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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. \((0,1),(1,1),(2,2),(3,2)\)

Short Answer

Expert verified
The equation of the least-squares line is \(y = 0.9 + 0.4x\).

Step by step solution

01

Calculate the Means

First, calculate the means of the x-values and y-values. The x-values are \(0, 1, 2, 3\). Thus, \( \bar{x} = \frac{0 + 1 + 2 + 3}{4} = 1.5 \). The y-values are \(1, 1, 2, 2\). Thus, \( \bar{y} = \frac{1 + 1 + 2 + 2}{4} = 1.5 \).
02

Calculate Slope (\(\beta_1\))

Use the formula for \(\beta_1\): \( \beta_1 = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \). Calculate \((x_i - \bar{x})\) and \((y_i - \bar{y})\) for each point, multiply them, and sum them up. Similarly, compute \((x_i - \bar{x})^2\) and sum it up:- \((0-1.5)(1-1.5) = 0.75\)- \((1-1.5)(1-1.5) = 0.25\)- \((2-1.5)(2-1.5) = 0.25\)- \((3-1.5)(2-1.5) = 0.75\)Sum: \(0.75+0.25+0.25+0.75 = 2.0\)For \((x_i - \bar{x})^2\):- \((0-1.5)^2 = 2.25\)- \((1-1.5)^2 = 0.25\)- \((2-1.5)^2 = 0.25\)- \((3-1.5)^2 = 2.25\)Sum: \(2.25+0.25+0.25+2.25 = 5.0\)Thus, \(\beta_1 = \frac{2.0}{5.0} = 0.4\).
03

Calculate Intercept (\(\beta_0\))

The formula for \(\beta_0\) is: \(\beta_0 = \bar{y} - \beta_1 \times \bar{x} \). We have \(\beta_1 = 0.4\), \(\bar{y} = 1.5\), and \(\bar{x} = 1.5\). Substitute these values:\[ \beta_0 = 1.5 - 0.4 \times 1.5 = 1.5 - 0.6 = 0.9 \].
04

Write the Equation of the Line

Now write the equation using the calculated \(\beta_0\) and \(\beta_1\). The equation of the least-squares line is:\[ y = 0.9 + 0.4x \].

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Key Concepts

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

Linear Regression
Linear Regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. In simple terms, it's about finding the straight line that best fits a set of data points. For a simple linear regression, we represent this line with the equation \( y = \beta_0 + \beta_1 x \), where \( y \) is the dependent variable, \( x \) is the independent variable, \( \beta_0 \) is the y-intercept, and \( \beta_1 \) is the slope. The goal is to minimize the discrepancies between the actual data points and the values predicted by the line.

This approach assumes that there is a linear relationship between the variables. The closer the data points are to forming a straight line, the more accurate the linear regression analysis will be. Despite its simplicity, linear regression can be a powerful tool for making predictions and understanding relationships within data.
Slope and Intercept Calculation
Calculating the slope and intercept of a line is essential in linear regression analysis. Let's break down how each is determined.

First, we calculate the slope, \( \beta_1 \). The formula for this is \( \beta_1 = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \), where \( x_i \) and \( y_i \) are the data points, and \( \bar{x} \) and \( \bar{y} \) are the means of the x-values and y-values, respectively.

For the given data points \((0,1),(1,1),(2,2),(3,2)\), we compute \( \bar{x} = 1.5 \) and \( \bar{y} = 1.5 \). Then, substitute these into the formula to calculate \( \beta_1 \), which results as 0.4.

Next, we find the intercept, \( \beta_0 \), using the equation \( \beta_0 = \bar{y} - \beta_1 \times \bar{x} \). By plugging in \( \beta_1 = 0.4 \), \( \bar{x} = 1.5 \), and \( \bar{y} = 1.5 \), we find \( \beta_0 = 0.9 \).

These calculations lead to the formation of the line equation \( y = 0.9 + 0.4x \), which best represents the linear relationship between the variables in the data set.
Statistical Data Analysis
Statistical data analysis involves examining data to uncover patterns and insights, often using statistical models like linear regression. It's crucial because it provides a systematic approach to understanding data relationships and making predictions.
  • Firstly, data collection is essential. Gather accurate and relevant data points for effective analysis.
  • Then, visualize the data using graphs and plots to observe potential relationships and outliers.
  • With linear regression, identify how well the line fits the data by examining the residuals — the differences between observed and predicted values.
Data analysis using techniques like least-squares regression allows us to determine the strength and direction of relationships between variables. It helps quantify variability, make predictions, and ultimately, draw significant conclusions that can guide decision-making processes. Overall, statistical data analysis is a powerful tool, enabling the understanding and articulation of complex quantitative scenarios.

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Most popular questions from this chapter

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 17 and \(18, A\) is an \(m \times n\) matrix and \(b\) is in \(\mathbb{R}^{m} .\) Mark each statement True or False. Justify each answer. a. If \(\mathbf{b}\) is in the column space of \(A,\) then every solution of \(A \mathbf{x}=\mathbf{b}\) is a least-squares solution. b. The least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is the point in the column space of \(A\) closest to \(\mathbf{b}\) . c. A least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is a list of weights that, when applied to the columns of \(A,\) produces the orthogonal projection of \(\mathbf{b}\) onto \(\operatorname{Col} A .\) d. If \(\hat{\mathbf{x}}\) is a least-squares solution of \(A \mathbf{x}=\mathbf{b},\) then \(\hat{\mathbf{x}}=\left(A^{T} A\right)^{-1} A^{T} \mathbf{b}\) e. The normal equations always provide a reliable method for computing least- squares solutions. f. If \(A\) has a QR factorization, say \(A=Q R\) , then the best way to find the least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is to compute \(\hat{\mathbf{x}}=R^{-1} Q^{T} \mathbf{b} .\)

In Exercises 13 and \(14,\) find the best approximation to \(z\) by vectors of the form \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}\) $$ \mathbf{z}=\left[\begin{array}{r}{3} \\ {-7} \\ {2} \\\ {3}\end{array}\right], \mathbf{v}_{1}=\left[\begin{array}{r}{2} \\ {-1} \\\ {-3} \\ {1}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{1} \\\ {1} \\ {0} \\ {-1}\end{array}\right] $$

In Excrcises \(1-6,\) the given set is a basis for a subspace \(W .\) Use the Gram-Schmidt process to produce an orthogonal basis for \(W\) . $$ \left[\begin{array}{r}{1} \\ {-4} \\ {0} \\\ {1}\end{array}\right],\left[\begin{array}{r}{7} \\ {-7} \\ {-4} \\\ {1}\end{array}\right] $$

In Exercises 17 and \(18,\) all vectors and subspaces are in \(\mathbb{R}^{n} .\) Mark each statement True or False. Justify each answer. a. If \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is an orthogonal basis for \(W,\) then mul- tiplying \(\mathbf{v}_{3}\) by a scalar \(c\) gives a new orthogonal basis \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, c \mathbf{v}_{3}\right\\} .\) b. The Gram-Schmidt process produces from a linearly in- dependent set \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{p}\right\\}\) an orthogonal set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) with the property that for each \(k,\) the vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\) span the same subspace as that spanned by \(\mathbf{x}_{1}, \ldots, \mathbf{x}_{k}\) c. If \(A=Q R,\) where \(Q\) has orthonormal columns, then \(R=Q^{T} A\)

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