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.

Answer 1

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 Σx_i = 30, Σy_i = 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.