A study was performed to investigate the shear strength of...

A study was performed to investigate the shear strength of soil $(y)$ as it related to depth in feet $\left(x_{1}\right)$ and $\%$ moisture content $\left(x_{2}\right)$. Ten observations were collected, and the following summary quantities obtained: $n=10, \sum x_{i 1}=223,$ $\sum x_{2}=553, \Sigma y_{i}=1,916, \sum x_{i 1}^{2}=5,200.9, \sum x_{R}^{2}=31,729$ $\sum x_{i 1} x_{i 2}=12,352, \sum x_{i 1} y_{i}=43,550.8, \sum x_{i 2} y_{i}=104,736.8$ and $\sum y_{i}^{2}=371,595.6$ (a) Set up the least squares normal equations for the model $Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\epsilon$ (b) Estimate the parameters in the model in part (a). (c) What is the predicted strength when $x_{1}=18$ feet and $x_{2}=43 \% ?$

A study was performed to investigate the shear strength of soil $(y)$ as it related to depth in feet $\left(x_{1}\right)$ and $\%$ moisture content $\left(x_{2}\right)$. Ten observations were collected, and the following summary quantities obtained: $n=10, \sum x_{i 1}=223,$ $\sum x_{2}=553, \Sigma y_{i}=1,916, \sum x_{i 1}^{2}=5,200.9, \sum x_{R}^{2}=31,729$ $\sum x_{i 1} x_{i 2}=12,352, \sum x_{i 1} y_{i}=43,550.8, \sum x_{i 2} y_{i}=104,736.8$ and $\sum y_{i}^{2}=371,595.6$
(a) Set up the least squares normal equations for the model $Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\epsilon$
(b) Estimate the parameters in the model in part (a).
(c) What is the predicted strength when $x_{1}=18$ feet and $x_{2}=43 \% ?$

Key Concepts

Parameter Estimation

Parameter estimation in the context of regression involves calculating estimates for the intercept and slope coefficients based on the observed data. These estimates quantify the relationship between the predictor variables and the response variable, and are essential for interpreting the influence of each independent variable on the dependent variable.

Normal Equations

Normal equations are derived from setting the partial derivatives of the sum of squared residuals with respect to each regression parameter to zero. The resulting system of equations is used to calculate the estimates of the regression coefficients, ensuring that the residual sum of squares is minimized. This concept is critical in understanding the mathematical mechanics behind the least squares method.

Prediction in Regression

Prediction in regression involves using the estimated regression model to compute the expected value of the dependent variable for given values of the independent variables. This process allows for forecasting and making informed decisions based on the relationship captured by the regression model.

Multiple Linear Regression

Multiple linear regression is a statistical method used to model the relationship between a dependent variable and two or more independent variables by fitting a linear equation to observed data. It extends simple linear regression by allowing for multiple explanatory variables, enabling a more comprehensive understanding of how various factors influence the outcome variable.

Least Squares Estimation

The least squares estimation method is a standard approach used in regression analysis to determine the coefficients that minimize the sum of the squared differences (residuals) between the observed and predicted values. This method provides the best linear unbiased estimates (BLUE) under the Gauss-Markov assumptions, making it a foundational concept in regression modeling.

Transcript

00:01 Here's a solution to 11 -1, and we're looking at strength and permeability, and i think these are concrete mixes.

00:06 So we're given some summary stats.

00:07 Sample size, summation y, summation y, squared, summation of x, summation of x squared, and summation of xy.

00:13 So that was all given to you.

00:14 And before we do anything, let's just get the regression equation.

00:18 The way to get to regression equation, we need a y intercept, we need a slope.

00:22 So the quickest way is to find the slope first, and that's just ssxy over ssx, which you use this shortcut formula here where you just plug.

00:30 In the numbers here.

00:31 So summation of xy is 1697 .8, not that number there, and minus summation of x, which is the 43, times summation of y, which is 572, and then that's divided by the sample size, 14, and then we divide all of that by a summation of x squared, which is 157 .42, minus summation of x quantity squared, so that would be 43 squared.

01:03 And then we divide that little piece by 14.

01:06 So that should give you negative 59 .057 over 25 .349.

01:17 And whenever you simplify that, you get about negative 2 .33.

01:21 We'll go to a couple decimal places here.

01:23 So that's your slope, negative 2 .33.

01:25 And then the way to find the y intercept, the b0, it's the summation of y, which is 572, minus b1.

01:35 So that becomes a plus because it's minus a negative.

01:37 So plus 2 .33 times summation of x, which was 43, and then all of that divided by the 14, the sample size.

01:45 And that gives us 48 .013.

01:49 So that is the y intercept.

01:51 And that's really, you know, basically all the work that we need mostly.

01:57 Is just write what the regression line is.

02:00 So here's what it is.

02:01 It's the y intercept 48 .013 and then minus 2 .33x.

02:07 So that's the regression line.

02:10 Okay, and we're also supposed to find the standard error, or i should say the estimate for sigma squared.

02:19 And what that is, that's sse over n minus 2.

02:24 Well, sse, if you have this formula, you should have it, but it's the summation of y squared minus beta not summation y minus beta 1 summation xy.

02:41 So the sse in this case would be 23 ,000, the summation of y squared is 23 ,530, and then minus that 48 .013 times summation of y, which is 572, and then plus because it's minus beta at 1, so plus 2 .33 times summation of xy, which is that 1697 .8.

03:07 So that sse is about 22 .438...

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