0
$\begingroup$

Someone either has disease X or they are healthy. A test is used which is 90% sensitive and 80% specific (81.81% precise). The test is performed 8 times with different samples from the same person.

1) What is the rule to calculate the probability that someone has the disease given that 5 out of 8 tests came back positive?

2) More generally: that N out of M samples were positive.

3) Is there a way of calculating the confidence?

Kind regards

$\endgroup$
  • $\begingroup$ Hello jamesj629, welcome to the site. If this is a homework question please add the self-study tag. $\endgroup$ – Andy Oct 16 '14 at 11:26
  • $\begingroup$ Thanks Andy - its not homework, just learning. $\endgroup$ – jamesj629 Oct 16 '14 at 11:32
0
$\begingroup$

You need the disease prevalence in order to calculate this probability.

Let $T$ be the indicator variable with $T=1$ denoting the event of a positive test. Similarly let $D$ be the indicator of the disease. You have $P(T=1|D=1)=0.9$ and $P(T=0|D=0)=0.8$. Define $A$ to be the event of N positive result from M tests. Then

$$P(A|D=1)=\binom{M}{N}0.9^N 0.1^{M-N}$$

What you want is $P(D=1|A)$. Using Bayes' rule $$P(D=1|A)= \frac{P(A|D=1)P(D=1)}{P(A)}=\frac{P(A|D=1)P(D=1)}{P(A|D=1)P(D=1) +P(A|D=0)P(D=0)}$$

$\endgroup$
  • $\begingroup$ I take it P(D=1) on its own is disease prevalence? and P(D=0)=1-P(D=1) ? $\endgroup$ – jamesj629 Oct 16 '14 at 12:59
  • $\begingroup$ The prevalence is (say) 1 in 100000, but for people taking the test, about 90% actually have the disease, because the test is only done on those presenting with suspect symptoms - does this invalidate the probability? $\endgroup$ – jamesj629 Oct 16 '14 at 13:05
  • $\begingroup$ @jamesj629 It will increase the probability. If you let $S$ be the indicator of symptom onset. With what you described in the comment, now you are calculating $P(D=1|A,S=1)$ instead of $P(D=1|A)$. You numerator becomes $P(A|D=1,S=1)P(D=1|S=1)=P(A|D=1)P(D=1|S=1)$ if test result is independent of symptom. Note $P(D=1|S=1) > P(D=1)$ if disease is positively dependent with symptom. $\endgroup$ – Peter Oct 16 '14 at 21:19
0
$\begingroup$

You can use the information to calculate the base rate fo the disease:

accuracy: $A$

sensitivity: $s^+$

specificity: $s^-$

base rate: $p$

positive test result: $T^+$

disease: $D$

$A=p\cdot s^++(1-p)\cdot s^-$

$0.8181= p\cdot0.9+ (1-p)\cdot 0.8$

$0.0181= 0.1p$

$p=.181$

For one test positive out of one:

using Bayes' theorem: $P(D|T^+)=\frac{p \cdot s^+}{p \cdot s^+ + (1-p) \cdot (1-s^-)}$

$P(D|T^+)=\frac{.181 \cdot .9}{.181 \cdot .9 + (1-.181) \cdot (1-0.8)}$

$P(D|T^+)=\frac{.1629}{.1629 + .1638}=.4986$

now for k tests out of m: as in Peter's answer, the probabilities are now drawn from a binomial to generate likelihoods:

$L^+$~$B(n,s^+)$ and $L^-$~$B(n,s^-)$ with inverted number of successes for the second. You can probably calculate confidence intervals based on the knwon properties (variance) of bionomial distributions...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.