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?

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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
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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