A regression model is to be developed for predicting the...

A regression model is to be developed for predicting the ability of soil to absorb chemical contaminants. Ten observations have been taken on a soil absorption index $(y)$ and two regressors: $x_{1}=$ amount of extractable iron ore and $x_{2}=$ amount of bauxite. We wish to fit the model $Y=\beta_{0}+\beta_{1} x_{1}+$ $\beta_{2} x_{2}+\epsilon$. Some necessary quantities are. $$ \begin{array}{c} \left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}=\left[\begin{array}{ccc} 1.17991 & -7.30982 \mathrm{E}-3 & 7.3006 \mathrm{E}-4 \\ -7.30982 \mathrm{E}-3 & 7.9799 \mathrm{E}-5 & -1.23713 \mathrm{E}-4 \\ 7.3006 \mathrm{E}-4 & -1.23713 \mathrm{E}-4 & 4.6576 \mathrm{E}-4 \end{array}\right] \\ \mathbf{X}^{\prime} \mathbf{y}=\left[\begin{array}{r} 220 \\ 36,768 \\ 9,965 \end{array}\right] \end{array} $$ (a) Estimate the regression coefficients in the model specified above. (b) What is the predicted value of the absorption index $y$ when $x_{1}=200$ and $x_{2}=50 ?$

A regression model is to be developed for predicting the ability of soil to absorb chemical contaminants. Ten observations have been taken on a soil absorption index $(y)$ and two regressors: $x_{1}=$ amount of extractable iron ore and $x_{2}=$ amount of bauxite. We wish to fit the model $Y=\beta_{0}+\beta_{1} x_{1}+$ $\beta_{2} x_{2}+\epsilon$. Some necessary quantities are.
$$ \begin{array}{c} \left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}=\left[\begin{array}{ccc} 1.17991 & -7.30982 \mathrm{E}-3 & 7.3006 \mathrm{E}-4 \\ -7.30982 \mathrm{E}-3 & 7.9799 \mathrm{E}-5 & -1.23713 \mathrm{E}-4 \\ 7.3006 \mathrm{E}-4 & -1.23713 \mathrm{E}-4 & 4.6576 \mathrm{E}-4 \end{array}\right] \\ \mathbf{X}^{\prime} \mathbf{y}=\left[\begin{array}{r} 220 \\ 36,768 \\ 9,965 \end{array}\right] \end{array} $$
(a) Estimate the regression coefficients in the model specified above.
(b) What is the predicted value of the absorption index $y$ when $x_{1}=200$ and $x_{2}=50 ?$

Key Concepts

Prediction in Regression Models

Prediction in regression involves using the estimated regression equation to forecast the value of the dependent variable for given values of the independent variables. By substituting the new values into the regression equation, one obtains the predicted outcome, which is essential for forecasting and decision-making in applied settings.

Ordinary Least Squares (OLS) Estimation

OLS estimation is a method used to determine the coefficients of a linear regression model by minimizing the sum of the squared differences (residuals) between the observed values and those predicted by the linear model. It provides the best linear unbiased estimators under the Gauss-Markov assumptions.

Multiple Linear Regression

Multiple linear regression is a statistical technique that models the relationship between a continuous dependent variable and two or more independent variables. This method assumes that the effect of the independent variables on the dependent variable is linear, and it is often used to understand how different factors collectively influence a response.

Matrix Algebra in Regression

Matrix algebra provides a compact and efficient way to express the regression model and compute estimates. By representing the model in the form y = X? + ?, where X is the design matrix and ? is the vector of coefficients, one can use matrix operations, such as the inversion of X'X and its multiplication with X'y, to obtain the OLS estimates in a systematic way.

Transcript

00:02 In this problem we're using regression and we're going to do almost the entire problem on the graphing calculator.

00:09 So the first thing we want to do is grab a calculator and then go to stat and then go to edit.

00:17 And we're going to type our x values into list one and our y values into list two.

00:27 Now that we have all the data in the lists, our job is to find the power regression model.

00:33 So we go to stat and then over to calculate.

00:37 And then go down until you find power regression it's just under exponential regression if you're using the ti graphing calculator press enter and then we're using list one we're using list two you don't need to put anything in frequency list and here in store regression equation i'm going to store my equation in my y equals menu and so if i go to alpha trace that's a shortcut to find the y1 then select y1 that tells the calculator where to store the equation and then we press enter and this is our regression equation now if we go to y equals we'll see that was pasted in and just for the sake of writing down the answer we can round those numbers a little bit and write that down so that model was about y equals 2 .75 x to the fifth in the next part of the problem we're making a prediction.

01:57 So if x is 7 .1, let's find its y value.

02:01 So i'm going to do this graphically, and i already have the function in y equals, and now i'm going to go to zoom stat, which is zoom 9, because it creates a good window for a statistics graph.

02:16 And i'm interested in what's happening when x is 7 .1.

02:21 So if i press trace and type in 7 .1, well, i get a value that's off my screen apparently so now i need to fiddle with that and see what's going on okay so the problem is that to use zoom stat we need to turn on our scatter plot so i'm going to go up to plot one turn it on now the calculator knows to use zoom stat with list 1 and list 2 zoom 9 and there's our power curve so now if we press trace and make sure the cursor is on the curve and we type in 7 .1 we can see the y value there at the bottom of the screen 49 ,616 .3.

03:15 All right so we'll write down this answer when x is 7 .1, y is approximately 49 ,16...

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