Question 1181343: In a study of 1003 community college students, 521 indicated that they believe television is a luxury that they can live without.
a) (15 points) Construct a 95% confidence interval for the population proportion of community college students who believe television is a luxury that they can live without.
b) (5 points) Explain what the confidence interval you constructed means.
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! Here's how to construct and interpret the confidence interval:
**a) Constructing a 95% Confidence Interval:**
1. **Calculate the sample proportion (p̂):**
p̂ = (Number of students who believe TV is a luxury) / (Total number of students)
p̂ = 521 / 1003
p̂ ≈ 0.52
2. **Find the critical z-value:**
For a 95% confidence level, the alpha (α) level is 1 - 0.95 = 0.05. We need to find the z-score that corresponds to an area of 0.975 (1 - α/2) in the standard normal distribution. This z-value is approximately 1.96. You can find this using a z-table or calculator.
3. **Calculate the margin of error (E):**
E = z * √(p̂(1 - p̂) / n)
E = 1.96 * √(0.52 * (1 - 0.52) / 1003)
E ≈ 1.96 * √(0.2496 / 1003)
E ≈ 1.96 * 0.0157
E ≈ 0.0308
4. **Construct the confidence interval:**
Lower Bound = p̂ - E = 0.52 - 0.0308 ≈ 0.4892
Upper Bound = p̂ + E = 0.52 + 0.0308 ≈ 0.5508
Therefore, the 95% confidence interval is approximately (0.4892, 0.5508).
**b) Interpretation of the Confidence Interval:**
We are 95% confident that the true population proportion of community college students who believe television is a luxury they can live without is between 48.92% and 55.08%. This means that if we were to repeat this study many times and calculate a 95% confidence interval each time, 95% of those intervals would contain the true population proportion. It's important to remember that this interval is an estimate, and the true proportion may or may not fall within this range.
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