The regression equation is y = 0.89678X + 34.47178

Question

If two institutions have budgets that differ by 11 million dollars in 2006, by how much would you predict their budgets to differ last year? Round the answer to two decimal places as needed.

Answer 1

To determine the predicted difference in budgets for last year based on the given regression equation \( y = 0.89678X + 34.47178 \), we need to analyze how a difference in the independent variable \( X \) (the budget in millions of dollars for the year 2006) affects the predicted dependent variable \( y \).

The slope of the regression equation, \( 0.89678 \), indicates how much \( y \) changes for a one-unit change in \( X \).

Given that the budgets of the two institutions differ by 11 million dollars in 2006, we can apply the slope to find the predicted difference in budget:

\[ \text{Predicted difference} = \text{Slope} \times \text{Difference in } X \] \[ \text{Predicted difference} = 0.89678 \times 11 \]

Calculating this gives:

\[ \text{Predicted difference} = 0.89678 \times 11 = 9.86458 \]

Rounding this to two decimal places:

\[ \text{Predicted difference} \approx 9.86 \]

Thus, the predicted difference in budgets for last year would be $9.86 million.

Answer 2

Predict last year's budget for an organization whose 2006 budget was 80 million. Round the answer to two decimal places as needed.

Answer 3

To predict last year's budget for an organization whose 2006 budget was 80 million, we can use the provided regression equation:

\[ y = 0.89678X + 34.47178 \]

where \( X \) is the budget in millions for 2006. Substituting \( X = 80 \):

\[ y = 0.89678 \cdot 80 + 34.47178 \]

Calculating the first part:

\[ 0.89678 \cdot 80 = 71.7424 \]

Now, add this value to the intercept:

\[ y = 71.7424 + 34.47178 = 106.21418 \]

Rounding this to two decimal places:

\[ y \approx 106.21 \]

Thus, the predicted budget for last year for an organization with a 2006 budget of 80 million is $106.21 million.