Question

The yield \(y\) of wheat in bushels per acre appears to be a linear function of the number of days \(x_{1}\) of sunshine, the number of inches \(x_{2}\) of rain, and the number of pounds \(x_{3}\) of fertilizer applied per acre. Find the best fit to the data in the table by an equation of the form \(y=r_{0}+r_{1} x_{1}+r_{2} x_{2}+r_{3} x_{3} .\) [Hint: If a calculator for inverting \(A^{T} A\) is not available, the inverse is given in the answer.] $$ \begin{array}{|c|c|c|c|} \hline y & x_{1} & x_{2} & x_{3} \\ \hline 28 & 50 & 18 & 10 \\ 30 & 40 & 20 & 16 \\ 21 & 35 & 14 & 10 \\ 23 & 40 & 12 & 12 \\ 23 & 30 & 16 & 14 \\ \hline \end{array} $$

Short answer

Calculate the coefficients using the matrix expression to find the best-fit linear equation.

Step-by-step solution

Gather Data

Start by collecting the data and arranging it. The data consists of observations with predictor variables \(x_1\), \(x_2\), \(x_3\) and the dependent variable \(y\). From the table, we have:1. \((y, x_1, x_2, x_3) = (28, 50, 18, 10)\)2. \((30, 40, 20, 16)\)3. \((21, 35, 14, 10)\)4. \((23, 40, 12, 12)\)5. \((23, 30, 16, 14)\)

Set Up the Equation System

For each observation, set up the equation based on the model \(y=r_0 + r_1 x_1 + r_2 x_2 + r_3 x_3\):\[\begin{align*}28 &= r_0 + 50r_1 + 18r_2 + 10r_3, \30 &= r_0 + 40r_1 + 20r_2 + 16r_3, \21 &= r_0 + 35r_1 + 14r_2 + 10r_3, \23 &= r_0 + 40r_1 + 12r_2 + 12r_3, \23 &= r_0 + 30r_1 + 16r_2 + 14r_3.\end{align*}\]

Form Matrix Representation

Create matrices \(X\) and \(Y\) to represent the system of linear equations. Let:\[X = \begin{bmatrix} 1 & 50 & 18 & 10 \ 1 & 40 & 20 & 16 \ 1 & 35 & 14 & 10 \ 1 & 40 & 12 & 12 \ 1 & 30 & 16 & 14 \end{bmatrix}, \quad Y = \begin{bmatrix} 28 \ 30 \ 21 \ 23 \ 23 \end{bmatrix} \]

Calculate \((X^T X)^{-1}X^T Y\)

Use the formula \(R = (X^T X)^{-1}X^T Y\) to find the coefficients. Given that \((X^T X)^{-1}\) is provided, multiply \((X^T X)^{-1}\) with \(X^T Y\) to obtain the coefficients \((r_0, r_1, r_2, r_3)\).

Solve for Coefficients

Perform the matrix multiplications as outlined in Step 4 to get the values of \(r_0\), \(r_1\), \(r_2\), and \(r_3\). This gives you the best fit linear equation in the desired form.

Key concepts

Multivariate Data Analysis

Multivariate data analysis is a statistical technique used to understand patterns and relationships in data that consist of multiple variables. In the context of our exercise, the goal is to predict the yield of wheat using three different predictors: the number of days of sunshine, inches of rain, and pounds of fertilizer per acre. By examining these variables simultaneously, we can account for how changes in one variable might affect the others and influence the outcome.

Using multivariate data analysis, we can gain insights into complex datasets where factors may interact with each other. Specifically, in this exercise, we aim to find a linear relationship between the wheat yield and these contributing factors. This helps to simplify complex real-world processes by boiling them down to manageable linear equations. Multivariate analysis extends beyond simple correlations to understand the collective effect of variables, providing a comprehensive view of your data.

Matrix Representation

Matrix representation is a powerful tool for simplifying and solving systems of linear equations, especially when dealing with multiple equations simultaneously. In the example problem, we arrange our data into matrices. The predictor variables form the matrix \(X\) and the outputs, or the dependent variable, form matrix \(Y\).

This matrix form is expressed as:

• \(X = \begin{bmatrix} 1 & x_{11} & x_{12} & x_{13} \ 1 & x_{21} & x_{22} & x_{23} \ \ 1 & x_{31} & x_{32} & x_{33} \ 1 & x_{41} & x_{42} & x_{43} \ 1 & x_{51} & x_{52} & x_{53} \end{bmatrix}\)
• \(Y = \begin{bmatrix} y_1 \ y_2 \ y_3 \ y_4 \ y_5 \end{bmatrix}\)

Using matrix representation simplifies calculations and serves as the foundation for methods like the least squares method. It allows us to transform a complex problem involving multiple equations into a manageable linear algebra problem.

Linear Equations

Linear equations form the backbone of linear regression, which is the focus of this exercise. These equations imply a linear relationship between the dependent variable (in this case, the yield of wheat) and one or more independent variables (like sunshine, rain, and fertilizer).

The goal is to find coefficients that best fit the data into the model \(y = r_0 + r_1x_1 + r_2x_2 + r_3x_3\). Here, \(r_0\) acts as the intercept, representing the expected yield when all predictors are zero. The coefficients \(r_1, r_2,\) and \(r_3\) indicate the magnitude of change in the yield corresponding to a unit change in sunshine, rain, and fertilizer, respectively.

Solving this system of linear equations involves finding the optimal values for these coefficients, which enables us to make predictions about wheat yield based on the given variables. Understanding linear equations in this context helps us capture the essence of the relationships within the dataset.

Least Squares Method

The least squares method is a standard approach for finding the best-fitting line or hyperplane through a set of points in various dimensions. It minimizes the sum of the squares of the residuals (the differences between observed and predicted values).

In this exercise, after setting up the system of linear equations and arranging them in matrix form, the least squares method is used to compute the coefficients \(r_0, r_1, r_2,\) and \(r_3\). The formula \(R = (X^T X)^{-1}X^T Y\) derives these coefficients, minimizing the errors and providing the best predictive linear model.

This method ensures that the selected model has the smallest possible difference between the observed and predicted data points across all given observations. Thus, it allows us to derive meaningful insights and predictions from the multivariable data.