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Find the equation of the least-squares line for the given data. Graph the line and data points on the same graph. $$\begin{array}{l|r|r|r|r|r|r}x & 20 & 26 & 30 & 38 & 48 & 60 \\\\\hline y & 160 & 145 & 135 & 120 & 100 & 90\end{array}$$

Short Answer

Expert verified
Equation: \( y = -1.7748x + 190.6516 \). Graph with data points and this line.

Step by step solution

01

Calculate Mean Values

To find the equation of the least-squares line, first calculate the mean of the x-values and the y-values. The given data is as follows: \( x = \{20, 26, 30, 38, 48, 60\} \) and \( y = \{160, 145, 135, 120, 100, 90\} \).Calculate the mean of x: \[ \bar{x} = \frac{20 + 26 + 30 + 38 + 48 + 60}{6} = 37 \]Calculate the mean of y:\[ \bar{y} = \frac{160 + 145 + 135 + 120 + 100 + 90}{6} = 125 \]
02

Calculate the Slope (m)

The slope \( m \) of the least-squares line is given by the formula:\[ m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \]Substitute the values:\[ m = \frac{(20-37)(160-125) + (26-37)(145-125) + (30-37)(135-125) + (38-37)(120-125) + (48-37)(100-125) + (60-37)(90-125)}{(20-37)^2 + (26-37)^2 + (30-37)^2 + (38-37)^2 + (48-37)^2 + (60-37)^2} \]Calculate:\[ m = \frac{(-17)(35) + (-11)(20) + (-7)(10) + (1)(-5) + (11)(-25) + (23)(-35)}{289 + 121 + 49 + 1 + 121 + 529} \]Simplifying gives:\[ m = \frac{-595 - 220 - 70 - 5 - 275 - 805}{1110} = \frac{-1970}{1110} \approx -1.7748 \]
03

Calculate the Intercept (b)

The intercept \( b \) of the line is calculated using the formula:\[ b = \bar{y} - m \bar{x} \]Substitute the values:\[ b = 125 - (-1.7748)(37) \]Calculate:\[ b = 125 + 65.6516 \approx 190.6516 \]
04

Write the Equation of the Line

Now we can write the equation of the least-squares line in the form \( y = mx + b \).Substitute \( m \approx -1.7748 \) and \( b \approx 190.6516 \) into the equation:\[ y = -1.7748x + 190.6516 \]
05

Graphing the Line and Data Points

To graph the data points and the least-squares line, plot the original data points \((x_i, y_i)\) on a coordinate plane.Next, use the equation \( y = -1.7748x + 190.6516 \) to draw the least-squares line. Start at the y-intercept \( b \approx 190.65 \) on the y-axis and use the slope \( -1.7748 \) to find another point by moving down 1.7748 units in y for every 1 unit you move to the right in x.

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Key Concepts

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

Understanding Linear Regression
Linear regression is a powerful statistical method used to model the relationship between two variables by fitting a linear equation to observed data. This technique helps predict the value of a dependent variable based on the value of an independent variable.
The popularity of linear regression stems from its simplicity and applicability to a wide range of scientific and business-related inquiries.

Key features of a linear regression model include:
  • The line of best fit, or the least-squares line, minimizes the sum of the squared differences between observed and predicted values.
  • The model assumes a linear relationship between the variables, as seen in the equation of a straight line, typically represented as \(y = mx + b\), where \(m\) is the slope and \(b\) is the intercept.
  • It is useful for both predicting trends and analyzing data trends over time.
Linear regression models provide a straightforward method to estimate relationships and make predictions based on existing data.
The Role of Data Analysis in Linear Regression
Data analysis is a critical process that involves examining, cleaning, transforming, and modeling data to discover useful information.
In the context of linear regression, data analysis consists of preparing and summarizing data to establish a coherent relationship between the variables.

Here's how it supports the regression analysis:
  • Data Summary: Begin by computing the mean or average of your variables, which gives a central tendency of the data. In our exercise, the mean of x-values \(\bar{x}\) is 37, and for y-values \(\bar{y}\) is 125.
  • Data Visualization: Plotting the data points on a graph before performing regression often provides visual insight into potential relationships.
  • Identifying Patterns: By exploring relationships and patterns in the dataset, data analysts can determine the feasibility of applying linear regression and set realistic expectations for the model.
Data analysis ensures that the data is ready for regression, maximizing the accuracy of predictions and insights gathered from the model.
How Slope Calculation Influences the Model
The slope calculation in linear regression is vital, as it quantifies the change in the dependent variable for every one-unit increase in the independent variable. Put simply, it indicates the steepness and direction of the line.
In our example, the slope \(m\) was found to be approximately -1.7748, revealing a negative relationship between the variables, meaning that as the x-values increase, the y-values decrease.

To calculate the slope:
  • Subtract the mean from each data point for both the x and y values.
  • Multiply these deviations for x and y together, then sum all these products.
  • Divide this sum by the sum of the squared deviations of the x-values from their mean.
The formula for the slope is:\[m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}\]This calculated slope provides crucial information about the relationship between the variables, helping depict how changes in one variable are expected to influence the other.
The Significance of Intercept Calculation
The intercept in a linear regression model signifies the value of the dependent variable when the independent variable is zero.
It is the point where the line crosses the y-axis. In mathematical terms, the intercept \(b\) is calculated using the formula:\[b = \bar{y} - m \bar{x}\]Where \(\bar{y}\) is the mean of the y-values, \(m\) is the slope, and \(\bar{x}\) is the mean of the x-values.

In our data, the intercept was calculated to be approximately 190.6516, indicating the expected value of y when x is zero. This translates to the starting point of the line on the y-axis, providing a baseline prediction when there's no presence of the independent variable.

The intercept is essential because:
  • It provides a reference point for the relationship modeled by the regression line.
  • Understanding the intercept helps in interpreting the overall linear relationship.
  • It plays a critical role in the complete equation of the straight line, \(y = mx + b\).
Thus, accurately calculating the intercept is key for precision in predictions and clarity in data interpretation when using linear regression.

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Most popular questions from this chapter

Find the equation of the least-squares line for the given data. Graph the line and data points on the same graph. The heat loss \(L\) per hour through various thicknesses of a particular type of insulation was measured as shown in the table. Find the least-squares line for \(L\) as a function of \(t .\) Check the values and line with a calculator. $$\begin{array}{l|r|r|r|r|r} t \text { (in.) } & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 \\ \hline L \text { (Btu) } & 5900 & 4800 & 3900 & 3100 & 2450 \end{array}$$

Find the equation of the least-squares line for the given data. Graph the line and data points on the same graph. The speed \(\nu\) (in \(\mathrm{m} / \mathrm{s}\) ) of sound was measured as a function of the temperature \(T\) (in \(^{\circ} \mathrm{C}\) ) with the following results. Find \(\nu\) as a function of \(T\). $$\begin{array}{l|r|r|r|r|r|r|r}T\left(^{\circ} \mathrm{C}\right) & 0 & 10 & 20 & 30 & 40 & 50 & 60 \\\\\hline v(\mathrm{m} / \mathrm{s}) & 331 & 337 & 344 & 350 & 356 & 363 & 369\end{array}$$

Use the following sets of numbers. They are the same as those used in Exercise 22.2. A: 3,6,4,2,5,4,7,6,3,4,6,4,5,7,3 B: 25,26,23,24,25,28,26,27,23,28,25 C: 0.48,0.53,0.49,0.45,0.55,0.49,0.47,0.55,0.48,0.57,0.51,0.46,0.53,0.50,0.49,0.53 D: 105, 108, 103, 108, 106, 104, 109, 104, 110, 108, 108, 104, 113,106,107,106,107,109,105,111,109,108 to find the standard deviation s for the indicated sets of numbers. Set \(D\)

Find the equation of the indicated least squares curve. Sketch the curve and plot the data points on the same graph. The average daily temperatures \(T\) (in \(^{\circ} \mathrm{F}\) ) for each month in Minneapolis (National Weather Service records) are given in the following table. $$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c}t & \mathrm{J} & \mathrm{F} & \mathrm{M} & \mathrm{A} & \mathrm{M} & \mathrm{J} & \mathrm{J} & \mathrm{A} & \mathrm{S} & \mathrm{O} & \mathrm{N} & \mathrm{D} \\\\\hline T\left(^{\circ} \mathrm{F}\right) & 11 & 18 & 29 & 46 & 57 & 68 & 73 & 71 & 61 & 50 & 33 & 19\end{array}$$ Find the least-squares curve \(T=m \cos \left[\frac{\pi}{6}(t-0.5)\right]+b .\) Assume the average temperature is for the 15 th of each month. Then the values of \(t\) (in months) are \(0.5,1.5, \ldots, 11.5 .\) (The fit is fairly good.) Compare the equation using the sinusoidal regression feature of a calculator. What are the main reasons for the differences in the equations?

Find the equation of the indicated least squares curve. Sketch the curve and plot the data points on the same graph. The output torque (in \(\mathrm{J}\) ) of a certain engine was measured at various frequencies (un \(\mathrm{r} / \mathrm{min}\) ) with the following results. Find the least squares curve \(T=1 / f+b\) $$\begin{array}{l|c|c|c|c|c|c} f(\mathrm{r} / \mathrm{min}) & 500 & 1000 & 1500 & 2000 & 2500 & 3000 \\ \hline T(\mathrm{J}) & 220 & 102 & 77 & 50 & 43 & 30 \end{array}$$

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