Q: (I shortened the question to only include the important details) Sample size/n = 1974, 354 respondents..... construct a 99.9% confidence interval for the proportion of people with a Master's degree.

Question

My work:
354/1974=.179
1-.179=.821
sqrt (.179) (.821) / 1974 = .0086282858

margin of error = (3.29) x (.0086) =.028294

Calculation of intervals:
Lower limit = .179-.028294=.150706
Upper limit= .179-.028294= .207294
(.150706,.207294)