/*! 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 115 Does the airline you choose affe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Does the airline you choose affect when you'll arrive at your destination? The dataset DecemberFlights contains the difference between actual and scheduled arrival time from 1000 randomly sampled December flights for two of the major North American airlines, Delta Air Lines and United Air Lines. A negative difference indicates a flight arrived early. We are interested in testing whether the average difference between actual and scheduled arrival time is different between the two airlines. (a) Define any relevant parameter(s) and state the null and alternative hypotheses. (b) Find the sample mean of each group, and calculate the difference in sample means. (c) Use StatKey or other technology to create a randomization distribution and find the p-value. (d) At a significance level of \(\alpha=0.01\), what is the conclusion of the test? Interpret the conclusion in context.

Short Answer

Expert verified
Ultimately, you would need to calculate the p-value and compare it to the significance level (0.01) to determine the conclusion of your hypothesis test. Unfortunately, without the actual data this exercise cannot be completed in full. In general, if the p-value is less than 0.01, reject H_0: the average difference between actual and scheduled arrival time in United Air Lines is not significantly the same as Delta Air Lines. Otherwise, fail to reject H_0.

Step by step solution

01

Define Parameters and State Hypotheses

This study’s parameters would be the actual differences in arrival times for the two airlines, denoted as µ_1 for Delta Air Lines and µ_2 for United Air Lines. The null hypothesis (H_0) is that the difference in the average actual vs. scheduled arrival times (µ_1-µ_2) is the same for both airlines. People often state the null hypothesis as µ_1-µ_2 = 0 (which implies the same thing). The alternative hypothesis (H_a) is that µ_1-µ_2 ≠ 0 meaning there is a difference between the two airlines.
02

Find Sample Means and their Difference

Calculate the sample mean difference in arrival times for each airline. Then subtract the Delta Air Lines mean from the United Air Lines mean to get the observed difference in sample means, denoted as d_obs.
03

Carry Out a Permutation Test

Use statistical software to perform a permutation test and generate a randomization distribution. This will simulate the distribution under the null hypothesis and allows the calculation of the p-value.
04

Find the p-value

By comparing the observed mean difference (d_obs) to the randomization distribution, the p-value can be computed. The p-value shows the probability of getting a mean difference as extreme as d_obs or more under the null hypothesis
05

Interpret and Make Conclusion

At a significance level of \(\alpha=0.01\), if the p-value is less than \(\alpha\), reject the null hypothesis. This suggests that there is a significant difference in the flight arrival times between the two airlines. If the p-value is greater than \(\alpha\), fail to reject the null hypothesis, which would mean there is no significant difference between the flight arrival times of the two airlines.

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.

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that there is no effect or no difference. It is the assumption that any observed difference in a dataset is due to chance. In the context of the flight arrival times for Delta and United Airlines, the null hypothesis (\(H_0\)) asserts that the average difference in arrival times is the same for both airlines. In simpler terms, it means that both airlines arrive around the same time on average. We express this mathematically as \(\mu_1 - \mu_2 = 0\), where \(\mu_1\) and \(\mu_2\) are the mean arrival differences for Delta and United Airlines, respectively.
Alternative Hypothesis
The alternative hypothesis provides a statement that contradicts the null hypothesis. It suggests that there is a significant difference between groups being compared, indicating an effect exists. For our airline example, the alternative hypothesis (\(H_a\)) is that there is a significant difference between the average arrival times of Delta and United Airlines. This is expressed as \(\mu_1 - \mu_2 eq 0\). Under the alternative hypothesis, we suppose that factors other than random variation are responsible for any observed difference between the airlines.
Randomization Distribution
A randomization distribution is a crucial concept in hypothesis testing. It represents the possible outcomes of a test statistic (like mean difference) under the null hypothesis. For example, in a permutation test, the data is shuffled many times to simulate the process of random assignment. This creates a distribution of outcomes that might happen purely by chance if the null hypothesis were true. For the airline data, we would repeatedly sample the arrival time differences randomly between Delta and United Airlines. By examining the randomization distribution, we can see how unusual our observed sample difference is if no real difference exists between the airlines.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold chosen by the researcher to decide whether to reject the null hypothesis. It is the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. Common significance levels are 0.05 or 0.01, with lower levels indicating stricter criteria for finding evidence against \(H_0\). In our scenario, \(\alpha = 0.01\) means we are using a strict criterion. We require strong evidence from the data to conclude there’s a difference in the airlines’ arrival times, ensuring our conclusion is less likely a result of random chance.
P-value
A p-value measures the strength of evidence against a null hypothesis. It tells us the probability of observing data as extreme, or more extreme, than what was actually observed, assuming the null hypothesis is true. When we perform our statistical test on the flight data, we calculate a p-value by comparing the observed difference (the mean arrival time difference) to the randomization distribution. If the p-value is less than the significance level (\(\alpha = 0.01\) in our case), we reject the null hypothesis. A small p-value suggests that the observed difference is unlikely to have occurred by random chance alone, supporting the idea that there is a true difference between Delta and United Airlines arrival times.

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

Do You Own a Smartphone? A study \(^{19}\) conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and the study shows that 688 of the 989 men own a smartphone and 671 of the 1012 women own a smartphone. We want to test whether the survey results provide evidence of a difference in the proportion owning a smartphone between men and women. (a) State the null and alternative hypotheses, and define the parameters. (b) Give the notation and value of the sample statistic. In the sample, which group has higher smartphone ownership: men or women? (c) Use StatKey or other technology to find the pvalue.

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{52}\) The study reviewed the records of all 1,050 skin cancer patients referred to the St. Louis University Cancer Center in 2004\. Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

Indicate whether it is best assessed by using a confidence interval or a hypothesis test or whether statistical inference is not relevant to answer it. (a) What proportion of people using a public restroom wash their hands after going to the bathroom? (b) On average, how much more do adults who played sports in high school exercise than adults who did not play sports in high school? (c) In \(2010,\) what percent of the US Senate voted to confirm Elena Kagan as a member of the Supreme Court? (d) What is the average daily calorie intake of 20 year-old males?

Mating Choice and Offspring Fitness Does the ability to choose a mate improve offspring fitness in fruit flies? Researchers have studied this by taking female fruit flies and randomly dividing them into two groups; one group is put into a cage with a large number of males and able to freely choose who to mate with, while flies in the other group are each put into individual vials, each with only one male, giving no choice in who to mate with. Females are then put into egg laying chambers, and a certain number of larvae collected. Do the larvae from the mate choice group exhibit higher survival rates? A study \(^{44}\) published in Nature found that mate choice does increase offspring fitness in fruit flies (with p-value \(<0.02\) ), yet this result went against conventional wisdom in genetics and was quite controversial. Researchers attempted to replicate this result with a series of related experiments, \({ }^{45}\) with data provided in MateChoice. (a) In the first replication experiment, using the same species of fruit fly as the original Nature study, 6067 of the 10000 larvae from the mate choice group survived and 5976 of the 10000 larvae from the no mate choice group survived. Calculate the p-value. (b) Using a significance level of \(\alpha=0.05\) and \(\mathrm{p}\) -value from (a), state the conclusion in context. (c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions dif- fer from run to run, and thus it makes sense to treat each \(\mathrm{run}\) as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice \(-\) NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than \(0 .\) (Hint: this is a single quantitative variable, so the corresponding test would be for a single mean.) (d) Using a significance level of \(\alpha=0.05\) and the p-value from (c), state the conclusion in context. (e) The experiment being tested in parts (a)-(d) was designed to mimic the experiment from the original study, yet the original study yielded significant results while this study did not. If mate choice really does improve offspring fitness in fruit flies, did the follow-up study being analyzed in parts (a)-(d) make a Type I, Type II, or no error? (f) If mate choice really does not improve offspring fitness in fruit flies, did the original Nature study make a Type I, Type II, or no error?

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Using a sample of 10 games each to see if your average score at Wii bowling is significantly more than your friend's average score.

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.