What is the quadratic regression equation for the data set?
x y 3 3.45 5 6.9 6 8.79 8 12.91 10 17.48 12 22.49 15 30.85
A. $\widehat{y}=0.056{{x}^{2}}+1.278x$
B. $\widehat{y}=0.056{{x}^{2}}-1.278x-0.886$
C. $\widehat{y}=0.056{{x}^{2}}+1.278$
D. $\widehat{y}=0.056{{x}^{2}}+1.278x-0.886$
| x | y |
| 3 | 3.45 |
| 5 | 6.9 |
| 6 | 8.79 |
| 8 | 12.91 |
| 10 | 17.48 |
| 12 | 22.49 |
| 15 | 30.85 |
Answer
Verified508.2k+ views
Hint: We need to solve the following set of equations for the values of a, b and c which will determine the coefficients of the quadratic equation. These equations are given as follows:
$\begin{align}
&\Rightarrow a\sum{{{x}_{i}}^{4}}+b\sum{{{x}_{i}}^{3}}+c\sum{{{x}_{i}}^{2}}=\sum{{{x}_{i}}^{2}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{3}}+b\sum{{{x}_{i}}^{2}}+c\sum{{{x}_{i}}}=\sum{{{x}_{i}}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{2}}+b\sum{{{x}_{i}}}+cn=\sum{{{y}_{i}}} \\
\end{align}$
Here, n is the number of data in the data set. The summation of the terms needs to be calculated from the given data for x and y. Using these in the above equations and by solving simultaneously, we get the values of a, b and c and using these coefficients, we create a quadratic equation.
Complete step-by-step solution:
In order to solve this question, let us first write down the equations required. They are given as follows:
$\begin{align}
& \Rightarrow a\sum{{{x}_{i}}^{4}}+b\sum{{{x}_{i}}^{3}}+c\sum{{{x}_{i}}^{2}}=\sum{{{x}_{i}}^{2}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{3}}+b\sum{{{x}_{i}}^{2}}+c\sum{{{x}_{i}}}=\sum{{{x}_{i}}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{2}}+b\sum{{{x}_{i}}}+cn=\sum{{{y}_{i}}} \\
\end{align}$
We find the value of $\sum{{{x}_{i}}^{4}}$ for the given data x as given below,
$\Rightarrow {{\sum{x}}_{i}}^{4}={{3}^{4}}+{{5}^{4}}+{{6}^{4}}+{{8}^{4}}+{{10}^{4}}+{{12}^{4}}+{{15}^{4}}$
Expanding each of the terms by taking the powers of 4,
$\Rightarrow {{\sum{x}}_{i}}^{4}=81+625+1296+4096+10000+20736+50625$
Adding all the terms,
$\Rightarrow {{\sum{x}}_{i}}^{4}=87459$
Similarly, we need to calculate for the other summations as shown,
For ${{\sum{x}}_{i}}^{3}$ we take the cubed values of each of the x values and find their summation as,
$\Rightarrow {{\sum{x}}_{i}}^{3}={{3}^{3}}+{{5}^{3}}+{{6}^{3}}+{{8}^{3}}+{{10}^{3}}+{{12}^{3}}+{{15}^{3}}=6983$
Similarly, for ${{\sum{x}}_{i}}^{2}$ we take the squared values of each of the x values and find their summation as,
$\Rightarrow {{\sum{x}}_{i}}^{2}={{3}^{2}}+{{5}^{2}}+{{6}^{2}}+{{8}^{2}}+{{10}^{2}}+{{12}^{2}}+{{15}^{2}}=603$
We find ${{\sum{x}}_{i}}$ and ${{\sum{y}}_{i}}$ by adding the data values in the x and y columns,
$\Rightarrow {{\sum{x}}_{i}}=3+5+6+8+10+12+15=59$
$\Rightarrow {{\sum{y}}_{i}}=3.45+6.9+8.79+12.91+17.48+22.49+30.85=102.87$
We also need the values of the terms $\sum{{{x}_{i}}^{2}{{y}_{i}}}$ and $\sum{{{x}_{i}}{{y}_{i}}}.$ This can be done by taking the square of the x term multiplied by the corresponding y terms and taking their sum for the first summation. For the second summation, we take the product of corresponding x and y terms and add them all together.
$\begin{align}
& \Rightarrow {{\sum{{{x}_{i}}}}^{2}}{{y}_{i}}={{3}^{2}}\times 3.45+{{5}^{2}}\times 6.9+{{6}^{2}}\times 8.79+{{8}^{2}}\times 12.91+{{10}^{2}}\times 17.48+{{12}^{2}}\times 22.49+{{15}^{2}}\times 30.85 \\
& \Rightarrow {{\sum{{{x}_{i}}}}^{2}}{{y}_{i}}=13274.04 \\
\end{align}$
Similarly for the second summation term,
$\begin{align}
& \Rightarrow \sum{{{x}_{i}}}{{y}_{i}}=3\times 3.45+5\times 6.9+6\times 8.79+8\times 12.91+10\times 17.48+12\times 22.49+15\times 30.85 \\
& \Rightarrow \sum{{{x}_{i}}}{{y}_{i}}=1108.3 \\
\end{align}$
Substituting all these in the given set of equations, we get the equations as,
$\begin{align}
& \Rightarrow 87459a+6983b+603c=13274.04 \\
& \Rightarrow 6983a+603b+59c=1108.3 \\
& \Rightarrow 603a+59b+7c=102.87 \\
\end{align}$
Solving for all these values of a, b and c using the method of simultaneous equations, which can be done by taking 2 equations at a time. Take the first two equations. And multiply the second equation by a fraction $\dfrac{603}{59}$ so as to make the coefficient of c same in both the equations.
$\Rightarrow 87459a+6983b+603c=13274.04\ldots \left( 1 \right)$
$\Rightarrow 71368.6271a+6162.8644b+603c=11327.2017\ldots \left( 2 \right)$
Subtracting the above two equations,
$\begin{align}
& \Rightarrow \text{ }87459a+\text{ }6983b+603c=13274.04 \\
& \Rightarrow -71368.6271a-6162.8644b-603c=-11327.2017 \\
\end{align}$
This gives us the equation with the c term eliminated.
$\Rightarrow 16090.3729a+820.1356b=1946.8383\ldots \left( 3 \right)$
Next, we consider equations
$\begin{align}
& \Rightarrow 6983a+603b+59c=1108.3 \\
& \Rightarrow 603a+59b+7c=102.87 \\
\end{align}$
We now multiply the second equation here by a fraction $\dfrac{59}{7}$ to make the coefficients of c terms equal.
$\begin{align}
& \Rightarrow 6983a+603b+59c=1108.3 \\
& \Rightarrow 5082.4286a+497.2857b+59c=867.0471 \\
\end{align}$
Now, we subtract the two equations
$\begin{align}
& \Rightarrow \text{ }6983a+\text{ }603b+59c=1108.3 \\
& \Rightarrow -5082.4286a-497.2857b-59c=-867.0471 \\
& \text{ }\overline{\text{ 1900}\text{.5714}a+105.7143b+0c\text{ = 241}\text{.2529 }\ldots \left( 4 \right)\text{ }} \\
\end{align}$
Now we consider the equations 3 and 4 and multiply the equation 4 by a fraction $\dfrac{820.1356}{105.7143}$ to make the coefficient of b same and subtract the two equations as,
$\begin{align}
& \Rightarrow 16090.3729a+820.1356b=1946.8383 \\
& \Rightarrow -\text{14744}\text{.7059}a-820.1356b\text{ = }-1871.6492 \\
& \text{ }\overline{\text{ 1345}\text{.667}a\text{ =+75}\text{.1891 }} \\
\end{align}$
Now, we divide both sides of the resulting equation by 1345.667 to obtain the value of a.
$\Rightarrow a=\dfrac{75.1891}{1345.667}=0.05587$
Substituting this in the equation 3,
$\Rightarrow 16090.3729\times 0.05587+820.1356b=1946.8383$
Multiplying the terms,
$\Rightarrow 898.9691+820.1356b=1946.8383$
Subtracting both sides by 898.9691 and dividing both sides by 820.1356,
$\Rightarrow 820.1356b=1047.8692$
$\Rightarrow b=\dfrac{1047.8692}{820.1356}=1.2776$
Now, we take the third equation in the set of simultaneous equations and substitute the values of a and b to calculate c.
$\Rightarrow 603\times 0.05587+59\times 1.2776+7c=102.87$
Multiplying and adding the terms on the left-hand side,
$\Rightarrow 109.0680+7c=102.87$
Subtracting both sides by 109.0680 and dividing both sides by 102.87,
$\Rightarrow 7c=102.87-109.0680$
$\Rightarrow c=\dfrac{-6.1980}{7}=-0.8854$
We have the values of a, b and c as,
$\Rightarrow a=0.05587,b=1.2776,c=-0.8854$
Rounding them all to 3 decimal digits, we get
$\Rightarrow a=0.056,b=1.278,c=-0.886$
Forming a quadratic equation with these values as $a{{x}^{2}}+bx+c,$
$\Rightarrow \widehat{y}=0.056{{x}^{2}}+1.278x-0.886$
Hence, the correct option is D.
Note: We need to know the concept of regression equation and its formation for a given set of data. It is important to know the method of solving simultaneous equations in order to obtain a solution for such problems. Students are required to know how to solve simultaneous equations.
$\begin{align}
&\Rightarrow a\sum{{{x}_{i}}^{4}}+b\sum{{{x}_{i}}^{3}}+c\sum{{{x}_{i}}^{2}}=\sum{{{x}_{i}}^{2}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{3}}+b\sum{{{x}_{i}}^{2}}+c\sum{{{x}_{i}}}=\sum{{{x}_{i}}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{2}}+b\sum{{{x}_{i}}}+cn=\sum{{{y}_{i}}} \\
\end{align}$
Here, n is the number of data in the data set. The summation of the terms needs to be calculated from the given data for x and y. Using these in the above equations and by solving simultaneously, we get the values of a, b and c and using these coefficients, we create a quadratic equation.
Complete step-by-step solution:
In order to solve this question, let us first write down the equations required. They are given as follows:
$\begin{align}
& \Rightarrow a\sum{{{x}_{i}}^{4}}+b\sum{{{x}_{i}}^{3}}+c\sum{{{x}_{i}}^{2}}=\sum{{{x}_{i}}^{2}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{3}}+b\sum{{{x}_{i}}^{2}}+c\sum{{{x}_{i}}}=\sum{{{x}_{i}}{{y}_{i}}} \\
& \Rightarrow a\sum{{{x}_{i}}^{2}}+b\sum{{{x}_{i}}}+cn=\sum{{{y}_{i}}} \\
\end{align}$
We find the value of $\sum{{{x}_{i}}^{4}}$ for the given data x as given below,
$\Rightarrow {{\sum{x}}_{i}}^{4}={{3}^{4}}+{{5}^{4}}+{{6}^{4}}+{{8}^{4}}+{{10}^{4}}+{{12}^{4}}+{{15}^{4}}$
Expanding each of the terms by taking the powers of 4,
$\Rightarrow {{\sum{x}}_{i}}^{4}=81+625+1296+4096+10000+20736+50625$
Adding all the terms,
$\Rightarrow {{\sum{x}}_{i}}^{4}=87459$
Similarly, we need to calculate for the other summations as shown,
For ${{\sum{x}}_{i}}^{3}$ we take the cubed values of each of the x values and find their summation as,
$\Rightarrow {{\sum{x}}_{i}}^{3}={{3}^{3}}+{{5}^{3}}+{{6}^{3}}+{{8}^{3}}+{{10}^{3}}+{{12}^{3}}+{{15}^{3}}=6983$
Similarly, for ${{\sum{x}}_{i}}^{2}$ we take the squared values of each of the x values and find their summation as,
$\Rightarrow {{\sum{x}}_{i}}^{2}={{3}^{2}}+{{5}^{2}}+{{6}^{2}}+{{8}^{2}}+{{10}^{2}}+{{12}^{2}}+{{15}^{2}}=603$
We find ${{\sum{x}}_{i}}$ and ${{\sum{y}}_{i}}$ by adding the data values in the x and y columns,
$\Rightarrow {{\sum{x}}_{i}}=3+5+6+8+10+12+15=59$
$\Rightarrow {{\sum{y}}_{i}}=3.45+6.9+8.79+12.91+17.48+22.49+30.85=102.87$
We also need the values of the terms $\sum{{{x}_{i}}^{2}{{y}_{i}}}$ and $\sum{{{x}_{i}}{{y}_{i}}}.$ This can be done by taking the square of the x term multiplied by the corresponding y terms and taking their sum for the first summation. For the second summation, we take the product of corresponding x and y terms and add them all together.
$\begin{align}
& \Rightarrow {{\sum{{{x}_{i}}}}^{2}}{{y}_{i}}={{3}^{2}}\times 3.45+{{5}^{2}}\times 6.9+{{6}^{2}}\times 8.79+{{8}^{2}}\times 12.91+{{10}^{2}}\times 17.48+{{12}^{2}}\times 22.49+{{15}^{2}}\times 30.85 \\
& \Rightarrow {{\sum{{{x}_{i}}}}^{2}}{{y}_{i}}=13274.04 \\
\end{align}$
Similarly for the second summation term,
$\begin{align}
& \Rightarrow \sum{{{x}_{i}}}{{y}_{i}}=3\times 3.45+5\times 6.9+6\times 8.79+8\times 12.91+10\times 17.48+12\times 22.49+15\times 30.85 \\
& \Rightarrow \sum{{{x}_{i}}}{{y}_{i}}=1108.3 \\
\end{align}$
Substituting all these in the given set of equations, we get the equations as,
$\begin{align}
& \Rightarrow 87459a+6983b+603c=13274.04 \\
& \Rightarrow 6983a+603b+59c=1108.3 \\
& \Rightarrow 603a+59b+7c=102.87 \\
\end{align}$
Solving for all these values of a, b and c using the method of simultaneous equations, which can be done by taking 2 equations at a time. Take the first two equations. And multiply the second equation by a fraction $\dfrac{603}{59}$ so as to make the coefficient of c same in both the equations.
$\Rightarrow 87459a+6983b+603c=13274.04\ldots \left( 1 \right)$
$\Rightarrow 71368.6271a+6162.8644b+603c=11327.2017\ldots \left( 2 \right)$
Subtracting the above two equations,
$\begin{align}
& \Rightarrow \text{ }87459a+\text{ }6983b+603c=13274.04 \\
& \Rightarrow -71368.6271a-6162.8644b-603c=-11327.2017 \\
\end{align}$
This gives us the equation with the c term eliminated.
$\Rightarrow 16090.3729a+820.1356b=1946.8383\ldots \left( 3 \right)$
Next, we consider equations
$\begin{align}
& \Rightarrow 6983a+603b+59c=1108.3 \\
& \Rightarrow 603a+59b+7c=102.87 \\
\end{align}$
We now multiply the second equation here by a fraction $\dfrac{59}{7}$ to make the coefficients of c terms equal.
$\begin{align}
& \Rightarrow 6983a+603b+59c=1108.3 \\
& \Rightarrow 5082.4286a+497.2857b+59c=867.0471 \\
\end{align}$
Now, we subtract the two equations
$\begin{align}
& \Rightarrow \text{ }6983a+\text{ }603b+59c=1108.3 \\
& \Rightarrow -5082.4286a-497.2857b-59c=-867.0471 \\
& \text{ }\overline{\text{ 1900}\text{.5714}a+105.7143b+0c\text{ = 241}\text{.2529 }\ldots \left( 4 \right)\text{ }} \\
\end{align}$
Now we consider the equations 3 and 4 and multiply the equation 4 by a fraction $\dfrac{820.1356}{105.7143}$ to make the coefficient of b same and subtract the two equations as,
$\begin{align}
& \Rightarrow 16090.3729a+820.1356b=1946.8383 \\
& \Rightarrow -\text{14744}\text{.7059}a-820.1356b\text{ = }-1871.6492 \\
& \text{ }\overline{\text{ 1345}\text{.667}a\text{ =+75}\text{.1891 }} \\
\end{align}$
Now, we divide both sides of the resulting equation by 1345.667 to obtain the value of a.
$\Rightarrow a=\dfrac{75.1891}{1345.667}=0.05587$
Substituting this in the equation 3,
$\Rightarrow 16090.3729\times 0.05587+820.1356b=1946.8383$
Multiplying the terms,
$\Rightarrow 898.9691+820.1356b=1946.8383$
Subtracting both sides by 898.9691 and dividing both sides by 820.1356,
$\Rightarrow 820.1356b=1047.8692$
$\Rightarrow b=\dfrac{1047.8692}{820.1356}=1.2776$
Now, we take the third equation in the set of simultaneous equations and substitute the values of a and b to calculate c.
$\Rightarrow 603\times 0.05587+59\times 1.2776+7c=102.87$
Multiplying and adding the terms on the left-hand side,
$\Rightarrow 109.0680+7c=102.87$
Subtracting both sides by 109.0680 and dividing both sides by 102.87,
$\Rightarrow 7c=102.87-109.0680$
$\Rightarrow c=\dfrac{-6.1980}{7}=-0.8854$
We have the values of a, b and c as,
$\Rightarrow a=0.05587,b=1.2776,c=-0.8854$
Rounding them all to 3 decimal digits, we get
$\Rightarrow a=0.056,b=1.278,c=-0.886$
Forming a quadratic equation with these values as $a{{x}^{2}}+bx+c,$
$\Rightarrow \widehat{y}=0.056{{x}^{2}}+1.278x-0.886$
Hence, the correct option is D.
Note: We need to know the concept of regression equation and its formation for a given set of data. It is important to know the method of solving simultaneous equations in order to obtain a solution for such problems. Students are required to know how to solve simultaneous equations.
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