The mean number of hours of flying time for pilots at Continental Airlines is 49 hours per month (The Wall Street Journal, February 25, 2003). Assume that this mean was based on actual flying times for a sample of 100 Continental pilots and that the sample standard deviation was 8.5 hours. a. \(\quad\) At \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval estimate of the population mean flying time for the pilots? c. The mean number of hours of flying time for pilots at United Airlines is 36 hours per month. Use your results from part (b) to discuss differences between the flying times for the pilots at the two airlines. The Wall Street Journal reported United Airlines as having the highest labor cost among all airlines. Does the information in this exercise provide insight as to why United Airlines might expect higher labor costs?

Short Answer

Expert verified
The 95% confidence interval for the population mean flying time of pilots at Continental Airlines is approximately (47.342, 50.658) hours, and the margin of error is approximately 1.658 hours. Comparing this to the mean flying time of 36 hours for pilots at United Airlines, there is a significant difference in mean flying times between the two airlines. This difference may contribute to higher labor costs for United Airlines, as they might need to employ more pilots to cover the same routes as Continental Airlines. However, other factors also need to be considered in evaluating labor costs.

Step by step solution

01

Identify the parameters

We are given the following values: - Sample mean flying time (\(\overline{x}\)): 49 hours - Sample size (n): 100 pilots - Sample standard deviation (s): 8.5 hours - Confidence level: 95%
02

Calculate the margin of error

To calculate the margin of error at a 95% confidence level, we need to find the critical z-value for a two-tailed test. We'll use the z-distribution since the sample size is greater than 30. The critical z-value for a 95% confidence level is 1.96. Margin of error = z-score × (standard deviation / sqrt(sample size)) Margin of error = 1.96 × (8.5 / sqrt(100)) Margin of error ≈ 1.658
03

Calculate the 95% confidence interval

Now, we'll use the sample mean and margin of error to calculate the 95% confidence interval for the population mean flying time. Lower limit = Sample mean - Margin of error Lower limit = 49 - 1.658 Lower limit ≈ 47.342 Upper limit = Sample mean + Margin of error Upper limit = 49 + 1.658 Upper limit ≈ 50.658 The 95% confidence interval for the population mean flying time is approximately (47.342, 50.658) hours.
04

Evaluate the differences in flying times

We now compare the mean flying time for pilots at Continental Airlines with that at United Airlines, which is 36 hours per month. The 95% confidence interval for Continental pilots' mean flying time does not include 36 hours, indicating that there is a significant difference in the mean flying times between the two airlines.
05

Analyze the results in the context of labor costs

As mentioned in the exercise, United Airlines has the highest labor cost among all airlines. The fact that the mean flying time for pilots at United Airlines is 36 hours per month might be one reason for the higher labor costs. Since the pilots at United Airlines fly fewer hours, the airline may need to employ more pilots to cover the same routes as airlines with pilots flying more hours, which could increase labor costs. The information in this exercise may provide some insight into why United Airlines might expect higher labor costs, but it is not the only factor to consider.

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!

Key Concepts

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

Sample Mean
Understanding the 'sample mean' lies at the heart of statistical analysis. The sample mean, represented as \( \overline{x} \), is the average of all the observations in a sample. It serves as an estimate of the population mean when it's not feasible to measure every individual in a population. For example, if a study looks at the flying hours of a group of pilots from an airline and finds an average flying time of 49 hours a month, this figure is the sample mean. It’s critical to ensure the sample is representative of the entire population to minimize bias and have a reliable estimate of the true average flying time across all pilots at the airline.
Margin of Error
The 'margin of error' provides a range of uncertainty around a sample mean. It is influenced by the level of confidence we wish to have in our estimate and the variability within our sample's data. A higher margin of error indicates less certainty about the precision of the sample mean's reflection of the true population mean. Mathematically, it's calculated using the sample's standard deviation, the size of the sample, and the critical value (usually a z-score for large samples) from the distribution that corresponds to the desired confidence level. For instance, with a z-score of 1.96, a standard deviation of 8.5, and a sample size of 100 pilots, the margin of error is approximately 1.658 hours. This means we can be 95% confident that the population mean flying time lies within 1.658 hours of the sample mean.
Standard Deviation
The 'standard deviation' is a measure that tells us how much the values in a sample deviate from the sample mean. A small standard deviation means that the data points are clustered closely around the mean, indicating similar scores. In contrast, a large standard deviation shows that the values are spread out over a wider range. In the context of flying times, a standard deviation of 8.5 hours suggests that the individual flight times of the pilots vary by an average of 8.5 hours around the mean. The standard deviation is a crucial piece in calculating the margin of error and confidence interval, as it reflects the dispersal of the data, which impacts the precision of these estimates.
Z-Score
A 'z-score' in statistics is synonymous with a 'standard score' and gives us an idea of how many standard deviations an element is from the mean. When constructing confidence intervals, the z-score helps us grasp the range within which the population mean is likely to fall, given our sample data. The z-score is particular to the confidence level we select, such as 95%. The higher the confidence level, the larger the z-score and consequently, the wider the confidence interval. A 95% confidence level typically corresponds to a z-score of 1.96, meaning the sample mean is 1.96 standard deviations away from the population mean for such a large proportion of different possible samples.
Population Mean
The 'population mean' is the average of an entire population's values for a certain variable. It is often unknown and thus estimated through the analysis of a random sample from that population. The population mean is the target figure that surveys and studies aim to estimate through sample means and confidence intervals. In the context of our example, while 49 hours is the sample mean flying time for a group of pilots, the population mean would be the average flying time of all the pilots at this airline. The confidence interval estimate provides a range in which we expect the true population mean to lie, which is crucial for making informed decisions at a business or policy level.

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

A simple random sample of 50 items from a population with \(\sigma=6\) resulted in a sample mean of 32 a. Provide a \(90 \%\) confidence interval for the population mean. b. Provide a \(95 \%\) confidence interval for the population mean. c. Provide a \(99 \%\) confidence interval for the population mean.

A simple random sample of 60 items resulted in a sample mean of \(80 .\) The population standard deviation is \(\sigma=15\) a. Compute the \(95 \%\) confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a \(95 \%\) confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?

According to Thomson Financial, through January \(25,2006,\) the majority of companies reporting profits had beaten estimates (Business Week, February 6, 2006). A sample of 162 companies showed that 104 beat estimates, 29 matched estimates, and 29 fell short. a. What is the point estimate of the proportion that fell short of estimates? b. Determine the margin of error and provide a \(95 \%\) confidence interval for the proportion that beat estimates. c. How large a sample is needed if the desired margin of error is \(.05 ?\)

A simple random sample with \(n=54\) provided a sample mean of 22.5 and a sample standard deviation of 4.4 a. Develop a \(90 \%\) confidence interval for the population mean. b. Develop a \(95 \%\) confidence interval for the population mean. c. Develop a \(99 \%\) confidence interval for the population mean. d. What happens to the margin of error and the confidence interval as the confidence level is increased?

The National Center for Education Statistics reported that \(47 \%\) of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a \(95 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. b. Provide a \(99 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. c. What happens to the margin of error as the confidence is increased from \(95 \%\) to \(99 \% ?\)

See all solutions

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