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Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of thóse whò have the disease, \(95 \%\) test possitive whèn a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P\) (has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease| positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

Short Answer

Expert verified
a. The tree has been correctly constructed. b. P(has disease and positive test) = 0.00095. c. P(positive test) = 0.10085. d. P(has disease | positive test) = 0.0094. The low probability of having the disease given a positive test is due to the rareness of the disease and higher possibility of false positives.

Step by step solution

01

Construct the Tree Diagram

First generate two branches from a root, one for 'Has Disease' and the other 'Doesn't Have Disease'. The probabilities for these branches are 0.001 and 0.999, respectively. Now, from each of these branches sprout two new branches, 'Positive Test' and 'Negative Test'. For the 'Has Disease' branch, assign a probability of 0.95 to 'Positive Test' and 0.05 to 'Negative Test'. For the 'Doesn't Have Disease' branch, the probability of 'Positive Test' is 0.10 and 'Negative Test' is 0.90.
02

Calculate P(has disease and positive test)

To solve this, multiply the probability of the 'Has Disease' branch with the 'Positive Test' sub-branch under it. P(has disease and positive test) = P(has disease) * P(positive test | has disease) = 0.001 * 0.95 = 0.00095.
03

Calculate P(positive test)

P(positive test) can be calculated by adding together: the probability of having a positive test when the disease is present and the probability of having a positive test when the disease is not present. P(Positive Test) = P(has disease and positive test) + P(no disease and positive test) = 0.00095 + (0.999 * 0.10) = 0.00095 + 0.0999 = 0.10085.
04

Calculate P(has disease | positive test)

Use the formula for conditional probability, P(A|B) = P(A and B) / P(B), to calculate the probability of someone having the disease given that they tested positive. P(has disease | positive test) = P(has disease and positive test) / P(positive test) = 0.00095 / 0.10085 = 0.0094.
05

Explanation of the Result

Though it may be surprising, only around 0.94% of individuals who test positive actually have the disease. This is because the disease is so rare in the population (0.1%). Thus, even though the test is fairly accurate, many of the positive results it generates will be false positive, leading to a relatively low conditional probability of having the disease given a positive test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tree Diagram
A tree diagram is a visual representation often used in probability theory to outline the possible outcomes of a series of related events. It resembles an upside-down tree with branches that split off, representing different scenarios or choices at each stage. For the diagnostic test accuracy problem, imagine drawing a trunk labeled 'Individual Tested'. From there, two branches represent 'Has Disease' (with a probability of 0.001) and 'Doesn't Have Disease' (with a probability of 0.999). Further branching occurs with 'Positive Test' and 'Negative Test' outcomes from each disease status.

Each branch is assigned a probability, indicating how likely that step is to occur. By tracing pathways from the trunk to the end of each branch, you can compute the overall probability of any sequence of events—key in understanding how likely an individual is to have the disease given a positive or negative test result.
Bayes' Theorem
Bayes' theorem is a powerful tool in probability theory that relates the conditional and marginal probabilities of statistical quantities. It's named after Thomas Bayes and provides a way to update our belief about the likelihood of an event based on new evidence.

Using the context of a diagnostic test, Bayes' theorem helps answer the question, 'What's the probability that a person has the disease, given that they tested positive?' In mathematical terms, it's expressed as:

\[\begin{equation}P(\text{has disease} \,|\, \text{positive test}) = \frac{P(\text{positive test} \,|\, \text{has disease})P(\text{has disease})}{P(\text{positive test})}\end{equation}\]
This equation uses prior probability (the initial likelihood of having the disease), the likelihood of the test result given the disease state, and the overall probability of a positive test result to calculate the posterior probability, which is the revised likelihood after gaining new evidence (the test result).
Probability Theory
Probability theory is a branch of mathematics that deals with quantifying uncertainty. Every event has a probability, ranging from 0 (impossible) to 1 (certain), that measures its likelihood of occurrence.

When dealing with medical diagnostics, probability theory enables us to predict the frequency of an outcome (like testing positive for a disease) and is crucial when assessing the effectiveness and reliability of a diagnostic test. It incorporates concepts such as conditional probability—calculating the chance of an event (like having a disease) given that another event has occurred (like a positive test result). With diagnostic tests, understanding false positives and negatives is integral, which is deeply rooted in probability theory.
Diagnostic Test Accuracy
The accuracy of a diagnostic test is measured by how well the test can distinguish between the presence and absence of a condition in patients. There are four main outcomes from a diagnostic test: true positives, true negatives, false positives, and false negatives. The terms sensitivity and specificity are often used to describe test accuracy. Sensitivity, or the true positive rate, is the proportion of sick individuals who correctly test positive. Specificity, or the true negative rate, is the proportion of healthy individuals who correctly test negative.

In the given exercise, even a diagnostic test with high sensitivity (95%) and high specificity (90%) can yield a low probability of actually having the disease after a positive test result. This counterintuitive outcome is due to the low base rate (prevalence) of the disease and highlights why it's important to consider both the test's accuracy and the pre-test likelihood of the disease when interpreting results.

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Most popular questions from this chapter

Five hundred first-year students at a state university were classified according to both high school GPA and whether they were on academic probation at the end of their first semester. The data are $$ \begin{array}{lcccc} && {\text { High School GPA }} \\ & 2.5 \text { to } & 3.0 \text { to } & 3.5 \text { and } & \\ \text { Probation } & <3.0 & <3.5 & \text { Above } & \text { Total } \\ \hline \text { Yes } & 50 & 55 & 30 & 135 \\ \text { No } & 45 & 135 & 185 & 365 \\ \text { Total } & 95 & 190 & 215 & 500 \\ \hline \end{array} $$ a. Construct a table of the estimated probabilities for each GPA-probation combination. b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of \(3.5\) or above? d. Are the two outcomes selected student has a bigh school GPA of \(3.5\) or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between \(2.5\) and \(3.0\) who are on academic probation at the end of the first semester. f. Estimate the proportion of those first-year students with high school GPAs \(3.5\) and above who are on academic probation at the end of the first semester.

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) -belongs to a medical clinic that always has a physician at each of stations 1,2, and \(3 .\) During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1, \mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station \(1 .\) a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1, then the nine in which \(\mathrm{P}_{1}\) goes to station 2, and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station 2 . e. Identify outcomes in each of the following events: \(B^{C}, C^{C}, A \cup B, A \cap B, A \cap C\).

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