Questions: For questions #47-50, refer to the following: A possible linear correlation between (x) hours of exercise per week and (y) a person's cholesterol level was examined with the following results: (x): 5 12 3 1 4 7 (y): 212 165 178 195 210 145 #47) What is the correlation coefficient? a) .43 b) .87 c) .53 d) .63 e) -27 Question 48 (Mandatory) (2 points) What percentage of a person's cholesterol level is directly related to their hours of weekly exercise? a) .28 b) 45.25 c) 28.09 d) 3.67 e) 53 Question 49 (Mandatory) (2 points) What is the equation of the regression line? a) y=203.75-3.67 x b) y=154.87-5.24 x c) y=203.75+3.67 x d) y=198.55-125 x e) y=-53+210 x

What is the correlation coefficient?

Calculation of Covariance and Standard Deviations.

The covariance \( \text{Cov}(X,Y) \) is calculated as \( -53.87 \). The standard deviation of \( X \) is \( \sigma_X = 3.83 \) and the standard deviation of \( Y \) is \( \sigma_Y = 26.45 \).

Calculation of the Correlation Coefficient.

The correlation coefficient \( r \) is given by the formula: \[ r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} = \frac{-53.87}{3.83 \cdot 26.45} = -0.53 \]

The answer is \(\boxed{-0.53}\).

What percentage of a person's cholesterol level is directly related to their hours of weekly exercise?

Calculation of the Coefficient of Determination.

The coefficient of determination \( r^2 \) is calculated as: \[ r^2 = (-0.53)^2 = 0.2809 \]

Conversion to Percentage.

The percentage of cholesterol level related to exercise is: \[ \text{Percentage} = r^2 \cdot 100 = 28.09\% \]

The answer is \(\boxed{28.09}\).

What is the equation of the regression line?

Calculation of Means.

The mean of \( X \) is: \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 5.33 \] The mean of \( Y \) is: \[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 184.17 \]

Calculation of Slope and Intercept.

The numerator for \( \beta \) is: \[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 5624 - 6 \cdot 5.33 \cdot 184.17 = -269.33 \] The denominator for \( \beta \) is: \[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 244 - 6 \cdot (5.33)^2 = 73.33 \] Thus, the slope \( \beta \) is: \[ \beta = \frac{-269.33}{73.33} = -3.67 \] The intercept \( \alpha \) is: \[ \alpha = \bar{y} - \beta \bar{x} = 184.17 - (-3.67) \cdot 5.33 = 203.75 \] The equation of the regression line is: \[ y = 203.75 - 3.67x \]

The answer is \(\boxed{y = 203.75 - 3.67x}\).

The correlation coefficient is \(\boxed{-0.53}\).
The percentage of cholesterol level related to exercise is \(\boxed{28.09}\).
The equation of the regression line is \(\boxed{y = 203.75 - 3.67x}\).