MA: Data analysis

Answer: D. Quota sampling divides the population into groups and sets a target number (quota) to interview from each group. Crucially, the selection within each group is not random — researchers interview whoever is available until the quota is met. This distinguishes it from stratified random sampling, where members within each stratum are selected randomly. Systematic sampling selects every nth item from a list. Cluster sampling selects entire subgroups (clusters) at random.

Answer: C. High-low method: Variable cost per unit = (Highest cost − Lowest cost) ÷ (Highest activity − Lowest activity). Variable cost per unit = (£46,000 − £28,000) ÷ (10,000 − 4,000) = £18,000 ÷ 6,000 = £3.00 per unit. Fixed costs can then be found: £46,000 − (10,000 × £3.00) = £16,000. Option A (£2.50) divides cost by highest volume only. Option B (£4.60) divides total cost at high level by units. Option D (£7.00) divides the cost difference by the low volume figure.

Answer: B. This involves stepped fixed costs which must be stripped out before applying the high-low method. Supervisor cost at 1,500 units = 2 × £3,000 = £6,000. Supervisor cost at 2,500 units = 3 × £3,000 = £9,000. Variable costs only: Low = £18,000 − £6,000 = £12,000. High = £25,000 − £9,000 = £16,000. Variable cost per unit = (£16,000 − £12,000) ÷ (2,500 − 1,500) = £4,000 ÷ 1,000 = £4.00. Option A is the error of not stripping out supervisor costs. This question type is flagged as a common exam weakness in the examiner's report.

Answer: B. Index number = (Current year value ÷ Base year value) × 100. Index = (£300,000 ÷ £240,000) × 100 = 125.0. This means revenue in year 3 is 125% of the base year value — an increase of 25%. Option A (80.0) divides base by current (the inverse). Option C (60.0) calculates the increase as a percentage of current revenue. Option D (140.0) uses an incorrect calculation.

Answer: C. Seasonal variation is the regular, predictable fluctuation that repeats within a year (or other consistent cycle). A consistently higher Q3 is a seasonal pattern. Cyclical variation (A) operates over multiple years linked to economic cycles, not regular quarters. Random variation (B) is unpredictable by definition. The trend (D) is the overall long-run direction, not the regular quarterly deviation from it.

Answer: D. Positive correlation means both variables tend to increase or decrease together. The scatter shows this pattern. Negative correlation (A) means one rises as the other falls. Zero correlation (B) means no discernible relationship. Causation (C) is a critical distinction — correlation does not prove causation. Higher advertising may cause higher sales, but other factors (economic conditions, competitor actions) may also be at work. This is a classic exam point.

Answer: A. Expected value = Σ(outcome × probability). EV = (£60,000 × 0.3) + (£20,000 × 0.5) + (−£10,000 × 0.2) = £18,000 + £10,000 − £2,000 = £26,000. Check probabilities sum to 1: 0.3 + 0.5 + 0.2 = 1.0 ✓. Option B divides the simple average by 3. Option C sums the outcomes without weighting. Option D is a common error — forgetting to deduct the loss scenario.

Answer: A. Step 1 — contribution per machine hour: X = £30 ÷ 4 = £7.50/hr; Y = £20 ÷ 2 = £10.00/hr. Rank Y first (higher contribution per limiting factor). Step 2 — allocate hours to Y at maximum demand: 300 units × 2 hrs = 600 hrs. Step 3 — remaining hours for X: 1,200 − 600 = 600 hrs ÷ 4 hrs = 150 units of X. Optimal plan: 150 X and 300 Y — option A. Contribution = (150 × £30) + (300 × £20) = £4,500 + £6,000 = £10,500. Option B (200X, 100Y) prioritises X, which has the lower contribution per machine hour. Option C (200X, 200Y) uses 800 + 400 = 1,200 hrs but yields only £10,000 contribution — feasible but not optimal. Option D exceeds maximum demand for X (300 > 200).

Answer: A. When a resource is scarce, rank by contribution per unit of the scarce resource, not by contribution per unit of product. Product A = £24 ÷ 4kg = £6.00/kg. Product B = £15 ÷ 2kg = £7.50/kg. Product B earns more contribution per kg of material, so it should be prioritised — option A, which gives both the right product and the right reasoning. Option B reaches the right product but for the wrong reason ('less waste' is irrelevant; the limiting-factor principle is about contribution per unit of the constraint). Option C applies total contribution per unit, the classic error when a constraint exists. Option D uses selling price, which ignores variable costs and the constraint entirely.

Answer: C. Y = 120 + 8(15) = 120 + 120 = 240. Sales = £240,000 (in £'000). The regression equation extrapolates the trend. Option A uses X=14 (120+112=232). Option B uses only the constant term (ignoring X). Option D uses X=16 (120+128=248). Always substitute the correct time period value — a common exam error is using the wrong X value.

Answer: B. A scatter diagram plots the raw data points to allow visual inspection of the relationship. If the points appear randomly scattered, a linear model is inappropriate. If they suggest a non-linear pattern, regression will produce a misleading equation. Inspection confirms linearity before computation. The regression equation is calculated separately (A). The choice of dependent variable is determined by the analysis objective, not the scatter diagram (C). Identifying fixed costs uses the high-low method, not a scatter diagram (D).

Answer: A. A 4-point moving average must be centred because 4 is an even number — each MA falls between quarters, not on one. First 4-point MA (Q1–Q4): (120 + 140 + 180 + 160) ÷ 4 = 150.0, centred between Q2 and Q3. Second 4-point MA (Q2–Q1 next year): (140 + 180 + 160 + 130) ÷ 4 = 152.5, centred between Q3 and Q4. Centred MA for Q3 = (150.0 + 152.5) ÷ 2 = 151.25 — option A. Option B (150.0) is the first uncentred moving average; forgetting the centring step is the most common error in time series questions.

Answer: C. Sensitivity analysis identifies the percentage change in a variable that would make NPV zero — the margin of safety for that variable. A 15% margin on sales volume means the project remains viable unless sales fall more than 15% below forecast. Whether to proceed depends on management's assessment of how likely that fall is — sensitivity analysis does not make the decision, it informs it. Options A, B and D either misapply the finding or draw unwarranted conclusions.

Answer: B. Tables present data in a structured grid format that facilitates comparison — the eye can quickly scan rows and columns to spot patterns, outliers, and relationships. Tables do not automatically calculate (A) — that requires formulas or manual work. Tables can present any quantitative or qualitative data (C). Tables complement but do not replace commentary (D) — the meaning of the data still needs to be explained.

Answer: B. In linear programming, the optimal solution (maximum or minimum of the objective function) always lies at a corner point of the feasible region — this is a fundamental property of linear programming. The feasible region is bounded by the constraint lines, and the optimal corner is found by substituting corner coordinates into the objective function and selecting the best value. The midpoint (A), origin (C), and axis intersection (D) are not generally optimal.

Answer: C. The Delphi method aggregates expert opinion through anonymous, iterative rounds — experts respond, see a summary of others' views, and revise their estimates, converging toward consensus without groupthink. Regression analysis (A) is a quantitative statistical technique. Monte Carlo simulation (B) generates random scenarios. Simple moving average (D) averages recent data. The Delphi method is qualitative and relies on expert judgement rather than historical data.

Answer: B. In an additive model: Actual = Trend + Seasonal variation. Therefore: Trend = Actual − Seasonal variation = £180,000 − £15,000 = £165,000. The positive seasonal variation means Q2 is typically above trend — removing it gives the underlying trend. Option A adds the variation (incorrect direction). Option C ignores the adjustment. Option D is just the seasonal variation. Seasonal adjustment reveals the underlying trend by removing the predictable seasonal component.

Answer: C. Weighted average contribution = (0.60 × £30) + (0.40 × £20) = £18 + £8 = £26 per unit. Breakeven = Fixed costs ÷ Weighted average contribution = £130,000 ÷ £26 = 5,000 units total. Of these: X = 5,000 × 60% = 3,000 units; Y = 5,000 × 40% = 2,000 units. Option A divides by £30 (X contribution only). Option B divides by £20 (Y contribution only). Option D is half the correct answer.

Answer: D. The chi-squared test assesses whether two categorical variables are associated — product type (categorical) and sales region (categorical) is exactly this situation. Regression (A) is for continuous variables with a predictive relationship. T-tests (B) compare means of continuous variables between groups. Standard deviation (C) describes spread within a data set — not a test of association. The MA syllabus includes awareness of when different statistical tools are appropriate.

Answer: B. In Y=a+bX: 'a' (50) is the intercept (fixed element); 'b' (3.5) is the slope — for each unit increase in X, Y increases by 3.5. The correlation coefficient r=0.92 indicates a strong positive relationship (r near +1 = strong positive, near −1 = strong negative, near 0 = no relationship). Option A misidentifies the slope and intercept. Option C reverses a and b and confuses r with r². Option D — r=0.92 means strong but not perfect correlation; r=1.0 would be perfect.

Answer: C. Relative risk is measured by the coefficient of variation (CV) = standard deviation ÷ expected value, which scales risk to the size of the return. CV(A) = £10,000 ÷ £50,000 = 0.20; CV(B) = £20,000 ÷ £80,000 = 0.25. The higher CV means higher relative risk, so Project B is riskier in relative terms — option C. Option A looks only at the absolute standard deviation (£20,000 > £10,000), ignoring that B also offers a much larger return. Option B reasons loosely from the expected return alone. Option D wrongly claims the ratios are equal; 0.25 does not equal 0.20.

Answer: A. 8,400 = 10,000 − 2×800 (two standard deviations below mean). 11,600 = 10,000 + 2×800 (two standard deviations above mean). The empirical rule for normal distributions: ±1 SD = ~68% of values; ±2 SD = ~95% of values; ±3 SD = ~99.7%. The range 8,400–11,600 is ±2 standard deviations, covering approximately 95% of observations. Option B (68%) covers ±1 SD. Option C (99.7%) covers ±3 SD. Option D (50%) covers half the distribution.