/*! 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 108 A consumer group is interested i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A consumer group is interested in comparing operating costs for two different types of automobile engines. The group is able to find 15 owners whose cars have engine type \(A\) and 15 who have engine type B. All 30 owners bought their cars at roughly the same time and all have kept good records for a certain 12 month period. In addition, owners were found that drove roughly the same mileage. The cost statistics are \(y_{A}=\$ 87.00 / 1,000\) miles, \(y_{B}=\$ 75.00 / 1,000\) miles, \(s_{A}=\$ 5.99,\) and \(S B-\$ 4.85 .\) Compute a \(95 \%\) confidence interval to estimate \(\mu_{A}-\mu_{B_{1}}\) the difference in the mean operating costs. Assume normality and equal variance.

Short Answer

Expert verified
The estimated 95% confidence interval for the difference in mean operating costs is between \$7.39 and \$16.61.

Step by step solution

01

Calculate Sample Mean Difference

Start by finding the difference between the two mean costs. This can be obtained by subtracting the mean cost for type B from that of A: \(\mu_{A}-\mu_{B} = y_{A} - y_{B} = 87 - 75 = 12\) dollars.
02

Calculate Standard Error

Since we assume equal variances, we use pooled standard error to find the standard error. The pooled standard error formula is \(SE = \sqrt{s_{A}^2/n_{A} + s_{B}^2/n_{B}}\), where \(s_{A}\) and \(s_{B}\) are the respective standard deviations, and \(n_{A}\) and \(n_{B}\) are the respective sample sizes. In this case both \(n_{A}=n_{B}=15\). Substituting the given values into the formula, we get: \(SE = \sqrt{(5.99)^2/15 + (4.85)^2/15} = 2.2583001\) dollars.
03

Determine the Quantile

We need to get the quantile value for a 95% confidence interval from the t-distribution table since the sample size is small (<30) and the population standard deviation is unknown. This value is commonly denoted by \(t\) . For a 95% confidence level and degrees of freedom = \(n_{A} + n_{B} - 2 = 15 + 15 - 2 = 28\), \(t = 2.048409007\).
04

Calculate Confidence Interval

The confidence interval is determined by \(Mean Difference ± t* SE\). Substituting our calculated values, the interval = \(12 ± 2.048409007*2.2583001 = [7.387310381, 16.61268962]\) dollars. This is the 95% confidence interval for the difference in mean running costs.

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 91Ó°ÊÓ!

Key Concepts

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

Pooled Standard Error
Understanding the pooled standard error is critical when working with two independent samples from populations with equal variances. In statistical analysis, particularly when comparing means, it's essential to assess the sampling variability. The pooled standard error is a measure of this variability that combines the standard deviations of the two samples.
The formula for the pooled standard error when comparing two sample means is:

\[SE_{pooled} = \sqrt{ \frac{s_{A}^2}{n_{A}} + \frac{s_{B}^2}{n_{B}} }\]
where \(s_{A}\) and \(s_{B}\) are the sample standard deviations, and \(n_{A}\) and \(n_{B}\) represent the sample sizes.
  • The assumption of equal variance is crucial; if the variances vary significantly, a different method should be used.
  • For the example problem, after plugging in the given values, the pooled standard error assists in determining the precision of the estimated difference in operating costs between engine types A and B.
To reiterate, the pooled standard error is the foundation upon which the confidence interval is built, ensuring that the estimate of the mean difference is made with a definable degree of certainty.
T-Distribution
The t-distribution is a critical concept in statistics, especially when dealing with small sample sizes or unknown population variances. As compared to the normal distribution, the t-distribution is slightly flatter and has heavier tails.
This distribution is characterized by its degrees of freedom (df), which is linked to sample size. The degrees of freedom for two independent samples is calculated as:

\[df = n_{A} + n_{B} - 2\]
In our consumer group problem, where each group has 15 members, the df equals 28. With degrees of freedom and the desired confidence level, we obtain the quantile value from the t-distribution table which is necessary for constructing the confidence interval. The further apart the true mean is from zero, the more likely it is to fall within the confidence interval, a consideration reflected in the width of the t-distribution tails.
  • For small samples, the t-distribution is the best guide, while the normal distribution is more appropriate for larger samples.
  • In our exercise, using the t-distribution accounts for the uncertainty due to a smaller sample size, giving a more conservative estimate for the confidence interval.
Mean Difference
When comparing two groups, statisticians often focus on the mean difference. This value simply reflects the arithmetic difference between the mean values of two distinct samples. It is usually represented as:

\[Mean\ Difference = \bar{y}_{A} - \bar{y}_{B}\]
In our context, the consumer group is assessing the difference in operating costs between two different engine types, A and B. Calculating the mean difference is straightforward:
  • The mean operating cost per 1,000 miles for engine type A is subtracted from that of engine type B, yielding a positive or negative value.
  • This figure represents how much more, on average, one engine costs to operate than the other over a similar distance.
This computed mean difference forms the basis for the confidence interval calculation, guiding us to understand not just if there's a difference, but how large that difference might be, within a specified level of confidence.
Statistical Hypothesis Testing
Finally, statistical hypothesis testing provides a framework for decision-making about a population parameter based on sample data. In the exercise, we're indirectly performing a hypothesis test by computing a confidence interval:
  • The null hypothesis (\(H_0\)) typically states that there's no effect or no difference -- in this case, that the mean operating costs are the same.
  • The alternative hypothesis (\(H_A\)) suggests that there is an effect or a difference -- here, that the mean operating costs differ between engine types.
The confidence interval offers a range of plausible values for the mean difference between engine costs. If this interval does not contain zero, we'd have evidence to reject the null hypothesis at our chosen confidence level. By calculating a 95% confidence interval, we're saying we're 95% confident the true mean difference lies within this range, assuming the null hypothesis is true. This integrates the pooled standard error and the t-distribution to reach a conclusion about the hypotheses, all while considering the variability and sampling distribution of the data.

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

Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 12 students in the section with labs made an average examination grade of 84 with a standard deviation of \(4,\) and 18 students in the section without labs made an average: grade of 77 with a standard deviation of \(6,\) find a \(99 \%\). confidence interval for the difference between the average grades for the two courses. Assume the populations to be approximately normally distributed with equal variances.

A study was conducted to determine if a certain metal treatment has any effect on the amount of metal removed in a pickling operation. A random sample of 100 pieces was immersed in a bath for 24 hours without the treatment, yielding an average of 12.2 millimeters of metal removed and a sample standard deviation of 1.1 millimeters. A second sample of 200 pieces was exposed to the treatment, followed by the 24 -hour immersion in the bath, resulting in an average: removal of 9.1 millimeters of metal with a sample standard deviation of 0.9 millimeter. Compute a \(98 \%\) confidence interval estimate for the difference between the population means. Docs the treatment appear to reduce the mean amount of metal removed?

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming a normal distribution for the number of words typed per minute, find a \(95 \%\) confidence interval for the average number of words typed by all graduates of this school.

In a batch chemical process, two catalysts arc being compared for their effect on the output of the process reaction. A sample of \(\lfloor 2\) batches was prepared using catalyst 1 and a sample of 10 batches was obtained using catalyst \(2 .\) The 12 batches for which catalyst 1 was used gave an average yield of 85 with a sample standard deviation of \(4,\) and the second sample gave an average of 81 and a sample standard deviation of \(5 .\) Find a \(90 \%\) confidence interval for the difference between the population means, assuming that the: populations art: approximately normally distributed with equal variances.

Many cardiac patients wear implanted pacemakers to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 and an approximate normal distribution, find a \(95 \%\) confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of 0.310 inch.

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.