4 Statistics and probability

A population is the full group a question concerns; a sample is the subset actually observed. Data may be qualitative or quantitative, discrete or continuous, and the type determines which summaries and graphs make sense.

Sampling method changes the route by which evidence enters the study. A random or stratified sample can reduce selection bias, while an opportunity sample is faster but may overrepresent whoever is available and willing.

If 40 of 200 sampled students prefer option A, the 20% is a sample proportion—not a fact about every student. Its usefulness depends on how the 200 were selected and whether the measurements were reliable.

A large sample does not repair a biased selection process. State the target population, sampling frame and likely source of error before generalising.

Sampling map: simple random gives every member an equal selection chance; systematic uses every $k$th member after a random start; stratified samples each subgroup in population proportion; quota fills category targets non-randomly; convenience uses readily available participants. Audit missing values and recording errors before analysis. An outlier lies more than $1.5\times IQR$ below $Q_1$ or above $Q_3$; investigate it rather than deleting it automatically.

Use a frequency table for discrete or grouped continuous data, an equal-class-width histogram for grouped continuous data, and a cumulative-frequency graph to read median, quartiles and percentiles. Frequency density histograms are not required at SL.

A box plot shows minimum, $Q_1$, median, $Q_3$ and maximum, with outliers marked separately by crosses. Compare two distributions using median, IQR, range and symmetry; a roughly symmetric box and whiskers can support, but not prove, possible normality.

Example

If a cumulative-frequency graph for 80 observations gives $Q_1=18$, median $=25$ and $Q_3=31$, then $IQR=13$. The outlier fences are $18-1.5(13)=-1.5$ and $31+1.5(13)=50.5$, so a value of 54 is plotted as an outlier cross.

Do not use bar gaps for continuous histogram classes or confuse cumulative frequency with class frequency. State units, preserve inequality class boundaries and compare both centre and spread.

The mean is the arithmetic average, the median is the middle ordered value, and the mode is the most frequent value. Range, variance and standard deviation describe how far scores vary around a centre.

The mean uses every value but is pulled by outliers; the median is more resistant. Standard deviation is small when scores cluster near the mean, but neither statistic alone tells you whether the measurement is valid.

For 2, 3, 3, 4 and 18, mean=6 while median=3. Reporting both exposes the outlier and prevents a reader from mistaking the arithmetic average for a typical observation.

Do not choose a statistic by habit. Match it to the measurement scale and distribution, and never delete an extreme value without explaining how it arose.

For grouped data, estimate the mean with class midpoints: $\bar x\approx\sum fm/\sum f$; the modal class has greatest frequency when class widths are equal. If every value becomes $aX+b$, the mean and median become $a$ times the old value plus $b$, while standard deviation becomes $|a|$ times the old standard deviation and variance becomes $a^2$ times the old variance. Quartile conventions can differ, so report the technology method when relevant.

A correlation coefficient summarises the direction and strength of an association between paired variables. A regression model uses an explanatory variable to estimate a response, with residuals showing what the model misses.

A positive correlation means larger values tend to occur together; a negative one means one tends to fall as the other rises. Strength concerns consistency around a pattern, not the slope's units or a causal mechanism.

If study time and score have r=0.82, a fitted line may predict scores within the observed range. It cannot show that study time alone caused the result: prior attainment, teaching and selection may also matter.

r=0 does not prove independence, and a strong r does not prove causation. Check outliers, range restriction, residual pattern and whether extrapolation leaves the observed data.

A by-eye best-fit line should pass through the mean point $(\bar x,\bar y)$. For $y=ax+b$, $a$ is the predicted change in $y$ per unit increase in $x$ and $b$ is predicted $y$ at $x=0$ when meaningful. A y-on-x regression predicts $y$ from $x$; rearranging it to predict $x$ is not generally valid. Pearson's $r$ measures only linear association.

A trial produces an outcome; the sample space $U$ lists all possible outcomes, and an event $A$ is a subset of $U$. For equally likely outcomes, $P(A)=n(A)/n(U)$ and $P(A')=1-P(A)$.

Represent $U$ with a complete list or outcome table. Relative frequency estimates probability from repeated trials and may differ from the theoretical value; with more trials it can become more stable but is not guaranteed to equal it.

Example

For two fair coins, $U=\{HH,HT,TH,TT\}$. The event of exactly one head is $\{HT,TH\}$, so its probability is $2/4=1/2$. If this probability applies to 128 trials, the expected number is $128(1/2)=64$.

The formula $n(A)/n(U)$ requires equally likely outcomes. An expected count such as 12.8 is a long-run average, not a claim that 0.8 of a person will occur in one trial.

Mutually exclusive events cannot occur together, so P(A∩B)=0. Independent events do not change one another's probability, so P(A∩B)=P(A)P(B) and P(A|B)=P(A) when defined.

Two non-impossible mutually exclusive events are automatically dependent: learning that A occurred makes B impossible. Independent events can occur together, such as separate results from a fair coin and die.

For a fair coin and die, P(head and six)=1/2×1/6=1/12. For one die roll, ‘odd’ and ‘even’ are mutually exclusive and their intersection is zero; they are not independent because P(odd|even)=0.

‘Independent’ does not mean unrelated in every philosophical sense, and ‘exclusive’ does not mean equally likely. Use the intersection or conditional rule that matches the claim.

Use a Venn diagram for overlapping sets, a tree for sequential conditions, and a table or sample-space diagram for paired outcomes. Without replacement, branch probabilities change because the remaining total and composition change; with replacement they reset. Independence must be checked from $P(A\cap B)=P(A)P(B)$ or $P(A|B)=P(A)$, not assumed from the wording.

A discrete random variable $X$ takes countable numerical values with probabilities $P(X=x)$ satisfying $0\le P(X=x)\le1$ and $\sum P(X=x)=1$. Its expected value is $E(X)=\sum xP(X=x)$.

Complete a missing probability by making the total equal 1, then multiply each outcome by its probability and add. In a gain model, positive values are winnings and negative values are losses; $E(X)=0$ describes a fair game in the long run.

Example

A game pays \4$with probability$0.2$and loses$$1$with probability$0.8$. Then$E(X)=4(0.2)-1(0.8)=0$, so it is fair by expected gain, although any individual play still wins$4 or loses $1.

Expected value need not be an attainable outcome and does not predict one play. At this SL objective, do not import variance transformations from the separate AHL random-variable content.

X~B(n,p) models the number of successes in n trials when each trial has two outcomes, trials are independent, n is fixed and the success probability p stays constant. P(X=r)=C(n,r)p^r(1−p)^(n−r).

The modelling decision comes before the calculator command. If trials affect one another, have more than two outcomes, or p changes, the binomial assumptions fail even if the question uses the word ‘success’.

If a sensor fails with probability 0.1 independently on each of 20 checks, the number of failures is B(20,0.1). ‘At least two’ means P(X≥2), not P(X=2), so the event wording controls the calculation.

A binomial distribution is not any count between 0 and n. State the trial, success definition, independence and constant-p assumption explicitly.

For $X\sim B(n,p)$, $E(X)=np$ and $Var(X)=np(1-p)$, so $SD(X)=\sqrt{np(1-p)}$. In the sensor example, $E(X)=20(0.1)=2$ failures and $Var(X)=20(0.1)(0.9)=1.8$. Use technology for exact, cumulative or interval probabilities and translate phrases such as 'at most' and 'more than' into inequalities before entering values.

A normal distribution $N(\mu,\sigma^2)$ is continuous, bell-shaped and symmetric about mean $\mu$; mean, median and mode coincide, and $\sigma$ controls spread. Total area under the curve is 1 and an exact point has probability 0.

Approximately 68%, 95% and 99.7% of values lie within $1\sigma$, $2\sigma$ and $3\sigma$ of $\mu$. Use technology for lower-tail, upper-tail and interval probabilities, and inverse normal to find a boundary from a given cumulative probability.

Example

If $X\sim N(300,40^2)$, then about 68% of observations lie from $260$ to $340$. An inverse-normal calculation for cumulative probability $0.95$ returns the value $x$ satisfying $P(X\le x)=0.95$.

At this SL objective, standardized $z$ transformation and continuity correction are not required. Check that a normal model is contextually reasonable, and enter standard deviation $\sigma$, not variance $\sigma^2$, when technology requests it.

Spearman's rank correlation converts paired observations to ranks and measures whether high ranks tend to accompany high or low ranks. With no ties, r_s=1−6Σd²/[n(n²−1)], where d is the rank difference.

Rank ties need an agreed average-rank procedure, and a significance decision depends on the sample size and test rule. The coefficient describes a monotonic pattern, not necessarily a straight-line relationship.

If study hours and score have r_s=0.80, students with higher hours generally rank higher in score. A third variable, restricted range or a few influential pairs could still explain or distort the pattern.

Ranking removes units but not bias. A high r_s is not proof of causation, and a low value can hide a curved relationship that a rank coefficient does not capture.

State $H_0$ and $H_1$, choose significance level $\alpha$, and calculate a test statistic or p-value. Reject $H_0$ when $p\le\alpha$ (or the statistic enters the supplied critical region); otherwise fail to reject $H_0$, always in context.

For a $\chi^2$ independence test, expected frequency is $(\text{row total})(\text{column total})/(\text{grand total})$ and $df=(r-1)(c-1)$. For goodness of fit at SL, $df=n-1$. Expected frequencies exceed 5 in examinations; technology gives $\chi^2$ and the upper-tail p-value.

For the SL t-test, compare two unpaired population means with unknown variance, assume normal underlying variables and equal group variances, and use the pooled two-sample test. Choose one- or two-tailed $H_1$ from the claim, not after viewing the result.

Do not say 'accept $H_0$' or treat significance as practical importance. Poisson-mean tests and Type I/II error calculations belong to AHL 4.18, not this SL Objective.

Data collection is the planned process of defining variables, selecting observations and recording measurements. Validity asks whether the method measures the intended construct; reliability asks whether it is consistent enough to repeat.

Sampling frame, wording, response options and recording conditions can introduce bias or random error. A questionnaire may be efficient, but leading questions and non-response can change who is represented and what is reported.

To estimate sleep in a school, define sleep time, choose a sampling method, pilot the question and record missing answers. A precise question about the previous night is more valid than an unbounded question about ‘usual’ sleep.

A reliable measurement can be consistently wrong, and a valid idea can be measured unreliably. Name the source of error and its direction before judging the data.

Question design should be unbiased, structured with consistent choices, precise and limited to relevant variables. For a $\chi^2$ table, justify category boundaries, keep expected frequencies above 5 and reduce degrees of freedom when parameters are estimated from data. Reliability checks include test–retest and parallel forms; validity checks include content and criterion-related validity.

Regression estimates a relationship between explanatory and response variables. If a straight line leaves a systematic curve in the residuals, a non-linear model or transformation may represent the pattern better.

Choose a model from theory, residual behaviour and the observed range. A transformed linear model can simplify calculation, but the fitted relationship must be converted back before interpreting predictions in the original units.

A growth dataset may follow y=ab^x; plotting ln y against x can make the pattern linear and estimate a and b. A good fit inside the measured range still says little about values far outside it.

A high R² does not prove the model is causal or valid everywhere. Check residuals, outliers, units and extrapolation before trusting a forecast.

Technology may fit linear, quadratic, cubic, exponential, power and sine regressions. Compare models using residual plots, $SS_{res}=\sum(y-\hat y)^2$ and $R^2$, the proportion of response variability accounted for by that model. Smaller $SS_{res}$ or larger $R^2$ helps only for the same response data; neither overrides mechanism, residual pattern or extrapolation limits.

A transformation maps a random variable to a new variable, such as Y=aX+b. For expectation and variance, E(Y)=aE(X)+b and Var(Y)=a²Var(X); a shift changes location while a scale changes spread.

For a discrete variable, transform each possible value and keep its probability. For a continuous variable, monotonic changes also alter the density and interval interpretation, so do not transform only the mean and ignore the distribution.

If E(X)=10 and Var(X)=4, then Y=3X−2 has E(Y)=28 and Var(Y)=36. The subtraction moves every value down by two; it does not reduce the variance.

Variance is squared in the scale factor and is not measured in X's units. Check whether the transformation reverses order or changes the support before interpreting probabilities.

For $Y=\sum a_iX_i+b$, $E(Y)=\sum a_iE(X_i)+b$; if the $X_i$ are independent, $Var(Y)=\sum a_i^2Var(X_i)$. The sample mean $\bar X$ is an unbiased estimator of $\mu$, and $s_{n-1}^2=\sum f_i(x_i-\bar x)^2/(n-1)$ is an unbiased estimator of $\sigma^2$. Independence is required for adding variances without covariance terms.

A sampling distribution is the probability distribution of a statistic, such as a sample mean, over repeated samples of the same size. It is not the distribution of individual observations.

For independent observations with population mean μ and standard deviation σ, the sample mean has mean μ and standard deviation σ/√n. The central limit theorem makes the shape approximately normal when n is sufficiently large under suitable conditions.

If μ=50, σ=12 and n=36, the standard error of the mean is 2. A sample mean of 54 is two standard errors above μ; that statement concerns sampling variation, not one student's score.

Increasing n reduces the standard error, not the population spread σ. The CLT does not erase dependence, extreme bias or a badly defined sampling process.

A linear combination of independent normal variables is exactly normal. If the population is normal, $\bar X\sim N(\mu,\sigma^2/n)$ for any $n$; for a general population the central limit theorem makes $\bar X$ approximately normal as $n$ grows, with $n>30$ treated as sufficient in examinations. Use variance $\sigma^2/n$ or standard error $\sigma/\sqrt n$ consistently.

A confidence interval combines a sample estimate with a margin of error to give a range of plausible values for a population parameter under a stated procedure and confidence level.

The interval widens when variability increases or the sample becomes smaller, and narrows when the sample is larger. Its interpretation is about the long-run performance of the method, not a probability assigned to a fixed parameter after the interval is calculated.

If a mean estimate is 72 with margin 4, a 95% interval is (68,76) under the model used. Repeating the sampling procedure would produce intervals that capture the true mean about 95% of the time in the long run.

A confidence level is not the chance that this already-fixed interval contains the parameter. Check assumptions, units and whether the target parameter matches the estimate.

For a normal population mean, use $\bar x\pm z^*\sigma/\sqrt n$ when population $\sigma$ is known and $\bar x\pm t^*s/\sqrt n$ with $n-1$ degrees of freedom when $\sigma$ is unknown, regardless of sample size. Report the confidence level, parameter, units and population in the contextual interpretation.

A Poisson random variable counts events in a time, length or area interval when events occur independently at a stable average rate λ. Its mean and variance are both λ, and disjoint intervals have independent counts under the model.

If the interval is multiplied by k, the mean rate becomes kλ. Use the model only when events are countable, the rate is approximately constant and one event does not make another more or less likely.

If calls arrive at an average rate of 3 per hour, the count in two hours is Poisson with mean 6. The probability of exactly four calls uses that interval's λ, not the original one-hour value.

Poisson is not any small count. Clustering, a changing rate or a maximum capacity can violate the assumptions even when the data look discrete.

If independent counts satisfy $X\sim Pois(\lambda_1)$ and $Y\sim Pois(\lambda_2)$, then $X+Y\sim Pois(\lambda_1+\lambda_2)$. Choose binomial for successes in fixed independent trials with constant $p$, Poisson for independent events at a uniform average rate, and normal for an appropriate continuous symmetric measurement model. Match $\lambda$ to the interval before using technology.

Choose the test from the parameter and information available: use a normal mean test when $\sigma$ is known, a t-test when $\sigma$ is unknown, a binomial test for a proportion, a Poisson test for a rate/mean, or a technology test of $H_0:\rho=0$ for bivariate normal data.

Samples may be paired or unpaired; matched pairs become one sample of differences. Normal, Poisson and binomial tests use the tail named by $H_1$; discrete critical regions maximize Type I error probability while keeping it below $\alpha$. t-test critical regions need not be calculated manually.

Error interpretation

A Type I error rejects a true $H_0$ and has probability controlled by the critical region. A Type II error fails to reject a false $H_0$; calculate its probability under the stated alternative parameter by finding the chance of landing outside the rejection region.

Do not select a test by distribution name alone: identify the population parameter, known or unknown variance, pairing, assumptions and direction. Poisson and binomial hypothesis tests are one-tailed in this syllabus.

A transition matrix stores probabilities of moving between states. With a column-vector convention, v_(n+1)=Pv_n; each column must sum to one. Powers of P give multi-step transition probabilities.

The orientation is part of the model: row-vector conventions transpose the multiplication order. A steady state is a distribution π with Pπ=π and entries summing to one, if the chain converges to one.

If 70% of users remain in state A and 30% move to B each month, the first column of P records those two destinations under the chosen convention. Multiply by the current state vector before interpreting the next month.

Matrix entries are not percentages to add across unrelated rows, and a stationary distribution is not guaranteed for every chain. State the convention, initial vector and convergence assumption.

A transition diagram labels the same probabilities encoded in $T$. A regular Markov chain has some power of $T$ with all positive entries and approaches a unique steady state independent of the initial state. Find long-run probabilities by repeated multiplication or solve $T\pi=\pi$ with entries summing to 1; $\pi$ is the eigenvector for eigenvalue 1 after normalization.