ANOVA Test for Flight Delay Times in U.S., Canada, and Great Britain

Math Problem Statement

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Topic 7 Homework (Nonadaptive) Question 15 of 16 (1 point)|Question Attempt: 1 of Unlimited

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Question 15 For the time being, the air traffic control system in the United States is run by the federal government. Several groups disagree with this and would like to privatize the air traffic control system. Among the arguments given for privatization is that flight delays are much too common in the United States. A first step in deciding if, indeed, delays in the U.S. are "too common" might be to compare the flight delay information from the U.S. and other nations. You have in front of you the following chart, which summarizes the information from three samples of flights from the past six weeks. One of the samples is composed of domestic U.S. flights, one is composed of domestic Canadian flights, and one is composed of domestic British flights. The information recorded for each flight is the time (in minutes) that the plane was late to the arrival gate. (If the plane was early, the time recorded was negative.)

Groups Sample size Sample mean Sample variance United States 208 18.7 99.1 Canada 206 16.1 85.9 Great Britain 235 16.6 110.0

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Send data to Excel You perform a one-way, independent-samples ANOVA test to decide if there are differences in the mean flight delay times among the three countries. Such a test uses the following statistic. =FVariation between the samplesVariation within the samples For the flight delay information you've been given ≈F 4.03. (a)Give the numerator degrees of freedom of this F statistic.

(b)Give the denominator degrees of freedom of this F statistic.

(c)Using the 0.05 level of significance, can you conclude from your ANOVA test that there are differences in the mean flight delay times among the three countries?

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Solution

For this problem, you are performing a one-way ANOVA test to determine if there are differences in the mean flight delay times among the three countries (United States, Canada, and Great Britain). The F-statistic for the test has already been provided as 4.03.

Now, let's address each part of the question:

(a) Numerator degrees of freedom:

The numerator degrees of freedom in an ANOVA test are calculated as: $\text{df}_{\text{between}} = k - 1$ where $k$ is the number of groups (countries in this case). Since we have three groups (U.S., Canada, and Great Britain), the degrees of freedom for the numerator are: $\text{df}_{\text{between}} = 3 - 1 = 2$

(b) Denominator degrees of freedom:

The denominator degrees of freedom are calculated as: $\text{df}_{\text{within}} = N - k$ where $N$ is the total number of observations, and $k$ is the number of groups.

The sample sizes for the three countries are: 208 (U.S.), 206 (Canada), and 235 (Great Britain). Therefore, the total sample size $N$ is: $N = 208 + 206 + 235 = 649$

Thus, the denominator degrees of freedom are: $\text{df}_{\text{within}} = 649 - 3 = 646$

(c) Conclusion of the ANOVA test:

To determine if there are differences in the mean flight delay times, we need to compare the F-statistic (4.03) with the critical value of F at a 0.05 significance level.

For $\text{df}_{\text{between}} = 2$ and $\text{df}_{\text{within}} = 646$ , we would look up the critical value of F in an F-distribution table or use statistical software. Assuming typical thresholds for a 0.05 significance level, if the F-statistic (4.03) exceeds the

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Math Problem Analysis

Mathematical Concepts

Statistics
ANOVA Test
F-statistic
Degrees of Freedom

Formulas

F = (Variation between the samples) / (Variation within the samples)
Numerator degrees of freedom = k - 1
Denominator degrees of freedom = N - k

Theorems

One-Way ANOVA

Suitable Grade Level

College Level Statistics

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