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Enter the values of X and Y variavles to calcualte the quadratic regression equation through this calculator.

Quadratic regression calculator helps you to determine the quadratic regression representing the parabola that best suits the data points. We have arranged a proper guide in this content so that you may not face any hurdle while doing such analysis.

In statistical analysis: “A particular operation that is performed on a set of data points to find the equation of the parabola is known as regression analysis”

You can work for the quadratic regression equations in the following form: $$y = ax^{2} + bx + c$$

#### Mean:

As we have x and y values in the points defined, so we have to determine the mean for both x and y values as follows: $$\bar{x} = \frac{1}{n}\sum_{i=1}^nx_{i}$$ $$\bar{x^{2}} = \frac{1}{n}\sum_{i=1}^nx_{i}^{2}$$ $$\bar{y} = \frac{1}{n}\sum_{i=1}^ny_{i}$$

#### Summations:

After doing so, we need to calculate a series of sums with the help of the following formulae:

$$S_{xx} = \sum_{i=1}^n \left(x_{i} - \bar{x}\right)^2$$

$$S_{xy} = \sum_{i=1}^n \left(x_{i} - \bar{x}\right) \left(y_{i} - \bar{y}\right)$$

$$S_{xx^{2}} = \sum_{i=1}^n \left(x_{i} - \bar{x}\right) \left(x_{i}^2 - \bar{x^{2}}\right)$$

$$S_{x^{2}x^{2}} = \sum_{i=1}^n \left(x_{i}^2 - \bar{x^{2}}\right)^2$$

$$S{x^{2}y} = \sum_{i=1}^n \left(x_{i}^2 - \bar{x^{2}}\right) \left(y_{i} - \bar{y}\right)$$

#### Coefficients:

Next, we need to determine the coefficients of the equation as follows:

$$a = \bar{y}-b\bar{x}-c\bar{x^2}$$

$$b = \dfrac{S_{xy}S_{x^2x^2}-S_{x^2y}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2}$$

$$c = \dfrac{S_{x^2y}S_{xx}-S_{xy}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2}$$

## How To Find Quadratic Regression?

Example:

Determine quadratic regression for the following data set of points:

$$(12, 13), (11, 17), (14, 11), (9, 12), (2, 11), (13, 10)$$

Solution:

From the data set given, we can separate the values of X and Y as follows:

$$X = 12, 11, 14, 9, 2, 13$$

$$y = 13, 17, 11, 12, 11, 10$$

First of all, we have to determine the mean of both X and Y values:

$$Mean X = \bar{x} = \frac{1}{n}\sum_{i=1}^nx_{i}$$

$$mean X = \bar{x} =\frac{1}{n} \left(12 + 11 + 14 + 9 + 2 + 13\right)$$

$$Mean X = \bar{x} = \frac{\left(12 + 11 + 14 + 9 + 2 + 13\right)}{6}$$

$$Mean X = \bar{x} = \frac{61}{6}$$

$$Mean X = \bar{x} = 10.166$$

Now we have:

$$Mean Y = \bar{y} = \frac{1}{n}\sum_{i=1}^ny_{i}$$

$$Mean Y = \bar{y} = \frac{1}{n} \left(13 + 17 + 11 + 12 + 11 + 10\right)$$

$$Mean Y = \bar{y} = \frac{\left(13 + 17 + 11 + 12 + 11 + 10\right)}{6}$$

$$Mean Y = \bar{y} = \frac{74}{6}$$

$$Mean Y = \bar{y} = 12.33$$

Also we have:

$$\bar{x^{2}} = \frac{1}{n}\sum_{i=1}^nx_{i}^{2}$$

$$\bar{x^{2}} = \frac{1}{n} \left(12 + 11 + 14 + 9 + 2 + 13\right)^2$$

$$\bar{x^{2}} = \frac{\left(61\right)^2}{6}$$

$$\bar{x^{2}} = \frac{3721}{6}$$

$$\bar{x^{2}} = 620.16$$

Now we need to calculate the following values and arrange them in the table just like below:

 $$\left(x_{i} - \bar{x}\right)^2$$ $$\left(x_{i} - \bar{x}\right)\left(y_{i} - \bar{y}\right)$$ $$\left(x_{i} -\bar{x}\right)\left({x_i}^2 - \bar{x^2}\right)$$ $$\left({x_i}^2 - \bar{x^2}\right)^2$$ $$\left({x_i}^2 - \bar{x^2}\right) \left(y_{i} - \bar{y}\right)$$ $$3.36$$ $$1.223$$ $$45.519$$ $$616.678$$ $$16.564$$ $$0.694$$ $$3.888$$ $$1.527$$ $$3.36$$ $$8.555$$ $$14.692$$ $$-5.109$$ $$294.501$$ $$5903.31$$ $$-102.418$$ $$1.362$$ $$0.389$$ $$44.541$$ $$1456.72$$ $$12.71$$ $$66.7$$ $$10.887$$ $$940.569$$ $$13263.438$$ $$153.518$$ $$8.026$$ $$-6.609$$ $$141.177$$ $$2483.328$$ $$-116.26$$

Calculating important summations as follows:

$$S_{xx} = \sum_{i=1}^n \left(x_{i} - \bar{x}\right)^2$$

$$S_{xx} = 3.36 + 0.694 + 14.692 + 1.3612 + 66.7 + 8.026$$ $$S_{xx} = 94.83$$

$$S_{xy} = \sum_{i=1}^n \left(x_{i} - \bar{x}\right) \left(y_{i} - \bar{y}\right)$$

$$S_{xy} = 1.223 + 3.888 + (-5.109) + 0.389 + 10.887 + (-6.609)$$

$$S_{xy} = 1.223 + 3.888 - 5.109 + 0.389 + 10.887 - 6.609$$

$$S_{xy} = 4.67$$

$$S_{xx^{2}} = \sum_{i=1}^n \left(x_{i} - \bar{x}\right) \left(x_{i}^2 - \bar{x^{2}}\right)$$

$$S_{xx^{2}} = 45.519 + 1.527 + 294.501 + 44.541 + 940.569 + 141.177$$

$$S_{xx^{2}} = 1467.83$$

$$S_{x^{2}x^{2}} = \sum_{i=1}^n \left(x_{i}^2 - \bar{x^{2}}\right)^2$$

$$S_{x^{2}x^{2}} = 616.678 + 3.36 + 5903.31 + 1456.72 + 13263.438 + 2483.328$$ $$S_{x^{2}x^{2}} = 23726.83$$

$$S{x^{2}y} = \sum_{i=1}^n \left(x_{i}^2 - \bar{x^{2}}\right) \left(y_{i} - \bar{y}\right)$$

$$S{x^{2}y} = 16.564 + 8.555 + (-102.418) + 12.71 + 153.518 + (-116.26)$$

$$S{x^{2}y} = 16.564 + 8.555 - 102.418 + 12.71 + 153.518 - 116.26$$

$$S{x^{2}y} = -27.33$$

Determining the coefficients of the equation:

$$b=\dfrac{S_{xy}S_{x^2x^2}-S_{x^2y}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2}$$

$$b = \frac{\left(4.67\right) \left(23726.83\right) + \left(27.33\right) \left(1467.83\right)}{\left(94.83\right) \left(23726.83\right) - \left(1467.83\right)^2}$$

$$b = \frac{110804.2961 + 40115.7939}{2250015.2889 - 2154524.9089}$$

$$b = \frac{150920.09}{95490.38}$$

$$b = 1.580$$

Now we have:

$$c = \dfrac{S_{x^2y}S_{xx}-S_{xy}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2}$$

$$c = \frac{\left(-27.33\right) \left(94.83\right) - \left(4.67\right) \left(1467.83\right)}{\left(94.83\right) \left(23726.83\right) - \left(1467.83\right)^2}$$

$$c = \frac{-2591.7039 - 6854.7661}{2250015.2889 - 2154524.9089}$$

$$c = \frac{-9446.47}{95490.38}$$

$$c = -0.098$$

Now we have:

$$a = \bar{y}-b\bar{x}-c\bar{x^2}$$

$$a = 12.33 - \left(1.580\right) \left(10.167\right) - \left(-0.098\right) \left(103.367889\right)$$ $$a = 12.33 - 16.06386 + 10.130053122$$

$$a = 8.05845$$

At last, we have to find correlation coefficient as follows:

$$\text{Correlation Coefficient} = r = \frac{n \left(\sum xy\right) - \left(\sum x\right) \left(\sum y\right)}{\sqrt([n\sum x^{2} - \left(\sum x\right)^2][n\sum y^{2} - \left(\sum y\right)^2])}$$

$$\text{Correlation Coefficient} = r = 0.3213$$ (for calculations, click Correlation Coefficient Calculator) Now the quadratic regression is as follows:

$$y = ax^{2} + bx + c$$

$$y = 8.05845x^{2} + 1.57855x - 0.09881$$