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For data set, find the regression line without using a calculator. $$\begin{array}{|ccccccccc|}\hline x & 1 & 2 & 3 & 5 \\\\\hline y & 0 & 1 & 3 & 4 \\\\\hline\end{array}$$

Short Answer

Expert verified
The regression line is \(y = 1.03x - 0.825\).

Step by step solution

01

Calculate the Mean of x and y

Calculate the mean values of the x and y data points. The mean is given by the sum of the values divided by the number of values.\[\bar{x} = \frac{1 + 2 + 3 + 5}{4} = \frac{11}{4} = 2.75\]\[\bar{y} = \frac{0 + 1 + 3 + 4}{4} = \frac{8}{4} = 2\]
02

Compute the Slope (m) of the Line

The slope \(m\) is calculated using the formula:\[m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}\]First, compute each product and square.\[\sum{(x_i - \bar{x})(y_i - \bar{y})} = (1 - 2.75)(0 - 2) + (2 - 2.75)(1 - 2) + (3 - 2.75)(3 - 2) + (5 - 2.75)(4 - 2)\]\[= -1.75(-2) + -0.75(-1) + 0.25(1) + 2.25(2)\]\[= 3.5 + 0.75 + 0.25 + 4.5 = 9\]Next, compute the denominator.\[\sum{(x_i - \bar{x})^2} = (1 - 2.75)^2 + (2 - 2.75)^2 + (3 - 2.75)^2 + (5 - 2.75)^2\]\[= 3.0625 + 0.5625 + 0.0625 + 5.0625 = 8.75\]Therefore, the slope \(m\) is:\[m = \frac{9}{8.75} \approx 1.03\]
03

Find the y-intercept (b)

The y-intercept \(b\) is calculated using the formula:\[b = \bar{y} - m\bar{x}\]Substitute the values calculated earlier:\[b = 2 - 1.03 \times 2.75 \approx -0.825\]
04

Write the Equation of the Line

The equation of the regression line is given by the slope-intercept form \(y = mx + b\). Using the calculated slope and y-intercept, the equation is:\[y = 1.03x - 0.825\]

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Key Concepts

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

Mean Calculation
To find the mean of a set of numbers, you simply add up all the numbers and divide the sum by the count of those numbers.
For example, given a set of x values: 1, 2, 3, and 5, the mean (or average) is calculated as follows:
\[\bar{x} = \frac{1 + 2 + 3 + 5}{4} = \frac{11}{4} = 2.75\]
Here, we added up all the x values to get 11, then divided by 4 (the number of x values). Likewise, for the y values: 0, 1, 3, and 4, the calculation is:
\[\bar{y} = \frac{0 + 1 + 3 + 4}{4} = \frac{8}{4} = 2\]
The mean helps to find the central tendency of the data, which is crucial for further calculations in linear regression.
Understanding the mean is important as it serves as a reference point for both the slope and the y-intercept in linear regression.
Slope Determination
To determine the slope of the regression line, we use all data points in combination with their mean values.
The formula for calculating the slope (m) is:
\[\text{slope} (m) = \frac{\sum{ (x_i - \bar{x})(y_i - \bar{y}) }}{ \sum{ (x_i - \bar{x})^2 }} \]
Here's a breakdown:
  • \( x_i \) and \( y_i \) are the individual data points.
  • \( \bar{x} \) and \( \bar{y} \) are the mean values we found earlier.
To understand better, let's break it into steps:
  • Calculate \( (x_i - \bar{x})(y_i - \bar{y}) \) for each data pair.
  • Add up these products to get the numerator.
  • Calculate \( (x_i - \bar{x})^2 \) for each x value.
  • Add these squared differences to get the denominator.
For our example:
\[\text{numerator} = (1 - 2.75)(0 - 2) + (2 - 2.75)(1 - 2) + (3 - 2.75)(3 - 2) + (5 - 2.75)(4 - 2)\] \[\text{denominator} = (1 - 2.75)^2 + (2 - 2.75)^2 + (3 - 2.75)^2 + (5 - 2.75)^2 \]
After calculation, you find:
\[\text{slope} (m) = \frac{9}{8.75} \approx 1.03 \] This slope indicates how much y increases for a unit increase in x.
y-intercept Calculation
The y-intercept (b) is the point where the regression line crosses the y-axis, i.e., the value of y when x is 0.
This is calculated using the mean values and the slope with the formula:
\[\text{y-intercept} (b) = \bar{y} - m\bar{x} \] Substituting the values from our example:
\[\bar{y} = 2 \] \[\bar{x} = 2.75 \] \[\text{slope} (m) \approx 1.03 \]
The calculation becomes:
\[\text{y-intercept} (b) = 2 - 1.03 \times 2.75 \approx -0.825 \]
Hence, our equation of the regression line is:
\[\text{line equation} = y = 1.03x - 0.825 \]
The y-intercept connects the dots between the slope and how the line starts on the graph. It ensures the line fits the data by adjusting how high or low it starts.

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