Questions: Question 31 of 40 For a linear regression, what is the r-value of the following data to three decimal places? x y ------ 1 14 6 16 9 11 20 7 22 4 A. -0.929 B. 0.863 C. -0.863 D. 0.929

Solution Steps

The covariance between \( X \) and \( Y \) is calculated using the formula: \[ \text{Cov}(X,Y) = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] For the given data, the covariance is found to be: \[ \text{Cov}(X,Y) = -41.55 \]

The standard deviation of \( X \) is calculated as: \[ \sigma_X = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})^2} \] The standard deviation of \( Y \) is calculated similarly: \[ \sigma_Y = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (Y_i - \bar{Y})^2} \] For the given data, the standard deviations are: \[ \sigma_X = 9.072 \] \[ \sigma_Y = 4.93 \]

The correlation coefficient \( r \) is calculated using the formula: \[ r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} \] Substituting the values obtained: \[ r = \frac{-41.55}{9.072 \cdot 4.93} \] This results in: \[ r = -0.929 \]

The correlation coefficient (r-value) for the given data is: \[ r = -0.929 \]

Final Answer