SOLUTION: 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 tw

Question 1192436: 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 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. Calculate the Difference in Proportions**
* Difference in proportions: p2 - p1 = 0.80 - 0.75 = 0.05
**6. Construct the Confidence Interval for the Difference in Proportions**
* Lower Limit: Difference in proportions - ME = 0.05 - 0.0306 = 0.0194
* Upper Limit: Difference in proportions + ME = 0.05 + 0.0306 = 0.0806
* **90% Confidence Interval for the Difference in Proportions: (0.0194, 0.0806)**
**Interpretation:**
* We are 90% confident that the true difference in the proportion of individuals experiencing pain relief between Drug 2 and Drug 1 lies between 1.94% and 8.06%.
* Since the interval does not include 0, this suggests that there is a statistically significant difference in the effectiveness of the two drugs. Drug 2 appears to be more effective than Drug 1.
**Note:**
* This analysis assumes that the samples are independent and that the conditions for using the normal approximation to the binomial distribution are met.
I hope this revised explanation is helpful!

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