Correlation: tree circumference and height

Let X = circumference and Y = height, n = 9.

a) Does a correlation exist?

Karl Pearson's r = Sxy / √(Sxx·Syy)

Sums (n = 9):

        x        y       xy       x²       y²
      1.8       21     37.8     3.24      441
      1.9     33.5    63.65     3.61  1122.25
      1.8     24.6    44.28     3.24   605.16
      2.4     40.7    97.68     5.76  1656.49
      5.1     73.2   373.32    26.01  5358.24
      3.1     24.4    75.64     9.61   595.36
      5.5     40.4    222.2    30.25  1632.16
      5.1     45.3   231.03    26.01  2052.09
      8.3     53.5   444.05    68.89  2862.25
---------------------------------------------
       35    356.6  1589.65   176.62    16325

Σx = 35, Σy = 356.6, Σxy = 1589.65, Σx² = 176.62, Σy² = 16325

Means: x̄ = 3.8889, ȳ = 39.6222

Sxx = Σx² − (Σx)²/n = 40.5089 Sxy = Σxy − (Σx)(Σy)/n = 202.8722 Syy = Σy² − (Σy)²/n = 2195.7156

Slope b = Sxy/Sxx = 202.8722/40.5089 = 5.0081 Intercept a = ȳ − bx̄ = 39.6222 − (5.0081)(3.8889) = 20.1463

r = 202.8722 / √(40.5089 × 2195.7156) = 0.6802

Test of significance of r

• H₀: ρ = 0 (no linear correlation in the population)
• H₁: ρ ≠ 0 (two-tailed)
• Level of significance α = 0.05
• Test statistic: t = r√(n − 2)/√(1 − r²) = 0.6802×√7/√(1 − 0.4627) = 2.455
• Critical value: t(0.025, df = 7) = ±2.365
• Decision: |t| = 2.455 > 2.365, so reject H₀ (p ≈ 0.044).

Conclusion: at the 5% level there is significant linear correlation between tree circumference and height.

b) Explanation

r = 0.680 is positive and moderately strong. Taller trees tend to have larger circumference: as circumference increases, height generally increases. The coefficient of determination r² = 0.463, so about 46.3% of the variation in height is explained by the linear relationship with circumference; the remaining 53.7% is due to other factors (age, species, soil, etc.). Correlation does not by itself prove that a larger circumference causes greater height.