/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Show that in a simple linear reg... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that in a simple linear regression model the point \((\bar{x}, \bar{y})\) lies exactly on the least squares regression line.

Short Answer

Expert verified
The point \((\bar{x}, \bar{y})\) satisfies the regression line equation, so it lies on the line.

Step by step solution

01

Identify the components of the regression line

The least squares regression line is given by the equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. The slope \( m \) can be calculated using the formula \( m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \). The y-intercept \( b \) is calculated as \( b = \bar{y} - m\bar{x} \).
02

Substitute the average points into the regression line

Substitute the point \((\bar{x}, \bar{y})\) into the regression line equation \( y = mx + b \). This yields \( \bar{y} = m\bar{x} + b \).
03

Show the satisfaction of the regression equation

To confirm \( \bar{y} = m\bar{x} + b \), recall from the intercept equation \( b = \bar{y} - m\bar{x} \). Substituting \( b \) in the regression line equation results in \( \bar{y} = m\bar{x} + (\bar{y} - m\bar{x}) \), which simplifies to \( \bar{y} = \bar{y} \).
04

Conclude the derivation

Since our simplification confirms \( \bar{y} = \bar{y} \), it proves that the point \((\bar{x}, \bar{y})\) satisfies the regression line equation, thus it lies on the least squares regression line.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Least Squares Method
The least squares method is a standard approach in regression analysis for finding the best-fitting line or curve to a set of data points. This method minimizes the sum of the squares of the residuals, which are the differences between the observed values and the values predicted by the model.
The main goal of using the least squares method is to find a line that would make these residuals as small as possible. This is why it is called the "least squares" method. The adjustments made to the data points ensure that the line is as close to the actual data points as feasible.

To apply the least squares method in simple linear regression, you start with an equation of the line, usually represented as \( y = mx + b \). Here:
  • \( y \) is the dependent variable
  • \( x \) is the independent variable
  • \( m \) represents the slope of the line
  • \( b \) is the intercept on the y-axis
By choosing \( m \) and \( b \) such that the residuals' squares are minimized, a best-fit line can be established, making it a crucial method in statistics and data analysis.
Regression Line
A regression line is a straight line that describes how a response variable \( y \) changes as an explanatory variable \( x \) changes. It allows you to predict \( y \) for any given \( x \). In the context of simple linear regression, the regression line helps summarize the relationship between two variables with a linear equation.
The equation for a regression line in simple linear regression is \( y = mx + b \). The regression line minimizes the distance between the points and the line, capturing the trend shown by the data points.

The regression line is significant because:
  • It reveals the nature of the relationship between \( x \) and \( y \).
  • It allows us to make predictions about the dependent variable \( y \) based on new values of \( x \).
  • It helps identify how changes in the independent variable \( x \) affect the dependent variable \( y \).
By plotting the regression line, you gain visual insight into the correlation and any potential trends or patterns within your data.
Slope and Intercept
The slope and intercept are key components of the regression line equation, representing its orientation and position. Understanding these terms is critical when analyzing the relationship between the variables in a regression model.
The **slope** \( m \) indicates how much \( y \) changes for a one-unit change in \( x \). It reveals the steepness or angle of the regression line. A positive slope means that as \( x \) increases, \( y \) also increases, and vice versa.

Conversely, the **intercept** \( b \) is the value of \( y \) when \( x \) equals zero. It gives the starting point of the line on the y-axis. The intercept shows where the line crosses the y-axis and is crucial for forming the complete equation of the line.
Some key points about slope and intercept are:
  • The slope determines the direction and the steepness of the line.
  • The intercept helps in understanding the baseline level of the dependent variable \( y \).
  • The combination of slope and intercept offers a full mathematical description of the line's behavior.
By grasping these two concepts, you can better interpret any linear models created using simple linear regression.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An article in Wood Science and Technology [ "Creep in Chipboard, Part 3 : Initial Assessment of the Influence of Moisture Content and Level of Stressing on Rate of Creep and Time to Failure" (1981, Vol. \(15,\) pp. \(125-144\) ) ] studied the deflection (mm) of particleboard from stress levels of relative humidity. Assume that the two variables are related according to the simple linear regression model. The data are shown below: \(\begin{array}{l}x=\text { Stress level }(\%): 54 & 54 & 61 & 61\end{array} \quad 68\) \(y=\) Deflection \((\mathrm{mm}): 16.473 \quad 18.693 \quad 14.305 \quad 15.121 \quad 13.505\) \(x=\) Stress level \((\%): 68 \quad 75 \quad 75 \quad 75\) \(y=\) Deflection \((\mathrm{mm}): 11.64011 .16812 .53411 .224\) (a) Calculate the least square estimates of the slope and intercept. What is the estimate of \(\sigma^{2}\) ? Graph the regression model and the data. (b) Find the estimate of the mean deflection if the stress level can be limited to \(65 \%\) (c) Estimate the change in the mean deflection associated with a \(5 \%\) increment in stress level. (d) To decrease the mean deflection by one millimeter, how much increase in stress level must be generated? (e) Given that the stress level is \(68 \%,\) find the fitted value of deflection and the corresponding residual.

Consider the computer output below. The regression equation is \(Y=12.9+2.34 x\) $$\begin{array}{lrrll}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\\\\text { Constant } & 12.857 & 1.032 & ? & ? \\\\\text { X } & 2.3445 & 0.1150 & ? & \text { ? }\end{array}$$ \(\begin{array}{ll}\mathrm{S}=1.48111 & \mathrm{R}-\mathrm{Sq}=98.1 \% & \mathrm{R}-\mathrm{Sq}(\mathrm{adj})=97.9 \%\end{array}\) Analysis of Variance $$\begin{array}{lrrrl}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\\\\text { Regression } & 1 & 912.43 & 912.43 & ? \\\\\text { Residual Error } & 8 & 17.55 & ? & \\\\\text { Total } & 9 & 929.98 & &\end{array}$$ (a) Fill in the missing information. You may use bounds for the \(P\) -values (b) Can you conclude that the model defines a useful linear relationship? (c) What is your estimate of \(\sigma^{2}\) ?

The following data gave \(X=\) the water content of snow on April 1 and \(Y=\) the yield from April to July (in inches) on the Snake River watershed in Wyoming for 1919 to 1935. (The data were taken from an article in Research Notes, Vol. \(61,1950,\) Pacific Northwest Forest Range Experiment Station, Oregon.) $$\begin{array}{cccc}\hline x & y & x & y \\\\\hline 23.1 & 10.5 & 37.9 & 22.8 \\\32.8 & 16.7 & 30.5 & 14.1 \\\31.8 & 18.2 & 25.1 & 12.9 \\ 32.0 & 17.0 & 12.4 & 8.8 \\\30.4 & 16.3 & 35.1 & 17.4 \\\24.0 & 10.5 & 31.5 & 14.9 \\\39.5 & 23.1 & 21.1 & 10.5 \\\24.2 & 12.4 & 27.6 & 16.1 \\\52.5 & 24.9 & & \\\\\hline\end{array}$$ (a) Estimate the correlation between \(Y\) and \(X\). (b) Test the hypothesis that \(\rho=0,\) using \(\alpha=0.05\). (c) Fit a simple linear regression model and test for significance of regression using \(\alpha=0.05 .\) What conclusions can you draw? How is the test for significance of regression related to the test on \(\rho\) in part (b)? (d) Analyze the residuals and comment on model adequacy.

Suppose that each value of \(x_{i}\) is multiplied by a positive constant \(a\), and each value of \(y_{i}\) is multiplied by another positive constant \(b\). Show that the \(t\) -statistic for testing \(H_{0}: \beta_{1}=0\) versus \(H_{1}: \beta_{1} \neq 0\) is unchanged in value.

A random sample of \(n=25\) observations was made on the time to failure of an electronic component and the temperature in the application environment in which the component was used. (a) Given that \(r=0.83,\) test the hypothesis that \(\rho=0\), using \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Find a \(95 \%\) confidence interval on \(\rho\). (c) Test the hypothesis \(H_{0}: \rho=0.8\) versus \(H_{1}: \rho \neq 0.8\), using \(\alpha=0.05 .\) Find the \(P\) -value for this test.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.