/*! 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 56 The manufacturer of a gasoline a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 56

The manufacturer of a gasoline additive claims that the use of this additive increases gasoline mileage. A random sample of six cars was selected, and these cars were driven for 1 week without the gasoline additive and then for 1 week with the gasoline additive. The following table gives the miles per gallon for these cars without and with the gasoline additive. $$ \begin{array}{l|cccccc} \hline \text { Without } & 24.6 & 28.3 & 18.9 & 23.7 & 15.4 & 29.5 \\ \hline \text { With } & 26.3 & 31.7 & 18.2 & 25.3 & 18.3 & 30.9 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is equal to the miles per gallon without the gasoline additive minus the miles per gallon with the gasoline additive. b. Using the \(2.5 \%\) significance level, can you conclude that the use of the gasoline additive increases the gasoline mileage?

Short Answer

Expert verified
The actual answer depends on the calculations made. After calculating the paired differences, their mean, and standard deviation, construct a 99% confidence interval for the mean difference. Then perform the t-test. If the P-value is less than 0.025, the conclusion is that the additive significantly increases the gasoline mileage.

Step by step solution

01

Calculate the paired differences

First, calculate the difference in miles per gallon for each car when driven without and with the additive. This is the paired difference for each car. The list of differences will be used in further steps.
02

Calculate mean and standard deviation

Next, calculate the mean and the standard deviation of the paired differences. The formula for the mean is \(\bar{x} = \frac{1}{n} \sum x_i\), while the formula for the standard deviation is \(s = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2}\). These values will be used to construct the confidence interval.
03

Construct a Confidence Interval

To construct a 99% confidence interval for the mean, we will use the formula for a confidence interval for a mean, which is \(\[\bar{x} \pm t_{\alpha/2,n-1}\frac{s}{\sqrt{n}}\]\). In this formula, \(t_{\alpha/2,n-1}\) is the t-value that cuts off the desired area in both tails of the t-distribution with \(n-1\) degrees of freedom. The result will be the interval within which we can be 99% confident that the population mean difference lies.
04

Conduct the t-test

Finally, conduct a one-sample t-test to see if the mean of the differences is significantly different from 0 at the 2.5% significance level. The null hypothesis is that the mean difference is 0, while the alternative hypothesis is that the mean difference is greater than 0. If the P-value for the t-test is less than 0.025, we reject the null hypothesis in favor of the alternative. If the P-value is greater than 0.025, we do not reject the null hypothesis.

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 91影视!

Key Concepts

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

Paired Difference
A paired difference is a critical concept in statistical hypothesis testing, especially when dealing with experiments involving the same subjects under two different treatments. In this scenario, we examine the effect of a gasoline additive on fuel efficiency.

The mileage of six cars was recorded without and then with the additive. To analyze these differences, we first compute the change for each car, which is simply the mileage without the additive minus the mileage with the additive.

This concept helps in focusing on the change per subject rather than comparing separate overall averages. It accounts for individual subject variability and provides a clearer picture of how each treatment affects the outcome.

It's like asking each subject, "How did your performance change with the treatment?" rather than "How did you perform compared to others?" This method is particularly useful when improvement is relative to a baseline for each subject.
Confidence Interval
Constructing a confidence interval provides us with a range within which we can be reasonably assured that the true mean of our population parameter resides. Here, we aim to determine a 99% confidence interval for the mean paired difference in mileage.

It uses the formula: \[\bar{x} \pm t_{\alpha/2,n-1}\frac{s}{\sqrt{n}}\]\ where
  • \(\bar{x}\) is the sample mean of the paired differences,
  • \(s\) is the sample standard deviation,
  • and \(n\) is the number of pairs (here, cars).
  • The \(t\)-value is determined based on our confidence level and the degrees of freedom \((n-1)\).
This calculation results in a specific interval that we are 99% confident includes the true mean paired difference.

A larger confidence level, like 99%, leads to a broader interval, reflecting greater certainty about where the true mean lies. This approach allows us to make informed decisions about the population parameter based on sample data.
t-test
The t-test is a statistical method used to determine whether there is a significant difference between the means of two groups. In this exercise, we employ a one-sample t-test, focusing on the mean of the paired differences in gasoline mileage.

Here鈥檚 how it works:
  • Null Hypothesis \((H_0)\): This posits that the mean paired difference is 0. In other words, the gasoline additive does not truly increase mileage.
  • Alternative Hypothesis \((H_a)\): This suggests that the mean paired difference is greater than 0, asserting that the additive boosts mileage.
Now, we compute the t-statistic and compare it against a critical t-value, determined from t-distribution tables for our specific significance level, here 2.5%.

  • If our calculated p-value is less than 0.025, the evidence against the null hypothesis is strong enough to conclude that the additive increases mileage.
  • If it is greater, we do not have enough evidence to reject the null hypothesis.
Thus, the t-test helps decide whether the observed effects are significant or might have occurred by random chance.

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

Gamma Corporation is considering the installation of governors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a \(90 \%\) confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

The management of a supermarket chain wanted to investigate if the percentages of men and women who prefer to buy national brand products over the store brand products are different. A sample of 600 men shoppers at the company's supermarkets showed that 246 of them prefer to buy national brand products over the store brand products. Another sample of 700 women shoppers at the company's supermarkets showed that 266 of them prefer to buy national brand products over the store brand products. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the proportions of all men and all women shoppers at these supermarkets who prefer to buy national brand products over the store brand products. c. Testing at the \(5 \%\) significance level, can you conclude that the proportions of all men and all women shoppers at these supermarkets who prefer to buy national brand products over the store brand products are different?

We wish to estimate the difference between the mean scores on a standardized test of students taught by Instructors \(\mathrm{A}\) and \(\mathrm{B}\). The scores of all students taught by Instructor A have a normal distribution with a standard deviation of 15, and the scores of all students taught by Instructor B have a normal distribution with a standard deviation of 10 . To estimate the difference between the two means, you decide that the same number of students from each instructor's class should be observed. a. Assuming that the sample size is the same for each instructor's class, how large a sample should be taken from each class to estimate the difference between the mean scores of the two populations to within 5 points with \(90 \%\) confidence? b. Suppose that samples of the size computed in part a will be selected in order to test for the difference between the two population mean scores using a \(.05\) level of significance. How large does the difference between the two sample means have to be for you to conclude that the two population means are different? c. Explain why a paired-samples design would be inappropriate for comparing the scores of Instructor A versus Instructor \(\mathrm{B}\).

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using the \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by male college students is larger than the average amount dispensed by female college students? Use both approaches to make this test.

Does the use of cellular telephones increase the risk of brain tumors? Suppose that a manufacturer of cell phones hires you to answer this question because of concern about public liability suits. How would you conduct an experiment to address this question? Be specific. Explain how you would observe, how many observations you would take, and how you would analyze the data once you collect them. What are your null and alternative hypotheses? Would you want to use a higher or a lower significance level for the test? Explain.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
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.