5 ms·

The survey sample was 200 RNs in the USA.

by monkpit 4y ago

spywaregorilla 4y ago
That's plenty so long as the sample is random
- deleted 4y ago
  [deleted]
degenerate 4y ago
I wish there was some law requiring surveying companies to fully disclose their method of contact and compensation. I can absolutely see an employee looking to leave their job being more receptive to taking a survey. Those happy with their jobs are not as likely to respond to third-party entities like "Hospital IQ" contacting them. Maybe the survey company masked this survey as a "pre-screen" to finding new jobs! We will never know. The process of recruiting survey participants has GREAT implications on the results.
- mikkergp 4y ago
  Nursing is a underappricated underpaid job, but we can't get 90% of people in this world to agree that the sky is blue.
PheonixPharts 4y ago
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 4y ago
  It’s good information to have in the comments, I was not making any judgement.
- seaman1921 4y 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 4y 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 4y 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 ?
- tqi 4y 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 4y 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 4y 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.
vmception 4y ago
And for just one year with no information on what they would have said in other years. "I plan on quitting my job if I had money" wow stop the presses.

The survey sample was 200 RNs in the USA.

spywaregorilla 4y ago

That's plenty so long as the sample is random

deleted 4y ago
[deleted]

degenerate 4y ago

I wish there was some law requiring surveying companies to fully disclose their method of contact and compensation. I can absolutely see an employee looking to leave their job being more receptive to taking a survey. Those happy with their jobs are not as likely to respond to third-party entities like "Hospital IQ" contacting them. Maybe the survey company masked this survey as a "pre-screen" to finding new jobs! We will never know. The process of recruiting survey participants has GREAT implications on the results.

mikkergp 4y ago
Nursing is a underappricated underpaid job, but we can't get 90% of people in this world to agree that the sky is blue.

PheonixPharts 4y ago

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 4y ago
It’s good information to have in the comments, I was not making any judgement.
seaman1921 4y 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 4y 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 4y 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 ?
tqi 4y 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 4y 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 4y 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.

vmception 4y ago

And for just one year with no information on what they would have said in other years. "I plan on quitting my job if I had money" wow stop the presses.