Question

given the data set.
perform a quadratic regression on the data and give the equation
y = \square x^2 + \square x + \square round your answers to the nearest thousandth.

Explanation:

Step1: Define quadratic regression form

A quadratic regression equation has the form:
$$y = ax^2 + bx + c$$
where $a$, $b$, $c$ are coefficients calculated using the least squares method.

Step2: List the full data set

The complete paired $(x,y)$ data points are:
$(5,99), (4,67), (3,40), (2,20), (1,9), (0,1), (-1,0), (-2,1), (-3,12), (-4,32), (-5,56)$

Step3: Calculate sum values

We first compute the necessary sums for the least squares formulas:

• $n = 11$ (number of data points)
• $\sum x = 5+4+3+2+1+0+(-1)+(-2)+(-3)+(-4)+(-5) = 0$
• $\sum y = 99+67+40+20+9+1+0+1+12+32+56 = 337$
• $\sum x^2 = 25+16+9+4+1+0+1+4+9+16+25 = 110$
• $\sum x^3 = 125+64+27+8+1+0+(-1)+(-8)+(-27)+(-64)+(-125) = 0$
• $\sum x^4 = 625+256+81+16+1+0+1+16+81+256+625 = 1968$
• $\sum xy = (5×99)+(4×67)+(3×40)+(2×20)+(1×9)+(0×1)+(-1×0)+(-2×1)+(-3×12)+(-4×32)+(-5×56)$

$= 495+268+120+40+9+0+0-2-36-128-280 = 486$

Step4: Solve for coefficients

Since $\sum x = 0$ and $\sum x^3 = 0$, the system of equations simplifies:

1. For $c$:

$$c = \frac{\sum y \cdot \sum x^4 - \sum x^2 \cdot \sum x^2 y}{n \cdot \sum x^4 - (\sum x^2)^2}$$
First calculate $\sum x^2 y = (25×99)+(16×67)+(9×40)+(4×20)+(1×9)+(0×1)+(1×0)+(4×1)+(9×12)+(16×32)+(25×56)$
$=2475+1072+360+80+9+0+0+4+108+512+1400 = 6020$
$$c = \frac{337×1968 - 110×6020}{11×1968 - 110^2} = \frac{663216 - 662200}{21648 - 12100} = \frac{1016}{9548} ≈ 0.106$$

1. For $b$:

$$b = \frac{\sum xy}{\sum x^2} = \frac{486}{110} ≈ 4.418$$

1. For $a$:

$$a = \frac{n \cdot \sum x^2 y - \sum x^2 \cdot \sum y}{n \cdot \sum x^4 - (\sum x^2)^2} = \frac{11×6020 - 110×337}{9548} = \frac{66220 - 37070}{9548} = \frac{29150}{9548} ≈ 3.053$$

Answer:

$y = 3.053x^2 + 4.418x + 0.106$