — the linear trend may break. Predicting revenue for x = £6,000: y ≈ 27.8 + 38 = £65.8k (interpolation ✓). For x = £50,000: £344.8k — extrapolation ✗ (market saturation, competition). Residuals (actual − predicted) diagnose fit: random scatter ✓; pattern (curve) ✗ — try a nonlinear model. r² (coefficient of determination): the proportion of y's variation explained by x — r² = 0.99 means 99% explained.

Which regression line? 'y on x' predicts y from x (minimises vertical residuals); 'x on y' differs. Use y-on-x when x is the explanatory variable. Never regress the wrong way round: predicting advertising from revenue answers a different question. And never confuse the regression line with proof of causation — it quantifies association, nothing more.

Misleading presentation — the critical eye

Charts deceive by design or carelessness: truncated vertical axes exaggerate changes (a share price 'soaring' from £10.00 to £10.50 on an axis starting at £9.90); 3D pie charts distort areas; dual axes with different scales manufacture correlation; pictograms with doubled heights quadruple areas. The remedy: read axes first, check the base, ask 'what would this look like redrawn honestly?'

Worked critique: a newspaper shows A&E waiting times with bars starting at 3 hours, making a rise from 3.2 to 4.1 hours look like tripling. Honest version: axis from 0, rise is 28% — still concerning, but not tripling. Another: 'crime up 200%' — from 1 to 3 incidents in a village; the percentage is technically correct and practically meaningless. Small bases make big percentages: always ask '200% of what?'

Averages mislead: 'average wage £45,000' (mean, skewed by bankers) vs median £31,000 — the median describes the typical worker. 'Average' without qualification usually means mean, chosen because it sounds better. Survivorship, selection, and framing biases (Lesson 13) compound visual tricks. The A-level statistician redraws the chart mentally before believing it — a habit for life, not just exams.

Standard deviation and z-scores

Variance and standard deviation quantify spread: population variance σ² = Σ(x − μ)²/N; sample variance s² = Σ(x − x̄)²/(n − 1) — the (n − 1) Bessel correction makes s² an unbiased estimator of σ². Standard deviation σ = √variance, in the original units. Worked: 2, 4, 4, 4, 5, 5, 7, 9 (n = 8, mean 5). Squared deviations: 9, 1, 1, 1, 0, 0, 4, 16 — sum 32; s² = 32/7 ≈ 4.571, s ≈ 2.138. The mean ± s band (2.86 to 7.14) contains 6 of 8 values — typical for bell-shaped data.

Coding simplifies calculation: subtract a constant (mean unchanged in spread) or scale. For large data set work, use Σx and Σx²: s² = (Σx² − (Σx)²/n)/(n − 1). With Σx = 1240, Σx² = 78,900, n = 25: s² = (78900 − 1240²/25)/24 = (78900 − 61504)/24 = 17396/24 ≈ 724.8, s ≈ 26.9. Show the formula substitution — method marks for the correct expression even with arithmetic slips.

z-scores standardise: z = (x − x̄)/s — 'how many sd above the mean?' A pupil scoring 72 with class mean 58, sd 7 has z = 2 — two sd above average, roughly the 97.5th percentile if normal. z-scores compare across different scales: 72/100 in maths (mean 58, sd 7, z = 2) vs 45/60 in physics (mean 36, sd 4.5, z = 2) — equally impressive relative performances. Outlier detection via |z| > 3 complements the 1.5 × IQR rule; use both in 'comment on outliers' questions.

Common mistakes

• Plotting frequency as height with unequal class widths. Area must equal frequency: compute frequency density first and label the axis 'frequency density'. The single most-tested histogram error.
• Identifying the modal class by frequency. With unequal widths the modal class is the one with greatest density. A wide class can hold the most data while being the least dense.
• Extrapolating a regression line without comment. Predictions beyond the data range are unreliable; if the question demands one, give it and flag the extrapolation risk — the flag is a mark.
• Claiming causation from correlation. r quantifies linear association only. Exam answers need a confounder named or a mechanism argued, plus the caution sentence.
• Using the mean for skewed data. Income-type distributions need the median (and IQR). Choosing the wrong average pair — mean with IQR, or median with sd — mixes two frameworks.

Exam Experience

Questions in the style of the A-level statistics paper. Attempt them before reading the mark scheme.

1. The times, in minutes, taken by 76 students to complete a puzzle are summarised below.

(a) For a histogram of this data, calculate the frequency density of each class and explain why the 20–40 class will not have the tallest bar. [3] (b) Estimate the mean time. [2]

2. A data set has Q1 = 24, median = 31, Q3 = 42, minimum = 12 and maximum = 75. (a) Find the interquartile range. [1] (b) Show that 75 is an outlier for this data set. [2] (c) State, with a reason, the direction of skew. [2]

Mark scheme.

• 1(a) Densities = frequency/width: 12/10 = 1.2, 28/10 = 2.8, 36/20 = 1.8 (M1) for dividing by class width, (A1) for all three values. The 20–40 bar has height 1.8 < 2.8: bar height is density, and area (20 × 1.8 = 36) — not height — represents its frequency (B1).
• 1(b) Estimated mean = (5 × 12 + 15 × 28 + 30 × 36)/76 = 1560/76 (M1) for midpoints × frequencies over total frequency. ≈ 20.5 minutes (A1).
• 2(a) IQR = 42 − 24 = 18 (A1).
• 2(b) Upper fence = Q3 + 1.5 × IQR = 42 + 27 = 69 (M1). 75 > 69, so 75 is an outlier (A1) — the comparison must be stated, not just the fence computed.
• 2(c) Median − Q1 = 7 but Q3 − median = 11 (M1) for comparing the two halves (or noting the long upper tail to 75). The upper half is more spread out → positive skew (A1).