Testing and False Positives

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Bill Foster

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7 years, 11 months ago

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Anne-Sophie Kaloghiros 5 years, 4 months ago

Gave some feedback: Has some problems

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Sonny Li 7 years, 4 months ago

Gave some feedback: Has some problems

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Bill Foster 7 years, 11 months ago

Gave some feedback: Ready to use

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Bill Foster 7 years, 11 months ago

Published this.

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Bill Foster 11 years ago

Created this as a copy of Bayes' Theorem and testing.

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Question statement

{test} is $\var{effective}\%$ {eff} in {finding} when {present}.

However, {it} also yields a "false positive" result for $\var{fal}\%$ of {notpresent} tested.

If $\var{user}\%$ of {something} {havecondition}, what percentage of the population has the condition, given that their test is positive?

(Thanks to Dr Bill Kirkup for providing the methodology and tabular format for this approach.)

Give any introductory information the student needs.

			Name	Type	Generated Value

thisprev	integer	1111
thatprev	integer	6
notpresent	string	healthy people
poswithout	number	3996.4
test	string	A laboratory test
it	string	the test
nus	number	0.9991
falseresult	string	a false positive result
percentprob	number	16.7
effective	integer	89
indef	string	a person
tol	number	0.001
eff	string	effective
used	string	has the disease
poswith	number	801
fal	number	0.4
effec	decimal	dec("0.89")
morethaneven	number	0.08
prob	number	0.167
havecondition	string	actually have the disease
pop	integer	1000000
something	string	the population
user	number	0.09
us	number	0.0009
totalwithout	number	999100
present	string	it is present
fa	number	0.004
finding	string	detecting a certain disease
totalwith	number	900
ans1	number	0.000801
testspositive	string	his test is positive
probpositive	number	0.0047974

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
round(100/user)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

1111

← Depends on:

This variable doesn't seem to be used anywhere.

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	With Condition	Without Condition	Totals
Test Positive	PosWith=Gap 0.	PosWithout=Gap 1.
Test Negative
Totals	TotalWith=Gap 2.		1000000

Suppose we are dealing with a population of $1,000,000.$

In the following enter all numbers in the table to one decimal place. Do not enter large numbers with commas i.e. enter 12,345.6 as 12345.6.

Given the percentage $\var{user}\%$ of this population with the condition find the expected total number, TotalWith, with the condition and enter this number in the appropriate answer box in the table above. Note that the number of the population without the condition is automatically filled in.

Now using the data on the effectiveness of the test enter the expected number, PosWith, of the TotalWith that test positively in the table.

Note that the expected number with the condition that test negatively is automatically filled in.

The last table entry is to fill in the number, PosWithout, of the population expected to test positively but do not have the condition. Use the data on false positives for this.

The population without the condition and that test negatively is filled in for you as well as totals for positive and negative tests.

Using this data, find the percentage of those that have the condition given that they have tested positively:

Percentage=Gap 3.% (Enter to 1 decimal place).

Advice

We present two solutions:

Solution a) uses the data in the table based on a population of $1,000,000$ in order to calculate the probability that a person has the condition given that the test is positive.

Solution b) uses conditional probability and Bayes' Theorem without reference to a specific population.

Solution a). Using a Table

We have that the number of people in a population of $1,000,000$ expected to have the condition is given by:

\[TotalWith =\frac{ \var{user}}{100} \times 1000000=\var{totalwith}.\]

Note that we have the expected number of the population without the condition to be $\var{totalwithout}$.

Of the $\var{totalwith}$ with the condition we can calculate the expected number whose condition will be detected by the test:

\[PosWith = \frac{\var{effective}}{100}\times \var{totalwith}=\var{poswith}.\]

Finally of the $\var{totalwithout}$ without the condition we are told to expect $\var{fal}$% to be wrongly diagnosed as with the condition i.e.

\[PosWithout = \frac{\var{fal}}{100}\times \var{totalwithout}=\var{poswithout}.\]

We now can calculate the number of the population which are expected to test positively:

\[TotalPos = \var{poswith}+\var{poswithout}=\var{poswith+poswithout}.\]

Of these $\var{poswith}$ actually have the condition so the percentage of those with the condition as against the total of those with positive tests is:

\[Percentage=100 \times\frac{\var{poswith}}{\var{poswith+poswithout}}=\var{percentprob}\%.\] to one decimal place.

This gives the percentage of those expected to have the condition after a positive test.

We conclude that if the results of a person's test is positive then there is a $\var{percentprob}\%$ chance that he/she has the disease i.e. the chances have increased from around 1 in $\var{thisprev}$ to about 1 in $\var{thatprev}$.

Note that if $PC\%$ is the percentage of people with the condition in the population, $FP\%$ is the percentage of false positives and $T\%$ is the effectiveness of the test then in order to for a positive test to indicate that there is more than an evens chance of having the condition we have:\[FP \lt \frac{PC\times T}{100-PC}.\]

If we take the above data for $PC$ and $T$ then for more than an evens chance of having the condition we would have to have:

\[FP \lt \frac{PC\times T}{100-PC}=\frac{\var{user}\times \var{effective}}{100-\var{user}}=\var{morethaneven}\%.\]

The above diagram shows the relationship between the percentage with the condition after testing positive and the percentage of false positives. This assumes that the test has a fixed $\var{effective}$% success rate of detecting the condition and that there is a fixed $\var{user}$% of the population with the condition.

You can drag the red point along the curve to see how the percentages change with the false positive percentage.

Solution b). Bayes' Theorem.

Consider the events

\[\begin{align} D &= \text{the person has the disease.}\\
ND &= \text{the person does not have the disease.}\\
PR&=\text{the test result is positive.}\end{align}\]

We are given that

\[P(PR|D)=\var{effec},\;P(PR|ND)=\var{fa},\;P(D)=\var{us}\]

Since the events $D$ and $ND$ form a partition of the sample space, we can use Bayes' Theorem to calculate

\[\begin{align}P(D|PR)&=\frac{P(PR|D)\times P(D)}{P(PR|D)\times P(D)+P(PR|ND)\times P(ND)}\\
&=\frac{\var{effec}\times \var{us}}{\var{effec}\times \var{us}+\var{fa}\times \var{nus}}\\
&=\var{percentprob/100}\;\;\text{to 3 decimal places}\end{align}\]

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This question is used in the following exam:

Middlesex Workshop by Bill Foster in Bill's workspace.

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Anne-Sophie Kaloghiros 5 years, 4 months ago

Sonny Li 7 years, 4 months ago

Bill Foster 7 years, 11 months ago

Bill Foster 7 years, 11 months ago

Bill Foster 11 years ago

Error in variable testing condition

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Generated value: `integer`

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Testing and False Positives

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Anne-Sophie Kaloghiros 5 years, 4 months ago

Sonny Li 7 years, 4 months ago

Bill Foster 7 years, 11 months ago

Bill Foster 7 years, 11 months ago

Bill Foster 11 years ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`