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Question
A sample of 5 items is taken from the production of a firm. Length and weight of 5 items are given below. [Given : `sqrt(0.8823)` = 0.9393]
| Length (cm) | 3 | 4 | 6 | 7 | 10 |
| Weight (gm.) | 9 | 11 | 14 | 15 | 16 |
Calculate the correlation coefficient between length and weight and interpret the result.
Solution
Let length = xi (in cm), Weight = yi (in gm)
| xi | yi | xi2 | yi2 | xiyi | |
| 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 the table, we have
n = 5, `sum"x"_"i"` = 30, `sum"y"_"i"` = 65, `sum"x"_"i"^2` = 210, `sum"y"_"i"^2` = 879, `sum"x"_"i""y"_"i"` = 420
`bar"x" = (sum"x"_"i")/"n" = 30/5` = 6
`bar"y" = (sum"y"_"i")/"n" = 65/5` = 13
Cov (X, Y) = `1/"n" sum"x"_"i"y_"i" - bar"x"bar"y"`
= `1/5 xx 420 - 6 xx 13`
= 84 – 78
= 6
`sigma_"x"^2 = (sum"x"_"i"^2)/"n" - (bar"x")^2`
= `210/5 - (6)^2`
= 42 – 36
∴ `sigma_"x"^2` = 6
∴ `sigma_"x" = sqrt(6)`
`sigma_"y"^2 = (sum"y"_"i"^2)/"n" - (bar"y")^2`
= `879/5 - (13)^2`
= 175.8 – 169
`sigma_"y"^2` = 6.8
∴ `sigma_"y" = sqrt(6.8)`
Thus, the coefficient of correlation between X and Y is
r = `("Cov(X, Y)")/(sigma_"x" sigma_"y")`
= `6/(sqrt(6)sqrt(6.8)`
= `sqrt(6/6.8)`
= `sqrt(60/68)`
= `sqrt(15/17)`
= `sqrt(0.8823)`
∴ r = 0.9393 ≈ 0.94
∴ The value of r indicates high degree of positive correlation between length and weight of items.
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