Question

Consider the following data. \begin{tabular}{c|cccc} \(x\) & 2 & 3 & 5 & 8 \\ \hline\(y\) & 3 & 4 & 4 & 5 \end{tabular} (a) Set up and solve the normal equations by hand to find the line of best fit, in \(y=m x+b\) form, for the given data. Check your answer using polyfit \((\mathrm{x}, \mathrm{y}, 1)\). (b) Compare to the solution found using Octave's left division operation directly on the relevant (inconsistent) system: $$ \left[\begin{array}{ll} 2 & 1 \\ 3 & 1 \\ 5 & 1 \\ 8 & 1 \end{array}\right] \cdot\left[\begin{array}{l} m \\ b \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \\ 4 \\ 5 \end{array}\right] $$ (c) Plot a graph showing the data points and the regression line.

Short answer

The line of best fit is approximately \(y = 0.3913x + 2.2392\).

Step-by-step solution

Calculate Means of x and y

First, calculate the mean of the "x" values and the "y" values.\[ \bar{x} = \frac{2+3+5+8}{4} = 4.5 \] \[ \bar{y} = \frac{3+4+4+5}{4} = 4 \] These means are used to compute the components of the normal equations.

Construct and Solve Normal Equations

Use the formulas for the components of the normal equations: sum of squares and cross-products.\[ S_{xx} = \sum (x_i - \bar{x})^2 = (2-4.5)^2 + (3-4.5)^2 + (5-4.5)^2 + (8-4.5)^2 = 23 \] \[ S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) = (2-4.5)(3-4) + (3-4.5)(4-4) + (5-4.5)(4-4) + (8-4.5)(5-4) = 9 \] Then solve for the slope \(m\) and intercept \(b\): \[ m = \frac{S_{xy}}{S_{xx}} = \frac{9}{23} \approx 0.3913 \] \[ b = \bar{y} - m\bar{x} = 4 - (0.3913 \times 4.5) \approx 2.2392 \] The equation of the line of best fit is \(y = 0.3913x + 2.2392\).

Comparing Using Polyfit

Use the polyfit functionality to automatically find the best-fit line. Polyfit for degree 1 (linear): "[m, b] = polyfit(x, y, 1)", calculated using tools like Python or Octave should yield coefficients close to our calculated values, confirming the correctness of our manual calculation.

Solve Using Octave's Left Division

Set up the matrix system: \(AX = B\), with \[ A = \begin{pmatrix} 2 & 1 \ 3 & 1 \ 5 & 1 \ 8 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 3 \ 4 \ 4 \ 5 \end{pmatrix} \]. Use Octave's left division '\' to solve for "m" and "b": \[ X = A \backslash B \]. This operation provides a line of best fit through least squares approximation, which should match the manually calculated line.

Plotting the Data and the Regression Line

Create a graph with the data points: (2,3), (3,4), (5,4), and (8,5). The regression line \(y = 0.3913x + 2.2392\) is plotted along with these points. The data points should lie near the line when plotted, showing how well it fits the data.

Key concepts

Least Squares Approximation

When we talk about least squares approximation, we're describing a method to find the best-fitting line to a set of data points. By "best-fitting," we mean the line that minimizes the total distance from all points to the line, when distances are measured vertically (up or down). This is particularly useful for data that doesn't perfectly follow a straight line pattern.

Here's how it works:

• First, you calculate the average of your x-values and y-values. These averages are your mean values, \( \bar{x} \) and \( \bar{y} \), respectively.
• Next, you determine how much each data point deviates from the mean. You do this by calculating \( S_{xx} \) and \( S_{xy} \). \( S_{xx} \) is the sum of squares of deviations of each x-value from the mean of x, and \( S_{xy} \) is the sum of the product of the deviations of x and y from their means.
• Using these sums, you compute the slope \( m \) of the line by dividing \( S_{xy} \) by \( S_{xx} \). Finally, the intercept \( b \) is found by rearranging the line equation \( y = mx + b \) and solving for \( b \), using the mean values.

Least squares approximation offers a simple way to statistically find a line that best represents the data, even if no specific data point lies on it precisely.

Normal Equations

Normal equations provide a systematic method to mathematically determine the optimal slope \( m \) and y-intercept \( b \) for our line of best fit. They derive from the principle of minimizing the sum of the squares of the residuals - the distances from each data point to the fitting line. This results in two core equations based on our dataset:

• \( S_{xx}m + S_{xy}b = \sum x_iy_i \)
• \( S_{xx}m + n\bar{b} = \sum y_i \)

In practice, solving the normal equations involves:

• Calculating \( S_{xx} \) (sum of squares of x) and \( S_{xy} \) (sum of cross-products of x and y).
• Using the already computed \( \bar{x} \) and \( \bar{y} \) to plug into the normal equations.
• Through substitution and algebraic manipulation, you can solve these equations to find the most accurate slope \( m \) and intercept \( b \).

The normal equations ensure that the resulting line minimizes the error across all data points, making it a cornerstone of linear regression.

Octave Polyfit Function

The Octave polyfit function simplifies the process of finding a linear fit to data points. Computers are fantastic at handling repetitive calculations, so functions like polyfit can quickly derive the best-fit line for any dataset without manual computation.

This function is particularly user-friendly:

• Polyfit requires the dataset variables along with the desired polynomial degree, typically 1 for linear regression.
• Using a simple command such as \( \text{[m, b] = polyfit(x, y, 1)} \), Octave computes values of \( m \) (slope) and \( b \) (intercept).
• The output provides coefficients that directly relate to the linear equation, confirming results you might have calculated manually through other methods.

Polyfit not only works effectively for linear regression but is also capable of fitting polynomials of higher degrees to more complex datasets, demonstrating its versatility and power in data analysis.