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Thread: Dichotomous test - posible to calculate the min number of 'positives' ?

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    Dichotomous test - posible to calculate the min number of 'positives' ?




    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

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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?

    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.

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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?

    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.
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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?

    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

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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?

    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."
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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?

    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."

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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?

    yes it means that at the number calculated you can say with 90% confidence that the percentage of passed tests is 80% or above.

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    Re: Dichotomous test - posible to calculate the min number of 'positives' ?


    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.

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    Linde (04-04-2017)

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