Chapter 6: Problem 4
Find the regression line for each data set. $$ \begin{array}{|c|c|c|c|} \hline x & 1 & 2 & 4 \\ \hline y & 3 & 5 & 8 \\ \hline \end{array} $$
Short Answer
Expert verified
The regression line is \( y = 1.642x + 1.503 \).
Step by step solution
01
Calculate Means
First, calculate the mean of the x-values and the mean of the y-values. The formula for the mean is \( \bar{x} = \frac{\sum x}{n} \) and \( \bar{y} = \frac{\sum y}{n} \), where \( n \) is the number of data points.Given \( x \) values: 1, 2, 4.\[ \bar{x} = \frac{1+2+4}{3} = \frac{7}{3} = 2.33 \]Given \( y \) values: 3, 5, 8.\[ \bar{y} = \frac{3+5+8}{3} = \frac{16}{3} = 5.33 \]
02
Compute Slopes
Next calculate the slope \( b \) using the formula \[ b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \]Calculate \( x_i - \bar{x} \) and \( y_i - \bar{y} \):\( x : (1-2.33), (2-2.33), (4-2.33) \rightarrow -1.33, -0.33, 1.67 \)\( y : (3-5.33), (5-5.33), (8-5.33) \rightarrow -2.33, -0.33, 2.67 \)\[ b = \frac{(-1.33)(-2.33) + (-0.33)(-0.33) + (1.67)(2.67)}{(-1.33)^2 + (-0.33)^2 + (1.67)^2} \]Calculate:Numerator: \[ (-1.33)(-2.33) + (-0.33)(-0.33) + (1.67)(2.67) = 3.099 + 0.109 + 4.457 = 7.665 \]Denominator: \[ (-1.33)^2 + (-0.33)^2 + (1.67)^2 = 1.7689 + 0.1089 + 2.7889 = 4.6667 \]\[ b = \frac{7.665}{4.6667} = 1.642 \]
03
Calculate Regression Line Intercept
Now, find the y-intercept \( a \) using the formula \( a = \bar{y} - b \cdot \bar{x} \).\[ a = 5.33 - (1.642 \times 2.33) \]Calculate:\[ 5.33 - 3.827 = 1.503 \]
04
Write the Equation
Finally, write the equation of the regression line in the form \( y = mx + c \), where \( m = b \) and \( c = a \).The regression line is:\[ y = 1.642x + 1.503 \]
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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 fundamental concept in statistics used to establish a relationship between two variables by fitting a linear equation to observed data. The key idea is to find a straight line that best predicts the dependent variable, often symbolized as \( y \), using the independent variable, noted as \( x \). Linear regression aims to simplify complex data into a model that is easy to interpret.
Here’s why it’s important:
Here’s why it’s important:
- It helps in trend prediction, such as forecasting future sales in a business.
- Provides a clear visual representation of how two variables are related.
- Enables hypothesis testing about variables, making it valuable in research.
Slope Calculation
The slope is a critical component of the linear regression equation. It represents the rate at which the dependent variable changes with respect to the independent variable. In our equation \( y = mx + b \), \( m \) is the slope.
Calculating the slope involves:
Calculating the slope involves:
- Determining how far the data points are from the mean in terms of both \( x \) and \( y \).
- Using the formula \( b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \), which measures the covariance between \( x \) and \( y \) over the variance of \( x \).
Y-intercept
The y-intercept is another vital component of the linear regression equation, represented by \( b \) in \( y = mx + b \). It indicates the point where the line crosses the y-axis. In practical terms, it predicts the value of \( y \) when \( x \) is zero.
To calculate the y-intercept, we use the formula \( a = \bar{y} - b \cdot \bar{x} \). This ensures that the line of best fit accurately represents the data trends around the mean values of \( x \) and \( y \).
For the exercise at hand, the y-intercept is calculated as approximately \( 1.503 \). This means that when \( x \) is zero, our predicted \( y \) value would be \( 1.503 \). Although \( x = 0 \) might not be realistic within the given data set, this value helps in constructing and plotting the regression line.
To calculate the y-intercept, we use the formula \( a = \bar{y} - b \cdot \bar{x} \). This ensures that the line of best fit accurately represents the data trends around the mean values of \( x \) and \( y \).
For the exercise at hand, the y-intercept is calculated as approximately \( 1.503 \). This means that when \( x \) is zero, our predicted \( y \) value would be \( 1.503 \). Although \( x = 0 \) might not be realistic within the given data set, this value helps in constructing and plotting the regression line.
Mean Calculation
Mean calculation is a simple yet powerful statistical tool that is foundational in linear regression analysis. The mean, denoted as \( \bar{x} \) for x-values and \( \bar{y} \) for y-values, is calculated by summing up all values and dividing by the number of values. This gives us the average value for a data set.
To compute the mean:
To compute the mean:
- For \( x \) values: \( \bar{x} = \frac{\sum x}{n} \)
- For \( y \) values: \( \bar{y} = \frac{\sum y}{n} \)