A transport company carries out a survey to gauge reaction to a proposed train timetable change. The survey finds that 122 out of a sample of 200 randomly selected passengers react positively to the new timetable. Infer a 95% confidence interval for the proportion of the train passenger population concerned who can be expected to react positively to the proposed timetable change.
I am confused as the SD is not given I know the formula to get calculate the answer but I am not sure if I am right
As e survey finds that 122 out of a sample of 200
Should I state N = 122?
The way I went about this question is using the formula but it is completely off as I am getting strange results
N = 200
P = 122/200 = .61
Interval at 95% = 1.96
√((.61)(1-.61)
= √((.61)(39)/200
= 0.0024 * 1.96
= 0.004