/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Data for this problem are based ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 11

Data for this problem are based on information taken from The Wall Street Journal. Let \(x\) be the age in years of a licensed automobile driver. Let \(y\) be the percentage of all fatal accidents (for a given age) due to speeding. For example, the first data pair indicates that \(36 \%\) of all fatal accidents of \(17-\) year-olds are due to speeding. $$ \begin{array}{l|rrrrrrr} \hline x & 17 & 27 & 37 & 47 & 57 & 67 & 77 \\ \hline y & 36 & 25 & 20 & 12 & 10 & 7 & 5 \\ \hline \end{array} $$ Complete parts (a) through (e), given \(\Sigma x=329, \Sigma y=115, \Sigma x^{2}=18,263\), \(\Sigma y^{2}=2639, \Sigma x y=4015\), and \(r \approx-0.959 .\) (f) Predict the percentage of all fatal accidents due to speeding for 25 -year-olds.

Short Answer

Expert verified
For a 25-year-old, the predicted percentage of fatal accidents due to speeding is approximately 26.84%.

Step by step solution

01

Determine the Linear Regression Equation

We need to use the formula for a linear regression line, which is given by \( y = mx + b \). Here, \( m \) is the slope of the line, and \( b \) is the y-intercept. The formulas related to these components are:1. Slope (\( m\)): \( m = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2} \)2. Intercept (\( b\)): \( b = \frac{(\Sigma y)(\Sigma x^2) - (\Sigma x)(\Sigma xy)}{n(\Sigma x^2) - (\Sigma x)^2} \)Substitute the given values into these formulas to find the slope and the intercept: Number of data points, \( n = 7 \)- \( m = \frac{7(4015) - (329)(115)}{7(18263)-(329)^2} \)- \( b = \frac{(115)(18263) - (329)(4015)}{7(18263) - (329)^2} \)
02

Calculate the Slope \( m \)

Solving for \( m \):\[ m = \frac{7 \cdot 4015 - 329 \cdot 115}{7 \cdot 18263 - 329^2} \]\[ = \frac{28105 - 37835}{127841 - 108241} \]\[ = \frac{-9730}{19600} \]\[ = -0.496 \]
03

Calculate the Intercept \( b \)

Solving for \( b \):\[ b = \frac{115 \cdot 18263 - 329 \cdot 4015}{127841 - 108241} \]\[ = \frac{2090245 - 1320955}{19600} \]\[ = \frac{769290}{19600} \]\[ = 39.24 \]
04

Formulate the Regression Equation

With the slope \( m = -0.496 \) and intercept \( b = 39.24 \), the equation of the line is:\[ y = -0.496x + 39.24 \]
05

Predict y for x = 25

Substitute \( x = 25 \) into the regression equation to predict \( y \).\[ y = -0.496(25) + 39.24 \]\[ = -12.4 + 39.24 \]\[ = 26.84 \]

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Key Concepts

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

Slope Calculation
Understanding how slope calculation works is a fundamental part of linear regression. The slope of a line in a linear regression model indicates the direction and steepness of the line. In simpler terms, it tells us how much the dependent variable (\(y\)) changes for a unit change in the independent variable (\(x\)).
The formula for the slope (\(m\)) is given by:\[m = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2}\]
Where:
  • \(n\) is the number of data points.
  • \(\Sigma xy\) is the sum of the product of \(x\) and \(y\).
  • \(\Sigma x\) and \(\Sigma y\) are the sums of \(x\) and \(y\) values respectively.
  • \(\Sigma x^2\) is the sum of the squares of \(x\) values.
To solve for the slope, plug the given values into the formula. In our exercise, with the calculations provided, the slope came out to be \(-0.496\). This negative value indicates a decreasing relationship, pointing out that as the age (or \(x\)) increases, the percentage of fatal accidents due to speeding (\(y\)) decreases.
Intercept Calculation
The intercept in a linear regression equation represents the expected value of \(y\) when \(x\) is zero. It's the point where the line crosses the \(y\)-axis. From a practical standpoint, it helps to provide an initial benchmark in many contexts, even if \(x = 0\) might not be a possible real-world value.
The formula to calculate the intercept (\(b\)) is:\[b = \frac{(\Sigma y)(\Sigma x^2) - (\Sigma x)(\Sigma xy)}{n(\Sigma x^2) - (\Sigma x)^2}\]
In our given exercise, after substituting the appropriate values, the intercept calculates to \(b = 39.24\). This means that when the age is theoretically zero, the percentage of fatal accidents due to speeding is projected to be around 39.24%. However, in the context of this exercise, this interpretation is mostly mathematical rather than practical, as newborn drivers aren't part of the scenario.
Regression Equation
The regression equation is a comprehensive representation that combines both the slope and intercept to predict outcomes. It follows the basic line formula:\[y = mx + b\]Where:
  • \(m\) is the slope showing how much \(y\) changes for each unit of \(x\).
  • \(b\) is the intercept indicating the \(y\) value when \(x\) is zero.
For this linear regression exercise, the complete equation to predict the percentage of fatal accidents due to speeding based on age is:\[y = -0.496x + 39.24\]
Using this equation, you can predict the \(y\) value for any given \(x\). For example, substituting \(x = 25\) gives a prediction of around 26.84% fatal accidents. This ability to predict outcomes is one of the key strengths of linear regression, allowing us to foresee potential trends and patterns quickly.

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Most popular questions from this chapter

In the least-squares line \(\hat{y}=5-2 x\), what is the value of the slope? When \(x\) changes by 1 unit, by how much does \(\hat{y}\) change?

Suppose you are interested in buying a new Lincoln Navigator or Town Car. You are standing on the sales lot looking at a model with different options. The list price is on the vehicle. As a salesperson approaches, you wonder what the dealer invoice price is for this model with its options. The following data are based on information taken from Consumer Guide (Vol. 677). Let \(x\) be the list price (in thousands of dollars) for a random selection of these cars of different models and options. Let \(y\) be the dealer invoice (in thousands of dollars) for the given vehicle. $$ \begin{array}{c|ccccc} \hline x & 32.1 & 33.5 & 36.1 & 44.0 & 47.8 \\ \hline y & 29.8 & 31.1 & 32.0 & 42.1 & 42.2 \\ \hline \end{array} $$ (a) Verify that \(\Sigma x=193.5, \Sigma y=177.2, \Sigma x^{2}=7676.71, \Sigma y^{2}=6432.5, \Sigma x y=\) 7023.19, and \(r \approx 0.977\). (b) Use a \(1 \%\) level of significance to test the claim that \(\rho>0\). (c) Verify that \(S_{e} \approx 1.5223, a \approx 1.4084\), and \(b \approx 0.8794\). (d) Find the predicted dealer invoice when the list price is \(x=40\) (thousand dollars). (e) Find a \(95 \%\) confidence interval for \(y\) when \(x=40\) (thousand dollars). (f) Use a \(1 \%\) level of significance to test the claim that \(\beta>0\). (g) Find a \(90 \%\) confidence interval for \(\beta\) and interpret its meaning.

Let \(x=\) day of observation and \(y=\) number of locusts per square meter during a locust infestation in a region of North Africa. $$ \begin{array}{r|rrrrr} \hline x & 2 & 3 & 5 & 8 & 10 \\ \hline y & 2 & 3 & 12 & 125 & 630 \\ \hline \end{array} $$ (a) Draw a scatter diagram of the \((x, y)\) data pairs. Do you think a straight line will be a good fit to these data? Do the \(y\) values almost seem to explode as time goes on? (b) Now consider a transformation \(y^{\prime}=\log y .\) We are using common logarithms of base \(10 .\) Draw a scatter diagram of the \(\left(x, y^{\prime}\right)\) data pairs and compare this diagram with the diagram of part (a). Which graph appears to better fit a straight line? (c) Use a calculator with regression keys to find the linear regression equation for the data pairs \(\left(x, y^{\prime}\right) .\) What is the correlation coefficient? (d) The exponential growth model is \(y=\alpha \beta^{x}\). Estimate \(\alpha\) and \(\beta\) and write the exponential growth equation. Hint: See Problem 22 .

If two variables have a negative linear correlation, is the slope of the least-squares line positive or negative?

Data for this problem are based on information from STATS Basketball Scoreboard. It is thought that basketball teams that make too many fouls in a game tend to lose the game even if they otherwise play well. Let \(x\) be the number of fouls more than (i.e., over and above) the opposing team. Let \(y\) be the percentage of times the team with the larger number of fouls wins the game. $$ \begin{array}{l|rrrr} \hline x & 0 & 2 & 5 & 6 \\ \hline y & 50 & 45 & 33 & 26 \\ \hline \end{array} $$ Complete parts (a) through (e), given \(\Sigma x=13, \quad \Sigma y=154, \quad \Sigma x^{2}=65\), \(\Sigma y^{2}=6290, \Sigma x y=411\), and \(r \approx-0.988 .\) (f) If a team had \(x=4\) fouls over and above the opposing team, what does the least-squares equation forecast for \(y\) ?
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