Question

5 16.8
6 14.2

regression equation:
use the regression equation to predict the next years attendance.
$r^2=0.776$ $24.2 + 6.01x + 2.2x^2 + -0.25x^3$

Explanation:

Step1: Determine the next year's x - value

From the table, the existing x - values are 5 and 6. So the next year's x - value (let's assume x represents the year index) will be \(x = 7\).

Step2: Substitute \(x = 7\) into the regression equation

The regression equation is \(y=24.2 + 6.01x+2.2x^{2}- 0.25x^{3}\) (assuming the equation is \(y = 24.2+6.01x + 2.2x^{2}-0.25x^{3}\) as the original writing has some formatting issues).

Substitute \(x = 7\) into the equation:

First, calculate each term:

• For the constant term: \(24.2\)
• For the \(6.01x\) term: \(6.01\times7=42.07\)
• For the \(2.2x^{2}\) term: \(2.2\times7^{2}=2.2\times49 = 107.8\)
• For the \(- 0.25x^{3}\) term: \(-0.25\times7^{3}=-0.25\times343=-85.75\)

Now, sum up all the terms:

\(y=24.2 + 42.07+107.8-85.75\)

First, add the positive terms: \(24.2+42.07 + 107.8=24.2+149.87 = 174.07\)

Then subtract the negative - contributing term: \(174.07-85.75 = 88.32\)

Answer:

The predicted next year's attendance is \(88.32\) (the unit of attendance is not specified in the problem, but based on the calculation from the regression equation with \(x = 7\)).