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Standard error in this case, assuming the 200 are randomly sampled is: sqrt((p(1-p)/n) sqrt((0.90.1)/200) = 0.021 So 95% confidence interval for this ~ 0.8

Standard error in this case, assuming the 200 are randomly sampled is: 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?

monkpit 5y ago
It’s good information to have in the comments, I was not making any judgement.
seaman1921 5y ago
lol.. what kind of calculation is this? n is 1 in this case, the survey was not repeated 200 times from which you derived 90% as the mean number of nurses quitting.
- PheonixPharts 5y ago
  It's a normal approximation of the expected variance (in terms of standard deviation) in the estimate of the mean of the sum of 200 Bernoulli random variables. Each nurses' response is considered an observation of a Bernoulli distributed random variable, and we trying to determine the rate of that variable. 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.
  - seaman1921 5y ago
    Thank you for taking the time to explain your modelling. Unfortunately I will need to read more on this topic, because I do not understand the intuition behind the priors "uniform prior of alpha_prior = 1, and beta_prior = 1". 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 ?
    - deleted 5y ago
      [deleted]
    - PheonixPharts 5y ago
      Simulations are fantastic, and often necessary for tricky statistics problems, however what you are describing is reinventing so much of the wheel using simulation that you are going to be spending multiple orders of magnitude extra computation to get an approximately correct solution. You also do have some conceptual errors in your plan. For example > 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%. You have just described the Binomial distribution [0], which is probably the most elementary distribution you learn about when studying probability and statistics (even the Bernoulli is just a special case of it). There's no need to run simulations to answer this particular question. There are also some fundamental misunderstandings with your approach: > increasing until I get >95% chance that a random sample of 200 nurses has E(quitting) > 90%. The probability of getting > 90% 'yes/quitting' (i.e. more than 180) if the true probability 'yes' is in fact 0.9 is only 0.46. You won't cross your threshold of 95% here until you reach X=0.933 If you wanted to construct the 95% CI from pure simulation, a better approach would be to sample 200 observations from a 0.9 Bernoulli random variable (just sample from a uniform, and check if it's less than 0.9), compute the mean of the samples, and repeat this 10,000 or so times. Then look at the empirical CDF [1] (fairly easy to implement in code) and look at the lower 2.5% and upper 2.5% values and you have your bounds (which will be the same as the ones I posted within some epsilon). I do recommend, if you're seriously interested in understanding this, picking up a basic probability/stats book and work your way through it. 0. https://en.wikipedia.org/wiki/Binomial_distribution https://en.wikipedia.org/wiki/Binomial_distribution 1. https://en.wikipedia.org/wiki/Empirical_distribution_function https://en.wikipedia.org/wiki/Empirical_distribution_functio...
- deleted 5y ago
  [deleted]
tqi 5y ago
That also assumes there is no response bias. The linked article doesn't seem to go too deep on methods but the source is a content marketing piece for Hospital IQ, so I'd take it with a grain of salt.
- jonshariat 5y ago
  Also my first thought when I saw n was 200, lots of surveys make mistakes in sample selection or question writing but it seems with the responses here and my own observations being close to nursing, the main point checks out. Lots of burn out and leaving the profession.
  - tqi 5y ago
    Agree, I'm definitely not making any statements as to whether or not the burnout is real. I just think that a lot times the assumptions we make when we apply statistical concepts (like standard error) to real world data don't hold up.

Standard error in this case, assuming the 200 are randomly sampled is: sqrt((p*(1-p)/n) sqrt((0.9*0.1)/200) = 0.021 So 95% confidence interval for this ~ 0.8

monkpit 5y ago

It’s good information to have in the comments, I was not making any judgement.

seaman1921 5y ago

lol.. what kind of calculation is this? n is 1 in this case, the survey was not repeated 200 times from which you derived 90% as the mean number of nurses quitting.

PheonixPharts 5y ago
It's a normal approximation of the expected variance (in terms of standard deviation) in the estimate of the mean of the sum of 200 Bernoulli random variables. Each nurses' response is considered an observation of a Bernoulli distributed random variable, and we trying to determine the rate of that variable. 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.
- seaman1921 5y ago
  Thank you for taking the time to explain your modelling. Unfortunately I will need to read more on this topic, because I do not understand the intuition behind the priors "uniform prior of alpha_prior = 1, and beta_prior = 1". 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 ?
  - deleted 5y ago
    [deleted]
  - PheonixPharts 5y ago
    Simulations are fantastic, and often necessary for tricky statistics problems, however what you are describing is reinventing so much of the wheel using simulation that you are going to be spending multiple orders of magnitude extra computation to get an approximately correct solution. You also do have some conceptual errors in your plan. For example > 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%. You have just described the Binomial distribution [0], which is probably the most elementary distribution you learn about when studying probability and statistics (even the Bernoulli is just a special case of it). There's no need to run simulations to answer this particular question. There are also some fundamental misunderstandings with your approach: > increasing until I get >95% chance that a random sample of 200 nurses has E(quitting) > 90%. The probability of getting > 90% 'yes/quitting' (i.e. more than 180) if the true probability 'yes' is in fact 0.9 is only 0.46. You won't cross your threshold of 95% here until you reach X=0.933 If you wanted to construct the 95% CI from pure simulation, a better approach would be to sample 200 observations from a 0.9 Bernoulli random variable (just sample from a uniform, and check if it's less than 0.9), compute the mean of the samples, and repeat this 10,000 or so times. Then look at the empirical CDF [1] (fairly easy to implement in code) and look at the lower 2.5% and upper 2.5% values and you have your bounds (which will be the same as the ones I posted within some epsilon). I do recommend, if you're seriously interested in understanding this, picking up a basic probability/stats book and work your way through it. 0. https://en.wikipedia.org/wiki/Binomial_distribution https://en.wikipedia.org/wiki/Binomial_distribution 1. https://en.wikipedia.org/wiki/Empirical_distribution_function https://en.wikipedia.org/wiki/Empirical_distribution_functio...
deleted 5y ago
[deleted]

tqi 5y ago

That also assumes there is no response bias. The linked article doesn't seem to go too deep on methods but the source is a content marketing piece for Hospital IQ, so I'd take it with a grain of salt.

jonshariat 5y ago
Also my first thought when I saw n was 200, lots of surveys make mistakes in sample selection or question writing but it seems with the responses here and my own observations being close to nursing, the main point checks out. Lots of burn out and leaving the profession.
- tqi 5y ago
  Agree, I'm definitely not making any statements as to whether or not the burnout is real. I just think that a lot times the assumptions we make when we apply statistical concepts (like standard error) to real world data don't hold up.