It sounds like it should be possible but it isn't clear (to me at least) exactly what the problem is. Can you elaborate a bit and give an example, perhaps.
Hi,
Is it possible to calculate the minimum number of positives (when I have 36 participants) for a dichotomous test at a 90% confidence level (one-sided) and the proportion of positives should be not less than 0.80?
Thanks
Linde
It sounds like it should be possible but it isn't clear (to me at least) exactly what the problem is. Can you elaborate a bit and give an example, perhaps.
Yeah, you need to provide much more information. It almost seems like you could be looking at a one-sample proportion test, but that is a guess based on your vague introduction in your post.
Stop cowardice, ban guns!
Thank you. yes, it concerns a one sample proportion test.
It concerns a task that people need to perform, and the outcome is scored as passed or failed
The issue is that a criterion was set that not less than 80% of the individuals performing the task should be able to pass the test, with a 90% confidence level, one-sided.
If 36 people would try to carry out this task, how many would need to pass to meet the requirement?
In case you need more information, what would it be?
Thanks
Linde
Is the below written correctly, it reads to me as, the upper 95% CI in a one-sided interval should be below 80% in the group that fails the test?
"The issue is that a criterion was set that not less than 80% of the individuals performing the task should be able to pass the test, with a 90% confidence level, one-sided."
Stop cowardice, ban guns!
So, for example, if the magic number you were looking for was say, 31/36 in one random experiment, then you could say "Because I saw 31/36 pass in this random experiment, I'm 90% sure that at least 80% of all individuals from the same population would pass if they sat."
yes it means that at the number calculated you can say with 90% confidence that the percentage of passed tests is 80% or above.
Because you can only have a whole number answer, there probably won't be an answer that satisfies both the 80% and 90%.
Using the binomial distribution, if the population proportion is in fact 80% then the probability of getting 32 or more is about 12%, and that of getting 33 or more is about 5%, so you need at least 33 to be at least 90% sure that the proportion is at least 80% passing. 32/36 will happen about 10% of the time if the population proportion is about 79%, so 32/36 would be pretty close to your 80% requirement.
Linde (04-04-2017)
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