sqrt((p*(1-p)/n)
sqrt((0.9*0.1)/200) = 0.021
So 95% confidence interval for this ~ 0.86 - 0.94
Does that radically change the message of this article?
You are incorrect that "n is 1" since, by that logic one survey talking to 100,000 nurses would be the same as one talking to 3.
If you would like an alternate, more Bayesian formulation we can use the Beta distribution which is parameterized by alpha (numbers of 'yes') and beta (number of 'no').
This approach is a bit more intuitive than the Frequentist method since it answers the question "what do we believe to be the expected rate of nurses answering 'yes'"
In this case alpha=180 and beta=20, we'll include uniform prior of alpha_prior = 1, and beta_prior = 1
For Betas the posterior is defined quite nicely as:
Beta(alpha_posterior, beta_posterior) = Beta(alpha_likelihood + alpha_prior, beta_likelihood + beta_prior)
In general for Beta distributions we can compute the expectation as:
E[Beta(alpha,beta)] = alpha/(alpha + beta)
In this case: 181/202 = ~0.9
And the variance of a Beta distributed random variable is:
Var[Beta(alpha, beta)] = (alpha*beta)/((alpha+beta)^2 * (alpha + beta + 1))
Which for our case is:
0.00046
and the standard deviation of this is just it's square root:
0.021
Which gives us the same answer as we get with the normal approximation.
The way I would generally approach such a problem is by running monte carlo simulations. Assuming the true rate of nurses quitting is X, what is the chance that a random sample of 200 nurses has the expectation of quitting >= 90%. To get the lower bound of the confidence interval, I will run this simulation for several values of X, starting at say X=60%, increasing until I get >95% chance that a random sample of 200 nurses has E(quitting) > 90%. Do you think this approach makes sense ?