Question

The coefficient of correlation of the above two data series will be equal to \(\underline{\hspace{1cm}}\)

\[\begin{array}{|c|c|} \hline X & Y \\ \hline -3 & 9 \\ -2 & 4 \\ -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ \hline \end{array}\]
 

• A. +1
• B. -1
• C. 0
• D. -0.5

Hint:

The Pearson correlation coefficient \(r\) only detects linear relationships. A value of \(r=0\) means there is no linear correlation, but there could still be a strong non-linear relationship, such as \(Y=X^2\).

Answer 1

The Correct Option is C

Step 1: Examine the relationship between X and Y. The table reveals that each Y value is the square of its corresponding X value, i.e., \(X^2\). This indicates a perfect, yet non-linear (quadratic), relationship.

Step 2: Define the coefficient of correlation (Pearson's r). The Pearson correlation coefficient quantifies the strength and direction of a linear association between two variables.

Step 3: Calculate the means of X and Y. \[ \bar{X} = \frac{-3-2-1+0+1+2+3}{7} = \frac{0}{7} = 0 \]\[ \bar{Y} = \frac{9+4+1+0+1+4+9}{7} = \frac{28}{7} = 4 \]

Step 4: Determine the covariance between X and Y, using the formula \( \text{Cov}(X,Y) = \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{n} \).\[ \sum(x_i - \bar{x})(y_i - \bar{y}) = \sum(x_i - 0)(y_i - 4) = \sum x_i(y_i-4) \]\[ = (-3)(9-4) + (-2)(4-4) + (-1)(1-4) + (0)(0-4) + (1)(1-4) + (2)(4-4) + (3)(9-4) \]\[ = (-3)(5) + (-2)(0) + (-1)(-3) + 0 + (1)(-3) + (2)(0) + (3)(5) \]\[ = -15 + 0 + 3 + 0 - 3 + 0 + 15 = 0 \]Thus, the covariance is 0.

Step 5: Calculate the correlation coefficient using \( r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} \). Since the covariance (numerator) is 0, the correlation coefficient is 0, assuming non-zero standard deviations. A value of 0 implies no linear correlation.