Methods of Demographic Analysis
What it is
Unit II is the toolkit of formal demography: comparable measures of events, growth, the location of events by age, period and cohort, projection and estimation, and the statistics underneath. UGC-NET items here are numeric (growth rate, doubling time, standardised rate, standard error) or turn on a definition (rate versus ratio, stable versus stationary), so exact formulas matter most.
Core concepts
Rates, ratios, proportions, density, incidence and prevalence. A rate relates events to the population at risk: events ÷ person-years of exposure, × 1,000 or 100,000. Person-years (total time lived by all members in the period) are usually approximated by mid-year population × length of period. A ratio compares two distinct quantities (sex ratio: females per 1,000 males in India, males per 100 females internationally); a proportion is a part over a whole containing it, and a percentage is a proportion × 100. Density is persons per sq km. Incidence is new cases in a period over the population at risk; prevalence is all existing cases at a point or over a period; in a steady state prevalence ≈ incidence × mean duration. The IMR (per 1,000 live births) and maternal mortality ratio (per 100,000 live births) are strictly ratios; the maternal mortality rate is per 100,000 women aged 15–49.
Rate of population growth. Arithmetic: r = (Pt − P0) / (P0 × t). Geometric: Pt = P0 (1 + r)^t, so r = (Pt/P0)^(1/t) − 1. Exponential: Pt = P0 e^(rt), so r = ln(Pt/P0) / t. For the same counts, arithmetic r > geometric r > exponential r: a doubling in 10 years gives 10%, 7.18% and 6.93%. The Indian Census reports decadal growth, (Pt − P0)/P0 × 100 over ten years. Doubling time under exponential growth is ln 2 / r = 0.693/r, about 70 divided by the percentage rate; under geometric growth it is ln 2 / ln(1 + r).
Population stabilisation and NRR unity. The NRR is the average number of daughters a woman would bear under current fertility and mortality; NRR = 1 means each generation replaces itself, a TFR of about 2.1 at low mortality. A stable population (Lotka) has constant age-specific fertility and mortality, hence a constant age structure and growth rate, with intrinsic rate r ≈ ln(NRR) / T, T being the mean length of generation. A stationary population is stable with r = 0: births equal deaths, and the age structure is the life table's. A young age structure keeps growth going for decades after NRR reaches 1: population momentum. The National Population Policy 2000 aimed at replacement-level TFR by 2010 (medium term) and a stable population by 2045 (long term).
Crude and standardised measures. The CBR and CDR (per 1,000 mid-year population) are confounded by age structure. Direct standardisation applies the study population's age-specific rates to a standard age structure; dividing by the standard's crude rate gives the comparative mortality figure. Indirect standardisation applies standard age-specific rates to the study population to get expected events: SMR = observed ÷ expected, and the indirectly standardised rate = SMR × the standard's crude rate. Use indirect when the study's age-specific rates are missing or rest on small numbers. Birth rates are standardised the same way.
Location of events in time: the Lexis diagram. Named after Wilhelm Lexis, it puts calendar time on the horizontal axis and age on the vertical, so each life is a 45-degree life line. A vertical line is a census count at one moment, a horizontal line marks persons reaching an exact age, and a diagonal band follows a birth cohort. A square (one age, one year) holds two cohorts, a parallelogram one cohort, and a triangle fixes all three, since cohort = period − age. Age last birthday is an interval; exact age is a point.
Mathematical projection methods. Linear: Pt = P0 (1 + rt). Exponential: Pt = P0 e^(rt). Polynomial: Pt = a + bt + ct^2, fitted to three or more censuses. The logistic (Verhulst 1838, revived by Pearl and Reed 1920), Pt = K / (1 + e^(a − bt)), is a symmetric S-curve with asymptote K and inflection at K/2. The Gompertz curve (1825, first a law of mortality), Pt = K × a^(b^t), is asymmetric, inflecting at K/e, about 37% of K. All project totals only and suit short horizons.
Component method and derived projections. The cohort-component method (Edwin Cannan 1895, then Whelpton; Leslie's matrix form 1945) carries each age–sex cohort forward with life-table survival ratios (5L(x+5) / 5Lx), adds births from projected ASFRs applied to women aged 15–49 and split by the sex ratio at birth, and adds net migrants. Sub-national projections extrapolate each state's share of the national total, or run components constrained to sum to it. Derived projections apply rates to the projected structure: labour force (participation rates), enrolment (enrolment ratios × school-age population), households (headship rates × adults).
Official projections. The UN's World Population Prospects, revised every two years, gives low, medium, high and constant-fertility variants; the World Bank issues its own. In India a Technical Group on Population Projections under the National Commission on Population reports after each census; its 2011–2036 report (2020) projects about 152 crore people in 2036 (as of 2026-10).
Population estimates and pyramids. Intercensal estimates interpolate between censuses, mathematically (geometric: Pt = P0 (P1/P0)^(t/n), n the intercensal interval) or by components from recorded births, deaths and migration; postcensal estimates extrapolate from the last census. The gap between a component estimate and the next census is the error of closure. A pyramid draws age groups as horizontal bars, males conventionally on the left: expansive (broad base, high fertility), constrictive (narrowing base, falling fertility) or near-rectangular (low fertility and mortality).
Sampling and statistical methods. A parameter describes the population (μ, σ), a statistic the sample. The sampling distribution of the mean centres on μ with standard error σ/√n (× √((N − n)/(N − 1)) without replacement from a finite population) and is near normal for large n by the central limit theorem; the SE of a proportion is √(pq/n). Descriptive statistics summarise (frequency distributions, averages, range, variance, SD, coefficient of variation = SD/mean × 100); inductive statistics generalise to the population. For moderately skewed data, Mode = 3 Median − 2 Mean (Karl Pearson). Pearson's r lies in −1 to +1, and b(yx) = r × σy/σx and b(yx) × b(xy) = r^2. A Type I error (probability α) rejects a true H0, a Type II error (β) accepts a false one, and power is 1 − β. Interpolation works within the observed range (Newton's forward-difference formula for equal intervals, Lagrange's for unequal); extrapolation goes beyond it.
Worked example
Direct and indirect standardisation of death rates. Two populations of 100,000; ASDRs per 1,000.
| Age group | Population A | Deaths A | ASDR A | Population B | Deaths B | ASDR B |
|---|---|---|---|---|---|---|
| 0–14 | 40,000 | 200 | 5.0 | 20,000 | 80 | 4.0 |
| 15–59 | 50,000 | 150 | 3.0 | 55,000 | 165 | 3.0 |
| 60+ | 10,000 | 500 | 50.0 | 25,000 | 1,000 | 40.0 |
| All ages | 100,000 | 850 | CDR 8.5 | 100,000 | 1,245 | CDR 12.45 |
B's ASDR is lower or equal at every age, yet its CDR is higher. Standard: A and B combined.
| Age group | Standard population | Expected deaths at A's rates | Expected deaths at B's rates |
|---|---|---|---|
| 0–14 | 60,000 | 300 | 240 |
| 15–59 | 105,000 | 315 | 315 |
| 60+ | 35,000 | 1,750 | 1,400 |
| All ages | 200,000 | 2,365 | 1,955 |
- Direct rates: A = 2,365 / 200 = 11.825; B = 1,955 / 200 = 9.775 per 1,000.
- Check with the standard's age shares (0.300, 0.525, 0.175): A = 1.5 + 1.575 + 8.75 = 11.825; B = 1.2 + 1.575 + 7.0 = 9.775.
- Indirect check, A's rates as standard: B's expected deaths = 20,000 × 0.005 + 55,000 × 0.003 + 25,000 × 0.050 = 100 + 165 + 1,250 = 1,515. SMR = 1,245 / 1,515 = 0.8218, so B's rate = 0.8218 × 8.5 = 6.99 per 1,000.
- Reading: the crude ranking (B above A) reverses once age is held constant (A 11.8, B 9.8): B's high CDR comes from its older age structure.
Common traps
- Calling the IMR or maternal mortality ratio a true rate: both divide by live births, not the exposed population.
- Mixing growth formulas: the rule of 70 assumes exponential growth; for the same counts arithmetic r > geometric r > exponential r.
- Treating stable and stationary as synonyms (stationary is stable with r = 0), or assuming NRR = 1 halts growth at once (momentum).
- Reversing the standardisations: direct needs the study's ASDRs; indirect needs its age structure and total deaths, and yields an SMR.
- Swapping inflection points: logistic at K/2, Gompertz at K/e.
Speed technique
- Doubling time: 70 ÷ percentage growth rate (2% gives about 35 years).
- Direction rule: direct sends the study's rates to the standard population; indirect brings the standard's rates to the study population.
- Intrinsic growth: r = ln(NRR) / T; NRR 1 gives r = 0.
- Regression check: r^2 = b(yx) × b(xy), with r taking the coefficients' common sign.
Check yourself
- A population grows exponentially at 2% a year. What is its doubling time?
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About 34.7 years — ln 2 / 0.02 = 0.693 / 0.02 = 34.66. - A population of 50,000 becomes 60,500 in two years. What is the geometric annual growth rate?
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10% — (60,500 / 50,000)^(1/2) = 1.21^(1/2) = 1.10. - Which standardisation needs the study population's age-specific death rates?
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Direct standardisation — indirect needs only the study's age structure and total deaths. - At what share of the upper asymptote K do the logistic and Gompertz curves inflect?
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K/2 and K/e — the logistic is symmetric; the Gompertz inflects at about 37% of K. - With σ = 12, a sample of 36 is drawn. What is the standard error of the mean?
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2 — 12 / √36 = 12 / 6.
Try it: Methods of Demographic Analysis questions
Real questions from the NET Population Studies bank on exactly this skill. Pick an answer to see the full solution — the intuition, the worked steps, the faster methods and the traps.
A district recorded 2,400 live births in a year, and its mid-year population was 80,000. Its crude birth rate is:
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Answer: option A
The crude birth rate is live births in the year ÷ mid-year population × 1,000.
CBR = 2,400 ÷ 80,000 × 1,000 = 0.03 × 1,000 = 30.0; as a check, 80,000 ÷ 1,000 = 80 thousands, and 2,400 ÷ 80 = 30.
So the crude birth rate is 30.0 per 1,000 population, option A.
Consider the following two statements: Statement I: In India, the sex ratio is expressed as the number of females per 1,000 males. Statement II: In international usage, the sex ratio is usually expressed as the number of males per 100 females. In the light of the above statements, choose the correct answer from the options.
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Answer: option A
Statement I: the Census of India reports the sex ratio as females per 1,000 males, as in the Census 2011 figure of 943; Statement I is true.
Statement II: the UN and most countries report males per 100 females, so a value above 100 means more males; Statement II is true.
So Statement I, "In India, the sex ratio is expressed as the number of females…", and Statement II are both true, option A.
A population pyramid with a very broad base that narrows quickly towards the top is typical of a population with:
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Answer: option D
In a pyramid, age groups are drawn as horizontal bars; a broad base means many children relative to adults.
This expansive shape comes from high fertility, which keeps adding large cohorts at the bottom, and gives a young age structure; a narrowing base marks falling fertility (constrictive) and a near-rectangle marks low fertility and mortality.
So a broad, fast-narrowing pyramid indicates high fertility and a young population, option D.
A town's population rises from 64,000 to 100,000 in two years. Its annual geometric growth rate is:
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Answer: option B
Geometric growth: Pt = P0 , so r = − 1.
Pt ÷ P0 = 100,000 ÷ 64,000 = 1.5625; its square root is 1.25 (since 1.25 × 1.25 = 1.5625), so r = 0.25 = 25.0%.
Check: 64,000 × 1.25 = 80,000 after one year and 80,000 × 1.25 = 100,000 after two years.
So the annual geometric growth rate is 25.0 per cent, option B.
A standard population has 30,000 persons aged 0–14, 60,000 aged 15–59 and 10,000 aged 60 and above. Population X has age-specific death rates of 4, 2 and 40 per 1,000 in these groups, and its own population is 20,000, 50,000 and 30,000 respectively. The directly standardised death rate of X, per 1,000, is:
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Answer: option D
Direct standardisation applies X's age-specific rates to the standard population: expected deaths = 30,000 × 0.004 + 60,000 × 0.002 + 10,000 × 0.040 = 120 + 120 + 400 = 640.
Standardised rate = 640 ÷ 100,000 × 1,000 = 6.4; check with the standard's age shares 0.3, 0.6 and 0.1: 0.3 × 4 + 0.6 × 2 + 0.1 × 40 = 1.2 + 1.2 + 4.0 = 6.4.
So X's directly standardised death rate is 6.4 per 1,000, option D.
Answer above — every one shows its working.