A travel analyst claims that the mean price of a round trip flight from New York City to Los Angeles is less than $507. In a random sample of 55 round trip flights from New York City to Los Angeles, the mean price is $502. Assume the population standard deviation is $111. At a = 0.05, is there enough evidence to support the travel analyst's claim?
A travel analyst claims that the mean price of a round trip flight from New York City to Los Angeles is less than $507. In a random sample of 55 round trip flights from New York City to Los Angeles, the mean price is $502. Assume the population standard deviation is $111. At a = 0.05, is there enough evidence to support the travel analyst's claim?
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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![**Question 28:** A travel analyst claims that the mean price of a round trip flight from New York City to Los Angeles is less than $507. In a random sample of 55 round trip flights from New York City to Los Angeles, the mean price is $502. Assume the population standard deviation is $111. At α = 0.05, is there enough evidence to support the travel analyst’s claim?
**Detailed Explanation:**
The problem presented is hypothesis testing to determine if we can support the travel analyst's claim about the mean price of round trip flights being less than $507, given the sample data.
**Steps to approach the problem:**
1. **State the Hypotheses:**
- Null hypothesis (H0): μ ≥ $507
- Alternative hypothesis (Ha): μ < $507
2. **Set the significance level (α):**
- α = 0.05
3. **Calculate the test statistic:**
Given:
- Sample mean (x̄) = $502
- Population mean (μ) = $507
- Population standard deviation (σ) = $111
- Sample size (n) = 55
The test statistic for a one-sample z-test is calculated using:
\[
z = \frac{x̄ - μ}{σ/√n}
\]
Substituting the given values:
\[
z = \frac{502 - 507}{111/√55} ≈ \frac{-5}{14.92} ≈ -0.34
\]
4. **Determine the critical value and rejection region:**
- For a one-tailed test at α = 0.05, the critical value (z-critical) from standard normal distribution tables is -1.645.
- Reject H0 if z < -1.645.
5. **Compare the test statistic to the critical value:**
- Our calculated z-value is -0.34, which is not less than -1.645.
**Conclusion:**
Since the calculated z-value (-0.34) does not fall in the rejection region (less than -1.645), we do not reject the null hypothesis. There is not enough evidence at the 0.05 significance level to support the travel analyst’s claim that the mean price of a round trip flight from New York City to Los Angeles](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69841bf7-b9f0-4fc9-8153-6ebe52f9298f%2Ff56c46eb-2387-4698-bd19-5061c3bc03f2%2Farqxafp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 28:** A travel analyst claims that the mean price of a round trip flight from New York City to Los Angeles is less than $507. In a random sample of 55 round trip flights from New York City to Los Angeles, the mean price is $502. Assume the population standard deviation is $111. At α = 0.05, is there enough evidence to support the travel analyst’s claim?
**Detailed Explanation:**
The problem presented is hypothesis testing to determine if we can support the travel analyst's claim about the mean price of round trip flights being less than $507, given the sample data.
**Steps to approach the problem:**
1. **State the Hypotheses:**
- Null hypothesis (H0): μ ≥ $507
- Alternative hypothesis (Ha): μ < $507
2. **Set the significance level (α):**
- α = 0.05
3. **Calculate the test statistic:**
Given:
- Sample mean (x̄) = $502
- Population mean (μ) = $507
- Population standard deviation (σ) = $111
- Sample size (n) = 55
The test statistic for a one-sample z-test is calculated using:
\[
z = \frac{x̄ - μ}{σ/√n}
\]
Substituting the given values:
\[
z = \frac{502 - 507}{111/√55} ≈ \frac{-5}{14.92} ≈ -0.34
\]
4. **Determine the critical value and rejection region:**
- For a one-tailed test at α = 0.05, the critical value (z-critical) from standard normal distribution tables is -1.645.
- Reject H0 if z < -1.645.
5. **Compare the test statistic to the critical value:**
- Our calculated z-value is -0.34, which is not less than -1.645.
**Conclusion:**
Since the calculated z-value (-0.34) does not fall in the rejection region (less than -1.645), we do not reject the null hypothesis. There is not enough evidence at the 0.05 significance level to support the travel analyst’s claim that the mean price of a round trip flight from New York City to Los Angeles
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