Question 1185470: According to the Bureau of Transportation Statistics, on-time performance by airlines is described as follows:
Action = % of time
On-Time = 68
National Aviation System Delay =8
Aircraft arriving late = 9
Other (weather and other conditions) =15
When a study was conducted it was found that of the 201 randomly selected flights, 145 were on time, 14 were a National Aviation System Delay, 18 arriving late, 24 were due to weather. Perform a test to see if there is sufficient evidence at α = 0.05 to see if these differ from the governments statistics.
The test value is_____
The p-value is_____
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to perform a Chi-Square Goodness-of-Fit test for this problem:
**1. State the Hypotheses:**
* **Null Hypothesis (H₀):** The observed distribution of flight delays matches the Bureau of Transportation Statistics' reported distribution.
* **Alternative Hypothesis (H₁):** The observed distribution of flight delays differs from the reported distribution.
**2. Calculate Expected Frequencies:**
Multiply each percentage from the Bureau of Transportation Statistics by the total number of flights in the study (201):
* **On-Time:** 0.68 * 201 = 136.68
* **National Aviation System Delay:** 0.08 * 201 = 16.08
* **Aircraft Arriving Late:** 0.09 * 201 = 18.09
* **Other (Weather):** 0.15 * 201 = 30.15
**3. Calculate the Chi-Square Statistic:**
χ² = Σ [(Observed - Expected)² / Expected]
χ² = (145 - 136.68)² / 136.68 + (14 - 16.08)² / 16.08 + (18 - 18.09)² / 18.09 + (24 - 30.15)² / 30.15
χ² ≈ 0.525 + 0.268 + 0.0002 + 1.258
χ² ≈ 2.051
**4. Determine Degrees of Freedom:**
df = Number of categories - 1 = 4 - 1 = 3
**5. Find the p-value:**
Consult a Chi-Square distribution table with df = 3. A χ² value of 2.051 is relatively low. You will find that the p-value is greater than 0.05. Statistical software can provide a more precise p-value which is approximately 0.561
**6. Make a Decision:**
* **Decision:** Since the p-value (0.561) is greater than the significance level (α = 0.05), we fail to reject the null hypothesis.
**7. Conclusion:**
There is not sufficient evidence at the α = 0.05 level to conclude that the observed distribution of flight delays differs significantly from the distribution reported by the Bureau of Transportation Statistics.
**Answers:**
* The test value (χ²) is approximately **2.051**.
* The p-value is approximately **0.561**.
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