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Question
A sample of 5 items is taken from the production of a firm. Length and weight of the five items arc given below :
| Length (cm) | 3 | 4 | 6 | 7 | 10 |
| Weight (gm) | 9 | 11 | 14 | 15 | 16 |
Calculate Karl Pearson's coefficient of correlation between the length and weight and interpret the result.
Solution
Let x = Length (cm), y = Weight (gm)
| x | y | x2 | y2 | xy | |
| 3 | 9 | 9 | 81 | 27 | |
| 4 | 11 | 16 | 121 | 44 | |
| 6 | 14 | 36 | 196 | 84 | |
| 7 | 15 | 49 | 225 | 105 | |
| 10 | 16 | 100 | 256 | 160 | |
| Total | 30 | 65 | 210 | 879 | 420 |
From table we have Σxi = 30, Σyi = 65.
`Σx_i^2 = 210, Σ_i^2 = 879, Σx_iΣy_i = 420, n = 5
Karl Pearson's coefficient of correlation is
r = `(nSigmax_iy_i - Sigmax_iSigmay_i)/(sqrtnSigmax_i^2 - (Sigmax_i)^2 sqrtnSigmay_i^2 - (Sigmay_i)^2`
= `(5(420) - 30 (65))/sqrt(5(210) - (30)^2 sqrt(5(879) - (65)^2`
= `(2100 - 1950)/sqrt(1050 - 900 sqrt(4395 - 4225)`
= `150/sqrt150 sqrt170`
= `(sqrt150 xx sqrt150)/(sqrt150 xx sqrt170`
= `sqrt(15/17)`
r = `sqrt0.8823`
r = 0.9393
Yhere is high degree positive correlation between x and y.
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