Probability practice questions: Flashcards
In a recent published study, the results of exercise tolerance tests were compared among subjects with
and without coronary artery disease (CAD). Of 1400 male subjects studied, 1000 were found to have
CAD. Of those, 800 had a positive exercise tolerance test. Of the 400 men who did not have CAD, 100
had a positive test.
- The specificity of the exercise tolerance test for CAD is:
A. 40% (200/500) B. 25% (100/400) C. 75% (300/400) D. 60% (300/500) E. None of the above
C
In a recent published study, the results of exercise tolerance tests were compared among subjects with
and without coronary artery disease (CAD). Of 1400 male subjects studied, 1000 were found to have
CAD. Of those, 800 had a positive exercise tolerance test. Of the 400 men who did not have CAD, 100
had a positive test.
The false positive rate is: A. 40% (200/500) B. 25% (100/400) C. 75% (300/400) D. 60% (300/500) E. None of the above
(B) False positive rate = P(Test+ | CAD-) = 100/400 = 25%. Or, FP rate = 1-specificity = 1-
.75=.25.
In a recent published study, the results of exercise tolerance tests were compared among subjects with
and without coronary artery disease (CAD). Of 1400 male subjects studied, 1000 were found to have
CAD. Of those, 800 had a positive exercise tolerance test. Of the 400 men who did not have CAD, 100
had a positive test.
The negative predictive value is: A. 40% (200/500) B. 25% (100/400) C. 75% (300/400) D. 60% (300/500) E. None of the above
(D) Negative predictive value = P(CAD- | Test-) = 300/500 = 60%
In a recent published study, the results of exercise tolerance tests were compared among subjects with
and without coronary artery disease (CAD). Of 1400 male subjects studied, 1000 were found to have
CAD. Of those, 800 had a positive exercise tolerance test. Of the 400 men who did not have CAD, 100
had a positive test.
The positive predictive value is: A. 40% (200/500) B. 25% (100/400) C. 75% (300/400) D. 60% (300/500) E. None of the above
(E) Positive predictive value = P(CAD+ | Test+) = 800 / 900 = 89%
You are considering using the exercise tolerance test to screen asymptomatic males for the
presence of CAD. If the above data on test sensitivity and specificity hold and the prevalence of CAD in
asymptomatic men is 4%, what percent of those with a positive exercise tolerance test are estimated to
have CAD?
A. 89%
B. 80%
C. 12%
D. 20%
E. 100%
(C) Specificity = 75% (from question 1). Sensitivity = P(T+ | D+) = 800/1000 = 80%.
Prevalence = .04. Use Bayes’ Theorem (not shown) or construct the following 2x2 table:
In a discussion about the use of serum prostatic acid phosphatase (P.A.P.) as an indicator of
cancer of the prostate, a clinician was concerned that too many men with a positive P.A.P test had no
signs, symptoms, or other evidence of disease. The clinician is concerned about the
A. specificity of the P.A.P. test
B. negative predictive value of the P.A.P. test
C. high prevalence of cancer of the prostate
D. positive predictive value of the of the P.A.P. test
(D) The problem is that many men with positive tests are found to not have disease. This
would imply that the positive predictive value of the test is low. Specificity and negative predictive value
deal with negative test results, and the prevalence of the disease is likely to be low in order to get such a
low positive predictive value.
The false positive rate associated with using a cutoff MANTRELS score of 5 or greater (score less
than 5 would be considered a “negative” result) is:
MANTRELS score/ Episodes with appendicitis/ Episodes without appendicitis ≤ 4 0 27 5 6 12 6 6 13 ≥7 104 21 Total 116 73
A. 63% (46/73) B. 72% (116/162) C. 61% (116/189) D. 28% (46/162) E. 100% (116/116)
(A) False positive rate is P(T+ | D-). There are 73 episodes without appendicitis (disease
negative). Of these, 12 + 13 + 21 = 46 have a positive test (MANTRELS score of 5 or greater). 46/73 =
63%.
In the Bond study, the prevalence of appendicitis was 61%. If the prevalence of appendicitis
were 80% (as in the original Alvarado study) and the cutoff remained at 5 or greater for deciding
whether to perform an appendectomy, what percent of appendectomies would be performed on
patients without appendicitis?
MANTRELS score/ Episodes with appendicitis/ Episodes without appendicitis ≤ 4 0 27 5 6 12 6 6 13 ≥7 104 21 Total 116 73
A. 37.0% B. 28.4% C. 0% D. 13.6% E. 22.0%
(D) The question asks for P(D- | T+), the probability of no appendicitis in a patient with a
positive test who was selected for appendectomy. P(D- | T+) = 1 – P(D+ | T+) or 1- positive predictive
value (PV+). You can calculate PV+ from sensitivity, specificity and prevalence by constructing a 2x2
table (not shown) or using Bayes’ theorem (Campbell p. 53). Prevalence is given as 0.8. Since all
episodes with appendicitis had scores of 5 or greater, sensitivity is 100%. Of 73 episodes without
appendicitis, 27 had a score below 5, so specificity = 27/73 = 37%. Using Bayes’ theorem, PV+ = (.8 x 1.0)
/ [(.8 x 1.0) + (1-.37)(1-.8) = 13.6%.
MANTRELS score/ Episodes with appendicitis/ Episodes without appendicitis ≤ 4 0 27 5 6 12 6 6 13 ≥7 104 21 Total 116 73
Which MANTRELS score would you use as a cutoff to define a negative result if you were attempting to “rule out” appendicitis? A. ≤ 4 B. ≤ 5 C. ≤ 6 D. ≤ 7 E. ≥ 7
(A) To rule out appendicitis, you want to make sure that there are very few or no false
negatives. Using a MANTRELS score of ≤ 4 to “exclude” appendicitis gives the fewest false negatives
It is generally accepted that in any decision regarding acceptable levels of test sensitivity and
specificity involves weighing the consequences of an undetected cases against erroneous classification
of healthy persons as diseased. With this in mind, which statement is (generally) true when the penalty
associated with missing a case is high?
A. Sensitivity should be decreased at the expense of specificity
B. Sensitivity should be increased at the expense of specificity
C. Specificity should be decreased at the expense of sensitivity
D. Specificity should be increased at the expense of sensitivity
E. Positive predictive value should be maximized
(B) If there is a desire to minimize missing cases, then one would like to minimize false
negatives. High sensitivity (P(T+ | D+)) means a low false negative rate (P(T- | D+)).
Pregnant women whose fetuses have certain severe defects (e.g. spina bifida) have particularly
high concentrations of alpha-fetoprotein (AFP) in their blood. Data indicate than of every 1000 women
tested, 50 will have elevated levels of AFP, but only 2 of those will have fetuses with severe defects.
From this information, one can determine:
A. negative predictive value = 48/950
B. specificity = 48/50
C. prevalence of birth defects = 2/1000
D. positive predictive value = 2/50
(D) Among the 50 who test positive, 2 will have the condition, so P(D+ | T+) = positive
predictive value = 2/50. Because there is no information about true negatives, negative predictive value
and specificity cannot be calculated. The prevalence cannot be calculated because the number of false
negatives is not provided.
Which of the following are influenced by the prevalence of the disease or condition under
study?
A. sensitivity
B. false negative rate
C. negative predictive value
D. test-retest reliability (reproducibility)
E. likelihood ratio
(C) Only predictive values are influenced by the prevalence of disease. The other answers
are properties of the test that do not depend on the prevalence.
Your patient has an abnormally high pathological finding: a high blood pressure or high blood
cholesterol concentration or severe rheumatic pain or long transit time in diverticulitis. You treat the
patient with a new drug, and are pleased when the BP, cholesterol level, pain, or transit time is reduced.
You treat the next patient with a similar condition with a placebo and observe a similar result. Which of
the following is the best explanation for these results?
A. The second patient’s improvement was probably due to the well known “placebo effect”.
B. The placebo and new drug are probably equally effective.
C. You may be dealing with the “regression to the mean” effect.
D. There may have been a laboratory error in the case of the placebo treated patient.
E. The intra-subject variability is too large to reliably compare the outcomes
(C) Regression-to-the-mean describes the phenomenon where extreme initial values of a
measurement tend on average to be closer to the mean upon repeat observations. In both patients, we
observed an abnormally high initial reading that returned closer to mean of all patients after a second
reading was done. While the other answers may be plausible, the regression-to-the-mean is the best
explanation for this change.
If a clinical lab uses the mean +/- 2 standard deviations to obtain the “reference interval”, then
which of the following statements is NOT TRUE?
A. the distribution should be approximately normal.
B. the selected sample should be representative of the reference population.
C. the sample size should be large enough so that the mean and standard deviation of the
reference population (estimated by the sample mean and sample standard deviation) are
reliably estimated.
D. 5% of persons should be diseased.
E. In general, the reference interval should apply only to results from that lab
(D) Only for distributions that have an approximately Gaussian (normal) distribution (and not
for all distributions in general) should you expect 95% of the values to fall within ± 2 standard deviations
of the mean. To get estimates of the mean and standard deviation of the reference population that make
sense, the sample ought to be representative of the reference population and large enough to give valid
estimates. Because of potential variability among different labs, reference intervals are often applicable
only to result from that particular lab.
In clinical medicine the 95% “reference interval” for a given measurement (e.g., cholesterol)
A. is determined so that only 5% of diseased persons will have a value within the given
range.
B. can be determined only if the measurement has a normal (Gaussian) distribution.
C. is determined so that 95% of diseased persons will have values outside the given range.
D. is the range in which it is expected that 95% of values (persons) from a specified healthy
reference population will fall.
E. is always equal to the mean ± 2 standard deviations
(D) The normal range is usually defined to be an interval (lower and upper values) in which
95% of the values of some reference (healthy) population would be expected to fall. Nothing can be said 8
specifically about diseased persons. The normal range can be computed using the 2.5 and 97.5
percentiles if the data are not Gaussian and the “mean ± 2 standard deviations” method is invalid
