Question 1192436
**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!