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Math Help - Bayes Rule?

  1. #1
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    Bayes Rule?

    Please help me with this problem.

    The sensitivity of a test is the probability that a person having the disease will test positive; the specificity of a test is the probability that a person not having the disease will have a negative test.
    Experience indicates that its sensitivity is .0934 and that its specificity is 0.968. Furthermore, it is know that roughly 1 in 500 people has the disease.

    Determine the probability that a person testing positive actually has the disease.
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  2. #2
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    Quote Originally Posted by Breeegzkm View Post
    Please help me with this problem.

    The sensitivity of a test is the probability that a person having the disease will test positive; the specificity of a test is the probability that a person not having the disease will have a negative test.
    Experience indicates that its sensitivity is .0934
    ;
    So:

    P(positive|disease)=0.0932

    and that its specificity is 0.968. Furthermore, it is know that roughly 1 in 500 people has the disease.
    So:

    P(negative| \overline{disease})=0.968

    Determine the probability that a person testing positive actually has the disease.
    You are asked to find:

    P(disease|positive)= \frac{P(positive|disease)P(disease)}{P(positive)}

    and we are told that:

    P(disease)=0.002

    and so:

    P(positive)=P(positive|disease)P(disease)+P(positi  ve|\overline{disease})P(\overline{disease})

    and you have sufficient data to evaluate allof these probbailities

    RonL
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  3. #3
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    Quote Originally Posted by Breeegzkm View Post
    Please help me with this problem.

    The sensitivity of a test is the probability that a person having the disease will test positive; the specificity of a test is the probability that a person not having the disease will have a negative test.
    Experience indicates that its sensitivity is .0934 and that its specificity is 0.968. Furthermore, it is know that roughly 1 in 500 people has the disease.

    Determine the probability that a person testing positive actually has the disease.
    I always find it easiest to try and show these things visually using, say, a karnaugh table:


    I start off with: \, \begin{tabular}{l | c | c | c}  & D & D$\, '$ & \\ \hline +ve & a & b & a + b \\ \hline -ve & c & d & c + d \\  \hline & a + c & b + d & 1 \\ \end{tabular}


    Note that the answer to the question will be \frac{a}{a + b} \, , so the values of a and a + b are needed.


    From the given data it's known that:

    a + c = 1/500 = 0.002.

     \frac{a}{a + c} = 0.0934 \Rightarrow \frac{a}{0.002} = 0.0934 \Rightarrow a = 0.0001868

    \frac{d}{b + d} = 0.968


    Update the table: \, \begin{tabular}{l | c | c | c}  & D & D$\, '$ & \\ \hline +ve & 0.0001868 & b & 0.0001868 + b \\ \hline -ve & c & d & c + d \\  \hline & 0.002 & b + d & 1 \\ \end{tabular}


    Clearly c = 0.002 - 0.0001868 = 0.0018132 and b + d = 1 - 0.002 = 0.998


    Update the table: \, \begin{tabular}{l | c | c | c}  & D & D$\, '$ & \\ \hline +ve & 0.0001868 & b & 0.0001868 + b \\ \hline -ve & 0.0018132  & d & 0.0018132  + d \\  \hline & 0.002 & 0.998 & 1 \\ \end{tabular}


    From ealier, \frac{d}{b + d} = 0.968. Substitute b + d = 0.998: d = (0.968)(0.998) = 0.966064.

    Therefore b = 0.998 - 0.966064 = 0.031936.


    Update the table: \, \begin{tabular}{l | c | c | c}  & D & D$\, '$ & \\ \hline +ve & 0.0001868 & 0.031936 & 0.0321228 \\ \hline -ve & 0.0018132  & 0.966064 & 0.9678772 \\  \hline & 0.002 & 0.998 & 1 \\ \end{tabular}


    As a check of this table, note that the elements in the far right column add up to 1.

    Now you can answer the question and any other question asked.
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