2
$\begingroup$

I have a measurement with two possible outcomes, let's say 0 and 1. If the outcome of the measurement is 1, the true value is always 1. If the outcome of the measurement is 0, there is a chance of 1% that the outcome of the measurement is false and the true value is 1 (and 99% that the outcome of the measurement was correct). For a sample of $n$ measurements, I want to test if the ratio of 1s, which we call $p$ is less than 0.05 with a significance level of $\alpha = 0.05$. First,let us assume that we ignore the measurement error. We can calculate if $$\sum_{i=0}^c B(i|p_0,n)\leq \alpha$$ where $c$ is the number of 1s measured and $p_0=0.05.$ So for 59 measurements where the outcome of all the measurements is 0, it is possible to conclude with a significance level of 0.05 that the probability for 1s is less than 0.05.

I had two ideas to include the measurement error, but these give different results:

  1. For the $n-c$ measurements with the outcome 0 we know that $B(l|0.01,n-c)$ is the probability for $l$ undetected 1s and therefore the probability for $c+l$ 1s. So $$\sum_{i=0}^{n-c} B(i|0.01,n-c) \sum_{j=0}^{c+l}B(j|p_0,n)$$ should be the probability for at most $c$ measured 1s under the assumption that $p=p_0$. Therefore, if this is less than $\alpha$ we conclude $p\leq p_0$.

  2. We do the test as if we ignore the mesurement error but with $p_0=0.0404$. In this case we know that the posibility of measuring a 1 is less than 0.0404 and the posibility of a measurements true value beinig a 1 is less than $$0.0404+(1-0.0404)*0.01 \approx 0.05.$$

But both ways lead to different results. For example when we have $n=74$ measurements where all the outcomes are 0s with the secound way we can conclude (with the significance $\alpha = 0.05$) that at most a ratio of 0.05 are 1s and with the first way we cannot. So at least one of them needs to be false, but which one?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ It's hard to justify (2). The likelihood is straightforward to write down. The maximum likelihood estimate for observing $x$ ones in a sample of size $n$ is $\hat p = x/n - q$ with $q = 0.01.$ You can go on from there. $\endgroup$ Commented Oct 30 at 15:55
  • $\begingroup$ @whuber Thank you for the hint, but I have problems writing down the likelihood. To connect the likelihood for observing $x$ ones with $p$ I need the probability for a one to be measured as a zero. But I only know the probability of a measured zero to be a one. These are connected via Bayes theorem but there I need the probability to measure a zero and that I don't know. $\endgroup$ Commented Nov 3 at 9:27
  • $\begingroup$ Those are good points, Adrian, and it's possible I originally misinterpreted your question. But it still looks like your assumptions will permit you to complete a $2\times 2$ table of the joint probabilities of (True value, measured value) in terms of $p.$ That, along with the assumption of independence of the observations, determines the likelihood. $\endgroup$ Commented Nov 3 at 12:58

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.