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A sample of small cars was selected, and the values of \(x=\) horsepower and \(y=\) fuel efficiency \((\mathrm{mpg})\) were determined for each car. Fitting the simple linear regression model gave the estimated regression equation \(\hat{y}=44.0-.150 x .\) a. How would you interpret \(b=-.150\) ? b. Substituting \(x=100\) gives \(\hat{y}=29.0\). Give two different interpretations of this number. c. What happens if you predict efficiency for a car with a 300-horsepower engine? Why do you think this has occurred? d. Interpret \(r^{2}=0.680\) in the context of this problem. e. Interpret \(s_{e}=3.0\) in the context of this problem.

Short Answer

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a) For each additional unit of horsepower, the model predicts a decrease of 0.150 mpg in fuel efficiency. b) At 100 horsepower, the predicted efficiency is 29.0 mpg; this is also the point on the regression line corresponding to 100 horsepower. c) Predicting for a 300-hp engine results in a nonsensical negative efficiency due to extrapolation. d) 68% of the change in fuel efficiency can be explained by the change in horsepower. e) The actual fuel efficiency values deviate from the predicted ones by approximately 3.0 mpg on average.

Step by step solution

01

Interpretation of \(b=-.150\)

The value \(b=-.150\) is the slope of the regression line. This means that for every increase by 1 unit in horsepower (x), we expect the fuel efficiency (y) to decrease by 0.150 mpg, assuming all other factors remain constant.
02

Interpretation of \(\hat{y}=29.0\)

The value of \(\hat{y}=29.0\) given \(x=100\) means two things: 1) When the car has 100 horsepower, the predicted fuel efficiency is 29.0 mpg. 2) It is also the point on the estimated regression line where \(x=100\).
03

Predicting for \(x=300\)

If we substitute \(x=300\) into the regression equation, the result will be \(\hat{y}<0\). In the real world, this doesn't make sense as fuel efficiency cannot be less than zero. This issue has occurred mainly due to extrapolation beyond the range of data that the model is built on.
04

Interpretation of \(r^{2}=0.680\)

The value of \(r^{2}=0.680\) means that 68% of the variation in fuel efficiency can be explained by horsepower. It provides a measure of how well observed outcomes are replicated by the model based on the proportion of total variation of outcomes explained by the model.
05

Interpretation of \(s_{e}=3.0\)

The value of \(s_{e}=3.0\) measures the standard error of the residuals. This means on average, the actual values of fuel efficiency deviate from the predicted values by approximately 3.0 mpg.

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Key Concepts

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

Slope Interpretation
In simple linear regression, the slope, often represented by the letter "b," is a key component in understanding the relationship between the independent and dependent variables. In this context, the slope is given as \(b = -0.150\), which means that for every 1 unit increase in horsepower, the car's fuel efficiency decreases by 0.150 miles per gallon (mpg).
This negative value indicates an inverse relationship between horsepower and fuel efficiency, suggesting that as cars become more powerful, they tend to become less fuel-efficient. The slope can show whether a relationship is positive or negative, but remember that this does not imply causation, only correlation.
Understanding the slope helps predict future values and also can guide decision-making, especially when considering factors like energy efficiency in car manufacturing.
R-squared
The R-squared value, also known as the coefficient of determination, is crucial in evaluating the effectiveness of the regression model. With an \(r^2 = 0.680\), it indicates that 68% of the variability in fuel efficiency can be explained by the horsepower.
This percentage shows how well our model fits the observed data. A higher \(r^2\) value generally indicates a better fit when more variability is explained by the model.
However, R-squared only reflects the fit of the model to the data used and should not be the only measure used to assess model quality. It's important to use it alongside other metrics and checks to ensure that the model is not merely capturing noise within the data.
Standard Error
The standard error of the residuals, symbolized as \(s_e\), provides insight into the accuracy of our predictions. In this case, \(s_e = 3.0\), implying that the actual fuel efficiency typically deviates by about 3 mpg from the predicted values.
This is a measure of the average size of the errors that arise from the regression line. The smaller the standard error, the closer the observed data points are to the fitted regression line.
Interpreting standard error is crucial for understanding the precision of the predictions. It also serves as an indication of the uncertainty surrounding the estimated coefficients in the regression equation.
Extrapolation Issues
Extrapolation occurs when we use a regression model to make predictions for values outside the range of the data set used to create the model. This is illustrated by predicting fuel efficiency for a car with 300 horsepower, which results in an unrealistic value.
This happens because the model was built from data that did not include such high horsepower values, leading to predictions that are not valid or reliable outside the data range.
When dealing with extrapolation, caution is required. Always make predictions within the range of the existing data set to avoid misleading results. Extrapolation can help in exploratory ideation, but it should be approached with skepticism and be well-supported by additional data or evidence when used for decision-making.

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Most popular questions from this chapter

A simple linear regression model was used to describe the relationship between sales revenue \(y\) (in thousands of dollars) and advertising expenditure \(x\) (also in thousands of dollars) for fast-food outlets during a 3 -month period. A sample of 15 outlets yielded the accompanying summary quantities. $$ \begin{aligned} &\sum x=14.10 \quad \sum y=1438.50 \quad \sum x^{2}=13.92 \\ &\sum y^{2}=140,354 \quad \sum x y=1387.20 \\ &\sum(y-\vec{y})^{2}=2401.85 \quad \sum(y-\hat{y})^{2}=561.46 \end{aligned} $$ a. What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? b. Calculate \(s_{e}\) and \(s_{b}\) c. Obtain a \(90 \%\) confidence interval for \(\beta\), the average change in revenue associated with a \(\$ 1000\) (that is, 1 -unit) increase in advertising expenditure.

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It seems plausible that higher rent for retail space could be justified only by a higher level of sales. A random sample of \(n=53\) specialty stores in a chain was selected, and the values of \(x=\) annual dollar rent per square foot and \(y=\) annual dollar sales per square foot were determined, resulting in \(r=.37\). Carry out a test at significance level \(.05\) to see whether there is in fact a positive linear association between \(x\) and \(y\) in the population of all such stores.

Television is regarded by many as a prime culprit for the difficulty many students have in performing well in school. The article reported that for a random sample of \(n=528\) college students, the sample correlation coefficient between time spent watching television \((x)\) and grade point average \((y)\) was \(r=-.26\). a. Does this suggest that there is a negative correlation between these two variables in the population from which the 528 students were selected? Use a test with significance level \(.01\). b. Would the simple linear regression model explain a substantial percentage of the observed variation in grade point average? Explain your reasoning.

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