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Chapter 9: Problem 10

Airline Occupancy Rates Suppose a scheduled airline flight must average at least \(60 \%\) occupancy in order to be profitable to the airline. An examination of the occupancy rate for 12010: 00 A.M. flights from Atlanta to Dallas showed a mean occupancy per flight of \(58 \%\) and a standard deviation of \(11 \%\). a. If \(\mu\) is the mean occupancy per flight and if the company wishes to determine whether or not this scheduled flight is unprofitable, give the alternative and the null hypotheses for the test. b. Does the alternative hypothesis in part a imply a one- or two-tailed test? Explain. c. Do the occupancy data for the 120 flights suggest that this scheduled flight is unprofitable? Test using \(\alpha=.05\)

Short Answer

Expert verified
Answer: Yes, based on the hypothesis testing, there is evidence to suggest that the flights from Atlanta to Dallas are unprofitable, as the mean occupancy per flight is less than 60%.

Step by step solution

01

a. Formulate the null and alternative hypotheses

We will test the hypothesis that the average occupancy rate is at least 60%. The null and alternative hypotheses are: - Null hypothesis (\(H_0\)): \(\mu \geq 60\%\) - Alternative hypothesis (\(H_1\)): \(\mu < 60\%\)
02

b. One- or two-tailed test?

The alternative hypothesis states that the mean occupancy per flight is less than 60%. Since we are only concerned with one direction (less than 60%), this is a one-tailed test.
03

c. Perform the hypothesis test

We will use a t-test to compare the sample mean with the population mean, since we have the sample size, sample mean, and sample standard deviation. Following the steps for a one-tailed t-test: 1. State the null and alternative hypotheses: \(H_0: \mu \geq 60\%\) \(H_1: \mu < 60\%\) 2. Identify the significance level, sample mean (\(\bar{x}\)), sample standard deviation (s), and sample size (n): \(\alpha = 0.05\) \(\bar{x} = 58\%\) \(s = 11\%\) \(n = 120\) 3. Calculate the t-value: \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\) \(t = \frac{58 - 60}{11/\sqrt{120}}\) \(t = -1.915\) 4. Find the critical t-value (one-tailed test) using the t-table. We have to look up the t-value for \(\alpha = 0.05\) and degrees of freedom (df), which is \(n - 1\): df = 120 - 1 = 119 From the t-table, we find that the critical t-value for a one-tailed test with \(\alpha = 0.05\) and df = 119 is approximately 1.657. 5. Compare the calculated t-value with the critical t-value: Our calculated t-value is -1.915, and the critical t-value is 1.657. 6. Make the conclusion: Since our calculated t-value is less than the critical t-value (-1.915 < 1.657), we reject the null hypothesis (\(H_0\)) and conclude that there is evidence to suggest that the flight is unprofitable, as the mean occupancy per flight is less than 60%.

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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 we aim to test. It usually represents a status quo or a claim of no effect or no difference. In our exercise on airline occupancy rates, the null hypothesis (\(H_0\)) suggests that the mean occupancy per flight is at least 60%. This implies that the flight is, on average, meeting the minimum requirement for profitability.
  • Symbolically represented as: \(\mu \geq 60\%\).
  • This hypothesis assumes that any observed deviation from the 60% occupancy rate is due to random chance.
  • It's what the airline hopes to confirm as true, maintaining the profitability status.
The burden of proof in statistical testing lies in disproving the null hypothesis, not proving it.
Alternative Hypothesis
The alternative hypothesis is the complement of the null hypothesis and represents a new claim or observation that contradicts the status quo. For the airline's occupancy rates, the alternative hypothesis (\(H_1\)) is that the mean occupancy per flight is less than 60%. This suggests that the flights may be operating below the threshold needed for profitability.
  • Expressed mathematically as: \(\mu < 60\%\).
  • Unlike the null hypothesis, the alternative hypothesis suggests a specific condition or effect.
  • The test's aim is to find enough statistical evidence to support the alternative hypothesis if the null hypothesis can be rejected.
By seeking evidence against the null hypothesis, we indirectly attempt to support the alternative hypothesis.
One-Tailed Test
A one-tailed test in hypothesis testing is used when we are interested in determining if there is a relationship in only one direction. In this exercise, the concern is whether the mean occupancy is less than 60%. Therefore, it is a one-tailed test.
  • This test is directional, focusing on values only in one tail of the distribution.
  • It gives more power to detect an effect in one direction by not testing the other direction.
  • A one-tailed test requires specifying the direction of the effect we are interested in before conducting the test.
In our scenario, testing unfolds only in one direction, allowing us to determine if occupancy truly falls below the required rate.
T-test
A t-test is a statistical method used to compare the means of two groups or a sample mean against a known population mean. It's particularly useful when dealing with small sample sizes or unknown population variances. In our exercise, the t-test evaluates if the average flight occupancy rate deviates from the 60% target.
  • The t-test formula: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\)
  • Where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
  • Critical values determine if we reject or accept the null hypothesis based on the t-test results.
This method effectively evaluates the significance of our sample data, confirming if the occupancy rates fall short in terms of profitability.

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