102 students at a college were asked whether they had completed their required English 101 course, and 66 students said "yes". Construct the $80 \%$ confidence interval for the proportion of students at the college who have completed their required English 101 course. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
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Step 1 ：First, we calculate the sample proportion (p̂), which is the number of students who have completed the course divided by the total number of students surveyed. In this case, we have 66 students who have completed the course out of a total of 102 students surveyed. So, \(p̂ = \frac{66}{102} = 0.647\)

Step 2 ：Next, we calculate the standard error of the proportion. The formula for this is \(SE = \sqrt{\frac{p̂(1 - p̂)}{n}}\), where n is the total number of students surveyed. Substituting the values we have, \(SE = \sqrt{\frac{0.647(1 - 0.647)}{102}} = 0.047\)

Step 3 ：We then calculate the margin of error (E) using the formula \(E = Z * SE\), where Z is the Z-score corresponding to the desired confidence level. For an 80% confidence level, Z = 1.28. So, \(E = 1.28 * 0.047 = 0.060\)

Step 4 ：Finally, we construct the confidence interval by subtracting and adding the margin of error from/to the sample proportion. The lower bound of the confidence interval is \(p̂ - E = 0.647 - 0.060 = 0.586\) and the upper bound is \(p̂ + E = 0.647 + 0.060 = 0.708\)

Step 5 ：\(\boxed{\text{Final Answer: The 80% confidence interval for the proportion of students at the college who have completed their required English 101 course is } 0.586 < p < 0.708}\)