With a trend line equation of y = -0.669x + 101.09, modify the table below showing the values as well as the residuals. Show your work and explain the steps you used to solve. X y ŷ e 106 28 83 51 DELL 61 39 56 76

MATLAB: An Introduction with Applications
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### Introduction
In statistical analysis, a trend line (or line of best fit) is a straight line that best represents the data on a scatter plot. This line can be used to predict future points. Given a trend line equation, we can modify the existing dataset to include the predicted values (\(\hat{y}\)) and the residuals (\(e\)), which represent the differences between the observed values and the predicted values.

### Problem Statement
Given the trend line equation:
\[ y = -0.669x + 101.09 \]

Modify the table below to show the \(\hat{y}\) values as well as the residuals (\(e\)).

### Original Table
\[
\begin{array}{c|c|c|c}
x & y & \hat{y} & e \\
\hline
106 & 28 &  &  \\
83 & 51 &  &  \\
61 & 56 &  &  \\
39 & 76 &  &  \\
\end{array}
\]

The columns in the table are defined as follows:
- \(x\): Independent variable
- \(y\): Actual dependent variable
- \(\hat{y}\): Predicted value from the trend line equation
- \(e\): Residual (difference between \(y\) and \(\hat{y}\))

### Steps to Solve
1. **Calculate \(\hat{y}\)**: Use the trend line equation \( y = -0.669x + 101.09 \) to find the predicted value for each \(x\).
2. **Calculate Residuals (\(e\))**: Subtract the predicted value (\(\hat{y}\)) from the actual value (\(y\)).

Let's go through each calculation step-by-step:

#### For \(x = 106\):
\[ \hat{y} = -0.669(106) + 101.09 = -70.914 + 101.09 = 30.176 \]
\[ e = y - \hat{y} = 28 - 30.176 = -2.176 \]

#### For \(x = 83\):
\[ \hat{y} = -0.669(83) + 101.09 = -55.527 + 101.09 = 45.563 \]
\[ e = y
Transcribed Image Text:### Introduction In statistical analysis, a trend line (or line of best fit) is a straight line that best represents the data on a scatter plot. This line can be used to predict future points. Given a trend line equation, we can modify the existing dataset to include the predicted values (\(\hat{y}\)) and the residuals (\(e\)), which represent the differences between the observed values and the predicted values. ### Problem Statement Given the trend line equation: \[ y = -0.669x + 101.09 \] Modify the table below to show the \(\hat{y}\) values as well as the residuals (\(e\)). ### Original Table \[ \begin{array}{c|c|c|c} x & y & \hat{y} & e \\ \hline 106 & 28 & & \\ 83 & 51 & & \\ 61 & 56 & & \\ 39 & 76 & & \\ \end{array} \] The columns in the table are defined as follows: - \(x\): Independent variable - \(y\): Actual dependent variable - \(\hat{y}\): Predicted value from the trend line equation - \(e\): Residual (difference between \(y\) and \(\hat{y}\)) ### Steps to Solve 1. **Calculate \(\hat{y}\)**: Use the trend line equation \( y = -0.669x + 101.09 \) to find the predicted value for each \(x\). 2. **Calculate Residuals (\(e\))**: Subtract the predicted value (\(\hat{y}\)) from the actual value (\(y\)). Let's go through each calculation step-by-step: #### For \(x = 106\): \[ \hat{y} = -0.669(106) + 101.09 = -70.914 + 101.09 = 30.176 \] \[ e = y - \hat{y} = 28 - 30.176 = -2.176 \] #### For \(x = 83\): \[ \hat{y} = -0.669(83) + 101.09 = -55.527 + 101.09 = 45.563 \] \[ e = y
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