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One of the authors travels regularly to Seattle, Washington. He uses either Delta or Alaska. Flight delays are sometimes unavoidable, but he would be willing to give most of his business to the airline with the best on-time arrival record. The number of minutes that his flight arrived late for the last six trips on each airline follows. Is there evidence that either airline has superior on-time arrival performance? Use \(\alpha=0.01\) and the Wilcoxon rank-sum test. $$ \begin{array}{l|l} \text { Delta: } & 13,10,1,-4, \quad 0,9 \text { (minutes late) } \\ \hline \text { Alaska: } & 15,8,3,-1,-2,4 \text { (minutes late) } \end{array} $$

Short Answer

Expert verified
There is no significant difference in on-time arrival performance between Delta and Alaska at \( \alpha = 0.01 \).

Step by step solution

01

Rank All Observations

First, we need to rank all the observations from both airlines from smallest to largest. The order of the data is: - Delta: 13, 10, 1, -4, 0, 9 - Alaska: 15, 8, 3, -1, -2, 4. Sorting and ranking these observations, while maintaining ties by assigning an average rank, results in: - -4: Rank 1 - -2: Rank 2 - -1: Rank 3 - 0: Rank 4.5 - 1: Rank 6 - 3: Rank 7 - 4: Rank 8 - 8: Rank 9 - 9: Rank 10 - 10: Rank 11 - 13: Rank 12 - 15: Rank 13 These ranks are calculated using (lowest value = rank 1), then ties are averaged.
02

Assign Ranks to Each Group

Once we have the ranks, we assign them to their respective airlines: - Delta: 13 (Rank 12), 10 (Rank 11), 1 (Rank 6), -4 (Rank 1), 0 (Rank 4.5), 9 (Rank 10) - Alaska: 15 (Rank 13), 8 (Rank 9), 3 (Rank 7), -1 (Rank 3), -2 (Rank 2), 4 (Rank 8).
03

Calculate Sum of Ranks

Sum up the ranks for each airline: - Delta: 12 + 11 + 6 + 1 + 4.5 + 10 = 44.5 - Alaska: 13 + 9 + 7 + 3 + 2 + 8 = 42.
04

Determine Test Statistic

The test statistic for the Wilcoxon rank-sum test is the smaller of the two rank sums. In this case: - Smaller rank sum (W) = 42 (Alaska).
05

Find Critical Value

For a two-tailed test with \( \alpha = 0.01 \) and sample sizes \( n1 = n2 = 6 \), refer to a Wilcoxon rank-sum table. The critical value is 21 for this test.
06

Conclude the Test

Since the smaller test statistic \( W = 42 \) is greater than the critical value of 21, we fail to reject the null hypothesis. This indicates that there is no significant difference in on-time arrival performance between the airlines at \( \alpha = 0.01 \).

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Key Concepts

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

Non-Parametric Statistics
When dealing with data that doesn't necessarily follow the normal distribution, like our airline delay times, traditional statistical tests might not be suitable. Here, we turn to non-parametric statistics. These are methods that do not rely on data belonging to any particular distribution.

One major advantage of non-parametric tests is that they are more flexible and can be used when we have small sample sizes or categorical data. They are particularly useful in situations where the data may have skewness or outliers, making them robust alternatives to their parametric counterparts.

The Wilcoxon rank-sum test, which was used in the exercise, is a popular non-parametric test. It compares the medians of two independent samples to check if they come from the same distribution. This test is not affected much by non-normal data distributions, making it perfect for analyzing our flight delay scenarios.
Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics, used to decide whether there is enough evidence in a dataset to support or reject a specific theory. In the flight delay problem, the hypothesis testing framework helps evaluate whether one airline has better on-time performance than the other.

In conducting hypothesis tests, we typically define two hypotheses:
  • Null Hypothesis (\( H_0 \)): Assumes no effect or no difference is present. For this exercise, it assumes that there is no difference in delay times between Delta and Alaska airlines.
  • Alternative Hypothesis (\( H_a \)): Suggests there is an effect or a difference. This would imply that one airline consistently has better on-time performance than the other.

Once we define these hypotheses, we use statistical tests to determine which is more likely given our data. The result of the Wilcoxon rank-sum test was that there was insufficient evidence to reject the null hypothesis, meaning there was no statistically significant difference at the \( \alpha = 0.01 \) level between the airlines' performances.
Statistical Analysis of Flight Data
Statistical analysis allows us to draw conclusions about data like flight delays, which can have variability and uncertainty. In the given exercise, the flight delay data was analyzed using the non-parametric Wilcoxon rank-sum test, which helped to understand the performances of Delta and Alaska airlines.

Flight delay times could be influenced by various unpredictable factors such as weather, airport congestion, or mechanical issues. By analyzing this data, we aim to identify patterns or differences in the performance of airlines.

Steps in analyzing flight data typically include:
  • Collecting data: Gathering flight delay times.
  • Ranking data: Organizing and assigning ranks to each observation.
  • Test calculation: Using statistical tests, like the Wilcoxon rank-sum, to evaluate hypotheses.
  • Interpretation: Analyzing results to make data-driven decisions, such as selecting a more punctual airline.
Analyzing flight delay data not only helps consumers make informed choices regarding airlines but also assists airlines in identifying areas for operational improvement.

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Most popular questions from this chapter

Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that \(\sigma_{1}=\sigma_{2}=1.0\) psi. From a random sample of size \(n_{1}=10\) and \(n_{2}=12,\) we obtain \(\bar{x}_{1}=162.5\) and \(\bar{x}_{2}=155.0 .\) The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. (a) Based on the sample information, should it use plastic \(1 ?\) Use \(\alpha=0.05\) in reaching a decision. Find the \(P\) -value. (b) Calculate a \(95 \%\) confidence interval on the difference in means. Suppose that the true difference in means is really 12 psi. (c) Find the power of the test assuming that \(\alpha=0.05\). (d) If it is really important to detect a difference of 12 psi, are the sample sizes employed in part (a) adequate, in your opinion?

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2} .\) Suppose that sample sizes are \(n_{1}=15\) and \(n_{2}=15,\) that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) for a true difference in means of \(3 ?\) (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if the true difference in means is \(-2 ?\) Assume that \(\alpha=0.05 .\)

The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process. The standard deviation of active concentration is known to be 3 grams per liter, regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data follow: $$\begin{aligned} \text { Catalyst } 1: & 57.9,66.2,65.4,65.4,65.2,62.6,67.6,63.7, 67.2,71.0\end{aligned}$$ $$\begin{array}{l} \text { Catalyst 2: } 66.4,71.7,70.3,69.3 ,64.8,69.6,68.6,69.4,65.3,68.8\end{array}$$ (a) Find a \(95 \%\) confidence interval on the difference in mean active concentrations for the two catalysts. Find the \(P\) -value. (b) Is there any evidence to indicate that the mean active concentrations depend on the choice of catalyst? Base your answer on the results of part (a). (c) Suppose that the true mean difference in active concentration is 5 grams per liter. What is the power of the test to detect this difference if \(\alpha=0.05 ?\) (d) If this difference of 5 grams per liter is really important, do you consider the sample sizes used by the experimenter to be adequate? Does the assumption of normality seem reasonable for both samples?

Presidential election, exit polls from the critical state of Ohio provided the following results: For respondents with college degrees, \(53 \%\) voted for Bush and \(46 \%\) voted for Kerry. There were 2020 respondents. (a) Is there a significant difference in these proportions? Use \(\alpha=0.05 .\) What is the \(P\) -value? (b) Calculate a \(95 \%\) confidence interval for the difference in the two proportions and comment on the use of this interval to answer the question in part (a).

Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in \(\bar{x}_{1}=290\) and \(s_{1}=12\), while another random sample of 16 gears from the second supplier results in \(\bar{x}_{2}=321\) and \(s_{2}=22\). (a) Is there evidence to support the claim that supplier 2 provides gears with higher mean impact strength? Use \(\alpha=0.05,\) and assume that both populations are normally distributed but the variances are not equal. What is the \(P\) -value for this test? (b) Do the data support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot-pounds higher than that of supplier \(1 ?\) Make the same assumptions as in part (a). (c) Construct a confidence interval estimate for the difference in mean impact strength, and explain how this interval could be used to answer the question posed regarding supplier-to-supplier differences.

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