PubH 6450 – Fall 2008 – Test 2 Review Problems  Answer Key
A.
1. Two random samples of parents/kids (participated in program, did not participate)
2. Binary outcome: fully vaccinated, yes or no
3. Large samples, all of the following
≥
5:
n
1
ˆ
p
1
=
X
1
= 73
n
1
(1

ˆ
p
1
) =
n
1

X
1
= 20
n
2
ˆ
p
2
=
X
2
= 30
n
2
(1

ˆ
p
2
) =
n
2

X
2
= 41
4. Twosample normal approximation to binomial,
H
0
:
p
1
=
p
2
vs.
H
a
:
p
1
> p
2
ˆ
p
1
= 73
/
93 = 0
.
785
ˆ
p
2
= 30
/
71 = 0
.
423
ˆ
p
= (73 + 30)
/
(93 + 71) = 0
.
628
z
=
0
.
785

0
.
423
p
(0
.
628)(0
.
372)(1
/
93 + 1
/
71)
= 4
.
76
p

value
= Pr(
Z >
4
.
76)
≈
0
We have substantial evidence showing greater vaccination rates for children of parents
who participated in the education program compared to children of parents who did not
participate in the education program.
1
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1. Sampled and matched on disease status and then exposure determined:
Case Control
Study
2. Binary outcome: exposure to county fair, yes or no
3. Large samples, all counts a, b, c and d
≥
10
4. 95% Conﬁdence Interval for the Odds Ratio
Odds
Exposed

Rash
= 25
/
50 = 0
.
50
Odds
Exposed

Norash
= 16
/
59 = 0
.
27
ˆ
OR
= 0
.
50
/
0
.
27 = 1
.
85
Ln
(
ˆ
OR
) =
ln
(1
.
85) = 0
.
6152
Ln
(
ˆ
OR
)
±
Z
*
p
1
/a
+ 1
/b
+ 1
/c
+ 1
/d
0
.
6152
±
1
.
96
p
1
/
25 + 1
/
16 + 1
/
50 + 1
/
59
0
.
6152
±
1
.
96(0
.
3734)
= (

0
.
1167
,
1
.
3471)
The conﬁdence interval above is on the log scale. Back transform to the original scale
by taking the exponential of the lower and upper bound.
[
exp
(

0
.
1167)
, exp
(1
.
3471)] = (0
.
89
,
3
.
85)
The estimate and 95% conﬁdence interval for the odds ratio are 1.85 (0.89, 3.85). This
conﬁdence interval contains the null value of 1. There is not suﬃcient evidence to
conclude exposure to the county fair is associated with the rash.
2
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 Fall '10
 AndyMugglin
 Normal Distribution, Binomial distribution, Simple random sample, Student's ttest, binary outcome

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