📌 1. Unit 1 · Collection, Classification and Presentation of Data
| Primary data | Secondary data | |
|---|---|---|
| Meaning | Collected first-hand by the investigator for the purpose in hand | Already collected by someone else, published or unpublished |
| Examples / sources | Surveys, interviews, questionnaires | Census reports, RBI bulletins, government publications, journals |
| Cost and time | More | Less (but check suitability and reliability) |
- Methods of collecting primary data: direct personal investigation, indirect oral investigation, information from local sources/correspondents, mailed questionnaire (filled by the informant), schedules through enumerators (filled by the enumerator).
- Census = every unit is studied; sample survey = only some units are studied.
- Bases of classification: geographical (place), chronological (time), qualitative (attribute), quantitative (class intervals).
- Class intervals: exclusive (10–20 excludes 20) and inclusive (10–19). Sturges’ rule: number of classes k = 1 + 3.322 log₁₀N.
- Parts of a table: table number, title, head note, column headings (caption), row headings (stub), body, footnote, source note.
- Presentation: histogram, frequency polygon, ogive (the ‘less than’ and ‘more than’ ogives meet at the median), pie chart (angle = value/total × 360°).
Frequency distributions and diagrams
- Discrete distribution: separate values (e.g. number of children in a family); continuous distribution: class intervals (e.g. height 150–155 cm).
- Cumulative frequency (cf): "less than" (add from the top) and "more than" (add from the bottom). Relative frequency = f/N.
- A histogram is only for continuous data; the area of each rectangle is proportional to the frequency. With unequal class widths, height = frequency density = f ÷ class width.
- One-dimensional diagrams: line and bar diagrams; two-dimensional: rectangles, squares, circles (pie). A Lorenz curve shows inequality of income or wealth.
→ Angle = (value ÷ total) × 360° = (6,000 ÷ 24,000) × 360°
→ = ¼ × 360° = 90°
→ Gap between classes = 20 − 19 = 1; half = 0.5. Subtract 0.5 from each lower limit and add 0.5 to each upper limit.
→ Exclusive classes: 9.5–19.5, 19.5–29.5, 29.5–39.5 (width 10).
→ Class mark = (10 + 19)/2 = (9.5 + 19.5)/2 = 14.5
→ k = 1 + 3.322 log₁₀100 = 1 + 3.322 × 2 = 7.644
→ Round up → about 8 classes
- A questionnaire is filled by the informant; a schedule is filled by the enumerator.
- Caption = column headings; stub = row headings.
📌 2. Unit 2 · Measures of Central Tendency and Partition Values
Data are classified into class intervals. Class mark (mid-value) = (lower limit + upper limit)/2. Presentation: histogram (locates mode), ogive (locates median and quartiles), frequency polygon, bar and pie diagrams.
- Mean (short-cut): x̄ = A + Σfd/N, where d = x − A
- Step-deviation: x̄ = A + (Σfu/N) × h, where u = (x − A)/h
- Median = L + [(N/2 − cf)/f] × h
- Mode = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
- Q₁ = L + [(N/4 − cf)/f] × h; for Q₃ use 3N/4; for Dₖ use kN/10; for Pₖ use kN/100
- Q₂ = D₅ = P₅₀ = Median
- For two numbers: GM² = AM × HM; and AM ≥ GM ≥ HM
- Weighted mean = Σwx/Σw; geometric mean GM = antilog(Σf log x / N); harmonic mean HM = N/Σ(f/x)
- Properties: Σ(x − x̄) = 0; Σ(x − A)² is least when A = x̄; Σ|x − A| is least when A = median
- Empirical relation (moderately skewed data): Mode ≈ 3 Median − 2 Mean
- Which average when: open-end classes → median; growth rates/ratios → GM; speeds/rates over equal distances → HM
→ Class marks x = 5, 15, 25, 35, 45; A = 25, h = 10 → u = −2, −1, 0, 1, 2
→ fu = −10, −8, 0, 7, 10 → Σfu = −1
→ x̄ = 25 + (−1/40) × 10 = 25 − 0.25 = 24.75
→ Cumulative frequencies: 5, 13, 28, 35, 40; N/2 = 20 → median class 20–30 (L = 20, cf = 13, f = 15)
→ Median = 20 + [(20 − 13)/15] × 10 = 20 + 4.67 = 24.67
→ Modal class 20–30: f₁ = 15, f₀ = 8, f₂ = 7 → Mode = 20 + [(15 − 8)/(30 − 8 − 7)] × 10 = 20 + 7/15 × 10 = 24.67
→ Mean ≈ median ≈ mode → the distribution is almost symmetric.
→ Q₁: N/4 = 10 → class 10–20 (cf = 5, f = 8): Q₁ = 10 + (10 − 5)/8 × 10 = 16.25
→ Q₃: 3N/4 = 30 → class 30–40 (cf = 28, f = 7): Q₃ = 30 + (30 − 28)/7 × 10 = 32.86
→ D₇: 7N/10 = 28 → class 20–30 (cf = 13, f = 15): D₇ = 20 + (28 − 13)/15 × 10 = 30
→ Equal distances at different speeds → harmonic mean
→ HM = 2 × 30 × 60/(30 + 60) = 3600/90 = 40 km/h (the arithmetic mean 45 would be wrong)
- L = lower limit of the median/modal class; cf = cumulative frequency of the class BEFORE it (not of that class).
- In the mode formula f₀ = frequency before the modal class, f₂ = frequency after it.
Partition values: deciles and percentiles
Deciles (D₁…D₉) split the data into 10 equal parts and percentiles (P₁…P₉₉) into 100 equal parts; quartiles (Q₁, Q₂, Q₃) into 4.
- Individual series (ascending order): Dₖ = k(n + 1)/10-th item; Pₖ = k(n + 1)/100-th item
- Grouped: Dₖ = L + [(kN/10 − cf)/f] × h; Pₖ = L + [(kN/100 − cf)/f] × h
- Relations: Q₁ = P₂₅, Q₂ = D₅ = P₅₀ = median, Q₃ = P₇₅, D₁ = P₁₀, D₉ = P₉₀
- Percentile range = P₉₀ − P₁₀; Kelly’s coefficient of skewness = (P₉₀ + P₁₀ − 2P₅₀)/(P₉₀ − P₁₀)
- Percentile rank 80 = 80% of the observations lie below that value
- In grouped data locate the class with kN/10 (not n + 1).
📌 3. Unit 3 · Measures of Dispersion
Dispersion shows how widely values are spread around the average. Absolute measures (range, QD, MD, SD) carry the unit of the data; relative measures (coefficients, CV) are unit-free and are used to compare two series.
- Quartile deviation QD = (Q₃ − Q₁)/2; coefficient of QD = (Q₃ − Q₁)/(Q₃ + Q₁)
- Mean deviation MD = Σ|x − A|/n; MD is least when taken about the median
- Coefficient of range = (L − S)/(L + S)
- Coefficient of variation CV = (σ/x̄) × 100%; lower CV = more consistent
- Combined mean x̄ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)
- Combined variance σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂), dᵢ = x̄ᵢ − x̄
- For a normal distribution, roughly QD : MD : SD = 10 : 12 : 15
- Variance σ² = Σx²/n − x̄² (short form); grouped: σ² = Σfx²/N − x̄²
- SD of the first n natural numbers = √[(n² − 1)/12]; SD of n terms of an AP with common difference d = d√[(n² − 1)/12]
- SD of two numbers a, b = |a − b|/2; range ≥ SD
→ Mean = 40/8 = 5
→ (x − 5)² = 9, 1, 1, 1, 0, 0, 4, 16 → sum = 32
→ σ² = 32/8 = 4 → σ = 2
→ Median = 8; |x − 8| = 5, 3, 0, 2, 6 → sum 16
→ MD = 16/5 = 3.2; coefficient of MD = 3.2/8 = 0.4
→ CV(A) = 10/50 × 100 = 20%; CV(B) = 6/40 × 100 = 15%
→ Lower CV = more consistent → B (even though A has the higher average)
→ Combined mean = (50 × 60 + 50 × 50)/100 = 55; d₁ = 5, d₂ = −5
→ σ² = [50(64 + 25) + 50(36 + 25)]/100 = (4450 + 3050)/100 = 75
→ σ = √75 ≈ 8.66
- A change of origin (x + k) leaves SD, QD, MD unchanged; a change of scale (kx) multiplies them by |k|.
📌 4. Unit 4 · Moments, Skewness and Kurtosis
- Raw moments about A: μ′ᵣ = Σf(x − A)ʳ/N; mean = A + μ′₁
- μ₂ = μ′₂ − μ′₁² (variance); μ₃ = μ′₃ − 3μ′₂μ′₁ + 2μ′₁³
- Karl Pearson's coefficient Sₖ = (Mean − Mode)/σ = 3(Mean − Median)/σ
- Bowley's coefficient = (Q₃ + Q₁ − 2Q₂)/(Q₃ − Q₁), lies between −1 and +1
- β₁ = μ₃²/μ₂³; β₂ = μ₄/μ₂²
- μ₄ = μ′₄ − 4μ′₃μ′₁ + 6μ′₂μ′₁² − 3μ′₁⁴
- About the mean: μ₀ = 1, μ₁ = 0, μ₂ = σ²; central moments do not change with a change of origin
- γ₁ = √β₁ (sign of μ₃), γ₂ = β₂ − 3; for a symmetric distribution μ₃ = 0 and β₁ = 0
- Karl Pearson’s coefficient usually lies between −3 and +3; Bowley’s between −1 and +1
| Value of β₂ | Curve |
|---|---|
| β₂ = 3 | Mesokurtic — normal curve |
| β₂ > 3 | Leptokurtic — more peaked |
| β₂ < 3 | Platykurtic — flatter |
→ Mean = 4 + 1 = 5; μ₂ = 4 − 1² = 3
→ μ₃ = 10 − 3(4)(1) + 2(1)³ = 0 → β₁ = 0 (symmetric)
→ μ₄ = 45 − 4(10)(1) + 6(4)(1) − 3(1) = 26 → β₂ = 26/3² = 2.89 (< 3, slightly platykurtic)
→ (Q₃ + Q₁ − 2Q₂)/(Q₃ − Q₁) = (50 + 20 − 60)/(50 − 20) = 10/30
→ = 0.33 → positive skewness
→ Sₖ = 3(Mean − Median)/σ = 3(45 − 42)/9
→ = 1 (positive skewness)
- Positive skew: Mean > Median > Mode (long right tail); reversed for negative skew.
📌 5. Unit 5 · Correlation and Regression
- Scatter diagram: points rising → positive, falling → negative, no pattern → r ≈ 0.
- Coefficient of determination r² = the share of the variance of y explained by x; probable error PE = 0.6745(1 − r²)/√n.
- At r = ±1 the two regression lines coincide; at r = 0 they are perpendicular (x = x̄ and y = ȳ).
- r = Cov(x, y)/(σₓσᵧ), −1 ≤ r ≤ 1
- Spearman's rank correlation: R = 1 − 6Σd²/[n(n² − 1)]
- Regression of y on x: y − ȳ = bᵧₓ(x − x̄), bᵧₓ = r·σᵧ/σₓ; bₓᵧ = r·σₓ/σᵧ
- r² = bᵧₓ × bₓᵧ; r, bᵧₓ and bₓᵧ always have the same sign
- Both regression lines intersect at (x̄, ȳ)
- If u = (x − a)/h, v = (y − b)/k, r is unchanged and bᵧₓ = (k/h)·bᵥᵤ
- Partial correlation r₁₂.₃ = (r₁₂ − r₁₃r₂₃)/√[(1 − r₁₃²)(1 − r₂₃²)]
- Association of attributes, Yule's Q = (ad − bc)/(ad + bc)
Multiple regression and partial/multiple correlation (three variables)
- Regression of X₁ on X₂ and X₃: X₁ = a + b₁₂.₃X₂ + b₁₃.₂X₃ (b₁₂.₃ = partial regression coefficient)
- Multiple correlation: R²₁.₂₃ = (r₁₂² + r₁₃² − 2r₁₂r₁₃r₂₃)/(1 − r₂₃²); 0 ≤ R₁.₂₃ ≤ 1
- Partial correlation: r₁₂.₃ = (r₁₂ − r₁₃r₂₃)/√[(1 − r₁₃²)(1 − r₂₃²)]; between −1 and +1
- R₁.₂₃ is never less than |r₁₂| or |r₁₃|
→ x̄ = 3, ȳ = 4; dx = −2, −1, 0, 1, 2; dy = −2, 0, 1, 0, 1
→ Σdx·dy = 4 + 0 + 0 + 0 + 2 = 6; Σdx² = 10; Σdy² = 6
→ r = 6/√(10 × 6) = 6/7.75 ≈ 0.77
→ bᵧₓ = 6/10 = 0.6 → y − 4 = 0.6(x − 3) → y = 0.6x + 2.2; at x = 6, y = 5.8
→ Solve together: first × 5 → 40x − 50y = −330; subtracting gives 32y = 544 → ȳ = 17, x̄ = 13
→ Take the first as y on x: y = 0.8x + 6.6 → bᵧₓ = 0.8; the second gives x = 0.45y + 5.35 → bₓᵧ = 0.45
→ r² = 0.8 × 0.45 = 0.36 → r = 0.6 (the other way round the product exceeds 1, which is impossible)
→ R = 1 − 6 × 14/[6(36 − 1)] = 1 − 84/210
→ = 0.6
→ Q = (ad − bc)/(ad + bc) = (30 × 40 − 10 × 20)/(30 × 40 + 10 × 20)
→ = 1000/1400 ≈ 0.71 → positive association
→ r₁₂.₃ = (0.7 − 0.5 × 0.5)/√[(1 − 0.25)(1 − 0.25)] = 0.45/0.75
→ = 0.6
- bᵧₓ and bₓᵧ cannot both exceed 1 (their product r² ≤ 1).
- r = 0 only means no LINEAR relation, not independence.
- Multiple correlation R is never negative; partial correlation can be negative.
📌 6. Unit 6 · Probability Theory
- Random experiment, sample space S, event (a subset of S). Mutually exclusive: A ∩ B = ∅; exhaustive: A ∪ B ∪ … = S; equally likely: all have equal probability.
- Definitions: classical P(A) = favourable/total equally likely outcomes; relative frequency (empirical) = limit of m/n over a very large number of trials; axiomatic (Kolmogorov): P(A) ≥ 0, P(S) = 1, and P(A ∪ B) = P(A) + P(B) for mutually exclusive events.
- Independent events: one happening does not change the probability of the other; then A and B′, and A′ and B′, are also independent.
- Odds in favour a : b → P = a/(a + b); odds against a : b → P = b/(a + b).
- Addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); three events: P(A ∪ B ∪ C) = ΣP(A) − ΣP(A ∩ B) + P(A ∩ B ∩ C)
- Multiplication theorem: P(A ∩ B) = P(A) · P(B|A) = P(B) · P(A|B)
- Conditional probability P(A|B) = P(A ∩ B)/P(B); if independent, P(A ∩ B) = P(A)P(B)
- Total probability: P(A) = Σ P(Eᵢ)P(A|Eᵢ); Bayes’ theorem: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ)
- At least one: 1 − P(none) = 1 − (1 − p₁)(1 − p₂)…
→ P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7
→ P(A|B) = 0.2/0.4 = 0.5 = P(A)
→ P(A)P(B) = 0.2 = P(A ∩ B) → independent
→ P = P(first ace) × P(second ace | first ace) = 4/52 × 3/51
→ = 12/2652 = 1/221 (or ⁴C₂/⁵²C₂ = 6/1326)
→ P(I) = P(II) = 1/2; P(W|I) = 3/5, P(W|II) = 2/5
→ Bayes: P(I|W) = (½ × 3/5)/(½ × 3/5 + ½ × 2/5) = (3/10)/(5/10)
→ = 3/5
→ P(nobody solves it) = (1/2)(2/3)(3/4) = 1/4
→ P(solved) = 1 − 1/4 = 3/4
- Mutually exclusive ≠ independent: if P(A), P(B) > 0 and A ∩ B = ∅, then P(A ∩ B) = 0 ≠ P(A)P(B) → such events are dependent.
- P(A|B) and P(B|A) are different — in Bayes questions the "given" part goes in the denominator.
📌 7. Unit 7 · Random Variables and Probability Distributions
- A random variable is a function that gives a number to every outcome of the sample space. Discrete (counts: 0, 1, 2…) → probability mass function p(x), Σp(x) = 1; continuous (measurements) → probability density function f(x), ∫f(x)dx = 1 and P(a < X < b) = area under the curve.
- Distribution function F(x) = P(X ≤ x): non-decreasing, F(−∞) = 0, F(∞) = 1. For a continuous variable P(X = a) = 0.
- Conditional probability P(A|B) = P(A ∩ B)/P(B); independent events: P(A ∩ B) = P(A)P(B)
- Bayes' theorem: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ)
- E(X) = Σx·p(x); Var(X) = E(X²) − [E(X)]²
- E(aX + b) = aE(X) + b; Var(aX + b) = a²Var(X)
| Distribution | Mean | Variance | Key point |
|---|---|---|---|
| Binomial B(n, p) | np | npq | Mean > variance; P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ |
| Poisson P(λ) | λ | λ | Mean = variance; P(X = r) = e⁻λλʳ/r! |
| Normal N(μ, σ²) | μ | σ² | Mean = median = mode; Z = (X − μ)/σ |
| Exponential (λ) | 1/λ | 1/λ² | f(x) = λe⁻λˣ, x ≥ 0; memoryless |
The four main distributions — which one when
- Binomial: n fixed independent trials, two outcomes each, constant success probability p. Symmetric when p = ½; positively skewed when p < ½. Variance npq ≤ n/4.
- Poisson: rare events (misprints on a page, calls in an hour). Limit of the binomial when n is large, p small and np = λ. P(X = 0) = e⁻λ. The sum of two independent Poisson variables is Poisson (λ₁ + λ₂).
- Normal: bell-shaped, symmetric (β₁ = 0, β₂ = 3); points of inflexion at μ ± σ; QD ≈ 0.6745σ (≈ ⅔σ), MD ≈ 0.7979σ (≈ ⅘σ). Standard normal Z = (X − μ)/σ ~ N(0, 1). Table: P(0 < Z < 1) = 0.3413, P(0 < Z < 1.96) = 0.4750, P(0 < Z < 2) = 0.4772.
- Exponential: waiting time / lifetime. P(X > t) = e⁻λᵗ; mean 1/λ and SD also 1/λ. Memoryless: P(X > s + t | X > s) = P(X > t).
- Continuous uniform U(a, b): mean (a + b)/2, variance (b − a)²/12.
→ E(X) = 0 + 1/2 + 2/4 = 1; E(X²) = 0 + 1/2 + 4/4 = 1.5
→ Var(X) = 1.5 − 1² = 0.5
→ Var(2X + 3) = 2² × 0.5 = 2 (the added 3 does not change the variance)
→ n = 6, p = q = ½: P(X = 4) = ⁶C₄ (½)⁶ = 15/64
→ Mean = np = 3; variance = npq = 1.5
→ P(0) = e⁻² ≈ 0.135
→ P(1) = 2e⁻² → P(X ≤ 1) = e⁻²(1 + 2) = 3 × 0.1353 ≈ 0.406
→ X = 80 → Z = (80 − 60)/10 = 2 → P(Z > 2) = 0.5 − 0.4772 = 0.0228 (≈ 2.3%)
→ 50 to 70 → Z = −1 to 1 → 2 × 0.3413 = 0.6826 (≈ 68%)
→ λ = 1/1000 → P(X > 1000) = e⁻¹ ≈ 0.368
→ By the memoryless property the second answer is also e⁻¹ ≈ 0.368 — the first 500 hours do not matter.
Joint distribution (two discrete random variables)
- Joint pmf p(x, y) = P(X = x, Y = y); p(x, y) ≥ 0 and ΣΣp(x, y) = 1
- Marginal: P(X = x) = Σᵧ p(x, y); conditional: P(X = x | Y = y) = p(x, y)/P(Y = y)
- Independent only if p(x, y) = P(X = x) × P(Y = y) for every (x, y)
- Cov(X, Y) = E(XY) − E(X)E(Y); Var(aX ± bY) = a²Var X + b²Var Y ± 2ab Cov(X, Y)
- For independent variables E(XY) = E(X)E(Y) and Cov = 0 (but Cov = 0 does not prove independence)
→ Marginals: P(X = 1) = 0.1 + 0.4 = 0.5; P(Y = 1) = 0.3 + 0.4 = 0.7
→ p(1, 1) = 0.4 ≠ 0.5 × 0.7 = 0.35 → not independent
→ E(XY) = 1 × 1 × 0.4 = 0.4 → Cov = 0.4 − 0.5 × 0.7 = 0.05
- Binomial: mean > variance; Poisson: mean = variance — a binomial with "mean 4, variance 5" is impossible.
- In a normal distribution divide by σ, not σ², when finding Z.
📌 8. Unit 8 · Sampling Theory
- Parameter = measure of the population (μ, σ); statistic = measure of the sample (x̄, s).
- Probability sampling: simple random, stratified, systematic, cluster, multistage, multiphase.
- Non-probability sampling: purposive, convenience, quota.
- Sampling error arises from taking a sample (falls as n grows); non-sampling error (wrong responses, processing mistakes) occurs even in a census.
- Properties of a good estimator: unbiased, consistent, efficient, sufficient.
- Methods of estimation: method of moments, maximum likelihood (MLE), least squares.
- Population: finite or infinite; sampling frame = the list of units from which the sample is drawn; sampling unit = the unit chosen (person, household, village).
- Advantages of a sample survey: lower cost, less time, more detailed study, and the only way for destructive testing (e.g. bulb life). Census: information on every unit, but costly.
- SRSWR (with replacement): Nⁿ possible samples; SRSWOR (without replacement): ᴺCₙ. Select by the lottery method or a random number table.
- Systematic: interval k = N/n, first unit random from 1 to k, then every k-th. Stratified: proportional allocation nₕ = n × Nₕ/N; Neyman allocation nₕ ∝ Nₕσₕ. Cluster: whole groups (e.g. villages) are chosen. Multistage: first districts, then villages, then households.
- Central limit theorem: for large n (above about 30) the sampling distribution of x̄ is approximately normal N(μ, σ²/n), whatever the population distribution.
- Standard error SE(x̄) = σ/√n
- 95% confidence interval for μ = x̄ ± 1.96·σ/√n; use 2.58 for 99%
- s² = Σ(x − x̄)²/(n − 1) is an unbiased estimator of the population variance
- Finite population (SRSWOR): SE(x̄) = (σ/√n) × √[(N − n)/(N − 1)] — finite population correction
- Standard error of a proportion SE(p) = √(PQ/n)
- Sample size to estimate a mean n = (Zσ/E)²; for a proportion n = Z²PQ/E² (E = permissible error)
→ SE = σ/√n = 12/6 = 2
→ n = (1.96 × 10/2)² = 9.8² = 96.04
→ Always round up → 97
→ nₕ = 50 × Nₕ/1000 → 50 × 0.6, 50 × 0.3, 50 × 0.1
→ = 30, 15, 5
→ k = 500/25 = 20
→ Units 7, 27, 47, 67, … , 487 (7 + 24 × 20)
→ σ/√n = 10/5 = 2; correction = √(75/99) ≈ 0.870
→ SE ≈ 2 × 0.870 = 1.74
- Sampling error falls as n grows; non-sampling error (wrong answers, enumerator mistakes, non-response) occurs even in a census and can grow with n.
- To halve the SE, n must be made four times as large (SE ∝ 1/√n).
📌 9. Unit 9 · Statistical Inference: Estimation and Testing of Hypotheses
Estimation
- Point estimate = a single number (e.g. x̄ = 52); interval estimate = a range that contains the parameter with a stated confidence (e.g. a 95% confidence interval).
- Unbiased: E(T) = θ. x̄ is an unbiased estimator of the population mean μ; Σ(x − x̄)²/n is biased, Σ(x − x̄)²/(n − 1) is unbiased. Consistent: T → θ as n grows. Efficient: least variance among unbiased estimators. Sufficient: keeps all the information in the sample.
- Maximum likelihood estimator (MLE): the value that maximises the likelihood L(θ). Examples: binomial p̂ = x/n; Poisson λ̂ = x̄; normal μ̂ = x̄, σ̂² = Σ(x − x̄)²/n (biased but consistent).
Testing of hypotheses
- Steps: (1) state the null hypothesis H₀ and the alternative H₁; (2) choose the level of significance α; (3) compute the test statistic; (4) compare with the critical value; (5) reject H₀ if |calculated| > critical.
- H₁: μ ≠ μ₀ → two-tailed; H₁: μ > μ₀ or μ < μ₀ → one-tailed. Reject H₀ if the p-value < α.
- Type-I error: rejecting a true H₀ (probability α = level of significance); Type-II error: accepting a false H₀ (β); power of a test = 1 − β.
- Large sample (n > 30) or known σ → Z-test; small sample with unknown σ → t-test; comparing variances → F; independence / goodness of fit → χ².
- At the 5% level the two-tailed critical Z = 1.96; one-tailed 1.645.
| Test | When | Statistic | Degrees of freedom (df) |
|---|---|---|---|
| Z (mean) | Large sample or σ known | Z = (x̄ − μ₀)/(σ/√n) | — |
| Z (proportion) | Large sample | Z = (p − P)/√(PQ/n) | — |
| t (one mean) | Small sample, σ unknown | t = (x̄ − μ₀)/(s/√n) | n − 1 |
| t (two independent means) | Two small samples | with the pooled s | n₁ + n₂ − 2 |
| Paired t | Before–after (same units) | t = d̄/(s_d/√n) | n − 1 |
| χ² | Goodness of fit / independence / variance | Σ(O − E)²/E | k − 1 − (parameters estimated); (r − 1)(c − 1) |
| F | Comparing two variances, ANOVA | F = s₁²/s₂² | (n₁ − 1, n₂ − 1) |
| Level of significance | Two-tailed Z | One-tailed Z |
|---|---|---|
| 5% | 1.96 | 1.645 |
| 1% | 2.58 | 2.33 |
- Z = (x̄ − μ₀)/(σ/√n)
- χ² = Σ(O − E)²/E; for an r × c table df = (r − 1)(c − 1)
- F = s₁²/s₂² (larger variance in the numerator)
- One-way ANOVA (k treatments, N observations): df treatments = k − 1, error = N − k, total = N − 1
- Two-way ANOVA (r rows, c columns): error df = (r − 1)(c − 1)
- χ²(n) distribution: mean = n, variance = 2n
→ SE = 8/√64 = 1
→ 50 ± 1.96 × 1 = (48.04, 51.96)
→ Z = (103 − 100)/(9/6) = 3/1.5 = 2
→ 5%: 2 > 1.96 → reject H₀; 1%: 2 < 2.58 → do not reject H₀
→ t = (52 − 50)/(8/4) = 1; df = 15
→ 1 < 2.131 → do not reject H₀
→ p = 0.55; SE = √(0.5 × 0.5/400) = 0.025
→ Z = 0.05/0.025 = 2 > 1.96 → claim rejected
→ E = 10 for each face; (O − E)² = 4, 4, 1, 1, 16, 16 → sum 42
→ χ² = 42/10 = 4.2; df = 6 − 1 = 5
→ 4.2 < 11.07 → the die is taken as fair
→ For the binomial p̂ = x/n = 7/10 = 0.7
- "Accept H₀" means "not enough evidence to reject", not that H₀ is proved.
- Lowering α (5% → 1%) reduces the Type-I error but raises the Type-II error (for the same n).
📌 10. Unit 10 · Analysis of Variance (ANOVA)
ANOVA (R. A. Fisher) tests the means of three or more groups at once. H₀: μ₁ = μ₂ = … = μₖ. The total variation is split into "between groups" and "within groups (error)" parts and F = MS(between)/MS(within) is computed; a large F means the means are not all equal.
- Assumptions: independent observations, each group from a normal population, equal variances in all groups (homoscedasticity), additive effects. Equal group sizes are not required.
- Link with design of experiments: completely randomised design (CRD) → one-way; randomised block design (RBD) → two-way; in an m × m Latin square (LSD) the error df = (m − 1)(m − 2).
- Correction factor CF = G²/N (G = total of all observations)
- Total SS (TSS) = Σx² − CF; treatment SS (SST) = Σ(Tᵢ²/nᵢ) − CF; error SS (SSE) = TSS − SST
- One-way df: treatments k − 1, error N − k, total N − 1; MST = SST/(k − 1), MSE = SSE/(N − k), F = MST/MSE
- Two-way (one observation per cell, r rows × c columns): TSS = SSR + SSC + SSE; df: (r − 1) + (c − 1) + (r − 1)(c − 1) = rc − 1
| Source of variation | SS | df | MS | F |
|---|---|---|---|---|
| Treatments (between groups) | SST | k − 1 | MST = SST/(k − 1) | MST/MSE |
| Error (within groups) | SSE | N − k | MSE = SSE/(N − k) | |
| Total | TSS | N − 1 |
→ Totals: T_A = 12, T_B = 18, T_C = 24; G = 54, N = 9 → CF = 54²/9 = 324
→ Σx² = 56 + 116 + 200 = 372 → TSS = 372 − 324 = 48
→ SST = (144 + 324 + 576)/3 − 324 = 348 − 324 = 24; SSE = 48 − 24 = 24
→ MST = 24/2 = 12; MSE = 24/6 = 4 → F = 12/4 = 3
→ 3 < 5.14 → do not reject H₀: no significant difference between the method means
→ N = 20 → df: treatments 3, error 16, total 19
→ MST = 60/3 = 20; MSE = 80/16 = 5 → F = 4
→ 4 > 3.24 → reject H₀: the treatment means are not all equal
→ Rows 4 − 1 = 3; columns 5 − 1 = 4
→ Error 3 × 4 = 12; total 20 − 1 = 19 (3 + 4 + 12 = 19 ✓)
- F is always ≥ 0; the ANOVA F-test is always right-tailed (one-tailed).
- Mind the df in MST/MSE: F(k − 1, N − k) — numerator df first.
📌 11. Unit 11 · Time Series
- Components of a time series: secular trend, seasonal (within a year), cyclical (business cycle, several years), irregular (floods, wars).
- Measuring trend: free-hand curve, semi-averages, moving averages, least squares.
- Models: additive Y = T + S + C + I; multiplicative Y = T × S × C × I (more common in practice).
- Examples: sales rising at festivals → seasonal; boom and recession → cyclical; steady population growth → secular trend; earthquake/strike → irregular.
- Moving averages: an odd period (3, 5 years) is placed against the middle year; an even period (4 years) must be centred. If the period equals the cycle length, the fluctuations are removed. Some values at the ends are lost.
- Least squares method: Y = a + bX; code X so that ΣX = 0, then a = ΣY/n and b = ΣXY/ΣX².
- Seasonal indices: simple averages, ratio-to-trend, ratio-to-moving-average, link relatives. Quarterly indices add up to 400, monthly ones to 1200. Deseasonalised value = actual ÷ seasonal index × 100.
→ 2020: (10 + 14 + 12)/3 = 12; 2021: (14 + 12 + 16)/3 = 14; 2022: (12 + 16 + 20)/3 = 16
→ Moving averages 12, 14, 16 (none for 2019 and 2023)
→ X = −2, −1, 0, 1, 2 (origin 2021); ΣY = 75, ΣXY = −20 − 12 + 0 + 17 + 42 = 27, ΣX² = 10
→ a = 75/5 = 15, b = 27/10 = 2.7 → Y = 15 + 2.7X
→ 2025 → X = 4: Y = 15 + 10.8 = 25.8
→ Grand average = 200/4 = 50 → index = quarter average/50 × 100 = 80, 100, 120, 100 (sum 400 ✓)
→ Deseasonalised = 66/120 × 100 = 55
→ First half: (20 + 24 + 22)/3 = 22 (against year 2); second: (28 + 32 + 30)/3 = 30 (against year 5)
→ Increase = (30 − 22)/(5 − 2) = 8/3 ≈ 2.67 per year
- Seasonal swings repeat within a year; cyclical ones over several years — "every summer" = seasonal, "every 7–8 years" = cyclical.
📌 12. Unit 12 · Index Numbers
- An index number measures the relative change in a group (prices, quantities) over time or place; base year = 100. Price, quantity and value (value = Σp₁q₁/Σp₀q₀ × 100) indices.
- Simple aggregative Σp₁/Σp₀ × 100; simple average of relatives Σ(p₁/p₀ × 100)/n; weighted: Laspeyres, Paasche, Fisher, Marshall–Edgeworth, Dorbish–Bowley (L + P)/2, Kelly (fixed weights).
- Tests: unit test; time reversal P₀₁ × P₁₀ = 1; factor reversal P₀₁ × Q₀₁ = V₀₁; circular test (met by the simple aggregative and fixed-weight indices).
- Chain base: each year uses the previous year = 100 (link relatives); multiplying them gives a fixed-base series.
- Laspeyres P₀₁ = Σp₁q₀/Σp₀q₀ × 100 (base-year quantities)
- Paasche P₀₁ = Σp₁q₁/Σp₀q₁ × 100 (current-year quantities)
- Fisher's ideal index = √(L × P)
- Marshall–Edgeworth = Σp₁(q₀ + q₁)/Σp₀(q₀ + q₁) × 100
- Fisher's index satisfies both the Time Reversal Test and the Factor Reversal Test, hence "ideal".
- Usually L > P (Laspeyres has an upward bias, Paasche a downward bias).
- Base shifting: new index = (old index ÷ old index of the new base year) × 100
- Splicing: new series to the old base = new index × (old index of the link year/100); old series to the new base = old × 100/(old index of the link year)
- Deflating: real wage = money wage ÷ price index × 100
- Purchasing power of money = 100/CPI (in rupees)
| Method of cost of living (consumer price) index | Formula |
|---|---|
| Aggregate expenditure method | Σp₁q₀/Σp₀q₀ × 100 (= Laspeyres) |
| Family budget method | ΣPW/ΣW; P = p₁/p₀ × 100, W = p₀q₀ (base-year expenditure) |
Both methods give the same answer. A CPI measures the change in the cost of living of a particular group (e.g. industrial workers).
→ Σp₁q₀ = 96 + 150 = 246; Σp₀q₀ = 80 + 120 = 200 → L = 123
→ Σp₁q₁ = 120 + 125 = 245; Σp₀q₁ = 100 + 100 = 200 → P = 122.5
→ Fisher = √(123 × 122.5) ≈ 122.75
→ New = 150/125 × 100 = 120
→ Real salary in 2024 = 16,000/160 × 100 = ₹10,000
→ The same as in 2015 → no real increase
→ ΣPW = 7500 + 2400 + 3300 = 13,200; ΣW = 100
→ CPI = 13,200/100 = 132
- In base shifting the new base year becomes 100 — divide by its old index.
📌 13. Practice MCQs
Solution: The x-coordinate where the 'less than' and 'more than' ogives intersect is the median. A histogram locates the mode.
Solution: Caption = column headings; stub = row headings; the body holds the figures.
Solution: The data were collected by someone else (the Census office), so for the researcher they are secondary.
Solution: Angle = (10,000/40,000) × 360° = ¼ × 360° = 90°.
Solution: Take class marks: Σfx = 1830, N = 40 → mean = 1830/40 = 45.75.
Solution: N/2 = 23; median class 110-120 (cf = 11, f = 12) → 110 + [(23 − 11)/12] × 10 = 120.
Solution: Modal class 40-50: f₁ = 14, f₀ = 8, f₂ = 10 → 40 + [6/10] × 10 = 46.
Solution: 75% of the observations lie below Q₃ → Q₃ = P₇₅.
Solution: 4N/10 = 20 → class 20–30 (cf 15 to 25). D₄ = 20 + [(20 − 15)/10] × 10 = 25.
Solution: (Q₃ − Q₁)/(Q₃ + Q₁) = (26 − 15)/(26 + 15) = 11/41.
Solution: CV = (σ/x̄) × 100 = (6/120) × 100 = 5%.
Solution: Combined mean = 53; d₁ = 6, d₂ = −6; σ² = [30(64 + 36) + 30(64 + 36)]/60 = 100 → σ = 10.
Solution: Mean = 40/5 = 8; |x − 8| = 4, 1, 0, 1, 4 → sum 10; MD = 10/5 = 2.
Solution: Sₖ = (Mean − Mode)/σ = (58 − 54)/8 = 1/2.
Solution: With positive skew the right tail is long and pulls the mean furthest right.
Solution: β₂ = μ₄/μ₂² = 40/16 = 2.5; β₂ < 3 → platykurtic.
Solution: Mean = A + μ′₁ = 5 + 2 = 7; variance μ₂ = μ′₂ − μ′₁² = 20 − 4 = 16.
Solution: r = Cov/(σₓσᵧ) = 6/(3 × 4) = 1/2.
Solution: R = 1 − 6Σd²/[n(n² − 1)] = 1 − 216/120 = −4/5.
Solution: Both lines pass through (x̄, ȳ); solving gives x = 8, y = 4 (check: 3·8 + 4·4 = 40, 6·8 + 5·4 = 68).
Solution: R² = (0.36 + 0.64 − 0)/(1 − 0) = 1 → R = 1. (R is never above 1.)
Solution: r² = bᵧₓ × bₓᵧ = 0.16 → r = 0.4 (both coefficients are positive, so r is positive).
Solution: Q = (ad − bc)/(ad + bc) = (1600 − 100)/(1600 + 100) = 1500/1700 = 15/17.
Solution: Bayes: P(A|D) = (30 × 4)/(30 × 4 + 30 × 2 + 40 × 5) = 120/380 = 6/19.
Solution: P(A ∩ B) = 0.6 + 0.5 − 0.8 = 0.3; P(A|B) = 0.3/0.5 = 0.6.
Solution: Independent → P(A ∩ B) = 0.4 × 0.5 = 0.2; P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.
Solution: P = 13/52 × 12/51 = 156/2652 = 1/17. (1/16 would be the answer with replacement.)
Solution: Bayes: P(girl ∩ fail) = 0.4 × 0.1 = 0.04; P(boy ∩ fail) = 0.6 × 0.05 = 0.03; P = 0.04/(0.04 + 0.03) = 4/7.
Solution: A ∩ B = ∅ → P(A ∩ B) = 0, while P(A)P(B) > 0; so the condition for independence fails — the events are dependent.
Solution: q = variance/mean = 3/12 = 1/4 → p = 3/4; n = mean ÷ p = 12 ÷ 3/4 = 16.
Solution: For Poisson, mean = variance = λ. Binomial has mean > variance; exponential has mean 1/λ, variance 1/λ².
Solution: Total is 1 → k = 0.4. P(X = 1) = p(1, 0) + p(1, 1) = 0.2 + 0.4 = 0.6.
Solution: Binomial, n = 4, p = ½: P(X = 2) = ⁴C₂ (½)⁴ = 6/16 = 3/8.
Solution: λe⁻λ = λ²e⁻λ/2 → 1 = λ/2 → λ = 2.
Solution: Z = (60 − 50)/5 = 2; P(Z > 2) = 0.5 − 0.4772 = 0.0228.
Solution: λ = 1/5 = 0.2; P(X > 10) = e⁻λᵗ = e⁻⁰·² ˣ ¹⁰ = e⁻².
Solution: E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1 + 0.9 = 2.1.
Solution: Var(aX + b) = a²Var(X) = 9 × 4 = 36; the −2 does not change the variance.
Solution: Strata are internally homogeneous and each is sampled at random — stratified sampling. In cluster sampling whole clusters are chosen.
Solution: SE = σ/√n = 15/√25 = 15/5 = 3.
Solution: n = (Zσ/E)² = (1.96 × 20/4)² = 9.8² = 96.04 → round up to 97.
Solution: N = 10,000; nₕ = 200 × Nₕ/N → 200 × 0.2 = 40, 200 × 0.3 = 60, 200 × 0.5 = 100.
Solution: k = N/n = 1200/60 = 20; the first unit is random from 1–20, then every 20th.
Solution: E = 25 (heads), 25 (tails); χ² = (37 − 25)²/25 + (13 − 25)²/25 = 5.76 + 5.76 = 11.52.
Solution: The probability of a Type-I error is α, the level of significance. Accepting a false H₀ is a Type-II error (β).
Solution: SE = 10/√100 = 1; 80 ± 1.96 × 1 = (78.04, 81.96). (2.58 gives the 99% interval.)
Solution: Z = (52 − 50)/(5/√25) = 2/1 = 2.
Solution: df = n₁ + n₂ − 2 = 12 + 10 − 2 = 20.
Solution: For the Poisson λ̂ = x̄ = (2 + 4 + 3 + 5 + 1)/5 = 15/5 = 3.
Solution: E[Σ(x − x̄)²] = (n − 1)σ², so dividing by n − 1 makes the estimator unbiased; dividing by n underestimates σ² (biased).
Solution: Error df = N − k = 30 − 5 = 25 (treatment df = 4, total df = 29).
Solution: MST = 48/(4 − 1) = 16; MSE = 60/(24 − 4) = 3; F = 16/3 ≈ 5.33.
Solution: TSS = SST + SSE → SSE = 120 − 45 = 75.
Solution: Error df = (r − 1)(c − 1) = 4 × 3 = 12; total df = 20 − 1 = 19.
Solution: The assumptions are independence, normality and equal variances; a one-way ANOVA can be done with unequal group sizes too.
Solution: Against the third term: (12 + 13 + 17)/3 = 42/3 = 14. (The first moving average, (8 + 12 + 13)/3 = 11, goes against the second term.)
Solution: A fluctuation that repeats regularly within a year = seasonal variation.
Solution: 2026 → X = 6; Y = 50 + 4 × 6 = 74.
Solution: Deseasonalised = actual ÷ index × 100 = 500 ÷ 125 × 100 = 400.
Solution: Each quarter averages 100, so the total is 4 × 100 = 400; monthly indices add up to 1200.
Solution: Fisher = √(L × P) satisfies both tests, hence 'ideal'. Laspeyres and Paasche satisfy neither.
Solution: 240/150 × 100 = 160.
Solution: 18,000 ÷ 150 × 100 = ₹12,000.
Solution: Fisher = √(L × P) = √(144 × 121) = 12 × 11 = 132. (132.5 is the arithmetic mean — Dorbish–Bowley.)