Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Automobile Depreciation For a random sample of 20 automobile models, we record the value of the model as a new car and the value after the car has been purchased and driven 10 miles. \({ }^{47}\) The difference between these two values is a measure of the depreciation on the car just by driving it off the lot. Depreciation values from our sample of 20 automobile models can be found in the dataset CarDepreciation. (a) Find the mean and standard deviation of the Depreciation amounts in CarDepreciation. (b) Use StatKey or other technology to create a bootstrap distribution of the sample mean of depreciations. Describe the shape, center, and spread of this distribution. (c) Use the standard error obtained in your bootstrap distribution to find and interpret a \(95 \%\) confidence interval for the mean amount a new car depreciates by driving it off the lot.

Short Answer

Expert verified
Part a: Mean and standard deviation are calculated using statistical formulas. Part b: Bootstrap distribution is created by resampling the data with replacement and noting the shape, center and spread of this distribution. Part c: The 95% confidence interval is found using the formula (sample mean ± (1.96 * bootstrap standard error)) which is then interpreted in the context of the problem.

Step by step solution

01

Compute for the Mean and Standard Deviation

To find the mean, calculate the sum of all depreciation amounts and divide it by the total number of observations. The standard deviation can be computed by finding the square root of the variance. Variance is the average of the squared differences from the Mean.
02

Create Bootstrap Distribution

A bootstrap distribution of the sample mean is created by resampling the data with replacement. This process is repeated many times, each time saving the sample mean. The collection of these resampled means are used to form the bootstrap distribution.
03

Describe The Shape, Center, And Spread Of Distribution

Note the shape of the bootstrap distribution. It might be symmetrical (normal), positively skewed, negatively skewed, or bimodal. Then find the center (mean or median) and spread or variability (standard deviation or inter-quartile range) of the distribution.
04

Compute for the Confidence Interval

A 95% confidence interval can be computed as the sample mean ± (1.96 * bootstrap standard error). The bootstrap standard error can be found as the standard deviation of the bootstrap distribution.
05

Interpret the Confidence Interval

Interpret the 95% confidence interval in the context of the problem. It indicates that we are 95% confident that the interval contains the true mean depreciation amount for new cars driven off the lot.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. $$ \hat{p}=0.32 \text { and the standard error is } 0.04 \text { . } $$

Have You Ever Been Arrested? According to a recent study of 7335 young people in the US, \(30 \%\) had been arrested \(^{28}\) for a crime other than a traffic violation by the age of 23. Crimes included such things as vandalism, underage drinking, drunken driving, shoplifting, and drug possession. (a) Is the \(30 \%\) a parameter or a statistic? Use the correct notation. (b) Use the information given to estimate a parameter, and clearly define the parameter being estimated. (c) The margin of error for the estimate in part (b) is \(0.01 .\) Use this information to give a range of plausible values for the parameter. (d) Given the margin of error in part (c), if we asked all young people in the US if they have ever been arrested, is it likely that the actual proportion is less than \(25 \% ?\)

Home Field Advantage Is there a home field advantage in soccer? We are specifically interested in the Football Association (FA) premier league, a football (soccer) league in Great Britain known for having especially passionate fans. We took a sample of 120 matches (excluding all ties) and found that the home team was victorious in 70 cases \(^{56}\) (a) What is the population of interest? What is the specific population parameter of interest? (b) Estimate the population parameter using the sample. (c) Using StatKey or other technology, construct and interpret a \(90 \%\) confidence interval. (d) Using StatKey or other technology, construct and interpret a \(99 \%\) confidence interval. (e) Based on this sample and the results in parts (c) and (d), are we \(90 \%\) confident a home field advantage exists? Are we \(99 \%\) confident?

Give information about the proportion of a sample that agrees with a certain statement. Use StatKey or other technology to estimate the standard error from a bootstrap distribution generated from the sample. Then use the standard error to give a \(95 \%\) confidence interval for the proportion of the population to agree with the statement. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. In a random sample of 400 people, 112 agree and 288 disagree.

Student Misinterpretations Suppose that a student is working on a statistics project using data on pulse rates collected from a random sample of 100 students from her college. She finds a \(95 \%\) confidence interval for mean pulse rate to be (65.5,71.8) . Discuss how each of the statements below would indicate an improper interpretation of this interval. (a) I am \(95 \%\) sure that all students will have pulse rates between 65.5 and 71.8 beats per minute. (b) I am \(95 \%\) sure that the mean pulse rate for this sample of students will fall between 65.5 and 71.8 beats per minute. (c) I am 95\% sure that the confidence interval for the average pulse rate of all students at this college goes from 65.5 to 71.8 beats per minute. (d) I am sure that \(95 \%\) of all students at this college will have pulse rates between 65.5 and 71.8 beats per minute. (e) I am \(95 \%\) sure that the mean pulse rate for all US college students is between 65.5 and 71.8 beats per minute. (f) Of the mean pulse rates for students at this college, \(95 \%\) will fall between 65.5 and 71.8 beats per minute. (g) Of random samples of this size taken from students at this college, \(95 \%\) will have mean pulse rates between 65.5 and 71.8 beats per minute.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free