Questions: Listed below are the heights ( cm ) of winning presidential candidates and their main opponents from several recent presidential elections. Find the regression equation, letting president be the predictor ( x ) variable. Find the best predicted height of an opponent given that the president had a height of 183 cm. How close is the result to the actual opponent height of 185 cm? Use a significance level of 0.05. President: 177, 188, 192, 188, 188, 183, 178, 183, 185 Opponent: 183, 173, 180, 188, 175, 185, 180, 182, 177 The regression equation is ŷ = + + + ( x. (Round the y-intercept to the nearest integer as needed. Round the slope to three decimal places as needed.)

Solution Steps

The means of the heights of presidents and their opponents are calculated as follows:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{1}{9} (177 + 188 + 192 + 188 + 188 + 183 + 178 + 183 + 185) = 184.6667 \]

\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{9} (183 + 173 + 180 + 188 + 175 + 185 + 180 + 182 + 177) = 180.3333 \]

The correlation coefficient \( r \) is computed as:

\[ r = -0.258 \]

The numerator for the slope \( \beta \) is given by:

\[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 299665 - 9 \cdot 184.6667 \cdot 180.3333 = -49.0 \]

The denominator for \( \beta \) is:

\[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 307112 - 9 \cdot 184.6667^2 = 196.0 \]

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

\[ \beta = \frac{-49.0}{196.0} = -0.25 \]

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

\[ \alpha = \bar{y} - \beta \bar{x} = 180.3333 - (-0.25) \cdot 184.6667 = 226.5 \]

The regression equation is expressed as:

\[ \hat{y} = 226 + -0.250x \]

To predict the height of an opponent given that the president has a height of \( 183 \) cm:

\[ \hat{y} = 226 + (-0.25) \cdot 183 = 180.75 \text{ cm} \]

The actual opponent height is \( 185 \) cm. The difference between the predicted and actual height is:

\[ \text{Difference} = |180.75 - 185| = 4.25 \text{ cm} \]

Final Answer