In statistical evaluation: “a selected operation that is achieved on a fixed of information factors to discover the equation of the parabola is called regression analysis”
you could work for the quadratic regression equations in the following form: $$ y = ax^{2} + bx + c $$
Mean:
As we've x and y values inside 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 assist 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:
subsequent, we need to decide the coefficients of the equation as follows:
$$ a = \bar{y}-b\bar{x}-c\bar{x^2} $$
$$ b = \dfrac{S_{xy}S_{x^2x^2}-S_{x^2y}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2} $$
$$ c = \dfrac{S_{x^2y}S_{xx}-S_{xy}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2} $$
Instance:
decide quadratic regression for the following facts set of factors:
$$ (12, 13), (11, 17), (14, 11), (9, 12), (2, 11), (13, 10) $$
Solution:
From the records set given, we will separate the values of X and Y as follows:
$$ X = 12, 11, 14, 9, 2, 13 $$
$$ y = 13, 17, 11, 12, 11, 10 $$
to start with, we have to determine the imply 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:
$$ \bar{x^{2}} = \frac{1}{n}\sum_{i=1}^nx_{i}^{2} $$
$$ \bar{x^{2}} = \frac{1}{n} \left(12 + 11 + 14 + 9 + 2 + 13\right)^2 $$
$$ \bar{x^{2}} = \frac{\left(61\right)^2}{6} $$
$$ \bar{x^{2}} = \frac{3721}{6} $$
$$ \bar{x^{2}} = 620.16 $$
Calculating crucial 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've got:
$$ c = \dfrac{S_{x^2y}S_{xx}-S_{xy}S_{xx^2}}{S_{xx}S_{x^2x^2}-(S_{xx^2})^2} $$
$$ c = \frac{\left(-27.33\right) \left(94.83\right) - \left(4.67\right) \left(1467.83\right)}{\left(94.83\right) \left(23726.83\right) - \left(1467.83\right)^2} $$
$$ c = \frac{-2591.7039 - 6854.7661}{2250015.2889 - 2154524.9089} $$
$$ c = \frac{-9446.47}{95490.38} $$
$$ c = -0.098 $$
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 closing, we must 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:
$$ y = ax^{2} + bx + c $$
$$ y = 8.05845x^{2} + 1.57855x - 0.09881 $$
