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Feb 17, 2022, 3:00:23 PMFeb 17

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Corona Update 24, Mathematical proof self-tests are only 3% correct, 97% false positives !

3% of the corona/covid 19 self-tests are correct, 97% are false positives !

The mathematical proof has been delivered (using Bayes' theorem):

Roche's corona self-test has a sensitivity of 96.52% and a specificity of 99.68%.

If "C19" stands for the presence of the disease COVID-19 ("corona") and

+ and − respectively for a positive and negative result of the test, this means:

sensitivity: P( + | C19 ): 0.9652

specificity: P( - | not C19): 0.9968

That seems very high. But if the prevalence is only 1 in 10,000, i.e

prevalence: P( C19 ): 0.0001

This implies:

P( + ) = P( + | C19 ) P( C19 ) + P( + | not C19 ) P ( not C19 ) =

= 0.9652 x 0.0001 + 0.0032 x 0.9999 = 0.0033

and

P( C19 | + ) = ( P( + | C19 ) P( C19 ) ) / P( + ) =

= ( 0.9652 x 0.0001 ) / 0.0033 = 0.03

If 10,000 people are tested with this test, including probably 1 infected person,

then the infected person will almost certainly get a positive result.

But of the 9,999 uninfected people, 32 will get a false positive result.

The 9967 people with a negative result are almost certain that they are not infected.

But of the 33 with a positive result, only 1 is infected,

only it is unknown who that is.

So the chance that an individual is actually infected after a positive result

is only slightly more than 3% in this scenario.

This self-test is therefore of little use to determine whether you are infected with corona,

unless the prevalence is around 1% or higher.

(This information has been known and published since 20 june 2021 !)

Bye,

Skybuck ! =D

3% of the corona/covid 19 self-tests are correct, 97% are false positives !

The mathematical proof has been delivered (using Bayes' theorem):

Roche's corona self-test has a sensitivity of 96.52% and a specificity of 99.68%.

If "C19" stands for the presence of the disease COVID-19 ("corona") and

+ and − respectively for a positive and negative result of the test, this means:

sensitivity: P( + | C19 ): 0.9652

specificity: P( - | not C19): 0.9968

That seems very high. But if the prevalence is only 1 in 10,000, i.e

prevalence: P( C19 ): 0.0001

This implies:

P( + ) = P( + | C19 ) P( C19 ) + P( + | not C19 ) P ( not C19 ) =

= 0.9652 x 0.0001 + 0.0032 x 0.9999 = 0.0033

and

P( C19 | + ) = ( P( + | C19 ) P( C19 ) ) / P( + ) =

= ( 0.9652 x 0.0001 ) / 0.0033 = 0.03

If 10,000 people are tested with this test, including probably 1 infected person,

then the infected person will almost certainly get a positive result.

But of the 9,999 uninfected people, 32 will get a false positive result.

The 9967 people with a negative result are almost certain that they are not infected.

But of the 33 with a positive result, only 1 is infected,

only it is unknown who that is.

So the chance that an individual is actually infected after a positive result

is only slightly more than 3% in this scenario.

This self-test is therefore of little use to determine whether you are infected with corona,

unless the prevalence is around 1% or higher.

(This information has been known and published since 20 june 2021 !)

Bye,

Skybuck ! =D

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