### Example Problem: Confidence Interval Calculation for Population Proportion#### Problem Statement:Assume that a sample is used to estimate a population proportion \( p \). Find the 99.9% confidence interval for a sample of size 269 with 52% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.\[ \boxed{} < p < \boxed{} \]**Note:**- Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to three decimal places.This type of problem is common in statistics, particularly in estimating population parameters based on sample data. The confidence interval provides a range within which the true population proportion is likely to lie with a specified level of confidence (99.9% in this case). To solve this, you can use the formula for the confidence interval for a population proportion:\[ \hat{p} \pm Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Where:- \( \hat{p} \) is the sample proportion (0.52 in this example).- \( Z_{\alpha/2} \) is the critical value for the specified confidence level (for 99.9%, this is typically around 3.291).- \( n \) is the sample size (269 in this case).Carefully compute this using the exact sample proportion and round only the critical value to three decimal places as per the instructions.

The following information has been given: The sample size is n=269. The sample proportion is p^=0.52

Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 269 with 52% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. < p < Answer should be obtained without any preliminary rounding. However, the critical value may be rounded al places