HookThe night every opinion poll in Britain was wrong
On the morning of 9 April 1992, Britain's newspapers were confident. Every major opinion poll of the campaign's final week had put Labour level with the Conservatives or narrowly ahead, and the consensus prediction was a hung parliament. Then the votes were counted. The Conservatives under John Major won a clear majority, taking roughly \(42\%\) of the vote to Labour's \(34\%\) — a gap of about eight points that not a single final poll had seen. The pollsters had each interviewed thousands of people. They were still wrong, all of them, in the same direction.
The Market Research Society ran a formal inquiry, and its verdict is the foundation of this whole section: the samples were not representative of the population. Quota methods had quietly over-sampled Labour-leaning voters, and some Conservative supporters would not admit their choice to an interviewer. Sample size had not saved them, because size fixes random noise, not bias built into who you ask. Statistics is the craft of inferring a truth about a whole population from a part of it, summarising that part honestly, drawing it clearly, and refusing to be fooled — by a biased sample, by an average that hides the spread, or by a coincidence dressed up as a cause. Higher tier hands you the sharper tools for all of that: frequency density, cumulative frequency, quartiles and box plots, and the discipline of comparing two distributions fairly.
ModelSampling — inferring a population from a part of it
A population is the entire group you want to know about: every voter, every battery off a production line, every tree in a wood. Measuring all of it is usually impossible, ruinously expensive, or destructive — you cannot flatten every battery to test its life and still sell them. So you take a sample, measure that, and infer a conclusion about the whole. That inference is only as good as the sample is fair.
A sample is trustworthy when it is representative — a faithful miniature of the population. The enemy is bias: any process that systematically over- or under-represents part of the group. A simple random sample, where every member has an equal chance of selection, is the standard defence, because it removes the human tendency to pick convenient or agreeable respondents. The 1992 polls failed precisely here — their selection method skewed who was asked before a single sum was done.
The limitation AQA wants you to name explicitly is this: a sample can mislead no matter how careful the arithmetic, and a bigger sample lowers random error but never removes bias. So the honest statistician always states two things — what the sample suggests about the population, and how much to trust it given how the sample was chosen. Bigger and fairer beats bigger alone.
A supermarket wants to estimate the mean weekly spend of its \(9{,}000\) loyalty-card holders and surveys \(300\) of them one Tuesday morning. Why might this mislead, and what is a better design? A Tuesday-morning sample over-represents shoppers who are retired or not in work and under-represents full-time workers, so it is biased — the \(300\) are not a fair miniature of all \(9{,}000\). A stronger design draws the \(300\) at random from the full membership list across all days and times, giving every card-holder an equal chance and letting the sample mean stand in for the population mean with far more confidence.
DataChoosing the chart — matching the picture to the data
Different data wants different pictures, and picking the wrong one is a genuine exam error. A bar chart compares separate categories, with gaps between bars and height showing frequency. A pictogram does the same using a symbol for a set number of items, so you must read the key and handle part-symbols. A vertical line chart is the same idea drawn with thin lines, suited to discrete numerical data like shoe sizes. A pie chart shows how a whole splits into parts, each slice's angle proportional to its share. A line graph joins readings taken over time and is the correct choice for time-series data — temperature through a day, monthly sales across a year — because the line shows the trend between points.
The rule of thumb: bar and pie charts for categories, line graphs for change over time. Pie charts carry their own arithmetic — because a full turn is \(360^\circ\), each slice's angle is its frequency as a fraction of the total multiplied by \(360^\circ\), and read backwards, a slice's angle over \(360\) gives its share of the whole. A useful check every time: the slice angles must sum to \(360^\circ\), or a frequency has gone astray.
In a survey of \(72\) commuters, \(30\) drive, \(24\) take the train, \(12\) cycle and \(6\) walk. Find the pie-chart angle for each. Each angle is \(\dfrac{\text{frequency}}{72}\times 360^\circ\), and since \(\dfrac{360}{72}=5\), every commuter is worth \(5^\circ\). Drive: \(30\times 5=150^\circ\). Train: \(24\times 5=120^\circ\). Cycle: \(12\times 5=60^\circ\). Walk: \(6\times 5=30^\circ\). Check: \(150+120+60+30=360^\circ\) ✓ — the angles close the circle, so the working is sound.
MechanismHistograms and frequency density — the Higher-tier trap
A histogram looks like a bar chart but obeys a completely different rule, and confusing the two is the classic Higher slip. In a histogram the class intervals can have unequal widths, and it is the area of each bar — not its height — that represents frequency. To keep areas honest, the vertical axis is not frequency but frequency density, defined as \[\text{frequency density}=\frac{\text{frequency}}{\text{class width}}.\] Because area equals height times width, multiplying frequency density by class width gives back the frequency: \(\text{frequency}=\text{frequency density}\times\text{class width}\).
This is why a wide class and a narrow class cannot be compared by height alone — a wide bar can be low yet contain the most data because it covers so much ground. Histograms are for grouped continuous data (times, heights, masses), which is why the bars touch with no gaps: the values flow continuously from one class into the next.
The two skills examined are drawing (work out each frequency density, then plot) and reading backwards (measure a bar's frequency density, multiply by its width to recover the frequency, and split a class proportionally to estimate part of it).
Reaction times (seconds) are grouped as \(0\le t<10\) with frequency \(8\); \(10\le t<20\) with \(15\); \(20\le t<40\) with \(24\); and \(40\le t<70\) with \(18\). Find the frequency densities and then estimate how many times fell between \(20\) and \(30\) seconds. Divide each frequency by its class width: \(8\div 10=0.8\); \(15\div 10=1.5\); \(24\div 20=1.2\); \(18\div 30=0.6\). Notice the \(20\le t<40\) class has the most data yet not the tallest bar — width matters. To estimate the \(20\)–\(30\) count, take that class of \(24\) spread over a width of \(20\) and assume it is even: \(\dfrac{30-20}{40-20}\times 24=\dfrac{10}{20}\times 24=12\) reaction times.
ModelAverages and the estimated mean from grouped data
Three averages each measure the middle differently. The mean is the total divided by how many values there are — it uses every value, which makes it powerful but vulnerable to outliers. The median is the middle value in order, unmoved by extremes, which is why the ONS reports median pay: a handful of very high earners would inflate the mean. The mode is the most common value, the only average that works for categories.
From a plain frequency table the mean is \(\dfrac{\Sigma fx}{\Sigma f}\): multiply each value by its frequency, total those, and divide by the total frequency — not by the number of rows, which is the single commonest table error. The median is found by position: the middle of \(n\) values sits at position \(\dfrac{n+1}{2}\), and you count through the frequencies to reach it.
With grouped data you no longer know the exact values, so you take the midpoint of each class as a best estimate and compute an estimated mean — and you must call it an estimate, because the midpoints are assumptions. For grouped data you also quote the modal class (the interval with the highest frequency) rather than a single mode.
Test marks are grouped as \(0\le m<20\) (\(4\) students), \(20\le m<40\) (\(9\)), \(40\le m<60\) (\(12\)) and \(60\le m<80\) (\(5\)). Estimate the mean and state the modal class. Use midpoints \(10, 30, 50, 70\). Then \(\Sigma fx=(10\times 4)+(30\times 9)+(50\times 12)+(70\times 5)=40+270+600+350=1260\), and \(\Sigma f=4+9+12+5=30\). Estimated mean \(=\dfrac{1260}{30}=42\) marks. The modal class is \(40\le m<60\), because its frequency of \(12\) is the highest. Dividing \(1260\) by \(4\) rows instead of \(30\) students would give a nonsensical \(315\) — always divide by \(\Sigma f\).
MechanismCumulative frequency, the median and quartiles
A cumulative frequency graph is the Higher tool for pulling a median and the spread out of grouped data. You build a running total — the number of values at or below each class boundary — and plot that total against the upper boundary of each class, joining the points with a smooth curve that rises in a stretched S. The height of the curve at any value tells you how many data points fall below it.
From the curve you read three landmarks by going across from the cumulative-frequency axis. The median sits at \(\dfrac{n}{2}\) up the axis; the lower quartile \(Q_1\) at \(\dfrac{n}{4}\); the upper quartile \(Q_3\) at \(\dfrac{3n}{4}\). Read across to the curve, then down to the value axis. The interquartile range is \(\text{IQR}=Q_3-Q_1\): the range of the middle half of the data, and a measure of spread that — unlike the plain range — ignores the extreme tails and so is not distorted by a single outlier.
Because a curve is a smoothed model of grouped data, every value you read is an estimate, and saying so is part of the mark.
A \(40\)-runner race gives cumulative frequencies (times in minutes): \(t\le 10\) is \(6\); \(t\le 20\) is \(16\); \(t\le 30\) is \(28\); \(t\le 40\) is \(36\); \(t\le 50\) is \(40\). Estimate the median and the IQR. Here \(n=40\), so read the median at \(\dfrac{40}{2}=20\): the running total passes \(20\) inside the \(20\)–\(30\) class (it is \(16\) at \(20\) and \(28\) at \(30\)), giving a median of about \(23\) minutes. Read \(Q_1\) at \(\dfrac{40}{4}=10\): inside \(10\)–\(20\), about \(14\) minutes. Read \(Q_3\) at \(\dfrac{3\times 40}{4}=30\): inside \(30\)–\(40\), about \(33\) minutes. So \(\text{IQR}=33-14=19\) minutes — the middle half of runners finished within a \(19\)-minute band.
CaseBox plots, outliers and comparing two distributions
A box plot is a five-number summary drawn to scale: minimum, lower quartile, median, upper quartile and maximum. The box spans the IQR with the median as a line inside it, and the whiskers reach out to the extremes. One glance shows both the centre (the median line) and the spread (the box width), which is exactly what you need to describe a population.
An outlier is a value far from the rest; a common rule flags anything more than \(1.5\times\text{IQR}\) beyond a quartile. Outliers matter because they inflate the range and drag the mean, and identifying one lets you decide whether it is a genuine extreme or a recording error.
The skill AQA rewards most is comparing two distributions fairly. Never compare on an average alone: quote one measure of location and one of spread, each interpreted in context. 'Group B's median is higher, so B typically scored more, and B's IQR is smaller, so B was also more consistent' is a full-mark comparison; 'B's median is higher' on its own is half of one. This is how a sample of numbers becomes an honest statement about the population behind it.
Two classes sit the same exam. Class A: min \(15\), \(Q_1=22\), median \(28\), \(Q_3=35\), max \(50\). Class B: min \(20\), \(Q_1=30\), median \(34\), \(Q_3=40\), max \(48\). Compare them, and check A for an outlier. Location: B's median \(34\) exceeds A's \(28\), so Class B typically scored higher. Spread: A's IQR is \(35-22=13\) while B's is \(40-30=10\), so B was more consistent. Outlier check for A's maximum: \(1.5\times 13=19.5\), and the upper boundary is \(Q_3+19.5=35+19.5=54.5\); since \(50<54.5\), A's top mark is a high score, not an outlier. Two measures, both in context — that is the answer the mark scheme is built around.
MechanismScatter graphs — correlation is not cause
A scatter graph plots two measurements for each individual — revision hours against exam mark, height against weight — to reveal whether they move together. Points trending upward show positive correlation; one rising as the other falls shows negative correlation; a shapeless cloud shows no correlation. The tighter the points hug a line, the stronger the correlation.
Where correlation exists you draw a line of best fit: a single ruled line following the trend with roughly equal points either side, ignoring any clear outlier. That line lets you predict — read across and up to estimate one variable from the other. Predicting within the data range (interpolation) is fairly safe; predicting beyond it (extrapolation) is unreliable, because you are assuming a pattern holds where you have no evidence.
The idea AQA tests hardest is that correlation does not prove causation. Ice-cream sales and drowning deaths climb together each summer, yet ice cream drowns no one — hot weather drives both, a lurking third variable. When a question asks you to comment on a relationship, name the correlation, but never claim one thing causes the other without evidence of a genuine mechanism.
A scatter of revision hours \((x)\) against test mark \((y)\) shows strong positive correlation, and a line of best fit passes through \((1,\ 38)\) and \((7,\ 74)\). Estimate the mark for \(4\) hours, and comment on estimating for \(20\) hours. The gradient is \(\dfrac{74-38}{7-1}=\dfrac{36}{6}=6\) marks per hour. From \((1,38)\), a student revising \(4\) hours is \(3\) hours further on: \(38+6\times 3=56\) marks — a safe interpolation, since \(4\) lies inside the data. Estimating for \(20\) hours would give \(38+6\times 19=152\) marks, which is impossible over \(100\): that is extrapolation far beyond the data, where the linear trend clearly breaks down, so any such prediction must be flagged as unreliable.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Statistics runs across all three Higher papers, with the calculator papers carrying the arithmetic and the marks concentrated in interpretation. For a mean from a frequency table, show \(\Sigma fx\) and divide by \(\Sigma f\) — the total frequency, never the number of rows — and for grouped data use class midpoints and label the answer an estimated mean, giving the modal class rather than a single mode.
Histograms are the Higher banker and the Higher trap: always work in frequency density (\(\text{frequency}\div\text{class width}\)), remember that area equals frequency, and to recover a count multiply a bar's frequency density by its width, splitting a class proportionally when only part of it is asked for. On cumulative frequency graphs, read the median at \(\tfrac{n}{2}\), \(Q_1\) at \(\tfrac{n}{4}\) and \(Q_3\) at \(\tfrac{3n}{4}\) up the cumulative axis, draw your reading lines so the examiner sees the method, and state that grouped-data readings are estimates.
When you compare two distributions — box plots, two samples, two classes — quote one measure of location and one of spread (median and IQR travel well together) and interpret each in the context of the question; a bare 'the median is higher' scores well below a comparison that also mentions consistency. Whenever a relationship is mentioned, write that correlation does not imply causation unless there is evidence of a genuine link — it is almost always worth a mark. For any prediction from a line of best fit, label it interpolation (safe) or extrapolation (unreliable). And on sampling questions, name a specific limitation — a small or biased sample may not represent the population — rather than a vague 'it might be wrong'.