Answer :
To find the equation of the line that best fits the data provided in the table, we will employ the method of linear regression. This involves finding the coefficients \( m \) (slope) and \( c \) (intercept) such that the line \( y = mx + c \) minimizes the sum of the squared differences between the observed values (GPA) and the predicted values on the line. Here are the steps to achieve this:
1. Organize the data points:
[tex]\[ \begin{array}{|c|c|} \hline IQ & GPA \\ \hline 117 & 3.9 \\ 93 & 2.7 \\ 102 & 2.9 \\ 110 & 3.1 \\ 88 & 2.4 \\ 75 & 1.9 \\ \hline \end{array} \][/tex]
2. Create the design matrix \( A \):
In linear regression, each row of matrix \( A \) should consist of an \((x, 1)\) pair from your data.
[tex]\[ A = \begin{pmatrix} 117 & 1 \\ 93 & 1 \\ 102 & 1 \\ 110 & 1 \\ 88 & 1 \\ 75 & 1 \\ \end{pmatrix} \][/tex]
3. Construct the \( y \) vector:
The \( y \) vector contains the GPA values corresponding to each IQ.
[tex]\[ y = \begin{pmatrix} 3.9 \\ 2.7 \\ 2.9 \\ 3.1 \\ 2.4 \\ 1.9 \\ \end{pmatrix} \][/tex]
4. Calculate the coefficients \( m \) and \( c \):
Using the least squares method, we solve for the coefficients that best fit the data. The least squares solution seeks to minimize \( \|Ax - y\| \), where \( x \) is a vector containing the coefficients \( m \) and \( c \).
5. Results:
After performing the calculations, the resulting values for the slope \( m \) and the intercept \( c \) are found to be:
[tex]\[ m = 0.0427354069024286 \\ c = -1.350035506320122 \][/tex]
6. Formulate the equation:
Substituting \( m \) and \( c \) into the linear equation \( y = mx + c \), we get:
[tex]\[ y = 0.0427354069024286 \cdot x - 1.350035506320122 \][/tex]
7. Compare with given choices:
We now compare the calculated equation with the given choices:
[tex]\[ \begin{align*} \text{Choice 1:} & \quad y = 1.09x - 2.55 \\ \text{Choice 2:} & \quad y = 0.09x - 1.67 \\ \text{Choice 3:} & \quad y = 21.89x - 2.04 \\ \text{Choice 4:} & \quad y = 0.04x - 1.08 \\ \end{align*} \][/tex]
Our calculated equation \( y = 0.0427354069024286x - 1.350035506320122 \) is closest to "Choice 4: \( y = 0.04x - 1.08 \)”.
Therefore, the equation of the line that best fits the data is [tex]\( y = 0.04x - 1.08 \)[/tex].
1. Organize the data points:
[tex]\[ \begin{array}{|c|c|} \hline IQ & GPA \\ \hline 117 & 3.9 \\ 93 & 2.7 \\ 102 & 2.9 \\ 110 & 3.1 \\ 88 & 2.4 \\ 75 & 1.9 \\ \hline \end{array} \][/tex]
2. Create the design matrix \( A \):
In linear regression, each row of matrix \( A \) should consist of an \((x, 1)\) pair from your data.
[tex]\[ A = \begin{pmatrix} 117 & 1 \\ 93 & 1 \\ 102 & 1 \\ 110 & 1 \\ 88 & 1 \\ 75 & 1 \\ \end{pmatrix} \][/tex]
3. Construct the \( y \) vector:
The \( y \) vector contains the GPA values corresponding to each IQ.
[tex]\[ y = \begin{pmatrix} 3.9 \\ 2.7 \\ 2.9 \\ 3.1 \\ 2.4 \\ 1.9 \\ \end{pmatrix} \][/tex]
4. Calculate the coefficients \( m \) and \( c \):
Using the least squares method, we solve for the coefficients that best fit the data. The least squares solution seeks to minimize \( \|Ax - y\| \), where \( x \) is a vector containing the coefficients \( m \) and \( c \).
5. Results:
After performing the calculations, the resulting values for the slope \( m \) and the intercept \( c \) are found to be:
[tex]\[ m = 0.0427354069024286 \\ c = -1.350035506320122 \][/tex]
6. Formulate the equation:
Substituting \( m \) and \( c \) into the linear equation \( y = mx + c \), we get:
[tex]\[ y = 0.0427354069024286 \cdot x - 1.350035506320122 \][/tex]
7. Compare with given choices:
We now compare the calculated equation with the given choices:
[tex]\[ \begin{align*} \text{Choice 1:} & \quad y = 1.09x - 2.55 \\ \text{Choice 2:} & \quad y = 0.09x - 1.67 \\ \text{Choice 3:} & \quad y = 21.89x - 2.04 \\ \text{Choice 4:} & \quad y = 0.04x - 1.08 \\ \end{align*} \][/tex]
Our calculated equation \( y = 0.0427354069024286x - 1.350035506320122 \) is closest to "Choice 4: \( y = 0.04x - 1.08 \)”.
Therefore, the equation of the line that best fits the data is [tex]\( y = 0.04x - 1.08 \)[/tex].
