Margin of Error, The rule for sample proportions?

OK so here is the question: 335 visitors were questioned as part of a survey done by the McDowell Group, 78 percent were "very satisfied" with their time in the town they were visiting.


This is what they wanted to know:





The reported margin of error was 5.5 percent. Does this agree with your calculation?





Construct a 95% confidence interval for the true proportion of visitors who are "very satisfied" with their visit. ( I came up with 3.1% - 6.1% am I wrong?)





Does it appear that the 'rule for sample proportions was used' to calculate the margin of error or the formula using the sample proportions?





I came up with a margin of error 2.68% am I wrong?





I don't understand how they came up with 5.5%!








Assuming the sample is representative of all multiday visitors to, can you reasonably conclude that more than 50% of all such visitors are "very satisfied" with their visit? Can you reasonably conclude that more than 75% of all such visitors are "very satisfied"?

Margin of Error, The rule for sample proportions?
5.5% sounds reasonable for an n=335 sample size.





The 95% confidence interval is probably 72.5%-83.5% of visitors very satisfied with their time. (Margin of error is the radius of a confidence interval for a particular statistic from a survey.)





It'd be impossible to get a margin of error of 2.86% on a sample size that small.





So, you could reasonably conclude that more than 50% of all the visitors were "very satisfied", but you couldn't say for sure that more than 75% of the visitors were "very satisfied".