Unfortunately, my ability to survey was eliminated for much of my time in the area by the record New York snowfall on October 29, and so I was only able to collect data for a few hours on Sunday, October 30. In that process I discovered that data collection in the park is difficult and slow. After conducting interviews, I was only able to find 23 people (including myself) who were in Zucotti Park on the days in question. Of those, none reported responding to a survey on October 10 or 11.
This number of responses is not very low. One other survey has been reported by Fordham professor Costas Panagopoulos. He reported that over the period from October 14 to October 18, five days, his survey only reached 301 respondents. He also reported that he had 15 survey takers, while I only had one.
ANALYSIS
Since there were no respondents who said they answered the Schoen survey, I cannot rule out the possibility that it was not conducted. In order to assess the probabilities, a Bayesian process can be used.
First, an idea of prior probabilities is required by the Bayesian analysis. Zuccotti Park has an area of only 3,100 square meters according to Wikipedia and thus the number of people in it can be at most several thousand. The Schoen survey was taken on a Monday (Columbus Day) and the following Tuesday, so attendance is probably off-peak. However, people can come in and out through the two days. I am fairly confident the number of people who attended on those two days is less than 10,000 and probably is significantly less. Thus, we expect that more than 2% of the sample responded to the Schoen survey.
For the Bayesian analysis, we consider four unmarked boxes. In this analysis, "yes" means that the person I surveyed responded to the Schoen survey. The four boxes are the trick box, which contains no yes responses, and the other three are boxes which contain 2% yes (corresponding to an attendance of 9900), 4% yes (4950 attendance), and 6% yes (3300 attendance). I randomly pick one of the four boxes with equal (25%) probability and draw 23 responses from the box. I can then calculate the probability that the box I selected was either the trick box or one of the other four boxes.
If I chose the trick box, there is a 100% chance I got 23 noes. For the 2% box, the probability of drawing 23 noes is 0.98 to the 23rd power, or 62.835%, for the 4% box, the probability is 39.106%, and for the 6% box, 24.096%. Overall, then, the probability that I drew 23 noes is the average of these four probabilities, or 56.509%.
The possibility that I drew the trick box is 25%. So, the probability knowing that I got 23 noes that I drew the trick box is 25% divided by 56.509%, or 44.2%. The probability I drew the 2% box and got 23 noes is 15.709%, which divided by 56.509% is 27.8%. The probability I drew the 4% box and got 23 noes is 9.777%, which divided by 56.509% is 17.3%, and the probability I drew the 6% box and got 23 noes is 6.024%, which divided by 56.509% is 10.7%.
So, in summary, the probabilities are:
Trick box-- 44.2%
2% box-- 27.8% (attendance is 9900)
4% box-- 17.3% (attendance is 4950)
6% box-- 10.7% (attendance is 3300)
More statistics could bring about much more certainty. In particular, a single respondent who answered the Schoen survey would entirely rule out the trick box. Having just a hundred responses would greatly improve the statistics. For this reason, I would like to repeat the survey on an upcoming weekend.
SUPPORT THIS WORK
The investigator is in need of funding for his travels from Ithaca to Wall Street. Any donations should go to the PayPal account connected to the email address hbowman108@hotmail.com.
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