Question 1179001: A study of 243 randomly selected occupants in passenger cars and 199 randomly selected occupants in pickup trucks shows that 91% of occupants in passenger cars and 88% of occupants in pickup trucks wear seat belts. At α=0.10, can you reject the claim that the proportion of occupants who wear seat belts is the different for passenger cars and pickup trucks?
Conditions:
Significant:
Hypothesis:
Conclusion:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to conduct a hypothesis test to determine if the proportions of seatbelt wearers in passenger cars and pickup trucks are different.
**1. Conditions:**
* **Random Samples:** Given (243 car occupants, 199 truck occupants)
* **Independence:** Assumed (samples are independent of each other)
* **Normality (Large Samples):**
* n₁p₁ = 243 * 0.91 = 221.13 > 10
* n₁(1-p₁) = 243 * 0.09 = 21.87 > 10
* n₂p₂ = 199 * 0.88 = 175.12 > 10
* n₂(1-p₂) = 199 * 0.12 = 23.88 > 10
* All conditions are met.
**2. Significance Level:**
* α = 0.10
**3. Hypothesis:**
* **Null Hypothesis (H₀):** p₁ = p₂ (The proportions are equal)
* **Alternative Hypothesis (H₁):** p₁ ≠ p₂ (The proportions are different)
* This is a two-tailed test.
**4. Calculate Sample Proportions:**
* p̂₁ (car) = 0.91
* p̂₂ (truck) = 0.88
* n₁ (car) = 243
* n₂ (truck) = 199
**5. Calculate the Pooled Proportion (p̂):**
* p̂ = (x₁ + x₂) / (n₁ + n₂) = (n₁p̂₁ + n₂p̂₂) / (n₁ + n₂)
* p̂ = (243 * 0.91 + 199 * 0.88) / (243 + 199)
* p̂ = (221.13 + 175.12) / 442
* p̂ = 396.25 / 442 ≈ 0.8965
**6. Calculate the Standard Error (SE):**
* SE = √[p̂(1 - p̂)(1/n₁ + 1/n₂)]
* SE = √[0.8965(1 - 0.8965)(1/243 + 1/199)]
* SE = √[0.8965(0.1035)(0.004115 + 0.005025)]
* SE = √[0.09278775(0.00914)]
* SE = √0.00084795
* SE ≈ 0.0291
**7. Calculate the Test Statistic (z-score):**
* z = (p̂₁ - p̂₂) / SE
* z = (0.91 - 0.88) / 0.0291
* z = 0.03 / 0.0291
* z ≈ 1.031
**8. Find the Critical Value:**
* For a two-tailed test with α = 0.10, the critical z-values are ±1.645.
**9. Make a Decision:**
* Compare the calculated z-score (1.031) to the critical z-values (±1.645).
* Since -1.645 < 1.031 < 1.645, the calculated z-score does not fall in the rejection region.
* Therefore, we fail to reject the null hypothesis.
**10. Conclusion:**
* There is not sufficient evidence at the 0.10 significance level to conclude that the proportion of occupants who wear seat belts is different for passenger cars and pickup trucks.
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