Questions: The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 43 inches. Is the result close to the actual weight of 247 pounds? Use a significance level of 0.05. Chest size (inches): 50, 54, 44, 54, 40, 36 Weight (pounds): 279, 339, 219, 292, 201, 116 Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? ŷ = + x (Round to one decimal place as needed.)

Solution Steps

The means of the chest size (\( \bar{x} \)) and weight (\( \bar{y} \)) are calculated as follows:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 46.3 \]

\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 241.0 \]

The correlation coefficient (\( r \)) is found to be:

\[ r = 1.0 \]

The numerator for the slope (\( \beta \)) is calculated as:

\[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 69876 - 6 \cdot 46.3 \cdot 241.0 = 2878.0 \]

The denominator for the slope (\( \beta \)) is:

\[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 13164 - 6 \cdot 46.3^2 = 283.3 \]

Thus, the slope (\( \beta \)) is:

\[ \beta = \frac{2878.0}{283.3} = 10.2 \]

The intercept (\( \alpha \)) is calculated using:

\[ \alpha = \bar{y} - \beta \bar{x} = 241.0 - 10.2 \cdot 46.3 = -229.6 \]

The regression equation is given by:

\[ \hat{y} = -229.6 + 10.2x \]

To predict the weight of a bear with a chest size of 43 inches, we substitute \( x = 43 \) into the regression equation:

\[ \hat{y} = -229.6 + 10.2 \cdot 43 = 209.0 \text{ pounds} \]

The actual weight of the bear is 247 pounds. The difference between the predicted and actual weight is:

\[ \text{Difference} = |209.0 - 247| = 38.0 \text{ pounds} \]

Final Answer