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प्रश्न
Calculate the correlation coefficient for the following data.
| X | 5 | 10 | 5 | 11 | 12 | 4 | 3 | 2 | 7 | 1 |
| Y | 1 | 6 | 2 | 8 | 5 | 1 | 4 | 6 | 5 | 2 |
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उत्तर
| X | Y | x = `"X" - bar"X"` = X − 6 |
y = `"Y" - bar"Y"` = Y − 4 |
x2 | y2 | xy |
| 5 | 1 | − 1 | − 3 | 1 | 9 | 3 |
| 10 | 6 | 4 | 2 | 16 | 4 | 8 |
| 5 | 2 | − 1 | − 2 | 1 | 4 | 2 |
| 11 | 8 | 5 | 4 | 25 | 16 | 20 |
| 12 | 5 | 6 | 1 | 36 | 1 | 6 |
| 4 | 1 | − 2 | − 3 | 4 | 9 | 6 |
| 3 | 4 | − 3 | 0 | 9 | 0 | 0 |
| 2 | 6 | − 4 | 2 | 16 | 4 | − 8 |
| 7 | 5 | 1 | 1 | 1 | 1 | 1 |
| 1 | 2 | − 5 | − 2 | 25 | 4 | 10 |
| 60 | 40 | 0 | 0 | 134 | 52 | 48 |
N = 10, ∑X = 60, ∑Y = 40, ∑x2 = 134, ∑y2 = 52, ∑xy = 48, `bar"X" = (sum"X")/"N" = 60/10` = 6, `bar"Y" = (sum"Y")/"N" = 40/10` = 4.
Coefficient of correlation
r = `(sum"xy")/(sqrt(sum"x"^2 sum"y"^2))`, Where x = `"X" - bar"X"`, y = `"Y" - bar"Y"`
r = `48/sqrt (134 xx 52)`
r = 0.575
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संबंधित प्रश्न
Calculate the coefficient of correlation between X and Y series from the following data.
| Description | X | Y |
| Number of pairs of observation | 15 | 15 |
| Arithmetic mean | 25 | 18 |
| Standard deviation | 3.01 | 3.03 |
| Sum of squares of deviation from the arithmetic mean | 136 | 138 |
Summation of product deviations of X and Y series from their respective arithmetic means is 122.
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If the values of two variables move in the opposite direction then the correlation is said to be
Correlation co-efficient lies between
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30 |
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