Correlation — Practice

Question 1

Identify the type of correlation (positive, negative, or zero) for each pair: (a) Price and quantity demanded. (b) Height and income. (c) Study hours and marks. (d) Temperature and sales of woollen clothes.

Accepted Answer

(a) Price and quantity demanded: NEGATIVE correlation — as price rises, quantity demanded falls (Law of Demand). (b) Height and income: ZERO (or very low) correlation — there is no systematic economic relationship between a person's height and their income. (c) Study hours and marks: POSITIVE correlation — more study time generally leads to better marks. (d) Temperature and sales of woollen clothes: NEGATIVE correlation — as temperature rises, demand for woollens falls.

Question 2

Calculate Spearman's Rank Correlation for the following data on students' ranks in Maths and Economics: Student: A, B, C, D, E, F. Maths rank: 1, 2, 3, 4, 5, 6. Economics rank: 2, 4, 1, 5, 3, 6.

Accepted Answer

Step 1: Calculate d = Maths rank − Economics rank for each student: A: 1−2 = −1. B: 2−4 = −2. C: 3−1 = 2. D: 4−5 = −1. E: 5−3 = 2. F: 6−6 = 0. Step 2: Calculate d²: 1, 4, 4, 1, 4, 0. Step 3: Σd² = 1+4+4+1+4+0 = 14. Step 4: N = 6. r = 1 − (6Σd²) / [N(N²−1)] = 1 − (6 × 14) / [6(36−1)] = 1 − 84/210 = 1 − 0.4 = 0.6. Interpretation: r = 0.6 indicates moderate positive correlation — students who rank well in Maths tend to rank well in Economics, but the relationship is not very strong.

Question 3

Calculate Karl Pearson's r using assumed mean method for: X: 10, 20, 30, 40, 50; Y: 25, 35, 55, 65, 70. Use assumed mean A = 30 for X and B = 55 for Y.

Accepted Answer

Step 1: Build the table. dx = X−30; dy = Y−55. X=10: dx=−20, dy=−30, dx²=400, dy²=900, dxdy=600. X=20: dx=−10, dy=−20, dx²=100, dy²=400, dxdy=200. X=30: dx=0, dy=0, dx²=0, dy²=0, dxdy=0. X=40: dx=10, dy=10, dx²=100, dy²=100, dxdy=100. X=50: dx=20, dy=15, dx²=400, dy²=225, dxdy=300. Step 2: Σdx = −20−10+0+10+20 = 0. Σdy = −30−20+0+10+15 = −25. Σdx² = 400+100+0+100+400 = 1000. Σdy² = 900+400+0+100+225 = 1625. Σdxdy = 600+200+0+100+300 = 1200. N = 5. Step 3: r = [NΣdxdy − ΣdxΣdy] / √[NΣdx²−(Σdx)²] × √[NΣdy²−(Σdy)²] = [5×1200 − 0×(−25)] / √[5×1000−0] × √[5×1625−(−25)²] = 6000 / √5000 × √(8125−625) = 6000 / √5000 × √7500 = 6000 / (70.71 × 86.60) = 6000 / 6123.7 ≈ 0.98. Interpretation: r ≈ 0.98 indicates a very high positive correlation between X and Y.

Practice — Correlation

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