Geographical Techniques III: Statistical Methods — Central Tendency, Dispersion, Inequality, Sampling, Hypothesis Tests, Correlation, Indices, PCA and Cluster Analysis — and Morphometric Analysis
2. Sampling and hypothesis testing: chi-square, t-test and ANOVA
A sample is studied to infer the characteristics of a population. Probability designs give every unit a known chance of selection — simple random, systematic (every kth unit), stratified (random within sub-groups, improving precision when strata differ) and cluster or multi-stage sampling; non-probability designs — purposive, quota, snowball — cannot support formal inference. Geographers add spatial sampling by points, lines (transects) and areas (quadrats), each chosen randomly, systematically or by strata. Hypothesis testing sets a null hypothesis (H₀: no difference or no association) against an alternative, chooses a significance level (usually α = 0.05), computes a test statistic and compares it with the critical value for the right degrees of freedom. Rejecting a true H₀ is a Type I error; failing to reject a false one is a Type II error.
| Test | Use | Statistic and degrees of freedom |
|---|---|---|
| Chi-square (χ²) | Frequencies (counts) in categories: goodness of fit to an expected distribution, or independence of two attributes in a contingency table | χ² = Σ(O − E)² ÷ E; df = k − 1 for goodness of fit, (r − 1)(c − 1) for a table; expected counts should be at least 5 |
| Student's t-test | Whether two means differ (or a sample mean differs from a population mean), especially with small samples; paired form for before–after data | t = difference of means ÷ its standard error; df = n₁ + n₂ − 2 for two independent samples; assumes roughly normal data |
| Analysis of variance (ANOVA) | Whether three or more group means differ | F = variance between groups ÷ variance within groups (R.A. Fisher); a large F means the groups differ more than chance allows |
A worked chi-square: 40 landslides are recorded, 30 on shale and 10 on granite, in two terrains of equal area. If rock type made no difference we would expect 20 on each, so χ² = (30 − 20)² ÷ 20 + (10 − 20)² ÷ 20 = 5 + 5 = 10, with df = 1. The critical value at the 0.05 level is 3.84, so H₀ is rejected: landslides are significantly associated with shale. Choosing the test is itself a common question: counts in categories call for χ²; two means for t; three or more means for ANOVA.
3. Time series, correlation and regression, indices, composite indices, PCA and cluster analysis
A time series is decomposed into secular trend (long-run direction), seasonal variation, cyclical variation and irregular fluctuation; the trend is isolated by moving averages, the semi-average method or a least-squares trend line. Correlation measures the strength and direction of a linear relationship: Pearson's r = covariance ÷ (σₓ σᵧ), from −1 to +1; Spearman's rank correlation, ρ = 1 − 6Σd² ÷ [n(n² − 1)], uses ranks and suits ordinal data — for 5 districts with Σd² = 10, ρ = 1 − 60 ÷ 120 = 0.5. Regression fits the line y = a + bx by least squares, where b is the change in y for a unit change in x; the coefficient of determination, r², is the share of the variance of y explained, and a map of residuals shows where places depart from the relationship — often the most geographical part of the analysis.
Development and well-being are measured by indices built from several indicators in different units — literacy in per cent, income in rupees, hospital beds per lakh. They must first be made scale-free. The z-score, z = (x − x̄) ÷ σ, expresses each value in standard deviations from the mean. Range (min–max) normalisation, (x − min) ÷ (max − min), rescales to 0–1: a literacy rate of 70 per cent, where the minimum is 40 and the maximum 90, becomes (70 − 40) ÷ (90 − 40) = 30 ÷ 50 = 0.6 — the method the HDI uses with fixed "goalposts". Other options are the ratio to the mean and ranking (Kendall). A composite index then combines the standardised indicators by a simple or weighted average — or, as the HDI has done since 2010, by a geometric mean, which limits substitution between dimensions. Principal component analysis (PCA) transforms many correlated variables into fewer uncorrelated components, each with an eigenvalue giving the variance it explains and loadings linking it to the original variables; components with eigenvalues above 1 are usually kept (the Kaiser criterion), and the first component is often used as a development index. Cluster analysis groups areal units by their similarity across many variables — hierarchically, as in Ward's method with its dendrogram, or non-hierarchically, as in k-means — and is the statistical basis of regionalisation.
4. Morphometric analysis
Stream ordering follows A.N. Strahler (1952, 1957), who modified R.E. Horton's scheme (1945): unbranched fingertip channels are first order; where two streams of order u meet, the stream below is of order u + 1; where a lower-order stream joins a higher-order one, the order does not increase. (Shreve's magnitude, by contrast, adds the numbers of first-order links.) Horton's law of stream numbers states that the numbers of streams of successive orders form a geometric series, whose ratio is the bifurcation ratio, Rb = Nᵤ ÷ Nᵤ₊₁. With 40 first-order and 10 second-order streams, Rb = 40 ÷ 10 = 4. In basins without strong structural control Rb usually lies between 3 and 5; much higher values indicate structural control — elongated basins guided by faults or folds.
| Measure | Formula | Worked value and meaning |
|---|---|---|
| Drainage density (Dd), Horton | Total stream length ÷ basin area | 150 km of streams in 50 km² gives Dd = 3 km/km². High on impermeable rock, sparse vegetation and steep slopes (badlands); low on permeable rock and under dense forest |
| Stream frequency (Fs) — the document's "drainage frequency" | Number of stream segments ÷ basin area | 60 segments in 50 km² gives Fs = 1.2 per km²; rises with drainage density |
| Circularity ratio (Rc), Miller (1953) | 4πA ÷ P², area over the square of the perimeter | A = 64 km², P = 40 km: 4 × 3.14 × 64 ÷ 1,600 ≈ 0.50. A circle scores 1; elongated basins about 0.2–0.4 |
| Form factor (Ff), Horton | Basin area ÷ (basin length)² | A = 50 km², L = 10 km: Ff = 50 ÷ 100 = 0.5. Low values mean elongated basins with flatter, longer flood peaks; high values circular basins with sharp peaks |
Profiles display relief along a line: serial profiles are drawn at intervals across an area; superimposed profiles are drawn on the same base to reveal accordant summit levels; projected profiles show only the parts not hidden by nearer ground, as seen from a distance; composite profiles show the highest points of several parallel profiles; and the longitudinal profile of a river shows its gradient, graded concavity and knick points. Slope analysis maps average slope from contour spacing — methods include Wentworth's (1930) count of contour crossings along a grid and Smith's (1935) relative relief (the difference between the highest and lowest points in a unit area). The clinographic curve (Hanson-Lowe, 1935) plots the average slope between successive contours against altitude, showing at which heights a region is steep or gentle. The hypsographic curve plots altitude against the area lying above each height; its percentage form, Strahler's hypsometric curve (1952), uses relative height against relative area, and its hypsometric integral indicates the stage of erosion — high values for youthful, low values for old, planed-down basins. The altimetric frequency graph is a histogram of the frequency of spot heights or summit heights in altitude classes; peaks in it reveal erosion (planation) surfaces at particular levels.
Key takeaways
- Mean, median and mode suit interval, skewed and nominal data; σ = √[Σ(x − x̄)²/n] (2 for 2, 4, 4, 4, 5, 5, 7, 9); CV = σ/x̄ × 100; the Gini coefficient runs 0 to 1; LQ = (eᵢ/e)/(Eᵢ/E), above 1 for concentration.
- Counts in categories: χ² = Σ(O − E)²/E; two means: t-test (df = n₁ + n₂ − 2); three or more means: ANOVA with F = between ÷ within variance; Type I error rejects a true H₀.
- Time series = trend, seasonal, cyclical and irregular; Pearson's r and Spearman's ρ = 1 − 6Σd²/[n(n² − 1)]; regression y = a + bx with r² the share explained.
- Scale-free: z = (x − x̄)/σ or (x − min)/(max − min); composite index by arithmetic or geometric mean; PCA condenses correlated variables (eigenvalue > 1); cluster analysis (Ward, k-means) regionalises.
- Strahler ordering; Rb = Nᵤ/Nᵤ₊₁ (usually 3–5); Dd = ΣL/A; Fs = ΣN/A; Rc = 4πA/P² (circle = 1); Ff = A/L²; clinographic curve = slope against altitude, hypsographic = altitude against area, altimetric frequency graph = frequency of heights revealing erosion surfaces.
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
20 per cent of a district's workers are in manufacturing, compared with 10 per cent of all workers in the country. What is the location quotient for manufacturing in the district? Type a number.
Numerical answer — type the value.
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Answer: 2
LQ = (district share) ÷ (national share) = 0.20 ÷ 0.10 = 2. A value above 1 shows that manufacturing is more concentrated in the district than in the country as a whole.In a test of whether landslides depend on rock type, 30 landslides occur on shale and 10 on granite, while 20 were expected on each. Calculate χ² = Σ(O − E)² ÷ E. Type a number.
Numerical answer — type the value.
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Answer: 10
χ² = (30 − 20)² ÷ 20 + (10 − 20)² ÷ 20 = 100 ÷ 20 + 100 ÷ 20 = 5 + 5 = 10. With 1 degree of freedom the critical value at the 0.05 level is 3.84, so the null hypothesis of no association is rejected.A researcher wants to test whether mean wheat yield differs among four districts, using samples of farms from each. The appropriate test is:
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Answer: B — Analysis of variance (ANOVA)
ANOVA compares three or more group means through the ratio of between-group to within-group variance. A t-test compares only two means, chi-square tests frequencies in categories, and Spearman's ρ measures association between two ranked variables.Five districts are ranked on literacy and on per-capita income, and the sum of squared rank differences Σd² is 10. Using ρ = 1 − 6Σd² ÷ [n(n² − 1)], find Spearman's rank correlation. Type a number.
Numerical answer — type the value.
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Answer: 0.5
ρ = 1 − (6 × 10) ÷ [5 × (25 − 1)] = 1 − 60 ÷ 120 = 1 − 0.5 = 0.5, a moderate positive rank correlation.A district's literacy rate is 70 per cent. Among all districts the minimum is 40 per cent and the maximum 90 per cent. What is its range-normalised (min–max) score? Type a number.
Numerical answer — type the value.
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Answer: 0.6
Score = (x − min) ÷ (max − min) = (70 − 40) ÷ (90 − 40) = 30 ÷ 50 = 0.6. Rescaling every indicator to 0–1 in this way makes them scale-free, so they can be combined in a composite index.In principal component analysis, the eigenvalue of a component indicates:
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Answer: D — The amount of the total variance that the component explains
Each component's eigenvalue measures the variance it captures; components with eigenvalues above 1 (the Kaiser criterion) explain more than a single original standardised variable and are usually retained. Loadings, not eigenvalues, link components to variables.Which of the following make indicators measured in different units scale-free before they are combined? Select all that apply.
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Answer: B — Min–max (range) normalisation; C — Converting values to z-scores
z-scores express values in standard deviations and min–max scaling rescales them to 0–1, removing units in both cases. Adding raw values mixes incompatible units, and the mode is an average, not a standardisation.A drainage basin has 40 first-order streams and 10 second-order streams. What is the bifurcation ratio between these orders? Type a number.
Numerical answer — type the value.
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Answer: 4
Rb = Nᵤ ÷ Nᵤ₊₁ = 40 ÷ 10 = 4, within the usual range of 3 to 5 for basins without strong structural control.A basin of 50 km² contains streams with a total length of 150 km. What is its drainage density in km per km²? Type a number.
Numerical answer — type the value.
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Answer: 3
Dd = total stream length ÷ area = 150 ÷ 50 = 3 km/km². High drainage density goes with impermeable surfaces, sparse vegetation and high relief.A basin has an area of 50 km² and a basin length of 10 km. What is its form factor, Ff = A ÷ L²? Type a number.
Numerical answer — type the value.
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Answer: 0.5
Ff = 50 ÷ 10² = 50 ÷ 100 = 0.5. Lower values indicate more elongated basins, which produce flatter, longer-lasting flood peaks than compact ones.In Strahler's method of stream ordering, when a first-order stream joins a third-order stream, the stream below the junction is of:
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Answer: B — Third order
Order increases only where two streams of the same order meet; a lower-order tributary joining a higher-order stream does not change its order. Adding tributaries in that way is Shreve's magnitude, not Strahler's order.A curve that plots the average slope between successive contours against altitude is the:
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Answer: A — Clinographic curve
The clinographic curve shows slope against height. The hypsographic curve shows the area above each altitude, the altimetric frequency graph the frequency of spot heights in altitude classes, and the Lorenz curve cumulative inequality.Assertion (A): A circularity ratio close to 1 indicates a nearly circular drainage basin. Reason (R): The circularity ratio compares the basin's area with the area of a circle having the same perimeter, and a circle encloses the greatest area for a given perimeter. Choose the correct option.
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Answer: A — Both A and R are true, and R is the correct explanation of A
Both are true and R explains A. Rc = 4πA ÷ P² is exactly the basin's area divided by that of a circle with the same perimeter (P² ÷ 4π), so only a circle reaches 1 and elongated basins fall to about 0.2–0.4.Peaks in an altimetric frequency graph of summit heights are used mainly to identify:
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Answer: C — Erosion (planation) surfaces at particular levels
If many summits cluster at similar heights, the graph shows a peak, suggesting that they are remnants of an old erosion surface later dissected — a key tool in denudation chronology.