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22 pairs: means, consistency, correlation

Short answer

x̄ = 55.5, ȳ = 36.0. X is more consistent (CV 13.0% vs 23.8%). SE of the difference of means = 2.39. Karl Pearson r = -0.234, a weak negative correlation.

Following data reveals the sample of 22 pairs of observation (X,Y) drawn from large population. X: 46 61 56 68 58 45 50 59 45 66 57 59 66 62 57 57 45 50 61 55 47 51 Y: 49 46 43 32 26 27 29 47 37 30 43 32 27 37 24 43 49 48 29 37 32 26 i) Find the sample mean for each variable X and Y. ii) Which series is more consistent and why? iii) Find the standard error of the difference of mean. iv) Find the coefficient of Karl Pearson correlation.

n = 22 pairs. Σx = 1221, Σy = 793, Σx² = 68857, Σy² = 30125, Σxy = 43708.

i) Sample means

x̄ = Σx/n = 1221/22 = 55.500 ȳ = Σy/n = 793/22 = 36.045

ii) Consistency (coefficient of variation)

Sample standard deviations: s = √[(Σx² − n·x̄²)/(n − 1)]

  • s_x = 7.209, s_y = 8.566
  • CV(X) = s_x/x̄ × 100 = 7.209/55.500 × 100 = 12.99%
  • CV(Y) = s_y/ȳ × 100 = 8.566/36.045 × 100 = 23.76%

Series X is more consistent, because its coefficient of variation is smaller (less relative variation).

iii) Standard error of the difference of means

For independent samples of size 22 each,

SE(x̄ − ȳ) = √(s_x²/n₁ + s_y²/n₂) = √(51.976/22 + 73.379/22) = √5.6980 = 2.387

(x̄ − ȳ = 19.455.)

iv) Karl Pearson's correlation coefficient

Sxx = 1091.500, Syy = 1540.955, Sxy = Σxy − n·x̄·ȳ = -303.500

r = Sxy/√(Sxx·Syy) = -303.500/√(1091.500 × 1540.955) = -0.2340

The two series have a weak-to-moderate negative linear correlation.

✓ Answer checked by Claude Sonnet 5.5

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