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Find the equation \(y=\beta_{0}+\beta_{1} x\) of the least-squares line that best fits the given data points. \((-1,0),(0,1),(1,2),(2,4)\)

Short Answer

Expert verified
The least-squares line equation is \(y = 1.2 + 1.1x\).

Step by step solution

01

Understanding Least-Squares

The equation of the least-squares line is of the form, \(y = \beta_0 + \beta_1 x\), where \(\beta_0\) is the y-intercept and \(\beta_1\) is the slope of the line. The goal is to minimize the sum of the squares of the vertical distances of the points from the line.
02

Compute Means

Calculate the mean of the x-values and y-values. For the x-values \((-1, 0, 1, 2)\), \(\bar{x} = \frac{-1 + 0 + 1 + 2}{4} = 0.5\). For the y-values \((0, 1, 2, 4)\), \(\bar{y} = \frac{0 + 1 + 2 + 4}{4} = 1.75\).
03

Calculate the Slope, \(\beta_1\)

Use the formula for the slope: \(\beta_1 = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}\). Substitute \(x_i = (-1, 0, 1, 2)\) and \(y_i = (0, 1, 2, 4)\) into the formula to calculate the sums. Compute each term: \((-1-0.5)(0-1.75) + (0-0.5)(1-1.75) + (1-0.5)(2-1.75) + (2-0.5)(4-1.75) = 5.5\) and \((-1-0.5)^2 + (0-0.5)^2 + (1-0.5)^2 + (2-0.5)^2 = 5\), hence, \(\beta_1 = \frac{5.5}{5} = 1.1\).
04

Calculate y-intercept, \(\beta_0\)

Use the means to find \(\beta_0\) with the formula: \(\beta_0 = \bar{y} - \beta_1\bar{x}\). Substituting the values gives \(\beta_0 = 1.75 - 1.1 \times 0.5 = 1.2\).
05

Write the Least-Squares Equation

With \(\beta_0 = 1.2\) and \(\beta_1 = 1.1\), the equation of the least-squares line is \(y = 1.2 + 1.1x\).

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Key Concepts

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

Linear Equations
Linear equations are fundamental in understanding how relationships work between two variables. These equations play a pivotal role in statistics, particularly when applying the method of least-squares regression.
In general, a linear equation can be expressed in the form of \( y = mx + b \). Here,
  • \( y \) is the dependent variable,
  • \( x \) is the independent variable,
  • \( m \) represents the slope of the line,
  • \( b \) is the y-intercept.

In the context of least-squares regression, linear equations like \( y = \beta_0 + \beta_1 x \) are used to model the best fitting line through a set of data points. The goal is to find the values of \( \beta_0 \) (y-intercept) and \( \beta_1 \) (slope) that minimize the differences between the observed data points and those predicted by the linear equation. This method helps in predicting outcomes and understanding relationships in data.
Slope Calculation
The slope of a line (\( \beta_1 \)) is a measure of how steep the line is, and it describes the rate at which the dependent variable changes with respect to the independent variable. To calculate the slope in a least-squares regression model, a specific formula is used:
  • \( \beta_1 = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \).
This formula considers the means of the x-values (\( \bar{x} \)) and y-values (\( \bar{y} \)). It sums up the product of the deviations of each x and y point from their mean, and divides it by the sum of squares of the deviations of each x from its mean.
To put it simply, the slope tells us how much \( y \) changes with a one-unit increase in \( x \). In the specific exercise, after computation, it shows that for every one unit increase in \( x \), \( y \) increases by 1.1. Understanding slope makes it easier to visualize and predict trends in data.
Statistical Analysis
Statistical analysis often involves creating models like least-squares regression to summarize and draw insights from data. The emphasis here is understanding how data points spread around a line, and how well that line captures the trends within the data.

When engaging in statistical analysis, especially with least-squares regression, several important aspects are considered:
  • Data Fit: It's crucial to see how well the line fits the data points. In practice, metrics such as R-squared can help measure how well the model explains the variability of the data. A higher R-squared value indicates a better fit.
  • Error Minimization: Least-squares regression works by minimizing the sum of the squares of the vertical distances (errors) from each data point to the regression line. This ensures that the line fits in a way that reduces discrepancies between observed and predicted values.
  • Predictive Power: With the linear equation derived from the regression, one can predict the dependent variable's outcome when the independent variable is known. This predictive ability is essential in various practical applications, ranging from economics to biology.
Studying statistical analysis through least-squares regression provides a robust method to interpret data, observe correlations, and draw conclusions based on the model.

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Most popular questions from this chapter

A simple curve that often makes a good model for the variable costs of a company, as a function of the sales level \(x\), has the form \(y=\beta_{1} x+\beta_{2} x^{2}+\beta_{3} x^{3} .\) There is no constant term because fixed costs are not included. a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\) b. [M] Find the least-squares curve of the form above to fit the data \((4,1.58),(6,2.08),(8,2.5),(10,2.8),(12,3.1)\) \((14,3.4),(16,3.8),\) and \((18,4.32),\) with values in thou- sands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

[M] Generate random vectors \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{v}\) in \(\mathbb{R}^{4}\) with integer entries (and \(\mathbf{v} \neq \mathbf{0} ),\) and compute the quantities \( \left(\frac{\mathbf{x} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}\right) \mathbf{v},\left(\frac{\mathbf{y} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}\right) \mathbf{v}, \frac{(\mathbf{x}+\mathbf{y}) \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}, \frac{(10 \mathbf{x}) \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \) Repeat the computations with new random vectors \(\mathbf{x}\) and \(\mathbf{y} .\) What do you conjecture about the mapping \(\mathbf{x} \mapsto T(\mathbf{x})=\) \(\left(\frac{\mathbf{x} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}\right) \mathbf{v}(\text { for } \mathbf{v} \neq \mathbf{0}) ?\) Verify your conjecture algebraically.

In Exercises 13 and \(14,\) the columns of \(Q\) were obtained by applying the Gram-Schmidt process to the columns of \(A .\) Find an upper triangular matrix \(R\) such that \(A=Q R .\) Check your work. $$ A=\left[\begin{array}{rr}{-2} & {3} \\ {5} & {7} \\ {2} & {-2} \\ {4} & {6}\end{array}\right], Q=\left[\begin{array}{rr}{-2 / 7} & {5 / 7} \\ {5 / 7} & {2 / 7} \\ {2 / 7} & {-4 / 7} \\ {4 / 7} & {2 / 7}\end{array}\right] $$

In Exercises 13 and \(14,\) the columns of \(Q\) were obtained by applying the Gram-Schmidt process to the columns of \(A .\) Find an upper triangular matrix \(R\) such that \(A=Q R .\) Check your work. $$ A=\left[\begin{array}{rr}{5} & {9} \\ {1} & {7} \\ {-3} & {-5} \\ {1} & {5}\end{array}\right], Q=\left[\begin{array}{rr}{5 / 6} & {-1 / 6} \\ {1 / 6} & {5 / 6} \\ {-3 / 6} & {1 / 6} \\ {1 / 6} & {3 / 6}\end{array}\right] $$

Verify the parallelogram law for vectors \(\mathbf{u}\) and \(\mathbf{v}\) in \(\mathbb{R}^{n} :\) \(\|\mathbf{u}+\mathbf{v}\|^{2}+\|\mathbf{u}-\mathbf{v}\|^{2}=2\|\mathbf{u}\|^{2}+2\|\mathbf{v}\|^{2}\)

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