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Question 1178026: An airline has surveyed a simple random sample of air travelers to find out whether they would be interested in paying a higher fare in order to have access to email during their flight:Of the 400 travelers surveyed, 80 said email access would be worth a slight extra cost. Construct a 95% confidence interval for the population proportion of air travelers to who are in favour of the airline's email idea.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's construct the 95% confidence interval for the population proportion of air travelers who are in favor of the airline's email idea.
**1. Calculate the Sample Proportion (p̂)**
* **Sample Size (n):** 400
* **Number in Favor (x):** 80
* **Sample Proportion (p̂):** p̂ = x / n = 80 / 400 = 0.20
**2. Calculate the Standard Error (SE)**
* **Standard Error (SE):** SE = √[p̂(1 - p̂) / n]
* SE = √[0.20(1 - 0.20) / 400]
* SE = √[0.20(0.80) / 400]
* SE = √(0.16 / 400)
* SE = √0.0004 = 0.02
**3. Find the Critical Value (z*)**
* **Confidence Level:** 95%
* **Alpha (α):** 1 - 0.95 = 0.05
* **Alpha/2:** α/2 = 0.025
* **Critical Value (z*):** For a 95% confidence interval, z* = 1.96 (from the standard normal distribution table)
**4. Calculate the Margin of Error (ME)**
* **Margin of Error (ME):** ME = z* * SE
* ME = 1.96 * 0.02 = 0.0392
**5. Construct the Confidence Interval**
* **Confidence Interval:** p̂ ± ME
* **Lower Bound:** p̂ - ME = 0.20 - 0.0392 = 0.1608
* **Upper Bound:** p̂ + ME = 0.20 + 0.0392 = 0.2392
**Therefore, the 95% confidence interval for the population proportion of air travelers who are in favor of the airline's email idea is (0.1608, 0.2392).**
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