3 votes
Accepted

confidence interval: A/B test - unable to decide what to use

I almost address how to do this here. The test statistic for a difference in proportions is $$ z=\frac{p_{1}-p_{2}}{\sqrt{\frac{p_{1}\left(1-p_{1}\right)}{n_{1}}+\frac{p_{2}\left(1-p_{2}\right)}{n_{2}}...
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2 votes

simple probability/proportion question

Can I say a 65+ year old person will have dementia with 0.1 chance? It depends. First, "randomly sampled from the population" could mean different things. How exactly would you sample? It ...
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1 vote
Accepted

Is there a test (in R) for predicting the proportion of "successes" to "failures"?

Logistic regression does both: the predicted probabilities of success are the same as the predicted proportion of individuals with that set of covariates who would be a "success". Eg, ...
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1 vote

Hypothesis test for difference-of-differences of proportions?

Let's say our model on the log odds scale, allowing for interactions between sex and condition, is $$ \log \left( \dfrac{p}{1-p} \right) = \beta_0 + \beta_1 I(\mbox{sex = Male}) + \beta_2I(\mbox{...
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