Download Quantitative Methods Midterm: Anthrax Cases' Probability & Statistical Analysis and more Exams Introduction to Public Administration in PDF only on Docsity!
PUAF 610 QUANTITATIVE METHODS Fall 2001
MIDTERM EXAM SOLUTIONS
- In the wake of the 11 September hijackings, some airports have instituted very thorough examinations of randomly selected passengers. Yesterday, when I boarded a plane at London’s Heathrow airport, police randomly selected about one quarter of the passengers for a thorough search.
A. If, on the morning of 11 September, this procedure had been in place at Dulles airport, what is the probability that at least one of the four hijackers would have been selected for a thorough search? (10 pts)
P(person is searched) = 0.25; P(person is not searched) = 0.
P(at least one of 4 is searched) = 1 – P(all 4 are not searched)
= 1 – (0.75) 4 = 1 – 0.316 = 0.684 ≅ 68%
B. What assumption(s) did you make in part B? (3 pts)
That the searches are independent random events. If the hijackers were able to detect and take advantage of some pattern in the searches, or if they were somehow able to circumvent the searches, then the probability would be lower.
C. At Dulles airport, it appeared that security was selecting every 20th^ person for a thorough search. What is this method of sampling called? (3 pts)
Systematic sampling.
- After Senator Daschle received an anthrax-contaminated letter, quick tests of 150 nearby people indicated that 33 had been exposed. A far more accurate test indicat- ed that only 30 had been exposed; 3 of the original positives were “false positives.”
A. What were the false positive and false negatives rates for the quick test? (6 pts)
If we assume that the final test was perfectly accurate, 30 were exposed and 120 were unexposed. The false positive rate is the fraction of unexposed people that the quick test incorrectly indicated were exposed: 3/120 = 0.025 = 2.5%. The false negative rate is the fraction of exposed people who are incorrectly said to be unexposed; this is 0/30 = 0%.
P(+|exposed) = 1 P(–|exposed) = 0
P(+|unexposed) = 0.025 P(–|unexposed) = 0.
B. Initially, the lab had claimed that the false positive rate of the quick test was less than 1 percent. If this is true, what is the likelihood the 3 or more false positives would have been observed? (6 pts; full credit for the correct Excel formula).
What is the probability of observing 3 more positives in a group of 120 unexposed people, if the probability of a positive test is 1 percent?
P(3 or more) = 1 – P(2 or fewer) = 1 – BINOMDIST(2,120,0.01,1) = 0.12 = 12%
So the actual experience was not inconsistent with the lab’s claim.
C. The FBI has investigated 100 letters credible enough to trigger tests for anthrax. In 5 of these cases, tests confirmed that nearby people were exposed to anthrax.
Your office receives a letter containing white powder. A quick test indicates that you have been exposed. Using the false positive and negative rates estimated in part A, what is the probability that the accurate test will show that you were, in fact, exposed to anthrax? (10 pts; if you did not answer part A, assume false positive and false negative rates of 5 percent).
( )
( ) ( ) ( ) (^ )^ ( ) (^ ) ( )( ) ( )( ) ( )( )
P exposed P exposed P exposed + P exposed P exposed P unexposed P unexposed
1 0.05 (^) 0. 0.678 68% 1 0.05 0.025 0.95 0.
Also with a tree: Test negative
exposed
P(–|E) = 0
P(E) = 0.
Test positive
letter p(+|E) = 1
received Test positive
unexposed
P(+|Ec^ ) = 0.
P(Ec^ ) = 0. Test negative P(–|Ec^ ) = 0.
If, lacking faith in your answers to part A, you assumed P(+|E) = 0.95 and P(+|Ec^ ) = 0.05, then P(E|+) = 0.5.
- The only clinical trial of the anthrax vaccine occurred at a mill where workers were at risk of cutaneous infection from anthrax spores in animal hides. Of the 1, workers, 720 workers received the vaccine; the remainder received a placebo. In the subsequent 3-year period, 26 cases of cutaneous anthrax were diagnosed among the workers; 5 of these cases were workers that had been vaccinated.
A. Did this clinical trial show that the vaccine was effective in preventing cutaneous anthrax infections? How powerful is the evidence? (20 pts)
( ) (^ )^ (^ )
p v
p v p v
p v p v
ˆ 21 ˆ^5
p 0.0362 p 0.00694 p 0. 580 720 1300
ˆ ˆ^1 1 1
SE p p p 1 p 0.02 0.98 0. n n 580 720
pˆ^ pˆ (^) 0.0362 0.00694 0. Z 3. SE pˆ^ pˆ^ 0.00781^ 0.
p NORMSDIST 3.75 0.00009 (one-tailed;
= − = two-tailed also fine)
I’d call this “convincing” evidence that the vaccine is efficacious.
B. Based on this data, what is the best estimate of the effectiveness of the vaccine? Effectiveness = (cases expected - cases observed)/(cases expected). (5 pts)
26.07 5 cases expected = (0.0362)(720) = 26.07; effectiveness 0.808 81%
Cases of inhalation anthrax are so rare that clinical trials, such as the one presented in question 2, cannot establish the efficacy of the vaccine. Deliberate exposure of humans to anthrax is out of the questions, so researchers have had to rely on animal tests to establish the effectiveness of the vaccine in preventing inhalation anthrax infections. Of 95 rhesus monkeys who were vaccinated and then exposed to lethal concentrations of spores, 5 died. Give a one-sided 95% confidence interval for the effectiveness of the vaccine, and state your result in plain English. (20 pts)
90 p 1( p^ ) p 1ˆ^ ( pˆ) ( 0.947^ )( 0.053)
pˆ^ 0.947 SE pˆ 0. 95 n n 95
For 95% confidence, one-tailed, Zc = NORMSINV(0.05) = –1.
p > pˆ^ − Z SE pc ( ˆ ) = 0.947 − 1.645 0.229( )= 0.947 − 0.0377 = 0.910 ≅91%
“We are 95% sure that the vaccine is at least 91% effective (i.e., at least 91% of monkeys exposed to lethal doses would survive).”
