/*! 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 34 Exercises 33 and 34 refer to the... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises 33 and 34 refer to the following setting. Thirty randomly selected seniors at Council High School were asked to report the age (in years) and mileage of their main vehicles. Here is a scatterplot of the data: We used Minitab to perform a least-squares regression analysis for these data. Part of the computer output from this regression is shown below. Predictor \(\quad\) coef \(\quad\) stdev \(\quad\) t-ratio \(\quad \mathrm{P}\) Constant \(-13832 \qquad 8773 \qquad-1.58 \qquad 0.126\) Age \(\quad 14954 \qquad 1546 \qquad 9.67 \quad 0.000\) \(s=22723 \qquad R-s q=77.08 \qquad R-s q(a d j)=76.18\) Drive my car (3.2, 4.3) (a) Explain what the value of r2 tells you about how well the least-squares line fits the data. (b) The mean age of the students’ cars in the sample was x 8 years. Find the mean mileage of the cars in the sample. Show your work. (c) Interpret the value of s in the context of this setting. (d) Would it be reasonable to use the least-squares line to predict a car’s mileage from its age for a Council High School teacher? Justify your answer.

Short Answer

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(a) The squared correlation (R-sq) is 77.08%, indicating a strong fit. (b) The mean mileage is 105,800 miles. (c) On average, actual mileage is 22,723 miles away from predicted. (d) Not reasonable; teachers' cars might differ greatly.

Step by step solution

01

Interpretation of R-Squared

The R-squared ( R-sq ) value is approximately 77.08%. R-squared represents the proportion of the variance in the dependent variable (mileage) that is predictable from the independent variable (age of cars). In this case, an R-squared value of 77.08% means that 77.08% of the variation in the car mileage can be explained by the age of the car. This indicates a strong relationship since a substantial portion of the variability in mileage is accounted for by the regression model.
02

Calculating Mean Mileage of Cars

To find the mean mileage of cars, we make use of the regression equation. The regression equation is derived from: \[\text{Mileage} = \text{Constant} + (\text{Slope} \times \text{Mean Age})\]Using the given coefficients, the equation becomes:\[\text{Mileage} = -13832 + 14954 \times 8\]Calculate the predicted mean mileage:\[\text{Mileage} = -13832 + 119632 = 105800 \]Thus, the mean mileage of the cars is 105,800 miles.
03

Interpretation of Standard Error s

The standard error (s) provided is 22723. This measure indicates the average distance that the observed values fall from the regression line. In this context, on average, each car's actual mileage is 22,723 miles away from the mileage predicted by the regression model. It gives us an idea of the expected variability of actual mileage around the predicted value based on their ages.
04

Application of the Regression Model to Teachers

The regression model was created based on data from students' cars at Council High School. Since we're considering applying this to teachers, whose cars may have different characteristics or usage patterns, it would be inappropriate without further validation. Teachers' cars might differ in terms of types, usage frequency, or maintenance, which could affect their mileage independently of age. Therefore, using this model for teachers without additional data could lead to erroneous predictions.

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Key Concepts

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

Least-Squares Regression
Least-squares regression is a common statistical method used to model the relationship between two variables. In our specific context, we're looking at how the age of a car relates to its mileage. The least-squares approach aims to find the line that best fits the data points by minimizing the sum of the squares of the vertical distances (also known as residuals) between the observed values and the values predicted by the line.
The line is represented by a regression equation of the form: \( y = a + bx \), where:
  • \( y \) is the dependent variable (mileage).
  • \( x \) is the independent variable (age of the car).
  • \( a \) is the y-intercept (constant term).
  • \( b \) is the slope of the line.
In the provided exercise, the constant term is \(-13832\), and the slope is \(14954\), indicating that for every additional year of a car's age, its mileage typically increases by \(14954\) miles. This calculated line provides the best estimate of the mileage based on the age of the car. Understanding and interpreting the coefficients in the regression equation is key to making predictions and drawing conclusions from your data.
R-squared Interpretation
The R-squared value, or \( R^2 \), serves as a statistical measure of how well the regression line explains the variance in the dependent variable. In simple terms, it tells us how much of the variation in mileage is explained by the age of the cars in our model. The closer \( R^2 \) is to 1, the better the model explains the variability of the response data around its mean.
In this exercise, the R-squared value is found to be approximately \(77.08\%\). This indicates that around 77% of the variability in the mileage can be explained by the age of the vehicles. It signifies a strong relationship, suggesting that age is a good predictor of car mileage for these students. This insight is crucial as it helps us gauge the model's reliability and assess whether other factors might need to be considered for a more comprehensive understanding.
Standard Error
The standard error (\( s \)) measures the average distance that the data points lie from the regression line. It is essentially a gauge of the accuracy of predictions made by the regression model. A smaller standard error represents more precise predictions.
In the given scenario, the standard error is \(22723\) miles. This means that on average, the cars' actual mileages deviate by about \(22723\) miles from the predicted mileage according to the regression line. While the standard error gives us a sense of variability, it also provides a tangible context for the predictions' reliability.
  • A higher standard error often indicates greater variability among data points not explained by the model.
  • Having this in mind helps in assessing the confidence we can place on the predictive power of our regression line.
Understanding the standard error is key in practical applications, especially when predictions involve crucial decisions.
Data Variability
Data variability refers to the extent to which the data points spread around the center. In the context of regression analysis, it reveals how consistent the relationship between variables is throughout the dataset.
High variability suggests that while our model may provide an average trend, individual data points can deviate notably from this trend. Such deviations indicate that factors other than the primary independent variable (in this case, car age) might influence the dependent variable (mileage).
It is essential to recognize and account for this variability when evaluating a regression model because:
  • It helps identify the limitations of the model — the more the variability, the lesser the model's accuracy in some situations.
  • Understanding variability assists in identifying if additional variables should be considered to better explain the dependent variable.
By addressing data variability directly, we gain insights into the data's structure and ensure that our analytical conclusions are well-founded and actionable.

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Most popular questions from this chapter

What’s wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Steroids in high school A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used to improve athletic performance.\(^{13}\) Is there a significant difference between the population proportions? State appropriate hypotheses for a significance test to answer this question. Define any parameters you use.

Who talks more—men or women? Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. (a) Do these data provide convincing evidence of a difference in the average number of words spoken in a day by male and female students at this university? Carry out an appropriate test to support your answer. (b) Interpret the P-value from part (a) in the context of this study.

Explain why the conditions for using two-sample z procedures to perform inference about \(p_{1}-p_{2}\) are not met in the settings of Exercises 7 through 10 . Don’t drink the water! The movie A Civil Action (Touchstone Pictures, 1998) tells the story of a major legal battle that took place in the small town of Woburn, Massachusetts. A town well that supplied water to eastern Woburn residents was contaminated by industrial chemicals. During the period that residents drank water from this well, 16 of the 414 babies born had birth defects. On the west side of Woburn, 3 of the 228 babies born during the same time period had birth defects.

Multiple choice: Select the best answer for Exercises 67 to 70. Exercises 69 and 70 refer to the following setting. A study of road rage asked samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of telephone numbers. The two-sample t statistic for the road rage study (male mean minus female mean) is \(t=3.18\). The \(P\)-value for testing the hypotheses from the previous exercise satisfies (a) \(0.001 < P < 0.005 . \quad\) (d) \(0.002 < P < 0.01\) (b) \(0.0005 < P < 0.001 . \quad(\mathrm{e}) P > 0.01\) (c) \(0.001 < P < 0.002\)

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