Question 1192440: consider two pain relieving drugs compared two independent samples of 1000 individuals each. suppose 750 of these individuals receiving drugs one and 800 of those receiving drug two construct 90% confidence interval for the percentage of 2 pain receiving pills?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! **1. Calculate the Sample Proportions**
* **Drug 1:**
* Sample proportion (p1) = 750 / 1000 = 0.75
* **Drug 2:**
* Sample proportion (p2) = 800 / 1000 = 0.80
**2. Calculate the Standard Error**
* **Pooled Proportion (p̂):**
* p̂ = (Number of successes in both samples) / (Total sample size)
* p̂ = (750 + 800) / (1000 + 1000) = 1550 / 2000 = 0.775
* **Standard Error (SE):**
* SE = √[p̂ * (1 - p̂) * (1/n1 + 1/n2)]
* SE = √[0.775 * (1 - 0.775) * (1/1000 + 1/1000)]
* SE = √[0.775 * 0.225 * (2/1000)]
* SE ≈ 0.0186
**3. Determine the Z-score for 90% Confidence Level**
* For a 90% confidence interval, the Z-score is 1.645.
**4. Calculate the Margin of Error**
* Margin of Error (ME) = Z-score * SE
* ME = 1.645 * 0.0186
* ME ≈ 0.0306
**5. Construct the Confidence Intervals**
* **Drug 1:**
* Lower limit: p1 - ME = 0.75 - 0.0306 = 0.7194
* Upper limit: p1 + ME = 0.75 + 0.0306 = 0.7806
* 90% CI for Drug 1: (0.7194, 0.7806)
* **Drug 2:**
* Lower limit: p2 - ME = 0.80 - 0.0306 = 0.7694
* Upper limit: p2 + ME = 0.80 + 0.0306 = 0.8306
* 90% CI for Drug 2: (0.7694, 0.8306)
**Interpretation:**
* We are 90% confident that the true proportion of individuals who experience pain relief with Drug 1 lies between 71.94% and 78.06%.
* We are 90% confident that the true proportion of individuals who experience pain relief with Drug 2 lies between 76.94% and 83.06%.
**Note:**
* These confidence intervals provide a range of plausible values for the true population proportions.
* The intervals do not overlap, suggesting that there might be a statistically significant difference in the effectiveness of the two drugs. However, further statistical analysis (such as a hypothesis test) would be needed to confirm this.
**Disclaimer:**
* This analysis provides a basic framework for constructing confidence intervals.
* In real-world scenarios, more sophisticated statistical methods might be necessary, especially when dealing with medical data.
* This information should not be considered medical advice.
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