Mark Scheme

General instructions to examiners. Method marks (M) are for a correct process; accuracy marks (A) require the correct value. Error carried forward (ECF) applies throughout: a candidate whose earlier value is wrong but who applies a valid method to it retains all method marks and any accuracy marks consistent with their own figure. Accept answers within reasonable rounding of tabulated values. Penalise premature rounding once only across the whole paper. In hypothesis-testing parts, the conclusion in context carries its own mark and is not awarded for a bare "reject H₀".

Section A — Module 1

Question 1

(a) (5) — 3 marks for the distinction: a population is the entire set of individuals or items about which information is sought and to which the conclusions will apply; a sample is a subset of the population, selected for study, from which inferences about the population are drawn. 1 mark each for any two reasons: cost; time; the population may be very large or its membership not fully known; testing may be destructive (a fireworks manufacturer cannot test every firework); a sample allows greater care and accuracy per item; access to the whole population may be physically impossible.

• Common error: defining a sample as "a small part" — size is not the discriminator; representativeness and the purpose of inference are.

(b) (6) — Up to 4 marks for the method, 2 for the advantage.

• Identify the strata: the five year groups, which are the natural non-overlapping subgroups (1).
• Determine the number in each stratum and calculate the sampling fraction 200/1500 = 2/15 ≈ 0.1333 (1).
• Apply the fraction to each stratum so that each year group is represented in proportion to its size in the school — for example, a year group of 300 contributes 300 × 2/15 = 40 students (1).
• Within each stratum, select the required number by simple random sampling, using a numbered list or register and random numbers, without replacement (1). This mark is frequently lost: candidates describe the proportional allocation but never say how individuals are chosen inside a stratum, and a stratified sample that is not random within strata is not a random sample.
• Advantage (2): the sample is guaranteed to reflect the year-group composition of the school, whereas a simple random sample could by chance over- or under-represent a year group; since the variable of interest (attainment, opinion, height) is likely to differ systematically by year group, this reduces sampling error and produces more precise estimates.

(c) (i) (4) — The data are already in ascending order; n = 10. Median = mean of the 5th and 6th values = (25 + 27)/2 = 26 (M1 A1). Lower quartile Q₁: the median of the lower five values (12, 15, 18, 22, 25) = 18 (A1). Upper quartile Q₃: the median of the upper five values (27, 30, 33, 35, 41) = 33 (A1).

• Accept the alternative convention giving Q₁ = 16.5 and Q₃ = 34 provided the method is stated and used consistently; adjust subsequent parts by ECF.
• Common error: taking Q₁ as the 2.5th value without interpolation and then using an inconsistent rule for Q₃.

(c) (ii) (4) — IQR = Q₃ − Q₁ = 33 − 18 = 15 (M1 A1). Outlier boundaries: Q₁ − 1.5 × IQR = 18 − 22.5 = −4.5 and Q₃ + 1.5 × IQR = 33 + 22.5 = 55.5 (M1). All ten values lie between −4.5 and 55.5, so there are no outliers; in particular 41, the largest value, is well below 55.5 (A1).

• The final mark requires an explicit conclusion. A candidate who computes both boundaries but does not state the verdict scores 3.
• Common error: using 1.5 × IQR measured from the median rather than from the quartiles.

(c) (iii) (6) — Marks are for the candidate's own drawing; no plot is supplied.

• (1) A horizontal (or vertical) scale drawn and labelled with units — "Waiting time (minutes)" — and a linear scale covering at least 10 to 41.
• (1) Box drawn with its left edge at Q₁ = 18 and right edge at Q₃ = 33.
• (1) Median line drawn inside the box at 26.
• (1) Whiskers extended from the box to the minimum 12 and the maximum 41. Since part (ii) established that there are no outliers, the whiskers correctly run to the extreme values; a candidate who found an outlier in (ii) and truncated a whisker accordingly earns this mark by ECF.
• (2) Comment on skewness: the median at 26 sits 8 units above Q₁ and 7 units below Q₃, and the lower whisker (12 to 18, length 6) is shorter than the upper whisker (33 to 41, length 8). The distribution is therefore approximately symmetric with a slight positive (right) skew, the longer upper whisker being the clearer indication. Award 1 mark for naming the skew and 1 for justifying it by reference to specific features of the candidate's own plot; "it is skewed because it looks skewed" earns nothing.
• Common errors: drawing a bar chart or histogram instead of a box plot; omitting the scale entirely, which forfeits the first mark and makes the plot unreadable; placing the median at the centre of the box regardless of its value; describing right skew as "negative".

Question 2

(a) (4) — 1 mark each for the definitions and 1 each for the examples. Discrete data can take only particular separated values, usually arising from counting — e.g. the number of vehicles passing a junction in an hour, the number of occupants in a car. Continuous data can take any value within a range, limited only by the precision of measurement, and arise from measuring — e.g. the speed of a vehicle, the time taken to travel a route, the distance between two junctions. Do not accept "discrete means whole numbers" alone, since shoe sizes progress in halves and remain discrete.

(b) (i) (4) — Using midpoints x = 5, 15, 25, 35, 45 (M1 — midpoints correctly identified).

Mean = Σfx/Σf = 1300/50 (M1) = 26 minutes (A1 A1 — one mark for the correct total Σfx, one for the final value).

• Common errors: using the lower class boundary instead of the midpoint, giving 22; dividing by 5 (the number of classes) instead of 50.

(b) (ii) (5) — Σfx² = 4(5²) + 11(15²) + 18(25²) + 10(35²) + 7(45²) (M1) = 100 + 2475 + 11250 + 12250 + 14175 = 40,250 (A1). Variance = Σfx²/Σf − (mean)² = 40250/50 − 26² (M1) = 805 − 676 = 129 (A1). Standard deviation = √129 = 11.4 minutes (11.358) (A1).

• Accept the divisor (n − 1) with the sample formula, giving variance 131.6 and s = 11.5, provided the formula used is stated.
• ECF from an incorrect mean in (i) is allowed in full.
• Common errors: computing Σ(fx)² instead of Σfx²; forgetting to subtract the square of the mean; giving the variance as the final answer when the standard deviation is asked for — this last accounts for more lost marks in this part than any genuine calculation error.

(b) (iii) (2) — Because the data have been grouped, the individual values within each class are unknown; the calculation assumes that every observation in a class lies at the class midpoint, and hence that the values are evenly distributed within each class (1). In reality they are not, so the computed mean and standard deviation are estimates of the true values that would be obtained from the raw data (1).

• Common error: answering "because we rounded" — rounding is not the source of the approximation here.

(b) (iv) (5) — Modal class is 20 ≤ t < 30, since it has the highest frequency, 18 (B1 — the class must be given, not the frequency 18 nor the midpoint 25). Median position: n/2 = 25th value (M1). Cumulative frequencies are 4, 15, 33, 43, 50, so the 25th value lies in the class 20 ≤ t < 30 (A1). Linear interpolation: median = 20 + [(25 − 15)/18] × 10 (M1) = 20 + (10/18) × 10 = 20 + 5.556 = 25.6 minutes (A1).

• Common errors: using (n + 1)/2 = 25.5 for grouped continuous data (condone, giving 25.83, but the convention for grouped data is n/2); dividing by the cumulative frequency rather than the class frequency; using a class width other than 10.

(c) (5) — Up to 3 marks for the principle and up to 2 for the illustration. The mean uses every value and is therefore sensitive to extreme values; a small number of very large or very small observations pulls it towards the tail. The median depends only on the position of the middle value and is resistant to such extremes. The median is therefore preferable when the distribution is markedly skewed or contains outliers, since the mean would then not be representative of a typical observation (3). Illustration: the distribution of household incomes is strongly positively skewed — the great majority of households cluster at the lower and middle end of the range, while a small number of very high incomes form a long right-hand tail. Those few high incomes pull the arithmetic mean well above the income of a typical household, so the mean overstates typical living standards; the median household income is the more honest summary, and is the measure reported by statistical offices for exactly this reason (2).

• Common error: asserting that the median is "always better", or describing income distribution as negatively skewed.

Section B — Module 2

Question 3

(a) (i) (3) — A and B are independent if and only if P(A ∩ B) = P(A) × P(B) (M1 — the criterion must be stated). P(A) × P(B) = 0.4 × 0.5 = 0.2 (A1). Since this equals the given P(A ∩ B) = 0.2, the events are independent (A1 — the conclusion is a separate mark).

• Common error: testing for mutual exclusivity (P(A ∩ B) = 0) and concluding "not independent" — the two concepts are distinct and are frequently confused.

(a) (ii) (2) — P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (M1) = 0.4 + 0.5 − 0.2 = 0.7 (A1).

(a) (iii) (5) — P(A | B) = P(A ∩ B)/P(B) (M1) = 0.2/0.5 = 0.4 (A1). Note that this equals P(A), which confirms the independence established in (i) — reward candidates who make this observation. P(B | A′) = P(B ∩ A′)/P(A′) (M1). P(B ∩ A′) = P(B) − P(A ∩ B) = 0.5 − 0.2 = 0.3; P(A′) = 1 − 0.4 = 0.6 (A1). So P(B | A′) = 0.3/0.6 = 0.5 (A1).

• Common errors: inverting the conditional to P(B)/P(A ∩ B); computing P(B ∩ A′) as P(B) × P(A′) without first establishing independence (here it happens to give the same answer, which conceals the error — award the marks but note that the reasoning is only valid because independence was proved in (i)).

(b) (i) (4) — Marks are for the candidate's own tree diagram; none is supplied.

• (1) First stage drawn with two branches, labelled Red and Blue, carrying probabilities 7/12 and 5/12.
• (1) Second stage drawn with two branches from each first-stage branch, correctly labelled — four end-points in total.
• (1) Second-stage probabilities correct for the Red-first path: R then R = 6/11, R then B = 5/11. Denominators of 11 are the crux: the draw is without replacement.
• (1) Second-stage probabilities correct for the Blue-first path: B then R = 7/11, B then B = 4/11.
• Common errors: using 7/12 and 5/12 at the second stage, i.e. treating the draws as with replacement — this single error typically costs two marks here and both marks in (ii); failing to label which branch is which colour.

(b) (ii) (4) — P(different colours) = P(R then B) + P(B then R) (M1 — both paths identified) = (7/12)(5/11) + (5/12)(7/11) (M1 — correct products, ECF from the tree) = 35/132 + 35/132 = 70/132 (A1) = 35/66 ≈ 0.530 (A1).

• Accept the complementary method: 1 − [P(RR) + P(BB)] = 1 − [42/132 + 20/132] = 1 − 62/132 = 70/132.
• Common error: computing only one path and giving 35/132.

(b) (iii) (4) — This is a conditional probability requiring Bayes' reasoning. P(1st R | 2nd B) = P(1st R ∩ 2nd B) / P(2nd B) (M1 — correct structure). P(1st R ∩ 2nd B) = (7/12)(5/11) = 35/132 (A1). P(2nd B) = P(R then B) + P(B then B) = 35/132 + (5/12)(4/11) = 35/132 + 20/132 = 55/132 (M1 — both routes to a blue second marble). P(1st R | 2nd B) = (35/132)/(55/132) = 35/55 = 7/11 ≈ 0.636 (A1).

• Common errors: answering 7/12, i.e. ignoring the conditioning information entirely; computing P(2nd B) from only the one path that appears in the numerator, which is the most frequent structural error in Bayes questions.

(c) (3) — 1 mark each for any three of: a fixed number of trials, n; each trial has exactly two outcomes, success and failure; the probability of success, p, is constant from trial to trial; the trials are independent of one another.

• Common error: "the data must be discrete" — true but not a defining condition, and not creditworthy.

Question 4

(a) (i) (2) — X ~ B(10, 0.2) (B1 for the binomial model, B1 for both parameters).

(a) (ii) (3) — P(X = 3) = C(10, 3)(0.2)³(0.8)⁷ (M1 for the structure, A1 for correct C(10,3) = 120 and correct powers) = 120 × 0.008 × 0.2097152 = 0.201 (A1).

• Common errors: using (0.8)⁶ or (0.2)⁷; omitting the binomial coefficient entirely, giving 0.00168.

(a) (iii) (3) — P(X ≥ 1) = 1 − P(X = 0) (M1 — the complement must be recognised) P(X = 0) = (0.8)¹⁰ = 0.10737 (A1) P(X ≥ 1) = 1 − 0.10737 = 0.893 (A1).

• Common error: attempting to sum P(X = 1) through P(X = 10) — award full marks if completed correctly, but note the inefficiency; more often the candidate computes 1 − P(X = 1), which is a different and incorrect quantity.

(a) (iv) (3) — E(X) = np = 10 × 0.2 = 2 (M1 A1). Var(X) = np(1 − p) = 10 × 0.2 × 0.8 = 1.6 (A1).

• Common error: giving the standard deviation 1.26 when the variance is asked for.

(b) (i) (3) — X ~ Po(3). P(X = 2) = e⁻³ × 3²/2! (M1 — correct Poisson formula) = 0.049787 × 4.5 (A1) = 0.224 (A1).

(b) (ii) (3) — P(X ≤ 1) = P(X = 0) + P(X = 1) (M1 — both terms required) = e⁻³ + e⁻³ × 3 = e⁻³(1 + 3) (A1) = 0.049787 × 4 = 0.199 (A1).

• Common error: interpreting "at most 1" as P(X = 1) only, giving 0.149.

(c) (i) (3) — Marks are for the candidate's own sketch; no curve is supplied.

• (1) A symmetric bell-shaped curve drawn with a labelled horizontal axis (mass, kg) and the curve approaching but not meeting the axis at both ends.
• (1) The mean marked on the axis at 50 and the line of symmetry through it indicated.
• (1) The value 60 marked on the axis to the right of the mean, and the region under the curve to the right of 60 shaded. The shaded region must be the tail, not the region between 50 and 60; this is the mark most often lost.

(c) (ii) (3) — Z = (60 − 50)/8 (M1) = 1.25 (A1). P(X > 60) = P(Z > 1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.106 (A1).

• Common error: giving 0.8944, i.e. reading the table value directly without subtracting from 1 — the candidate's own sketch from (i) is the safeguard against this, which is precisely why the sketch is examined.

(c) (iii) (2) — Z₁ = (45 − 50)/8 = −0.625 and Z₂ = (58 − 50)/8 = 1.0 (M1 — both standardisations). P(45 < X < 58) = Φ(1.0) − Φ(−0.625) = 0.8413 − (1 − 0.7340) = 0.8413 − 0.2660 = 0.575 (A1).

• Accept 0.575 to 0.576 depending on interpolation in the tables.
• Common error: adding the two tail probabilities instead of subtracting the cumulative values.

Section C — Module 3

Question 5

(a) (4) — Up to 2 marks for the statement and up to 2 for the significance. Statement: for a random sample of size n drawn from any population with mean μ and finite variance σ², the distribution of the sample mean X̄ approaches a normal distribution with mean μ and variance σ²/n as n becomes large (in practice n ≥ 30 is usually taken as sufficient). Importance: it permits inference about a population mean — confidence intervals and hypothesis tests using the normal distribution — without knowing or assuming the shape of the underlying population distribution, provided the sample is large enough. This is what makes the normal-based procedures in this Module applicable to real data, whose distributions are rarely known and often skewed.

• Common error: stating that the theorem says the population becomes normal, or that the sample data become normal, as n increases. It is the distribution of the sample mean that tends to normality.

(b) (i) (3) — X̄ ~ N(50, 8²/25) (M1). Mean = 50 kg (A1). Standard error = σ/√n = 8/√25 = 8/5 = 1.6 kg (A1).

• Common errors: giving the standard error as 8/25 = 0.32, or leaving it as 8; giving the variance 2.56 when the standard error is asked for.

(b) (ii) (4) — Z = (52 − 50)/1.6 (M1 — standardising with the standard error, not σ) = 2/1.6 = 1.25 (A1). P(X̄ > 52) = P(Z > 1.25) = 1 − 0.8944 = 0.106 (A1 A1).

• ECF from an incorrect standard error in (i).
• Common error: dividing by 8 rather than by 1.6, giving Z = 0.25 and a probability of 0.401. This is the defining error of the topic: it treats a question about a sample mean as though it concerned a single observation, and the fact that this paper asks the numerically identical question about one bag in 4(c)(ii) is deliberate — the two answers coincide here only by the arithmetic accident that 10/8 and 2/1.6 are both 1.25, and candidates should not read anything into it.

(c) (i) (5) — n = 100 is large, so the normal approximation applies and s may be used in place of σ (B1 — a statement justifying the use of z rather than t). Standard error = s/√n = 12/√100 = 1.2 (M1 A1). 95% CI = x̄ ± 1.96 × SE = 78 ± 1.96 × 1.2 (M1 — correct critical value 1.96) = 78 ± 2.352. Interval: (75.6, 80.4) (A1).

• Common errors: using 1.645 (the 90% two-tailed or 95% one-tailed value); using 2.576 (99%); dividing 12 by 100 rather than by 10; presenting the answer as "78 ± 2.35" without converting to interval form — condone, but the endpoints are safer.

(c) (ii) (4) — 2 marks for the contextual interpretation, 2 for the precise meaning. Interpretation: the interval provides a range of plausible values for the population mean examination mark of all students, based on this sample; it is plausible that the true mean mark for the whole population lies between about 75.6 and 80.4 (2). Precise meaning: "95% confident" refers to the procedure, not to this particular interval. If the sampling and interval-construction process were repeated many times, approximately 95% of the intervals so constructed would contain the true population mean. It is not correct to say that there is a 95% probability that μ lies in this specific interval: μ is a fixed constant, and this interval either contains it or does not (2).

• The second mark of the pair is reserved for candidates who explicitly reject the "95% probability that μ is in this interval" reading. This is the standard misinterpretation and its correction is what discriminates the top band.

(c) (iii) (5) — Up to 2 marks per change with its disadvantage, plus 1 for overall quality.

• Increase the sample size. The standard error is s/√n, so a larger n reduces the width. Disadvantage: quadrupling n only halves the width, so the gain is subject to sharply diminishing returns, and the additional data cost time and money.
• Reduce the confidence level, for example to 90%, which replaces 1.96 by 1.645. Disadvantage: the interval is narrower only because it is less reliable — a greater proportion of such intervals would fail to contain μ, so precision has been bought directly at the cost of confidence, not gained.
• Reduce the variability of the data, for example by sampling a more homogeneous subgroup or improving measurement precision. Disadvantage: restricting the population changes what is being estimated, so the interval no longer refers to the original population of interest.
• Do not credit "take a better sample" or "use a more accurate calculator" — neither affects the width.

Question 6

(a) (i) (2) — H₀: μ = 500 (B1); H₁: μ < 500 (B1). The alternative must be one-tailed and in the correct direction, since the consumer group is testing whether bags are underweight.

• Common errors: stating hypotheses about the sample mean x̄ rather than the population mean μ; a two-tailed alternative μ ≠ 500, which does not match the question.

(a) (ii) (4) — Standard error = σ/√n = 15/√40 (M1) = 15/6.3246 = 2.3717 (A1). z = (x̄ − μ)/SE = (494 − 500)/2.3717 (M1) = −6/2.3717 = −2.53 (A1).

• Common errors: dividing by 15 rather than by the standard error, giving −0.4; dropping the negative sign and then failing to compare correctly against a negative critical value.

(a) (iii) (4) — One-tailed test at the 5% level: critical value z = −1.645 (B1 — the sign and the one-tailed value are both required). Since −2.53 < −1.645, the test statistic lies in the critical region (M1), so we reject H₀ (A1). Conclusion in context: there is sufficient evidence at the 5% significance level to conclude that the mean mass of the bags of cement is less than the 500 g claimed by the manufacturer (A1).

• The final mark is for the contextual conclusion. "Reject H₀" alone scores 3 of 4. Equally, a conclusion stated with unwarranted certainty — "the manufacturer is lying" or "we have proved the bags are underweight" — loses the mark: a significance test provides evidence, not proof.
• Accept an equivalent p-value approach: p = P(Z < −2.53) = 0.0057 < 0.05, so reject H₀.

(b) (9) — A χ² goodness-of-fit test.

• (1) H₀: the die is fair, i.e. all six scores are equally likely, each with probability 1/6. H₁: the die is not fair, i.e. at least one score does not have probability 1/6.
• (1) Expected frequency for each score = 100 × 1/6 = 16.667. All expected frequencies exceed 5, so the test is valid — award credit for candidates who check this condition.
• (3) Calculation of Σ(O − E)²/E:

(M1 for the correct formula, M1 for at least four correct terms, A1 for the total 7.88.) Accept 7.87 to 7.89.

• (2) Degrees of freedom = 6 − 1 = 5 (B1). Critical value χ²(5, 0.05) = 11.07 (B1).
• (2) Since 7.88 < 11.07, the test statistic does not fall in the critical region, so we do not reject H₀ (A1). There is insufficient evidence at the 5% significance level to conclude that the die is unfair; the observed variation, including the relatively high frequency of 25 for the score 4, is consistent with what would arise by chance from a fair die in 100 rolls (A1 — contextual conclusion).
• Common errors: using degrees of freedom 6 rather than 5; computing Σ(O − E)² without dividing by E, giving 155.3; using the expected frequency 16 or 17 rather than 16.667 throughout; concluding "the die is fair" — the test can only fail to find evidence of unfairness, and a conclusion asserting that H₀ is true loses the final mark.

(c) (6) — 2 marks for each definition and 1 for each application to the cement scenario. Type I error: rejecting H₀ when H₀ is in fact true. Its probability is the significance level α, here 0.05. In context: concluding that the mean mass is less than 500 g when in fact the machine is filling correctly to 500 g — the consumer group wrongly accuses the manufacturer, on the basis of a sample that happened to be light. Type II error: failing to reject H₀ when H₀ is in fact false. Its probability is denoted β. In context: concluding that there is insufficient evidence of underfilling when the true mean mass really is below 500 g — the manufacturer's underfilling goes undetected and consumers continue to be short-changed.

• Reward candidates who note the trade-off: reducing α to guard against a Type I error increases β for a given sample size, and only an increase in n reduces both.
• Common errors: defining a Type I error as "accepting H₁ when it is false" (a circumlocution that is often wrong on inspection); reversing the two; giving definitions without the contextual application, which caps the part at 4.

Level descriptors (extended and multi-step parts)

• Level 4 (top band): Correct statistical model identified and stated with its parameters; every stage of the method shown, including formulae before substitution; correct use of tables with values quoted to full precision before rounding; diagrams drawn with labelled scales and correctly shaded regions; hypotheses stated in terms of population parameters; conclusions expressed in context with appropriate tentativeness; validity conditions checked where relevant.
• Level 3: Correct method throughout with isolated arithmetic or table-reading slips; model identified but parameters incompletely stated; diagrams present but incompletely labelled; conclusions correct but stated as "reject H₀" without context.
• Level 2: The correct procedure is identified but misapplied in a way that changes its structure — standardising with σ rather than the standard error, treating a without-replacement problem as with replacement, using the wrong degrees of freedom; working fragmentary; answers stated without support.
• Level 1: Isolated correct values with no coherent method; formulae quoted but not applied.

Sample Answers with Examiner Commentary

Question 6(a) — Sample Answers

Grade I (Distinction) answer

"Let μ be the population mean mass of the bags of cement, in grams.

H₀: μ = 500 H₁: μ < 500 (one-tailed, since the consumer group is testing specifically for underfilling rather than for any departure from 500 g)

The population standard deviation σ = 15 is known and n = 40 ≥ 30, so by the Central Limit Theorem the sample mean is approximately normally distributed regardless of the distribution of individual bag masses, and a z-test is appropriate.

Under H₀, X̄ ~ N(500, 15²/40), so the standard error is 15/√40 = 2.3717.

z = (494 − 500)/2.3717 = −6/2.3717 = −2.53

At the 5% significance level in a one-tailed test, the critical value is z = −1.645. Since −2.53 < −1.645, the test statistic falls in the critical region.

Equivalently, p = P(Z < −2.53) = 0.0057, which is less than 0.05.

We therefore reject H₀. There is sufficient evidence at the 5% significance level to conclude that the mean mass of the bags is less than the 500 g the manufacturer claims. This does not prove the claim false — a result this extreme would arise by chance from correctly filled bags about 6 times in 1,000 — but it is strong enough evidence to warrant further investigation of the filling process."

Mark: 10/10. Examiner commentary: This is a complete piece of statistical reasoning rather than a calculation with a verdict attached. The candidate defines μ before using it, justifies the one-tailed alternative from the wording of the question, and — unusually — explicitly checks the conditions for the z-test and names the theorem that licenses it. The standard error is computed as a separate step and labelled, which protects it from the commonest error in the topic. Offering the p-value alongside the critical-value comparison is good practice and provides a cross-check. The final paragraph earns the contextual mark several times over: the conclusion is stated in the language of the problem, the strength of evidence is quantified, and the candidate is careful to say what a significance test can and cannot establish.

Grade II (Merit) answer

"H₀: μ = 500 H₁: μ < 500

z = (494 − 500)/(15/√40) = −6/2.37 = −2.53

Critical value = −1.645

−2.53 < −1.645 so reject H₀."

Mark: 7/10. Examiner commentary: Every calculation is correct and efficiently executed, and the hypotheses are properly stated in terms of the population mean. The response secures all the method and accuracy marks. What it loses is the conclusion mark in each of the three parts where interpretation is required: the candidate stops at "reject H₀" and never says what that means about the bags of cement, which is what the question actually asked for. There is also no statement of the model or of why a z-test is valid here. This is a very common profile — technically sound candidates treating the conclusion as a formality — and it costs three marks on this question alone. The remedy takes one sentence: state, in the words of the problem, what has been concluded and at what significance level.

Grade IV (Weak pass) answer

"H₀: x̄ = 500 H₁: x̄ ≠ 500

z = (494 − 500)/15 = −0.4

−0.4 is bigger than −1.96 so accept H₀. The bags are 500 g."

Mark: 1/10. Examiner commentary: Four separate errors, each characteristic. First, the hypotheses are written about the sample mean x̄; hypotheses are always statements about a population parameter, and the sample mean is a known number that requires no hypothesis. Second, the alternative is two-tailed when the question asks specifically whether the mean is less than claimed. Third, and most seriously, the test statistic divides by σ rather than by the standard error σ/√n — this discards the sample size entirely and is the single most consequential error in the Module, here turning a clearly significant result into a clearly non-significant one. Fourth, the conclusion is stated as certainty: a test never establishes that H₀ is true, and the correct language is "there is insufficient evidence to reject". One mark is retained for the correct structure of the hypothesis pair.

Question 5(c)(ii) — Sample Answers

Grade I (Distinction) answer

"The interval (75.6, 80.4) is a range of plausible values for μ, the mean examination mark of the whole population of students from which this sample of 100 was drawn.

The phrase '95% confident' describes the reliability of the method, not of this particular interval. If we were to repeat the whole exercise many times — drawing a fresh sample of 100 students and constructing an interval each time — approximately 95% of the intervals produced would contain the true population mean, and about 5% would not. We have one such interval and do not know which kind it is.

It would be wrong to say that there is a 95% probability that μ lies between 75.6 and 80.4. The population mean is a fixed number, not a random variable; it either lies in this interval or it does not, and no probability attaches to that. The randomness is in the sampling, and therefore in the endpoints of the interval, not in μ."

Mark: 4/4. Examiner commentary: This is exactly what the question is testing and it is answered completely. The candidate distinguishes the long-run property of the procedure from a probability statement about a fixed parameter, and — crucially — explains why the common misreading is wrong by locating the randomness in the interval rather than in μ. The final sentence is the discriminator. Candidates who can reproduce the repeated-sampling formulation but cannot say why the probability formulation fails typically score 3.

Grade III (Pass) answer

"There is a 95% chance that the true mean mark is between 75.6 and 80.4. This means we are 95% sure that the average mark of all the students is in this range."

Mark: 1/4. Examiner commentary: This is the standard misinterpretation, and the question is set precisely to detect it. The candidate earns one mark for correctly identifying that the interval concerns the population mean rather than the sample mean, which is more than the weakest responses manage. But the probability statement is not a loose paraphrase of the correct answer — it is a different claim. The population mean is a fixed constant; it does not have a 95% chance of being anywhere. The confidence attaches to the procedure that generated the interval: in repeated sampling, 95% of such intervals would capture μ. The second sentence merely restates the first in different words and adds nothing.

How Unit 1 is assessed

Paper 02 carries 150 marks and contributes 50% of the Unit 1 external assessment mark, alongside Paper 01 (multiple choice, 30%) and the Internal Assessment (20%). All six questions are compulsory, so no Module can be left out — a candidate who has avoided the χ² test loses nine marks in Question 6 that nothing else can replace.

Four habits recover most of the marks that competent candidates lose. First, divide by the standard error, not by σ, whenever the question concerns a sample mean; this single distinction is worth marks in Questions 5 and 6 and is the most frequently fumbled idea in the Unit. Second, conclude in context. Every hypothesis test in this paper reserves a mark for a conclusion phrased in the language of the problem and with appropriate tentativeness — "there is insufficient evidence to conclude", never "we have proved". Third, draw the diagram you are asked for, with a labelled scale and the correct region shaded; the box plot, the tree diagram and the normal sketch carry thirteen marks between them, and a shaded normal tail is also the best available guard against reporting Φ(z) when 1 − Φ(z) was wanted. Fourth, state the model before you use it: naming the distribution and its parameters, and checking that its conditions hold, earns marks directly in Questions 4 and 5 and prevents the structural errors — with-replacement probabilities in a without-replacement problem, wrong degrees of freedom, one-tailed values used two-tailed — that cost far more than arithmetic slips ever do.