IDNLearn.com provides a user-friendly platform for finding and sharing knowledge. Join our knowledgeable community to find the answers you need for any topic or issue.
Sagot :
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.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.