Answer :
Sure, let's work through the problem step-by-step. We need to fit a quadratic model \( y = b_0 + b_1 x + b_2 x^2 \) to the given data:
[tex]\[ \begin{array}{c|c|c|c|c|c} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline y & 5 & 20 & 45 & 75 & 110 \\ \hline \end{array} \][/tex]
Given the data transformation \( v = x - 3 \) and \( w = \frac{y - 45}{5} \), let's first compute the transformed values:
[tex]\[ \begin{array}{c|c|c|c|c|c} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline v & -2 & -1 & 0 & 1 & 2 \\ \hline \end{array} \][/tex]
[tex]\[ \begin{array}{c|c|c|c|c|c} \hline y & 5 & 20 & 45 & 75 & 110 \\ \hline w & -8 & -5 & 0 & 6 & 13 \\ \hline \end{array} \][/tex]
Next, we compute the necessary sums for the transformed values \(v\) and \(w\). We need the following sums:
[tex]\[ \sum v, \quad \sum v^2, \quad \sum v^3, \quad \sum v^4, \quad \sum w, \quad \sum vw, \quad \sum v^2 w \][/tex]
From the problem statement and provided results:
1. \( n = 5 \) (number of data points)
2. \( \sum v = 0 \)
3. \( \sum v^2 = 10 \)
4. \( \sum v^3 = 0 \)
5. \( \sum v^4 = 34 \)
6. \( \sum w = 6.0 \)
7. \( \sum vw = 53.0 \)
8. \( \sum v^2 w = 21.0 \)
These values are confirmed as follows:
\begin{align}
\sum v & = -2 + (-1) + 0 + 1 + 2 = 0 \\
\sum v^2 & = (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 4 + 1 + 0 + 1 + 4 = 10 \\
\sum v^3 & = (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 = -8 - 1 + 0 + 1 + 8 = 0 \\
\sum v^4 & = (-2)^4 + (-1)^4 + 0^4 + 1^4 + 2^4 = 16 + 1 + 0 + 1 + 16 = 34 \\
\sum w & = -8 + (-5) + 0 + 6 + 13 = 6 \\
\sum vw & = (-2)(-8) + (-1)(-5) + (0)(0) + (1)(6) + (2)(13) = 16 + 5 + 0 + 6 + 26 = 53 \\
\sum v^2 w & = (-2)^2(-8) + (-1)^2(-5) + (0)^2(0) + (1)^2(6) + (2)^2(13) = 4(-8) + 1(-5) + 0 + 1(6) + 4(13) \\
& = -32 - 5 + 0 + 6 + 52 = 21
\end{align}
Using these sums, we can form the system of equations as follows:
[tex]\[ \begin{pmatrix} n & \sum v & \sum v^2 \\ \sum v & \sum v^2 & \sum v^3 \\ \sum v^2 & \sum v^3 & \sum v^4 \end{pmatrix} \begin{pmatrix} b_0 \\ b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} \sum w \\ \sum vw \\ \sum v^2 w \end{pmatrix} \][/tex]
Substituting the computed values gives:
[tex]\[ \begin{pmatrix} 5 & 0 & 10 \\ 0 & 10 & 0 \\ 10 & 0 & 34 \end{pmatrix} \begin{pmatrix} b_0 \\ b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} 6 \\ 53 \\ 21 \end{pmatrix} \][/tex]
So the system of equations is:
[tex]\[ \begin{aligned} 5b_0 + 0b_1 + 10b_2 &= 6 \\ 0b_0 + 10b_1 + 0b_2 &= 53 \\ 10b_0 + 0b_1 + 34b_2 &= 21 \end{aligned} \][/tex]
This is the system of equations that yields the least squares regression coefficients for the quadratic model after applying the given data transformations.
[tex]\[ \begin{array}{c|c|c|c|c|c} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline y & 5 & 20 & 45 & 75 & 110 \\ \hline \end{array} \][/tex]
Given the data transformation \( v = x - 3 \) and \( w = \frac{y - 45}{5} \), let's first compute the transformed values:
[tex]\[ \begin{array}{c|c|c|c|c|c} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline v & -2 & -1 & 0 & 1 & 2 \\ \hline \end{array} \][/tex]
[tex]\[ \begin{array}{c|c|c|c|c|c} \hline y & 5 & 20 & 45 & 75 & 110 \\ \hline w & -8 & -5 & 0 & 6 & 13 \\ \hline \end{array} \][/tex]
Next, we compute the necessary sums for the transformed values \(v\) and \(w\). We need the following sums:
[tex]\[ \sum v, \quad \sum v^2, \quad \sum v^3, \quad \sum v^4, \quad \sum w, \quad \sum vw, \quad \sum v^2 w \][/tex]
From the problem statement and provided results:
1. \( n = 5 \) (number of data points)
2. \( \sum v = 0 \)
3. \( \sum v^2 = 10 \)
4. \( \sum v^3 = 0 \)
5. \( \sum v^4 = 34 \)
6. \( \sum w = 6.0 \)
7. \( \sum vw = 53.0 \)
8. \( \sum v^2 w = 21.0 \)
These values are confirmed as follows:
\begin{align}
\sum v & = -2 + (-1) + 0 + 1 + 2 = 0 \\
\sum v^2 & = (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 4 + 1 + 0 + 1 + 4 = 10 \\
\sum v^3 & = (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 = -8 - 1 + 0 + 1 + 8 = 0 \\
\sum v^4 & = (-2)^4 + (-1)^4 + 0^4 + 1^4 + 2^4 = 16 + 1 + 0 + 1 + 16 = 34 \\
\sum w & = -8 + (-5) + 0 + 6 + 13 = 6 \\
\sum vw & = (-2)(-8) + (-1)(-5) + (0)(0) + (1)(6) + (2)(13) = 16 + 5 + 0 + 6 + 26 = 53 \\
\sum v^2 w & = (-2)^2(-8) + (-1)^2(-5) + (0)^2(0) + (1)^2(6) + (2)^2(13) = 4(-8) + 1(-5) + 0 + 1(6) + 4(13) \\
& = -32 - 5 + 0 + 6 + 52 = 21
\end{align}
Using these sums, we can form the system of equations as follows:
[tex]\[ \begin{pmatrix} n & \sum v & \sum v^2 \\ \sum v & \sum v^2 & \sum v^3 \\ \sum v^2 & \sum v^3 & \sum v^4 \end{pmatrix} \begin{pmatrix} b_0 \\ b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} \sum w \\ \sum vw \\ \sum v^2 w \end{pmatrix} \][/tex]
Substituting the computed values gives:
[tex]\[ \begin{pmatrix} 5 & 0 & 10 \\ 0 & 10 & 0 \\ 10 & 0 & 34 \end{pmatrix} \begin{pmatrix} b_0 \\ b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} 6 \\ 53 \\ 21 \end{pmatrix} \][/tex]
So the system of equations is:
[tex]\[ \begin{aligned} 5b_0 + 0b_1 + 10b_2 &= 6 \\ 0b_0 + 10b_1 + 0b_2 &= 53 \\ 10b_0 + 0b_1 + 34b_2 &= 21 \end{aligned} \][/tex]
This is the system of equations that yields the least squares regression coefficients for the quadratic model after applying the given data transformations.
