Question
6 observations on (X, Y) yielded the following data:
i) Determine the correlation coefficient between X and Y.
ii) Given , what will be the predicted value of Y?
iii) Given , what will be the predicted value of X?
Answer :
Word Count : 377
First, note the given data: [ n = 6, \quad \sum X_i = 30, \quad \sum Y_i = 180, \quad \sum X_i Y_i = 1000, \quad \sum X_i^2 = 200, \quad \sum Y_i^2 = 5642 ] Step i: Correlation coefficient (r) The formula for (r) is: [ r = \frac{n \sum X_i Y_i - \sum X_i \sum Y_i}{\sqrt{\left(n \sum X_i^2 - (\sum X_i)^2\right)\left(n \sum Y_i^2 - (\sum Y_i)^2\right)}} ] Substitute the values: [ r = \frac{6(1000) - (30)(180)}{\sqrt{(6(200) - 30^2)(6(5642) - 180^2)}} ] Compute step by step: [ 6 \cdot 1000 = 6000, \quad _______ ____ ___ _____ ___ _________ ___ ______ ___ _______.
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First, note the given data: [ n = 6, \quad \sum X_i = 30, \quad \sum Y_i = 180, \quad \sum X_i Y_i = 1000, \quad \sum X_i^2 = 200, \quad \sum Y_i^2 = 5642 ] Step i: Correlation coefficient (r) The formula for (r) is: [ r = \frac{n \sum X_i Y_i - \sum X_i \sum Y_i}{\sqrt{\left(n \sum X_i^2 - (\sum X_i)^2\right)\left(n \sum Y_i^2 - (\sum Y_i)^2\right)}} ] Substitute the values: [ r = \frac{6(1000) - (30)(180)}{\sqrt{(6(200) - 30^2)(6(5642) - 180^2)}} ] Compute step by step: [ 6 \cdot 1000 = 6000, \quad _______ ____ ___ _____ ___ _________ ___ ______ ___ _______.
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