Sunday 30 August 2026, 12:30 to 14:30. Four questions, 10 marks each.
24 questions covering all 13 topics on the faculty's syllabus list, every topic reported by students who sat the paper, and all four model paper questions. Everything else is in question-bank-exam-list.md to learn afterwards.
Questions are in course order, module by module, so the shared patterns sit side by side. Each module opens with "Before the questions", which builds every concept the module needs from zero. Read that first, then the questions. Every question has the same four parts: when to use it, the question as it would appear on the paper, the answer in numbered steps, and what loses marks. Symbols, table-reading, arithmetic, the plan and the recall card are all in the Appendix at the end.
| Module | Questions |
|---|---|
| Probability and distributions | C1 PMF against PDF · C2 Binomial and Poisson · C3 Normal distribution and z-scores |
| Estimation and hypothesis testing | C4 Z-interval and t-interval · C5 What a 95% confidence interval means · C6 One-sample z-test for a mean · C7 One-sample t-test for a mean · C8 One-proportion z-test |
| Chi-square | C9 Chi-square test of independence · C10 Cramér's V |
| ANOVA | C11 One-way ANOVA · C12 Two-way ANOVA |
| Regression | C13 Simple linear regression · C14 R-squared against adjusted R-squared · C15 Reading a regression output |
| Maximum likelihood and Bayes | C16 Maximum likelihood estimation · C17 Bayes, both updates |
| Multivariate: PCA and clustering | C18 PCA and Kaiser · C19 The dendrogram and where to cut · C20 Ward's linkage against single linkage · C21 Hierarchical against k-means · C22 K-means, one iteration by hand |
| Logistic regression and forecasting | C23 Logistic regression · C24 Forecasting |
Priority is on each question's first line. Priority 1 was named by people who sat the paper, priority 2 is on the lecturer's syllabus list but was not reported, priority 3 is model paper only.
| Priority | Questions |
|---|---|
| 1 · Reported | C1, C2, C3, C4, C5, C6, C7, C8, C9, C10, C14, C15, C19, C20, C21, C22 |
| 2 · Syllabus | C11, C13, C23, C24 |
| 3 · Model paper | C12, C16, C17, C18 |
Probability is a number between 0 and 1 saying how likely something is. Across all the things that could happen, the probabilities add to 1.
toss a coin : P(heads) = 0.5 P(tails) = 0.5 0.5 + 0.5 = 1A random variable is a rule that turns each outcome into a number. It is written with a capital letter, usually X.
toss TWO coins, let X = number of heads
outcome HH HT TH TT
X 2 1 1 0A distribution is the full list of what values X can take and how much probability each one gets. It is the thing every question in this module is really about.
| Discrete | Continuous | |
|---|---|---|
| X takes | Separate, countable values: 0, 1, 2, 3 … | Any value on a scale, with no gaps |
| The test | You could list every possible value | Between any two values there is always another |
| Examples | Heads in 10 tosses, defectives in a batch, calls in an hour, marks out of 100 | Height, weight, time, temperature, distance |
| The distribution is called a | PMF |
For a discrete variable, the PMF is simply the table of P(X = x) for every value x. You can draw it as bars. The height of each bar is the probability.
two coins, X = number of heads
x 0 1 2
P(X=x) 0.25 0.50 0.25 ← this table IS the PMF
P(X=x)
0.5 ┤ █
│ █
0.25 ┤ █ █ █
└───┴────┴────┴───► x
0 1 2
three rules that make it a valid PMF:
every P(X=x) is between 0 and 1
the probabilities add to 1 0.25 + 0.50 + 0.25 = 1 ✓
to get P(X ≤ 1), ADD the bars 0.25 + 0.50 = 0.75For a continuous variable there are infinitely many possible values, so you cannot list them and you cannot give each one a probability. Instead there is a curve f(x), and probability is the area under the curve between two limits.
f(x)
│ ╱▔▔▔╲
│ ╱▓▓▓▓▓▓▓╲ P(a < X < b) = the SHADED AREA
│ ╱▓▓▓▓▓▓▓▓▓▓▓╲
└──┴───────────┴────► x
a b
three things that follow, and all three get examined:
the TOTAL area under the curve is 1
P(X = one exact value) is ZERO, because a single line has no area
f(x) is a HEIGHT, not a probability, so it can be bigger than 1Every distribution has a mean (its balance point, also called the expected value E(X)) and a variance (how spread out it is). Standard deviation is the square root of the variance.
discrete: mean = Σ x · P(x) two coins: 0(0.25) + 1(0.50) + 2(0.25) = 1.0You will not compute these from scratch in the exam. The named distributions below have formulas for them, printed on the paper.
Most exam questions do not ask you to build a PMF. They describe a situation, and you recognise which named distribution fits, then use its formula.
| Distribution | Fits when | Its PMF or PDF | Mean, variance |
|---|---|---|---|
| Binomial | Counting successes out of a fixed n trials, each with the same probability p | P(X=k) = C(n,k) pᵏ (1−p)ⁿ⁻ᵏ |
np, np(1−p) |
| Poisson | Counting events in a fixed interval at an average rate λ, with no fixed n | P(X=k) = e^(−λ) λᵏ / k! |
λ, λ |
| Normal | A continuous measurement that piles up around a centre and tails off symmetrically | The bell curve, set by μ and σ | μ, σ² |
Two pieces of notation the formulas use:
C(n,k) "n choose k" the number of ways to pick k things from n
C(20,2) = (20 × 19) / (2 × 1) = 190 cancel, never compute 20!
k! "k factorial" k × (k−1) × … × 1
5! = 120 and 0! = 1There is a different bell curve for every μ and σ, so no table could print them all. The z-score converts any normal value into the one standard normal with mean 0 and sd 1:
x − μ
z = ─────── z = +1.5 means "1.5 standard deviations ABOVE the mean"
σ z = −1.0 means "1.0 standard deviation BELOW the mean"
the printed table then gives P(Z < z), the area to the LEFT of zThat is the whole reason the z-score exists: one table serves every normal problem.
Priority 1 · reported by students as "PMF and PDF".
Use it when: the words distinguish or explain the difference, or a function f(x) handed to you to verify and integrate.
The one idea: for a discrete variable a probability is a height you read off. For a continuous variable it is an area you measure.
DISCRETE · PMF CONTINUOUS · PDF
P(X=x) f(x)
│ ▄ │ ╱▔▔▔╲
│ █ ▄ │ ╱▓▓▓▓▓▓▓╲
│ █ █ ▄ │╱▓▓▓▓▓▓▓▓▓▓▓╲
└──┴──┴──┴───► x └──┴───────┴────► x
0 1 2 a b
bar HEIGHT = probability SHADED AREA = probability
P(X = 1) can be 0.3 P(X = 1.0) is EXACTLY 0
heights sum to 1 total area = 1
height ≤ 1 always HEIGHT CAN EXCEED 1(a) Distinguish between a discrete and a continuous random variable, with one example of each. [3 marks] (b) Explain the difference between a probability mass function and a probability density function. State clearly why f(x) for a continuous variable is not itself a probability. [4 marks] (c) A continuous random variable X has f(x) = 2x for 0 ≤ x ≤ 1 and zero elsewhere. Verify that f(x) is a valid density and find P(0.2 < X < 0.5). [3 marks]
(a) Define a random variable first, then split it
A random variable assigns a number to each outcome of a random experiment. Toss three coins and let X be the number of heads: the outcome HTH becomes 2.
| Discrete | Continuous | |
|---|---|---|
| Values | Countable: 0, 1, 2, 3, … | Any value within an interval |
| The test | The possible values can be listed | Between any two values there is always another |
| Example | Defective boards in a batch of 20 | The height of a randomly chosen adult |
| Described by | PMF, P(X = x) | PDF, f(x) |
(b) Define both first, then contrast. The definition is the first mark.
A probability mass function is the function for a discrete variable that gives the probability of each exact value, P(X = x). It can be written as a table or drawn as bars, and the bar heights add to 1.
A probability density function is the function f(x) for a continuous variable whose area between two limits gives the probability of X falling in that interval. The total area under the curve is 1.
| PMF | ||
|---|---|---|
| What it gives | An actual probability | A density, which is a height |
| P(single point) | Can be positive | Always exactly zero |
| Getting a probability | Read it off, or add several | Area under the curve |
| Total | Values sum to 1 | Area equals 1 |
| Can exceed 1 | No | Yes, if total area is still 1 |
Why f(x) is not a probability. There are infinitely many possible values, so the chance of landing on any exact one is zero. Only intervals carry probability, and that is the area.
Uniform on [0, 0.5] : f(x) = 2 everywhere
f(x)
│ ┌──────┐ area = height × width
2 │ │▓▓▓▓▓▓│ = 2 × 0.5 = 1 ✓
│ │▓▓▓▓▓▓│
└─┴──────┴──► x height of 2 is NOT a probability of 2
0 0.5The consequence, worth a mark: for a continuous variable P(X < a) and P(X ≤ a) are the same number. For a discrete variable they differ by P(X = a), which is why binomial questions must be read carefully for "at least", "more than" and "at most".
(c) Verify, then find the area
The only integration this needs: the area under 2x is x², from the rule that ∫ x dx = x²/2. Substitute the upper limit, substitute the lower limit, subtract.
Condition 1 : f(x) = 2x ≥ 0 on [0,1] ✓
1 1
Condition 2 : ∫ 2x dx = [ x² ] = 1 − 0 = 1 ✓
0 0
→ f(x) is a valid probability density function.
0.5 0.5
P(0.2 < X < 0.5) = ∫ 2x dx = [ x² ] = 0.25 − 0.04 = 0.21
0.2 0.2If it also asks for the mean:
1 1
E(X) = ∫ x · 2x dx = [ 2x³/3 ] = 2/3 = 0.667
0 0Loses marks: saying a PDF gives "the probability of x", which is exactly what (b) tests; saying P(X = a) = 0 means the value is impossible; skipping the validity check when (c) says "verify".
Priority 1 · reported by students under "PMF".
Use it when: binomial if there is a fixed n and a percentage per trial. Poisson if there is an average rate per unit of time or space and no n at all.
The one idea: binomial counts successes out of n tries. Poisson counts events in an interval, with no upper limit.
BINOMIAL POISSON
"20 items, 5% defective" "3 calls per hour"
"50 patients, 30% respond" "2 accidents per month"
↑ ↑ ↑
fixed n p per trial rate λ, no n
P(X=k) = C(n,k) p^k (1−p)^(n−k) P(X=k) = e^(−λ) λ^k / k!
mean = np, var = np(1−p) mean = var = λA machine produces items of which 5% are defective. A random sample of 20 items is selected.
Formulas provided:
P(X = k) = C(n,k) p^k (1−p)^(n−k)·mean = np·variance = np(1−p)(a) State the conditions under which the binomial distribution is appropriate and confirm they hold here. [2 marks] (b) Find the probability that exactly 2 items are defective. [3 marks] (c) Find the probability that at least one item is defective. [2 marks] (d) Find the mean and standard deviation of the number of defectives. [3 marks]
(a) The four conditions ← stating them earns the marks outright
1 FIXED number of trials n = 20 ✓
2 TWO outcomes per trial defective / not ✓
3 CONSTANT probability of success p = 0.05 throughout ✓
4 INDEPENDENT trials ✓
→ X ~ Binomial(n = 20, p = 0.05)"Success" just means the outcome being counted, so here a defect counts as a success.
(b) Exactly 2
P(X = 2) = C(20,2) × 0.05² × 0.95¹⁸
20 × 19
C(20,2) = ───────── = 190 ← cancel, never compute 20!
2 × 1
0.05² = 0.0025
0.95¹⁸ = 0.3972
P(X = 2) = 190 × 0.0025 × 0.3972 = 0.1887 about 18.9%(c) At least one → use the complement
Direct way : P(1) + P(2) + … + P(20) 20 terms
Complement : 1 − P(0) 1 term ← always do this
P(X = 0) = C(20,0) × 0.05⁰ × 0.95²⁰ = 1 × 1 × 0.3585 = 0.3585
P(X ≥ 1) = 1 − 0.3585 = 0.6415 about 64.2%(d) Mean and spread
mean = n p = 20 × 0.05 = 1.0
variance = n p (1−p) = 20 × 0.05 × 0.95 = 0.95
sd = √0.95 = 0.975On average 1 defective per sample of 20, exactly the 5% rate expected, and the typical departure from that is about 1 item.
Turning the wording into an inequality ← misread this and the whole part goes
"exactly 2" → P(X = 2)
"at least 2" → 1 − P(0) − P(1)
"at most 2" → P(0) + P(1) + P(2)
"more than 2" → 1 − P(X ≤ 2)
"fewer than 2" → P(0) + P(1)If the question is Poisson
A call centre averages 3 calls per hour.
e^(−λ) λ^k e^(−3) × 3⁵ 0.049787 × 243
P(X = 5) = ─────────── = ─────────────── = ───────────────── = 0.1008
k! 5! 120If the interval changes, rescale λ. That is the mark, not the arithmetic.
3 per hour → ONE hour : λ = 3
→ TWO hours : λ = 3 × 2 = 6 then P(X ≥ 8) = 1 − P(X ≤ 7)Remember 0! = 1. Distinctive property: for a Poisson, mean = variance = λ. If real count data
show variance much larger than the mean, the data are overdispersed and Poisson is the wrong model.
Loses marks: forgetting 0! = 1 in P(X = 0); writing 1 − P(1) for "at least one" when it is 1 − P(0); not rescaling λ when the interval changes; not stating the four conditions when (a) asks for them.
Priority 1 · reported by students.
Use it when: the phrase normally distributed with mean … and standard deviation …, then asked for a probability, a proportion of people, or a cut-off score.
The one idea: z says how many standard deviations from the mean a value sits. That is what lets one printed table serve every normal problem.
x − μ
z = ───────── z = +1.5 → 1.5 sd ABOVE the mean
σ z = −1.0 → 1.0 sd BELOW the mean
╱▔▔▔╲
╱ ╲
▁╱ ╲▁
▁▁▁▁▁▁▁▁▁ ▁▁▁▁▁▁▁▁▁
──┴─────┴─────┴─────┴─────┴─────┴─────┴──
200 300 400 500 600 700 800 ← original scale (μ=500, σ=100)
−3 −2 −1 0 +1 +2 +3 ← z scale
├─── 68% ───┤
├──────── 95% ────────┤
├───────────── 99.7% ─────────────┤The table always gives the area to the LEFT of z. Everything else is built from that.
P(Z < z) → read straight off
P(Z > z) → 1 − (table value)
P(a < Z < b) → (table at b) − (table at a)Scores on an aptitude test are normally distributed with mean 500 and standard deviation 100.
Formula provided:
z = (x − μ)/σ, with the standard normal table. Table values: P(Z < 1.5) = 0.9332 · P(Z < −1.0) = 0.1587 · P(Z < 1.645) = 0.95(a) Find the probability that a randomly chosen score exceeds 650. [3 marks] (b) Find the probability that a score lies between 400 and 650. [3 marks] (c) Find the score exceeded by only 5% of candidates. [4 marks]
(a) P(X > 650)
650 − 500 150
z = ─────────── = ────── = 1.5
100 100
P(X > 650) = P(Z > 1.5) = 1 − 0.9332 = 0.0668 about 6.7%(b) P(400 < X < 650) ← standardise both ends, subtract the PROBABILITIES not the z values
z₁ = (400 − 500)/100 = −1.0
z₂ = (650 − 500)/100 = +1.5
P = P(Z < 1.5) − P(Z < −1.0) = 0.9332 − 0.1587 = 0.7745 about 77.5%(c) This one runs BACKWARDS: given the probability, find the score
"only 5% above" → 95% below → find z with P(Z < z) = 0.95 → z = 1.645
Now invert the formula:
x = μ + z σ = 500 + 1.645 × 100 = 664.5A score of about 664.5 is exceeded by only 5% of candidates.
The empirical rule, often a mark on its own:
within 1 sd of μ → 68% 400 to 600
within 2 sd of μ → 95% 300 to 700
within 3 sd of μ → 99.7% 200 to 800Two properties if asked to describe the normal: symmetric about the mean so mean = median = mode, and completely determined by just two parameters, μ for location and σ for spread.
Loses marks: forgetting the table gives the left area, so "greater than" needs 1 minus; subtracting z values instead of probabilities in (b); inverting the formula wrongly in (c).
Module numbers follow the course, and there is no Module 2 section here on purpose: its content, sampling distributions and the CLT, has no exam question of its own and is built into this module's opening instead.
| Population | Sample | |
|---|---|---|
| What it is | Everyone or everything you care about | The part you actually measured |
| Its numbers are called | Parameters, written in Greek | Statistics, written in Roman |
| Mean | μ | x̄ |
| Standard deviation | σ | s |
| Proportion | p | p̂ |
| Do you know it | No, that is the whole problem | Yes, you computed it |
Everything in this module is one move: use the sample statistic to say something about the population parameter.
Take a sample of 64 packets and compute x̄. Take another 64 and x̄ comes out slightly different. Keep going and the x̄ values form their own distribution, called the sampling distribution.
population one sample another another
μ = 500, σ=80 x̄ = 520 x̄ = 497 x̄ = 508 ...
the x̄ values pile up around μ like this:
╱▔▔╲
╱ ╲ centred at μ
╱ ╲ spread = σ / √n ← the STANDARD ERROR
▁╱ ╲▁
──────────┬──────────
μThe standard error, SE = σ/√n, is how far a sample mean typically lands from the true mean. Bigger samples give a smaller SE, which is why every formula has √n underneath. You will see SE in every single question in this module.
The Central Limit Theorem says: for n of about 30 or more, this pile is approximately normal whatever shape the original population has. That is what makes the z table usable.
| Confidence interval | Hypothesis test | |
|---|---|---|
| The question | "What range of values is the true μ plausibly in?" | "Is a specific claimed value of μ believable?" |
| Starts from | The sample estimate | A claim to check, called H₀ |
| Output | A range: estimate ± margin | A decision: reject the claim or not |
| Shape | x̄ ± (critical value) × SE |
(x̄ − claimed) / SE, compared to a critical value |
A number read from a table that marks off the tail of the distribution. 1.96 cuts off 2.5% in each tail of the normal, leaving 95% in the middle. That is why 95% intervals use 1.96.
╱▔▔▔▔▔╲
╱ 95% ╲
2.5% ╱ ╲ 2.5%
▓▓▓▓▁▁▁▁╱ ╲▁▁▁▁▓▓▓▓
────────┬─────────┬────────┬────────
−1.96 0 +1.96| Use | When |
|---|---|
| z, and the normal table | The population σ is known |
| t, and the t table with df = n − 1 | σ is unknown and you are using the sample s instead |
t has fatter tails than z because estimating σ adds uncertainty. Its critical values are a little bigger, and they depend on df = n − 1. When σ is unknown the safe answer is t whatever the sample size; at large n the two give almost the same number anyway, which is why some books say "z for large n". On this paper, s given means say t.
1 ASSUME the claim is true. H₀: μ = 500 (H₀ always holds the "=")
2 Ask: if that were true, how surprising is my sample?
3 If it is TOO surprising, reject the claim.| Term | Meaning |
|---|---|
| H₀, null hypothesis | The claim of no change, no effect. Always contains the equality |
| H₁, alternative | What you would conclude instead. Direction comes from the wording |
| α, significance level | How surprising counts as "too surprising". Almost always 0.05 |
| Test statistic | z or t: how many standard errors the sample sits from the claim |
| Critical value | The cut-off. Beyond it, reject H₀ |
| p-value | The probability of a sample at least this extreme if H₀ were true. p < α means reject |
| Reject H₀ | The data are inconsistent with the claim |
| Fail to reject H₀ | Not enough evidence against it. Never "H₀ is proved" |
H₁ says "different from" → TWO-TAILED → α split, 0.025 each side → ±1.96
H₁ says "greater than" → RIGHT-TAILED → all α on the right → +1.645
H₁ says "less than" → LEFT-TAILED → all α on the left → −1.645
The tail is fixed by the WORDING of the question, before you compute anything.Priority 1 · reported by students.
Use it when: the words construct a confidence interval, estimate with 95% confidence, or margin of error. No hypothesis is being tested.
The one idea: every confidence interval ever built is the same three pieces.
┌─ how confident you want to be
│ ┌─ how much the estimate bounces sample to sample
↓ ↓
ESTIMATE ± (critical value) × (standard error)
↑
└─ x̄ for a mean, p̂ for a proportion
├──────────────────┼──────────────────┤
lower estimate upper
└────── margin ───┘ └─── margin ──────┘(a) A random sample of 64 items has a mean of 52. The population standard deviation is known to be 8. Construct a 95% confidence interval for the population mean. [3 marks] (b) A random sample of 16 items has a mean of 25 and a sample standard deviation of 4. Construct a 95% confidence interval. [3 marks] (c) Both samples give the same standard error. Explain why the two intervals differ in width. [2 marks]
Formulas provided:
x̄ ± z (σ/√n)andx̄ ± t (s/√n)Critical values: z = 1.96 · t(15, 0.025) = 2.131
(a) σ known and n large → z-interval
SE = σ / √n = 8 / 8 = 1.0
margin = 1.96 × 1.0 = 1.96
CI = 52 ± 1.96 = (50.04, 53.96)We are 95% confident that the interval (50.04, 53.96) contains the true population mean.
(b) σ unknown and n small → t-interval, df = 15
SE = s / √n = 4 / 4 = 1.0
margin = 2.131 × 1.0 = 2.131
CI = 25 ± 2.131 = (22.87, 27.13)(c) Why (b) is wider, which is where the 2 marks are
(a) SE = 1.0 → width ± 1.96 ├────────┤
(b) SE = 1.0 → width ± 2.131 ├──────────┤ WIDER
Same standard error. Different critical value. Why?In (a) σ was known, so only x̄ varies. In (b) σ was estimated from the same small sample, so x̄ varies and s varies too. The t distribution has fatter tails to pay for that extra uncertainty, pushing the critical value from 1.96 up to 2.131. As n grows, s becomes reliable, t converges to the normal, and by about n = 30 the two critical values are almost identical.
If it asks for a proportion instead
p̂ = 240/400 = 0.60
SE = √( 0.60 × 0.40 / 400 ) = √0.0006 = 0.0245
margin = 1.96 × 0.0245 = 0.048
CI = 0.60 ± 0.048 = (0.552, 0.648)The whole interval sits above 0.50, so a majority can be claimed with confidence. Note this uses p̂, unlike the test in C8 which uses p₀.
If it asks for a sample size
Here E is the margin of error you are willing to accept, ±0.03 meaning 3 percentage points. It is not the expected count from chi-square, which reuses the same letter.
z² p(1−p) 1.96² × 0.5 × 0.5 0.9604
n = ────────────── = ───────────────────── = ───────── = 1067.1 → n = 1068
E² 0.03² 0.0009Use p = 0.5 when you have no prior data, because p(1−p) is largest there and that gives the safest sample size. Always round up. This is why national polls survey about a thousand people for a 3 point margin: the population size never enters the formula.
Loses marks: using 1.645, since a confidence interval is always two-sided; not naming which interval and why; rounding a sample size down.
Priority 1 · reported by students. The cheapest mark in this file.
Use it when: an interval is handed to you and you are asked to interpret it, or a quoted student statement is offered for you to judge.
The one idea: the parameter never moves. The interval is what is random.
TRUE MEAN μ ────────────────┼────────────────── fixed. Never moves.
│
sample 1 ├─────────┼───┤ contains μ ✓
sample 2 ├─────┼────────┤ contains μ ✓
sample 3 ├───────────┼─┤ contains μ ✓
sample 4 │ ├────────┤ MISSES μ ✗
sample 5 ├───────┼──────┤ contains μ ✓
│
Over many samples, about 95 intervals in 100 cover μ.
Yours is ONE of them. You cannot tell which kind.A 95% confidence interval for the population mean is calculated as (50.04, 53.96). A student interprets this as "there is a 95% probability that the population mean lies between 50.04 and 53.96". State whether this is correct, justify your answer, and give the correct interpretation. [4 marks]
1 · The verdict
The interpretation is incorrect.
2 · Why it is wrong
μ is a FIXED CONSTANT, not a random variable.
So either μ ∈ (50.04, 53.96) → probability 1
or μ ∉ (50.04, 53.96) → probability 0
We do not know which. But it is never 0.95.No probability statement can attach to a fixed constant.
3 · Where the randomness actually is
random sample → random x̄ → RANDOM INTERVAL
│
μ stays putA different sample gives a different interval. The 95% is a property of the procedure that generates intervals, not of any one interval.
4 · The correct interpretation
We are 95% confident that the interval (50.04, 53.96) contains the true population mean.
Spelt out for the second mark: if many samples of the same size were drawn and an interval built from each in the same way, about 95% of those intervals would contain the true mean. This is one of them, and we do not know whether it is one of the 95% that succeed or the 5% that fail.
5 · The contrast that shows real understanding
| Confidence interval | Credible interval (Bayesian) | |
|---|---|---|
| Parameter is | A fixed constant | A random variable |
| 95% describes | The procedure | The parameter |
| "95% chance μ is in here" | ✗ Not allowed | ✓ Allowed |
The student's sentence is exactly what a Bayesian credible interval does say. The difference is not in the arithmetic but in what each framework entitles you to claim.
Two further misreadings, if asked for common errors:
✗ "it contains 95% of the DATA" → that is a PREDICTION interval, far wider
✗ "95% of sample means fall in it" → it is about the population parameterLoses marks: answering only "it is wrong" without the justification, which is 2 of the 4 marks; giving the correct wording without the repeated-sampling explanation.
Priority 1 · reported by students. This is the shape every other test on the paper copies.
Use it when: one sample, one mean, a target value, and the phrase population standard deviation is known, or σ given directly.
The one idea: sample means bounce around even when nothing has changed. The test asks whether 520 against a target of 500 is a real shift or ordinary bouncing.
THE SIX-LINE SHAPE · every test on this paper is this, with a different denominator
1 H₀ and H₁ what is claimed, and what would contradict it
2 α and decision rule written BEFORE computing
3 name the test and say why it fits
4 SE = spread / √n ← only the "spread" changes between tests
5 statistic = (estimate − claimed) / SE
6 conclude IN CONTEXT never stop at "reject H₀" Which "spread" goes on top?
σ known → z-test SE = σ / √n
s (sample) → t-test SE = s / √n df = n − 1
proportion → z-test SE = √(p₀(1−p₀)/n)A machine is set to fill packets with a mean weight of 500 g. From long production experience the population standard deviation is known to be 80 g. A quality inspector draws a random sample of 64 packets and finds a sample mean weight of 520 g.
Formula provided:
z = (x̄ − μ₀) / (σ / √n)Critical values: Z two-tailed ± 1.960 · Z one-tailed right 1.645Test at the 5% level whether the machine is off target. State your hypotheses, the decision rule, the test statistic and your conclusion. [5 marks]
1 · Hypotheses
H₀ : μ = 500 the machine is filling on target
H₁ : μ ≠ 500 the machine is off target → TWO-TAILED"Off target" covers both overfilling and underfilling, so both tails matter.
Reading the tail from the wording:
"differs / changed / off target" → two-tailed → ± 1.96
"greater / exceeds / improved" → right-tailed → 1.645
"less / reduced / below" → left-tailed → −1.6452 · α and the decision rule, written before computing
α = 0.05, two-tailed → REJECT H₀ if |z| > 1.96Writing the rule first stops you fitting the conclusion to the number, and it carries a mark on its own even if the arithmetic later goes wrong.
3 · Name the test
One sample, one mean, σ known, n = 64 large → one-sample z-test. Had σ been unknown this would be a t-test.
4 · Standard error
σ 80 80
SE = ──── = ────── = ──── = 10
√n √64 85 · The statistic
x̄ − μ₀ 520 − 500 20
z = ──────── = ─────────── = ────── = 2.00
SE 10 106 · Compare and conclude
|z| = 2.00 > 1.96 → REJECT H₀
two-tailed p ≈ 0.0455 < 0.05 same decisionAt the 5% level there is sufficient evidence that the mean fill weight differs from 500 g. The sample mean of 520 g indicates the machine is overfilling by about 20 g, so it should be recalibrated.
Loses marks: stopping at "reject H₀" without the direction and the action; not stating the decision rule; treating "off target" as one-tailed.
Priority 1 · reported by students. Identical to C6 except for two things, and the marks are in naming them.
Use it when: one mean again, but the spread is called a sample standard deviation, or σ is never mentioned. Usually a small n.
The one idea: estimating σ from the same small sample adds a second source of uncertainty, so a wider distribution than the normal is needed.
z-test : σ known one thing varies : x̄ → normal
t-test : σ estimated TWO things vary : x̄ and s → fatter tails
╱▔▔╲
╱ ╲ normal
╱ ╲ t: fatter tails push the cut-off further out
▁╱ ╲▁
─────────┬───────┬──┬─────
0 ↑ ↑
1.96 2.131
z crit t crit same α, bigger cut-offA supplier claims the mean tensile strength of its wire is 100 units. A buyer tests a random sample of 25 wires and obtains a sample mean of 104.5 units with a sample standard deviation of 9 units. The population standard deviation is not known.
Formula provided:
t = (x̄ − μ₀) / (s / √n), withdf = n − 1Critical value: t(24, 0.025) = 2.064Test the supplier's claim at α = 0.05. [5 marks]
1 · Hypotheses
H₀ : μ = 100 the supplier's claim is correct
H₁ : μ ≠ 100 → TWO-TAILED2 · α and the decision rule
α = 0.05, two-tailed, df = n − 1 = 24 → REJECT H₀ if |t| > 2.0643 · Why t and not z ← this is the mark that separates C7 from C6
The population standard deviation is unknown; the value 9 is the sample standard deviation s. Because σ was estimated from the same small sample there are two sources of uncertainty, not one, so the t distribution with 24 degrees of freedom is used.
4 · Standard error
s 9 9
SE = ──── = ───── = ─── = 1.8
√n √25 55 · The statistic
x̄ − μ₀ 104.5 − 100 4.5
t = ──────── = ───────────── = ─────── = 2.50
SE 1.8 1.86 · Compare and conclude
|t| = 2.50 > 2.064 → REJECT H₀At the 5% level there is evidence that mean tensile strength differs from the claimed 100 units. The sample points to it being higher, which favours the buyer but is still a departure from the stated specification.
7 · State the assumption a t-test carries and a z-test does not
The population should be approximately normal. With n = 25 the CLT is not fully doing the work, so with a strongly skewed population a non-parametric alternative such as the Wilcoxon signed-rank test would be preferred.
The follow-up they often attach.
95% CI = x̄ ± t × SE = 104.5 ± 2.064 × 1.8 = 104.5 ± 3.72 = (100.78, 108.22)
100 is OUTSIDE the interval → same information as rejecting H₀A two-tailed test at α = 0.05 and a 95% confidence interval always agree.
Loses marks: using z because n = 25 feels large; using df = 25 instead of 24; not stating the normality assumption.
Priority 1 · reported by students. Model paper Q1(c), 3 marks.
Use it when: one group, a yes/no outcome, a claimed percentage, data given as x out of n. Wording like more than 30% makes it one-tailed.
The one idea: the test assumes H₀ is true, so p₀ goes inside the standard error, never p̂.
TEST → p₀ inside the SE → because H₀ is assumed true, and it fixes the spread
INTERVAL → p̂ inside the SE → because there is no hypothesised value to assume
Getting these the wrong way round is a favourite one-mark trap.The marketing team claims that more than 30% of customers aged 18 to 30 prefer Brand Y. Using the observed data (80 out of 200 in the 18 to 30 group prefer Brand Y), test this claim at α = 0.05 using a one-tailed proportion test. State your hypotheses and conclusion. [3 marks]
Formula provided:
z = (p̂ − p₀) / √( p₀(1 − p₀) / n )Critical values: Z two-tailed ± 1.960 · Z one-tailed right 1.645
1 · Name the test
One group, one categorical variable, a claim about a single percentage → one-sample z-test for a proportion.
2 · Hypotheses and the tail
H₀ : p = 0.30 at most 30% prefer Brand Y
H₁ : p > 0.30 more than 30% do → RIGHT-TAILEDH₀ must always carry the equality, so a "more than" claim always becomes H₁.
3 · The sample proportion
p̂ = 80 / 200 = 0.404 · Standard error, using p₀
p₀(1 − p₀) 0.30 × 0.70 0.21
SE = √ ──────────── = √ ─────────────── = √ ──────── = √0.00105 = 0.0324
n 200 2005 · The statistic
p̂ − p₀ 0.40 − 0.30 0.10
z = ──────── = ───────────── = ──────── = 3.09
SE 0.0324 0.03246 · Compare against the right critical value
one-tailed at α = 0.05 → critical value = 1.645 ← NOT 1.96
z = 3.09 ≫ 1.645 → REJECT H₀7 · Conclude in context
There is strong evidence at the 5% level that more than 30% of customers aged 18 to 30 prefer Brand Y. The marketing team's claim is supported by the data.
Loses marks: using 1.96 instead of 1.645, the commonest error on this type and the reason the paper prints both; putting p̂ inside the standard error. If there is room, add the validity check: n·p₀ = 60 and n(1−p₀) = 140, both above 10.
Everything in Module 3 measured something: a weight, a strength. Chi-square is for data that sorts things into categories and counts them: brand chosen, age group, pass or fail. There is no mean to test. There are only counts.
Two categorical variables cross-tabulated, with the totals on the edges.
Brand X Brand Y Brand Z │ row total
18 to 30 60 80 60 │ 200
31 to 50 50 40 30 │ 120
51+ 30 20 30 │ 80
──────────────────────────────────────────┼───────────
column total 140 140 120 │ 400 ← grand totalTwo categorical variables are independent if knowing one tells you nothing about the other. If brand and age are independent, every age group chooses brands in the same proportions as everyone else does. The test checks whether the counts in the table look like that.
INDEPENDENT every row has the same shape → no association
ASSOCIATED some rows lean towards some columns → association| Symbol | Where it comes from | |
|---|---|---|
| Observed count | O | The real number in the table. Printed in the question |
| Expected count | E | The number that cell would hold if the variables were independent. You calculate it |
The whole test is: measure how far every O sits from its E, add it up, and ask whether the total is bigger than chance alone could produce.
Because the statistic adds up squared gaps, it can never be negative and it only grows as the table departs from independence. So the test is always right-tailed, and the table has a single column for each α.
density
│▚
│ ▚▖
│ ▝▚▄▖
│ ▝▀▚▄▄▖
│ ▝▀▀▚▄▄▄▓▓▓▓▓ ← rejection region, all of α on the right
└────────────────┬──────────────► χ²
9.488
(df = 4, α = 0.05)Its degrees of freedom come from the size of the table: df = (rows − 1)(columns − 1).
Chi-square answers "is there an association" and its size grows with the sample. A separate number, Cramér's V, puts the strength on a 0 to 1 scale so tables of different sizes can be compared. C10 covers it.
Priority 1 · reported by students. Model paper Q1(a), 5 marks.
Use it when: two categorical variables, data given as counts in a table, and words like associated with, related to, independent of, depends on.
The one idea the whole test rests on: compare what you actually observed against what you would expect if the two variables were completely unrelated.
O = OBSERVED counts → what really happened. Printed in the question
E = EXPECTED counts → what independence would predict. You calculate it
O and E close together → independence looks fine
O and E far apart → there is an associationA mobile phone manufacturer surveyed 400 customers to study whether preferred phone brand (Brand X, Brand Y, or Brand Z) is associated with the customer's age group. The observed frequencies are:
Brand X Brand Y Brand Z Total 18 to 30 years 60 80 60 200 31 to 50 years 50 40 30 120 51+ years 30 20 30 80 Total 140 140 120 400 Formulas provided:
E_ij = (Row total × Column total) / Grand total·χ² = Σ (O − E)² / E·df = (r−1)(c−1)Critical values at α = 0.05: χ² df=2 is 5.991 · χ² df=4 is 9.488
State the null and alternative hypotheses. Calculate all expected frequencies, compute the chi-square test statistic, and state your conclusion at α = 0.05. [5 marks]
1 · Name the test
Two categorical variables, counts in a contingency table → chi-square test of independence.
2 · Hypotheses
H₀ : brand preference is INDEPENDENT of age group
H₁ : brand preference is ASSOCIATED with age groupIndependent means knowing a customer's age group tells you nothing about which brand they prefer: every age group chooses brands in the same proportions.
Always this pair, and always right-tailed. There is no one-tail or two-tail decision to make.
3 · Expected counts ← same grid as the observed table, so each E sits where its O sits
row total × column total
E = ──────────────────────────
grand total| Expected | Brand X (140) | Brand Y (140) | Brand Z (120) | Row total |
|---|---|---|---|---|
| 18 to 30 (200) | 200×140/400 = 70 | 200×140/400 = 70 | 200×120/400 = 60 | 200 |
| 31 to 50 (120) | 120×140/400 = 42 | 120×140/400 = 42 | 120×120/400 = 36 | 120 |
| 51+ (80) | 80×140/400 = 28 | 80×140/400 = 28 | 80×120/400 = 24 | 80 |
| Column total | 140 | 140 | 120 | 400 |
The totals in brackets are the numbers you multiply. Each cell is its row bracket × its column bracket, over 400.
4 · Check before going on
rows: 70 + 70 + 60 = 200 ✓ 42 + 42 + 36 = 120 ✓ 28 + 28 + 24 = 80 ✓
columns: 70 + 42 + 28 = 140 ✓ 70 + 42 + 28 = 140 ✓ 60 + 36 + 24 = 120 ✓The expected counts must reproduce the same totals as the observed table. Ten seconds, and it catches a slip before it spreads through nine cells.
5 · The statistic ← build ONE working table and fill it column by column
(O − E)²
χ² = Σ ──────────
all E
cells| Cell | O | E | O − E | (O−E)² | (O−E)² / E |
|---|---|---|---|---|---|
| 18-30, X | 60 | 70 | −10 | 100 | 100/70 = 1.4286 |
| 18-30, Y | 80 | 70 | +10 | 100 | 100/70 = 1.4286 |
| 18-30, Z | 60 | 60 | 0 | 0 | 0/60 = 0.0000 |
| 31-50, X | 50 | 42 | +8 | 64 | 64/42 = 1.5238 |
| 31-50, Y | 40 | 42 | −2 | 4 | 4/42 = 0.0952 |
| 31-50, Z | 30 | 36 | −6 | 36 | 36/36 = 1.0000 |
| 51+, X | 30 | 28 | +2 | 4 | 4/28 = 0.1429 |
| 51+, Y | 20 | 28 | −8 | 64 | 64/28 = 2.2857 |
| 51+, Z | 30 | 24 | +6 | 36 | 36/24 = 1.5000 |
| Totals | 400 | 400 | 0 | χ² = 9.405 |
Two checks built into the table itself:
column O must total 400 ✓ the grand total
column E must total 400 ✓ expected counts always reproduce it
column O−E must total 0 ✓ the gaps always cancel outIf any of those three fails, an E is wrong and you can find it before computing χ².
Reading the biggest cell: 51+ customers gave Brand Y only 20 when independence predicted 28. That one cell contributes 2.2857 of the 9.405, about a quarter of the whole statistic, so older customers avoiding Brand Y is where most of the evidence sits.
6 · Degrees of freedom
df = (r − 1) × (c − 1) = (3−1) × (3−1) = 47 · Compare
χ² computed = 9.405
χ² critical = 9.488 (df = 4, α = 0.05, right tail)
9.405 < 9.488 → FAIL TO REJECT H₀8 · Conclude in context
At the 5% level there is not enough evidence to conclude that brand preference depends on age group. The statistic sits very close to the critical value, so a larger sample could easily change this conclusion.
Loses marks: rounding 9.405 up into significance, and not saying it is close; writing "the variables are independent" instead of "not enough evidence"; using df = 8; skipping the expected table when the question says "calculate all".
If it is goodness of fit instead: one variable against a claimed distribution, not a cross-table. Expected counts come from the claim (150 sales over 5 weekdays under a uniform claim expects 30 each), and df = k − 1.
Priority 1 · reported by students. Model paper Q1(b), 2 marks.
Use it when: a chi-square result is in hand and the words strength of association appear, usually with the interpretation bands printed for you.
The one idea: χ² answers is there an association. It cannot answer how big is it.
Double every count in the table → χ² DOUBLES
→ but the pattern is IDENTICAL
So χ² is contaminated by sample size.
Cramér's V divides n back out, landing on a fixed 0 to 1 scale.Compute Cramér's V for the chi-square test performed in Part (a). Interpret the strength of association using: V < 0.1 negligible, 0.1 ≤ V < 0.3 small, 0.3 ≤ V < 0.5 medium, V ≥ 0.5 large. [2 marks]
Formula provided:
V = √( χ² / (n × min(r−1, c−1)) )
1 · Say why an effect size is needed
With only 2 marks on offer, this sentence is worth one of them. Use the box above: χ² grows with sample size, so a big χ² can mean a strong pattern or just a big survey.
2 · Find min(r−1, c−1)
3 × 3 table → r−1 = 2 , c−1 = 2 → min = 23 · Substitute
χ² 9.405 9.405
V = √ ───────────────── = √ ─────────────── = √ ───────── = √0.01176 = 0.108
n × min(r−1,c−1) 400 × 2 8004 · Read the band and tie it back
V < 0.1 negligible
0.1 ≤ V < 0.3 SMALL ← 0.108 lands here
0.3 ≤ V < 0.5 medium
V ≥ 0.5 largeThe test found no significant association, and Cramér's V agrees that even if one existed it would be weak. The two results tell one consistent story rather than contradicting each other.
Loses marks: giving 0.108 without the word "small"; not linking back to part (a), which is what
the second mark is for. V can never exceed 1, so a value above 1 means the min term was dropped.
Module 3 compared one mean to a claim. ANOVA compares three or more group means at once and asks one question: are they all equal?
H₀ : μ_A = μ_B = μ_C all groups have the same mean
H₁ : at least one differs (NOT "they all differ")Statistics measures how spread out numbers are by taking each value's distance from the mean, squaring it, and adding up. That total is a sum of squares, SS.
values 12, 14, 11, 13, 15 mean = 13
gaps −1 +1 −2 0 +2
squares 1 1 4 0 4 SS = 10 SS_Total = SS_Between + SS_Within
all the how far each how far each observation
variation GROUP MEAN sits sits from ITS OWN group mean
there is from the grand
mean
the SIGNAL the NOISE
(real differences (scatter that would exist
between groups) even if the groups were identical)If the groups genuinely differ, the signal is large compared to the noise. If they do not, the two are about the same size.
The same three group means, drawn twice. Only the scatter inside the groups changes, and it completely changes the verdict.
Response
│ ●●● ● ● ●
│ ●●● ● ● ●
│ ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯ grand mean ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
│ ●●● ● ● ●
│ ● ● ●
└──────────────────── ────────────────────
A B C A B C
TIGHT within groups. SCATTERED within groups.
The gaps between the The same gaps now look like
means look REAL. they could be random luck.A gap between means only counts as evidence if it is big relative to the noise inside the groups. That is why a test about MEANS is run on VARIANCES, and why the method is named analysis of variance.
Sums of squares grow with the number of observations, so each is divided by its degrees of freedom to get a mean square, MS, which is variation per degree of freedom. Then take the ratio.
MS = SS / df
MS_Between signal per df
F = ──────────── = ───────────────
MS_Within noise per df
F near 1 → groups look the same
F large → at least one group differs (compare to the F table)Why F near 1 is the neutral point: if H₀ is true, MS_Between and MS_Within are two independent estimates of the same underlying variance, and two estimates of the same thing have a ratio near
| df | Formula | Why |
|---|---|---|
| Between | k − 1 | k groups, minus one for the grand mean |
| Within | N − k | N observations, minus one per group mean |
| Total | N − 1 | And the two above must add to this |
Like chi-square it is a ratio of squared quantities, so it is never negative and the test is always right-tailed. It has two df, numerator (effect) and denominator (error), and the order matters: F(2, 12) is not F(12, 2).
| Term | Meaning |
|---|---|
| Factor | A variable you group by. Packaging material, storage temperature |
| Level | One value of a factor. Plastic and Glass are the two levels of Packaging |
| Replication | Repeating each combination more than once, so noise can be estimated |
| Main effect | The effect of one factor on its own, averaged over the other |
| Interaction | When the effect of one factor depends on the level of the other |
| Post-hoc test | Run after a significant F to find which groups differ. Tukey's HSD |
Learning this fresh rather than revising? Do the first pass in Notes Module 5, which builds the idea up from the multiple-comparisons problem. This section is the drill layer.
Priority 2 · syllabus topic 8, not reported by anyone.
Use it when: three or more groups, one numeric outcome, one grouping factor. Raw data per group, or a partly filled table with Between and Within rows.
The one idea: split the total variation into the part between groups (signal) and the part within groups (noise), then take their ratio.
SS_Total = SS_Between + SS_Within
↑ ↑ ↑
all the how far each how far each
variation GROUP MEAN OBSERVATION sits
there is sits from the from ITS OWN
grand mean group mean
SIGNAL NOISE
MS_Between signal per df
F = ──────────── = ───────────── F near 1 → no group effect
MS_Within noise per df F large → real differencesA manufacturer tested three training methods on 15 employees, 5 per method. Output per shift:
Method A 12 14 11 13 15 Method B 18 20 17 19 16 Method C 22 25 24 21 23 Formulas provided:
SS_Between = Σ nⱼ(x̄ⱼ − x̄)²·SS_Within = ΣΣ(xᵢⱼ − x̄ⱼ)²·SS_Total = SS_Between + SS_Within·df1 = k−1·df2 = N−k·MS = SS/df·F = MS_Between / MS_WithinF critical at α = 0.05: F(2, 12) = 3.89(a) State the hypotheses and complete the ANOVA table. [6 marks] (b) State your conclusion at the 5% level and explain what further analysis is needed. [2 marks] (c) Explain why ANOVA is preferred to three separate two-sample t-tests. [2 marks]
(a) Hypotheses
H₀ : μ_A = μ_B = μ_C all three methods give the same mean output
H₁ : AT LEAST ONE mean differs (not "all three differ")Then the table, worked as seven steps. One operation per step, same order every time.
Step 1 — group means and grand mean
x̄_A = 65/5 = 13
x̄_B = 90/5 = 18
x̄_C = 115/5 = 23
x̄ = 270/15 = 18 ← grand mean
k = 3 groups, nⱼ = 5 each, N = 15Step 2 — SS Between: how far each GROUP MEAN sits from the grand mean. The signal
SS_B = Σ nⱼ(x̄ⱼ − x̄)²
= 5(13−18)² + 5(18−18)² + 5(23−18)²
= 5(25) + 5(0) + 5(25)
= 125 + 0 + 125 = 250Step 3 — SS Within: squared deviations INSIDE each group, from its own mean, added up. The noise
| Group | Values | Mean | Deviations | Squares | Sum |
|---|---|---|---|---|---|
| A | 12, 14, 11, 13, 15 | 13 | −1, +1, −2, 0, +2 | 1, 1, 4, 0, 4 | 10 |
| B | 18, 20, 17, 19, 16 | 18 | 0, +2, −1, +1, −2 | 0, 4, 1, 1, 4 | 10 |
| C | 22, 25, 24, 21, 23 | 23 | −1, +2, +1, −2, 0 | 1, 4, 1, 4, 0 | 10 |
| SS_Within | 30 |
Step 4 — check the decomposition before moving on
SS_Total = SS_Between + SS_Within = 250 + 30 = 280 ✓
and summing (x − 18)² across all 15 values also gives 280 ✓If the two disagree, something above is wrong. It is a free error-check, so do it every time.
Step 5 — mean squares: variation per degree of freedom
MS_Between = 250 / (k−1) = 250 / 2 = 125.0
MS_Within = 30 / (N−k) = 30 / 12 = 2.5Step 6 — F: signal per df over noise per df
F = 125.0 / 2.5 = 50.0 df = (2, 12)Step 7 — the finished table, ready to copy onto the paper
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Between groups | 250 | k−1 = 2 | 250/2 = 125.0 | 50.0 |
| Within groups (error) | 30 | N−k = 12 | 30/12 = 2.5 | |
| Total | 280 | N−1 = 14 |
(b) Compare and conclude
F = 50.0 vs F(2,12) critical = 3.89
50.0 ≫ 3.89 → REJECT H₀At the 5% level there is very strong evidence that mean output differs across the three training methods. A post-hoc test such as Tukey's HSD is needed to identify which pairs differ, since ANOVA only establishes that at least one does. The means of 13, 18 and 23 suggest all three do.
Look at why F came out so large. The group means 13, 18 and 23 sit far apart, while inside each group no value strays more than 2 from its own mean. Huge signal, tiny noise, ratio 50.
Effect size: η² = SS_Between / SS_Total = 250/280 = 0.893
→ 89.3% of the variation in output is explained by which method was used.
F says the effect is REAL. η² says it is also LARGE.(c) Why not three t-tests ← answer with the calculation, not the assertion
k groups → k(k−1)/2 pairwise comparisons
each test at α = 0.05 → 95% chance of avoiding a false positive
k = 3 : 3 tests → 1 − 0.95³ = 0.143 about 14%
k = 4 : 6 tests → 1 − 0.95⁶ = 0.265 about 27%
k = 5 : 10 tests → 1 − 0.95¹⁰ = 0.401 about 40%
↑
you INTENDED to work at 5% and are actually hereThis is inflation of the family-wise Type I error rate. ANOVA asks one question at one α, so the rate stays at 5% however many groups there are. It also pools the error term across all N observations, giving a more stable estimate and more power.
The three assumptions, if asked:
| Assumption | Check with | If it fails |
|---|---|---|
| Independence | The study design | Nothing statistical fixes it. Redesign |
| Normality of residuals | Q-Q plot, Shapiro-Wilk | Kruskal-Wallis, the non-parametric version |
| Equal variances | Levene's test | Welch's ANOVA |
Loses marks: writing H₁ as "all three differ"; using N − 1 or k for the within df; concluding "reject" without naming the post-hoc test; asserting Type I inflation without the 1 − 0.95ᵏ calculation.
Priority 3 · model paper only, Q2, 10 marks.
Use it when: two grouping factors, and almost always a partially completed ANOVA table with blanks to fill in.
The one idea: always work df first, because the df come from counting levels and need no arithmetic. Everything else follows from them.
THE FILL ORDER, with the reason each step sits where it does
1 count levels a, b, reps n. check N = a×b×n the df in step 2 need these counts
2 ALL the df a−1, b−1, (a−1)(b−1), N−ab first because they are pure counting,
and they must sum to N−1 no arithmetic yet to get wrong
3 the ODD ONE OUT read WHICH cell each row fills one row usually gives MS and hides SS,
before autopiloting breaking the pattern. Invert: SS = MS × df
4 every MS MS = SS / df
5 MS_error compute it early, keep visible EVERY F divides by this one number
6 every F F = MS_effect / MS_error effect over error, never the reverse
7 verify the SS must add to the Total SS free error-check, often asked by nameA food processing company studies the effect of two factors on shelf life (days): Packaging Material (Plastic, Glass) and Storage Temperature (Low, Medium, High). A balanced experiment with 3 replications per combination gives N = 18 observations.
Source SS df MS F Packaging 150.0 A D G Temperature 280.0 B E H Interaction C F 45.0 I Error 180.0 12 Total 700.0 17 Formulas provided:
MS = SS/df·F = MS_effect / MS_error·df_A = a−1·df_B = b−1·df_A×B = (a−1)(b−1)·df_error = N − ab·SS_Total = SS_A + SS_B + SS_A×B + SS_ErrorF critical at α = 0.05: F(1, 12) = 4.75 · F(2, 12) = 3.89(a) State the null hypotheses for the two main effects and the interaction. In practical terms, explain what a significant interaction would mean for the company. [3 marks] (b) Calculate A through I. Also verify the Total SS. Show your working. [5 marks] (c) Determine which effects are significant at α = 0.05. If the interaction is significant, explain why interpreting the main effects alone would be misleading. [2 marks]
(a) The three null hypotheses
| Effect | H₀ |
|---|---|
| Packaging | μ(Plastic) = μ(Glass). Packaging has no effect on mean shelf life |
| Temperature | μ(Low) = μ(Med) = μ(High). Temperature has no effect on mean shelf life |
| Interaction | No interaction: the effect of packaging is the same at every temperature |
What an interaction is, drawn
NO INTERACTION INTERACTION
shelf │ glass ──────── shelf │ glass ╲ ╱ plastic
life │ life │ ╲ ╱
│ plastic ─────── │ ╲ ╱
│ │ ╱ ╲
│ │ plastic ╱ ╲ glass
└──┬──────┬──────┬──► └──┬──────┬──────┬──►
Low Med High Low Med High
PARALLEL lines → no interaction CROSSING lines → interaction
the effects simply ADD the best packaging DEPENDS on temperatureThe practical answer, which is the mark most candidates miss
A significant interaction means there is no single best packaging material. Glass might extend shelf life at low temperature while plastic does better at high temperature, so the company cannot issue one blanket instruction and must specify packaging per storage condition. It also means a statement like "glass adds 5 days" is meaningless on its own, because that figure is an average across temperatures that may describe none of the actual conditions.
(b) Fill the table in the order above
1 LEVELS a = 2 packaging, b = 3 temperature, n = 3 reps
check N = 2 × 3 × 3 = 18 ✓
2 df A = a − 1 = 2 − 1 = 1
B = b − 1 = 3 − 1 = 2
F = (a−1)(b−1) = 1 × 2 = 2
N − ab = 18 − 6 = 12 matches the table
check 1 + 2 + 2 + 12 = 17 = df_Total ✓
3 ODD ONE OUT the table gives interaction MS and wants its SS, so INVERT:
C = MS × df = 45.0 × 2 = 90.0 ← what the question is really testing
4 MS D = 150/1 = 150.0
E = 280/2 = 140.0
MS_error = 180/12 = 15.0 ← the yardstick
5 F G = 150/15 = 10.00
H = 140/15 = 9.33
I = 45/15 = 3.00
6 VERIFY 150 + 280 + 90 + 180 = 700 = stated Total SS ✓| Source | SS | df | MS | F |
|---|---|---|---|---|
| Packaging | 150.0 | 1 | 150.0 | 10.00 |
| Temperature | 280.0 | 2 | 140.0 | 9.33 |
| Interaction | 90.0 | 2 | 45.0 | 3.00 |
| Error | 180.0 | 12 | 15.0 | |
| Total | 700.0 | 17 |
(c) Match each F to its OWN pair of df. They are not all the same.
| Effect | F | df | F critical | Decision |
|---|---|---|---|---|
| Packaging | 10.00 | 1, 12 | 4.75 | 10.00 > 4.75 → significant |
| Temperature | 9.33 | 2, 12 | 3.89 | 9.33 > 3.89 → significant |
| Interaction | 3.00 | 2, 12 | 3.89 | 3.00 < 3.89 → not significant |
3.00 against 3.89 is a near miss, and a near miss fails. The paper is testing whether you will round it up into significance. Do not.
The reading order rule, which earns the second mark
ALWAYS check the interaction FIRST.
interaction NOT significant → main effects can be read directly and separately
pick the best level of each independently ← this case
interaction significant → main effects are MISLEADING. report simple effects insteadThen answer the hypothetical anyway, since the question asked it
Had the interaction been significant, interpreting a main effect alone would mislead, because a main effect is an average across the other factor's levels that can describe none of the actual conditions. If glass added 10 days at low temperature but lost 4 at high, the main effect would report roughly +3 days, true of no real storage situation. You would report the simple effects instead: the effect of packaging separately at each temperature.
Loses marks: computing MS or F for the Total row; using one critical value for all three effects when packaging has 1 df and the others have 2; dividing error by effect; skipping the Total SS verification the question asks for by name.
Everything so far compared groups. Regression asks a different question: when x goes up, what happens to y? Advertising and sales. Hours studied and marks. One variable is used to explain or predict another.
x the PREDICTOR, the thing you control or know first (advertising)
y the RESPONSE, the thing you want to explain (sales)A single number from −1 to +1 saying how tightly the points follow a straight line, and which way.
r = +0.9 r = 0 r = −0.9
· · · · · · ·
· · · · · · · ·
· · · · · · ·
strong up no line strong downr says how strong. It does not say how much y changes per unit of x. That is the slope.
ŷ = b₀ + b₁ x
b₀ the INTERCEPT the predicted y when x = 0
b₁ the SLOPE how much ŷ changes for each one-unit rise in x ← the number that matters
ŷ "y hat" the PREDICTED y, as opposed to the actual yFor each data point, the residual is the vertical gap between the actual y and the line's ŷ. Least squares picks the one line that makes the sum of the squared residuals as small as possible. The formulas for b₀ and b₁ are printed on the paper; C13 shows the working table.
y │ ●
│ ● ╱│ residual = y − ŷ
│ ╱│ ╱ │
│ ╱ │ ╱ ●
│ ╱ ●
└────────────────► xStart from the dumbest possible prediction: ignore x completely and predict the mean ȳ for every point. That flat line misses badly. Fit the regression line instead and the misses shrink. R² is the fraction of the original miss that the line removed.
PREDICT WITH ȳ ALONE PREDICT WITH THE FITTED LINE
y │ ● ● y │ ● ● ╱
│ ────────────── ȳ │ ╱
│ ● ● │ ● ╱ ●
└──────────────► x └──────────────► x
each miss = y − ȳ each miss = y − ŷ
square, add up → SST square, add up → SSE, SMALLER:
the line ATE part of the miss.
the part it ate is SSR SST = SSR + SSE
miss with part the miss still left
ȳ alone line ate (the residuals)
SSR miss removed by the line
R² = ───── = ─────────────────────────
SST miss you started with
R² = 0 the line did nothing. it IS the flat ȳ line
R² = 1 every point exactly on the line, SSE = 0
R² = 0.60 the line removed 60% of the miss. the other 40% is other factorsKeep the letters straight: SSR is the Regression part, the explained bit. SSE is the Error part, the residuals. This is the ANOVA split again, total = explained + leftover, and R² plays exactly the role η² played there.
Same idea with several predictors at once.
ŷ = b₀ + b₁ x₁ + b₂ x₂ + b₃ x₃Each coefficient is now a partial effect: what that predictor adds holding the others constant. That phrase is the whole difference from simple regression, and it is marked on every coefficient you interpret.
| Test | Asks | Read it |
|---|---|---|
| Overall F | Does the model as a whole explain anything? | First. If it fails, stop |
| t for each coefficient | Does this one predictor earn its place? | p < 0.05 means yes |
Plain R² can never fall when you add a predictor, even a useless one, so it cannot compare models of different sizes. Adjusted R² subtracts a penalty per predictor and can fall. C14 covers it.
Priority 2 · syllabus topic 9, not reported by anyone.
Use it when: paired numeric x and y data, and the instruction fit the least squares line, interpret the slope, or predict.
The one idea: find the line that makes the total squared vertical miss as small as possible.
y │ ● residual = actual y − predicted ŷ
│ ● ╱│ = the VERTICAL gap
│ ╱│ ╱ │
│ ╱ │ ╱ ● least squares picks the line that
│ ╱ ● │ MINIMISES the sum of these gaps SQUARED
│╱ │ │
└──────┴─────┴──────► x
A company recorded advertising spend (lakhs) and sales (crores) over five months:
Advertising x 1 2 3 4 5 Sales y 2 4 5 4 5 Formulas provided:
b₁ = Σ(x−x̄)(y−ȳ) / Σ(x−x̄)²·b₀ = ȳ − b₁x̄·R² = SSR/SST(a) Fit the least squares regression line. [4 marks] (b) Interpret the slope and the intercept. [3 marks] (c) Compute R-squared and predict sales when advertising is 6. [3 marks]
(a) Means first, then ONE deviation table
x̄ = 15/5 = 3 ȳ = 20/5 = 4Build this table and every quantity you need falls out of the three totals.
| x | y | x − x̄ | y − ȳ | (x−x̄)(y−ȳ) | (x−x̄)² | (y−ȳ)² |
|---|---|---|---|---|---|---|
| 1 | 2 | −2 | −2 | 4 | 4 | 4 |
| 2 | 4 | −1 | 0 | 0 | 1 | 0 |
| 3 | 5 | 0 | +1 | 0 | 0 | 1 |
| 4 | 4 | +1 | 0 | 0 | 1 | 0 |
| 5 | 5 | +2 | +1 | 2 | 4 | 1 |
| Totals | 6 | 10 | 6 |
column 5 total → Σ(x−x̄)(y−ȳ) = 6 drives the slope
column 6 total → Σ(x−x̄)² = 10 denominator of the slope
column 7 total → Σ(y−ȳ)² = 6 this is SST, for R² Σ(x−x̄)(y−ȳ) 6
b₁ = ──────────────── = ──── = 0.6
Σ(x−x̄)² 10
b₀ = ȳ − b₁x̄ = 4 − 0.6 × 3 = 4 − 1.8 = 2.2
→ ŷ = 2.2 + 0.6x(b) Interpret both, in the units of the question
Slope 0.6: for every additional lakh spent on advertising, sales rise by an estimated 0.6 crore on average. Say "associated with", not "causes": this is observational data.
Intercept 2.2: predicted sales at zero advertising, so about 2.2 crore of baseline sales. Here x = 0 lies just outside the observed range of 1 to 5, so this is a mild extrapolation but not absurd.
Where x = 0 is far outside the data or physically impossible, call the intercept a mathematical anchor with no real interpretation.
(c) R-squared and the prediction
SST = Σ(y−ȳ)² = 6
SSR = b₁ × Σ(x−x̄)(y−ȳ) = 0.6 × 6 = 3.6
SSE = SST − SSR = 6 − 3.6 = 2.4
SSR 3.6
R² = ─────── = ───── = 0.60
SST 6Every number here was already on the deviation table: SST is the last column's total, and SSR is b₁ times the same cross-product total that built the slope. No new sums are needed. R² = 0.60 says the line removed 60% of the miss that predicting with ȳ alone would have made.
60% of the variation in sales is explained by advertising, leaving 40% to other factors. The correlation is r = √0.60 = 0.775, positive because the slope is.
ŷ at x = 6 = 2.2 + 0.6 × 6 = 5.8 crore
⚠ x = 6 is BEYOND the observed range of 1 to 5.
This is EXTRAPOLATION and the relationship is not guaranteed to continue.
A prediction at x = 50 would be indefensible. ← say this, it is a markIf it asks whether the slope is significant, answer honestly
MSE = SSE/(n−2) = 2.4/3 = 0.80 n−2 because the line
estimated TWO things, b₀ and b₁
SE(b₁) = √( MSE / Σ(x−x̄)² ) = √(0.80/10) = √0.08 = 0.283
t = b₁ / SE(b₁) = 0.6 / 0.283 = 2.12, df = n − 2 = 3
t critical (3, 0.025) = 3.182
2.12 < 3.182 → FAIL TO REJECT H₀: β₁ = 0With only five data points there is not enough evidence that the relationship is real, even though R² is 0.60. A reasonable R-squared on a tiny sample is not by itself evidence of anything.
The LINE assumptions, if asked:
L Linearity residuals vs fitted show NO pattern
I Independence comes from the study design
N Normality of RESIDUALS not of X or Y. Q-Q plot
E Equal variance constant-width band in the residual plot
The single most useful diagnostic, residuals against fitted values:
shapeless band ✓ good a CURVE → linearity failed
· · · ╱▔╲
· ·· · · ╱ ╲ a FUNNEL → equal variance failed
· · · ╱ ╲ · ·· ·······
·· ·······Loses marks: saying advertising causes sales; predicting at x = 6 without flagging extrapolation; treating R² = 0.60 as proof the slope is significant; interpreting the intercept without asking whether x = 0 is meaningful.
Priority 1 · reported by students.
Use it when: two models with different numbers of predictors to choose between, or the words explain the difference between R-squared and adjusted R-squared.
The one idea: plain R² can never fall when you add a predictor, even a useless one. So it cannot compare models of different size. Adjusted R² can fall, so it can.
add a predictor → model gets one more free parameter to fit with
→ fitted values CANNOT get worse
→ SSR cannot fall
→ R² ALWAYS rises or stays flat ← even for random noise
R² ▁▂▃▄▅▆▇█ always up
Adjusted R² ▁▂▃▄▅▄▃▂ turns DOWN once predictors stop earning their place
↑
best model(a) Explain the difference between R-squared and adjusted R-squared, and state when each should be used. [4 marks] (b) An analyst fits two models to the same dataset of n = 50 observations:
Model Predictors R-squared A 3 0.720 B 5 0.735 Formula provided:
Adjusted R² = 1 − (1 − R²)(n − 1)/(n − k − 1)Compute the adjusted R-squared for each and recommend which model to use, justifying your choice. [4 marks]
(a) Define both, then contrast
R² is the proportion of total variation in the response explained by the model, SSR/SST. Adjusted R² measures the same thing with a penalty for the number of predictors.
| R-squared | Adjusted R-squared | |
|---|---|---|
| Formula | SSR / SST | 1 − (1 − R²)(n−1)/(n−k−1) |
| Adding a predictor | Never falls | Can fall |
| Use for | Reporting what the current model explains | Comparing models of different size |
| Range | 0 to 1 | Can go slightly negative for a very poor model |
The mark that carries (a) is the mechanism in the box above: one more free parameter means the fit cannot get worse, so R² is structurally incapable of penalising a useless variable.
(b) Substitute for each model
MODEL A : n = 50, k = 3, R² = 0.720
adj = 1 − (1 − 0.720) × (50 − 1)/(50 − 3 − 1)
= 1 − 0.280 × (49/46)
= 1 − 0.280 × 1.0652
= 1 − 0.2983
= 0.702
MODEL B : n = 50, k = 5, R² = 0.735
adj = 1 − (1 − 0.735) × (49)/(50 − 5 − 1)
= 1 − 0.265 × (49/44)
= 1 − 0.265 × 1.1136
= 1 − 0.2951
= 0.705 R² Adjusted R²
Model A 0.720 → 0.702
Model B 0.735 → 0.705
▲ +0.015 ▲ +0.003
│ │
looks like a gain almost nothing, for TWO extra predictorsRecommend Model A. Plain R² rose from 0.720 to 0.735, as it always will when predictors are added. Adjusted rose only from 0.702 to 0.705, so the two extra predictors bought almost nothing. Model A explains nearly as much with fewer predictors, making it easier to interpret and less prone to overfitting.
Loses marks: saying adjusted R² is "more accurate", when it is penalised rather than more accurate; recommending Model B because 0.735 > 0.720, which is the trap; not explaining why R² cannot fall.
Priority 1 · reported by students, since R-squared lives here.
Use it when: a table of coefficients, standard errors, t and p values, with R² and an F-statistic underneath, and t