Statistics

Statistics turns data into reasoned conclusions. It asks three connected questions:

1. How were the data obtained? A biased sample cannot be repaired by elegant calculation.
2. What do the data show? Tables, diagrams and summary statistics reveal location, spread and association.
3. What can reasonably be inferred? Probability models quantify uncertainty, but conclusions must retain their context and limitations.

The central habit is to distinguish the observed sample from the wider population. A sample statistic such as $\bar{x}$ is known once the data are collected. A population parameter such as $\mu$ is generally unknown and is the subject of inference.

Prerequisites

You should be able to:

• calculate with fractions, percentages, powers and square roots;
• rearrange formulae and use sigma notation $\sum$;
• interpret inequalities, coordinates, gradients and areas;
• use combinations $\binom nr$ and the statistical functions on your calculator;
• communicate conclusions in complete sentences.

Review GCSE statistics and probability foundations if frequency diagrams, averages or basic probability rules are insecure. Calculator output is useful only when you know which quantity you requested and what it means.

The statistical investigation cycle

A statistical investigation is an iterative process:

$$\boxed{\text{question}\to\text{sample}\to\text{clean and present}\to\text{analyse}\to\text{conclude}}$$

The conclusion often exposes a weakness or suggests a new question, so the process may begin again.

1. Define the question and population

State what is being measured, in whom or what, and when. “Do pupils sleep enough?” is vague. “What is the mean number of hours slept on school nights by Year 12 pupils at this school this term?” identifies a variable, population and time period.

2. Choose a sample

A census observes every member of the population. A sample observes only some members. Sampling is usually quicker and cheaper, but introduces sampling variation and possible bias.

A simple random sample of size $n$ gives every possible sample of size $n$ an equal chance of selection. Opportunity and voluntary response samples are convenient, but can systematically overrepresent some groups.

Learn to identify sampling units, frames, strata and sources of bias in populations, samples and sampling.

3. Clean and present the data

Check impossible values, inconsistent units, duplicates and missing observations before calculating. An unusual value is not automatically an error. Removing a genuine extreme observation merely because it is inconvenient distorts the evidence.

Choose a display that preserves the important structure:

Data or purpose	Useful display
categorical frequencies	bar chart
discrete numerical data	vertical line chart
continuous grouped data	histogram
cumulative frequencies	cumulative frequency graph
compare distributions	box plots
relationship between two variables	scatter diagram

Study data presentation, histograms and outliers and cleaning data.

4. Analyse and interpret

Use measures of location and spread together. The mean alone does not describe variability, and the standard deviation alone does not locate the data.

For observations $x_1,x_2,\ldots,x_n$,

$$\bar{x}=\frac{\sum x}{n}, \qquad \operatorname{Var}(x)=\frac{\sum x^2}{n}-\bar{x}^{,2}, \qquad \sigma_x=\sqrt{\operatorname{Var}(x)}.$$

The median and interquartile range are resistant to extreme values. The mean and standard deviation use every value, so they are more sensitive to skew and outliers. See averages and measures of spread.

Worked example 1: compare two distributions

Two groups complete the same task. Their times, in minutes, are summarised below.

Group	Mean	Median	Standard deviation	Interquartile range
A	$18.4$	$17.1$	$5.8$	$6.2$
B	$19.0$	$19.2$	$2.1$	$2.8$

Group A has the lower mean and median, so it was generally faster. Group B has the smaller standard deviation and interquartile range, so its times were more consistent.

For A, $\bar{x}>\operatorname{median}$, which is evidence consistent with positive skew: a few long times may have pulled the mean upwards. This does not prove the exact shape without seeing the raw data or a diagram.

A strong comparison names the statistics, compares both location and spread, and interprets them in context.

Probability as a model of uncertainty

Probability assigns numbers from $0$ to $1$ to events. For an event $A$,

$$0\leq P(A)\leq1, \qquad P(A')=1-P(A).$$

If $A$ and $B$ are mutually exclusive,

$$P(A\cap B)=0.$$

If they are independent,

$$P(A\cap B)=P(A)P(B).$$

These statements describe different ideas. Mutually exclusive events cannot occur together. Independent events do not alter one another’s probabilities. Two events with positive probability cannot be both mutually exclusive and independent.

Conditional probability makes the available information explicit:

$$\boxed{P(A\mid B)=\frac{P(A\cap B)}{P(B)}} \qquad\text{provided }P(B)>0.$$

Build this strand through probability, conditional probability and probability modelling.

Worked example 2: conditional probability and independence

In a year group, $60$ pupils study physics, $45$ study chemistry, and $30$ study both. One of the $120$ pupils is selected at random. Let $P$ and $C$ denote studying physics and chemistry.

The probability that a chemistry pupil also studies physics is

$$P(P\mid C)=\frac{P(P\cap C)}{P(C)} =\frac{30/120}{45/120} =\frac{30}{45} =\frac23.$$

To test independence, compare $P(P\cap C)$ with $P(P)P(C)$:

$$P(P\cap C)=\frac{30}{120}=\frac14,$$

but

$$P(P)P(C)=\frac{60}{120}\times\frac{45}{120}=\frac{3}{16}.$$

Since $1/4\ne3/16$, the events are not independent. Notice that the denominator in $P(P\mid C)$ is the number studying chemistry because the condition restricts the relevant population to that group.

Self check 1

Suppose $P(A)=0.4$, $P(B)=0.5$ and $P(A\cap B)=0.12$. Find $P(A\cup B)$ and $P(A\mid B)$. Are $A$ and $B$ independent?

Answer

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.4+0.5-0.12=0.78.$$

Also,

$$P(A\mid B)=\frac{0.12}{0.5}=0.24.$$

The events are not independent because

$$P(A)P(B)=0.4(0.5)=0.20\ne0.12.$$

Random variables and distributions

A random variable assigns a numerical value to each outcome of a random process. Its probability distribution lists possible values and their probabilities. For a discrete random variable $X$,

$$\sum P(X=x)=1,$$ $$E(X)=\sum xP(X=x), \qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2.$$

The expected value is the long run mean over many repetitions. It need not be a possible individual outcome. Learn the underlying ideas in discrete random variables.

Two central A level models are:

$$X\sim\operatorname{B}(n,p)$$

for a count of successes in $n$ independent trials with constant success probability $p$, and

$$X\sim N(\mu,\sigma^2)$$

for a continuous, symmetric bell shaped model with mean $\mu$ and variance $\sigma^2$.

The model must fit the mechanism, not merely the appearance of the numbers. Use the binomial distribution, the normal distribution and choosing a distribution.

Worked example 3: choose and use a binomial model

A component is defective with probability $0.04$, independently of other components. Let $X$ be the number of defective components among the next $20$.

There is a fixed number of trials, two classifications per trial, independence and constant probability. Therefore

$$X\sim\operatorname{B}(20,0.04).$$

The probability of at least one defective component is most efficiently found by a complement:

$$\begin{aligned} P(X\geq1) &=1-P(X=0)\\ &=1-\binom{20}{0}(0.04)^0(0.96)^{20}\\ &=1-0.96^{20}\\ &\approx0.558. \end{aligned}$$

The expected number is

$$E(X)=np=20(0.04)=0.8.$$

This does not mean exactly $0.8$ defects will occur. It is the mean count over many groups of $20$ components.

Correlation and regression

A scatter diagram can show association between paired variables. The product moment correlation coefficient, $r$, measures the strength and direction of a linear relationship:

$$-1\leq r\leq1.$$

A value near $1$ indicates strong positive linear correlation, a value near $-1$ strong negative linear correlation, and a value near $0$ little linear correlation. It does not rule out a strong curved relationship.

A regression line predicts one variable from another. Use the correct regression direction, interpolate within the observed range where possible, and avoid treating association as causation. A lurking variable may influence both measured variables.

Study correlation and regression before correlation hypothesis tests.

Hypothesis testing

A hypothesis test asks whether sample evidence is unusually extreme under a stated null model. The general structure is:

1. State hypotheses about a population parameter.
2. Assume $H_0$ and identify the sampling distribution.
3. Calculate a tail probability or compare with a critical region.
4. Reject $H_0$ if the evidence is significant at the stated level.
5. Give a contextual conclusion with the correct degree of uncertainty.

The key logical rule is

$$\boxed{\text{do not reject }H_0\ne\text{prove }H_0.}$$

Worked example 4: interpret a p value

A test uses

$$H_0:p=0.30, \qquad H_1:p>0.30.$$

The calculated p value is $0.032$.

At the $5\%$ significance level,

$$0.032<0.05,$$

so reject $H_0$. There is sufficient evidence at the $5\%$ level that the population proportion exceeds $0.30$.

At the $1\%$ level,

$$0.032>0.01,$$

so do not reject $H_0$. There is insufficient evidence at the $1\%$ level that the proportion exceeds $0.30$.

The p value is not the probability that $H_0$ is true. It is a probability about results, calculated under the assumption that $H_0$ is true.

Begin with hypothesis testing language, then study binomial hypothesis tests and normal hypothesis tests.

Self check 2

A two tailed test gives p value $0.071$. State the conclusion at the $10\%$ and $5\%$ significance levels.

Answer

At $10\%$, $0.071<0.10$, so reject $H_0$. There is sufficient evidence for the alternative hypothesis at the $10\%$ level.

At $5\%$, $0.071>0.05$, so do not reject $H_0$. There is insufficient evidence for the alternative hypothesis at the $5\%$ level.

The contextual wording of the alternative hypothesis should replace the generic phrase “for the alternative hypothesis” in a full solution.

Common misconceptions

• A large sample removes bias. A large biased sample can estimate the wrong quantity very precisely.
• Correlation proves causation. Association alone cannot establish a causal mechanism.
• Mutually exclusive means independent. For events with positive probability, mutual exclusivity makes one event impossible when the other occurs.
• Expected value is the most likely outcome. It is a probability weighted long run mean and may not be attainable.
• A normal distribution has standard deviation $\sigma^2$. In $N(\mu,\sigma^2)$, the second parameter is the variance; the standard deviation is $\sigma$.
• A significant result proves the alternative. It provides evidence against $H_0$ at a stated error threshold.
• Calculator output is an interpretation. A decimal needs notation, method, context and an appropriately cautious conclusion.

A route through the topic

Follow this order if you are learning statistics for the first time:

1. Populations, samples and sampling
2. Data presentation, histograms and averages and measures of spread
3. Outliers and cleaning data and the large data set
4. Probability, conditional probability and probability modelling
5. Discrete random variables, the binomial distribution and the normal distribution
6. Correlation and regression
7. Hypothesis testing language, followed by the binomial, normal and correlation tests

When solving a mixed question, return to the investigation cycle. Identify the population and variables, inspect how the data were obtained, choose a justified model or summary, calculate accurately, and finish by answering the question in context.