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I'm building binomial proportion confidence intervals for a patient dataset containing the frequency of home nursing visits during the week prior to hospital admission. The freq. of home visit categories are integer values of 1, 2, 3,... and observations for a particular patient can be assumed to be independent from other patients.

It looks something like this for a dataset of 100 patients:

No. Visits:   Frequency:      % Total:
    1             10             10%
    2             50             50%
    3             30             30%
    5             10             10%
Total:           100            100% 

So in this example I'm building separate 95% CIs for $p_1 = 0.10$, $p_2 = 0.50$, etc. with $n=100$.

I'm using the Rule of Three to compute 95% CI $= [0, 3/n]$ for frequency of visits (in this case No. Visits $= 4$) which were not observed in the dataset. However it occurred to me I could mechanically use the Rule of Three to compute 95% CIs $= [0, 3/n]$ extending for No. Visits $= 6$ all the way up to infinity.

Is this reasonable?

kjetil b halvorsen
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RobertF
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    In case anyone is unfamiliar with this, OP means [this](http://en.wikipedia.org/wiki/Rule_of_three_(statistics)) "Rule of Three". – Glen_b Mar 19 '13 at 01:12
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    Does [this related question](https://stats.stackexchange.com/questions/497548/) assist you? I recommend the Wilson-score based "quasi-rule-of-3.84" which is $\text{CI}(0.95) = [0 , 3.841459/(n+3.841459) ]$. – Ben Jul 30 '21 at 23:25

1 Answers1

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This are number of home-nursing visits during one week prior to hospitalization, so if there is a maximum of one visit per day (probably unreasonable) the maximum would be seven. I think the way up to infinity is unreasonable!

You could summarize by saying that the events 4 visits and 6 or more visits where unobserved, and use the rule of three for those. For a descriptive analysis, that could be enough, but maybe think about smoothing the probability estimates via a model or otherwise.

kjetil b halvorsen
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