Statistical Evidence Evaluation, fall semester 2013
Solutions to Exercises 1
1. Correction: The number of cars is of course five and not six.
Pr ( A) = 3 5
Pr (B ) = 2 5
Pr (C ) = 4 5
Pr ( A, B ) = 1 5 ≠ Pr ( A) ⋅ Pr (B )(= (3 5) ⋅ (2 5) = 6 25)
⇒ A and B are non - independent events
Pr ( A C ) = 2 4 = 1 2 ; Pr (B C ) = 2 4 = 1 2
Pr ( A, B C ) = 1 4 = (1 2 ) ⋅ (1 2 ) = Pr ( A C )⋅ Pr (B C )
⇒ A and B are conditionally independent given C
2. Let A = Dye is present an B = Method gives positive detection
Pr ( A) = 0.001
Pr (B A) = 0.99 ⇒ Proportion of false negatives, p fn = 0.01
Pr (B A) = 0.02 = Proportion of false positives, p fp
Pr ( A B ) =
Pr (B A)⋅ Pr ( A)
=
Pr (B A)⋅ Pr ( A) + Pr (B A)⋅ Pr ( A)
(
1 − p )⋅ 0.001
=
(1 − p )⋅ 0.001 + p ⋅ 0.999
fn
fn
fp
0
0.2
1
0.8
0.4 0.6
1
0.8
0.6
0.4
Obviously it is the proportion of
false positives that is the most
crucial.
0.2
post prob
01
0.8
0.6
0.4
false pos 0.2
false neg
0
3. a) Gender: Two states, no numerical relationships them between
nominal scale
b) Temperature in F: Continuous variable but no well-defined zero
interval scale
c) Age: Even if usually represented with integral values it is still a
continuous variable and it has a well-defined zero
ratio scale
d) Acoustic intensity level: logarithmic scale
unequal distances
between consecutive values
ordinal scale
4. Let A = It has rained that day ; B = The parking place outside Mr
Johnson’s house is wet ; C = Somebody has washed their car up the
road
The information given is used to assign probabilities:
Pr ( A) = 17 30
Pr (B A) = 4 5
Pr (B C ) = 1 2
Pr (C A) = 0
Pr ( A B ) + Pr (C B ) = 1
17 out of 30 days
four out of five times
50 % of the cases
No car washing on rainy days
No other explanation than rain or car washing
Likelihood of rain when parking place is wet: Pr (B | A ) = 4/5
Pr ( A B ) Pr (B A) Pr ( A) 4 5 17 30 68 1
=
⋅
=
⋅
= ⋅
Pr (C B ) Pr (B C ) Pr (C ) 1 2 Pr (C ) 75 Pr (C )
Pr (A B ) + Pr (C B ) = 1 ⇒
Pr (A B )
68 1
= ⋅
1 − Pr (A B ) 75 Pr (C )
68 1
⋅
1
75 Pr (C )
⇒ Pr ( A B ) =
=
68 1
75
⋅
+ 1 1 + ⋅ Pr (C )
75 Pr (C )
68
We know that Pr(C | A) = 0
⇒ Pr (C ) ≤ Pr ( A) =
13
30
1
⇒ Pr ( A B ) ≥
≈ 0.67
75 13
1+ ⋅
68 30
5. Let π = The probability that the coin land “heads”
The hypotheses:
H0: “The coin is fair ⇔ π = 0.5
H1: π = 0.6
Experiment: Five tossings gives four heads
Model: Number of heads is Bin (5, π )
5
⇒ L(π Data ) =   ⋅ π 4 ⋅ (1 − π )
 4
 5
5
5
⇒ L(H 0 Data ) =   ⋅ 0.54 ⋅ 0.5 =   ⋅ 0.55 ; L(H1 Data ) =   ⋅ 0.6 4 ⋅ 0.4
 4
 4
 4
Simple hypothesis
The Bayes factor is the likelihood ratio
5
  ⋅ 0.55
L(H 0 Data )
4
0.55

LR =
=
=
≈ 0.60
4
L(H1 Data )  5 
0.6 ⋅ 0.4
  ⋅ 0.6 4 ⋅ 0.4
 4
Slight evidence against H0
6. Using π = The probability that the coin land “heads” the hypothesis
are
H0: π = 0.5
H1: π > 0.5
i.e. simple null hypothesis and composite alternative hypothesis
⇒B=
L(H 0 Data )
∫π
∈H1
L(π Data )⋅ p (π H1 ) dπ
=
Uniform prior
for π under H 1
=
5
  ⋅ 0.54 ⋅ (1 − 0.5)
4
0.55

=
=
=
1
4
1 5
 1 
4
2 ⋅ ∫ π ⋅ (1 − π )dπ


(
)
π
π
π
⋅
⋅
1
−
⋅
d


0.5
∫0.5  4 
1
−
0
.
5


=
0.55
2 ⋅ B(5,2) ⋅ ∫
1
0.5
π 4 ⋅ (1 − π )
B(5,2 )
dπ
0.55
=
2 ⋅ B(5,2) ⋅ (1 − Fbeta ( 5, 2) (0.5))
where B is the Beta function and Fbeta(5,2) is the cdf of a beta(5,2)distribution
Use R to find B(5,2) and Fbeta(5,2)(0.5):
B(5,2)
Fbeta(5,2)(0.5)
0 .5 5
⇒B≈
≈ 0.53
2 ⋅ 0.0333 ⋅ (1 − 0.1094)
Slight evidence against H0 (a little bit stronger than in task 5)
7. The hypotheses are
Hp : “The particular screwdriver was used to break the lock”
Hd : “Another screwdriver was used to break the lock”
The hypotheses are not formulated in terms of any parameter. Instead,
distinct probabilities are used to form the likelihoods.
The particular screwdriver would leave precisely those marks on the
lock that has been observed
(
)
⇒ Pr Marks observed H p = 1
⇒ L(H p Marks observed) = 1
Now, Hd = Hd1 ∪ Hd2 where
Hd1 = “Another screwdriver of the same type as the particular one
was used to break the lock”
Hd2 = “Another screwdriver of a different type than the particular
one was used to break the lock”
It is known that
Pr (H d 2 H d ) ≈ 0.99 and Pr (H d 1 H d ) ≈ 0.01
discarding that the particular screwdriver has been taken out of the
population of screwdrivers.
Now,
L(H d 1 Marks observed) ≈ L(H p Marks observed) = 1
since screwdrivers of the same type as the particular one are expected to
give the same type of marks.
Moreover,
L(H d 2 Marks observed ) = Pr (Marks observed H d2 ) ≈ 0.0002
⇒B=
=
L(H p Marks observed)
L(H d 1 Marks observed)⋅ Pr (H d 1 H d ) + L(H d 2 Marks observed)⋅ Pr (H d 2 H d )
1
≈ 98.0
1 ⋅ 0.01 + 0.0002 ⋅ 0.99
=