Stat 665 (Spring 2011)
Kaizar
Solutions 2
Exercises
40 points total
1. (8 points) Agresti 2.16
(a) Response: lung cancer
Explanatory: have smoked (or not)
(b) case-control study
(c) No. This would be looking at the probability of cancer
conditional on the smoking status
,
whereas we fixed the number of cancer/no cancer participants in our study. Therefore,
we can estimate the probability of smoking conditional on cancer, but not the other way
around. If we were to attempt to use Bayes rule to turn the conditioning around:
P
(
C
|
S
) =
P
(
S
|
C
)
P
(
C
)
P
(
S
)
we would need to know the probability of having cancer – a value that we can not estimate
with this design.
(d) The odds ratio is:
ˆ
θ
=
n
11
n
22
n
12
n
21
=
688
*
59
21
*
650
= 2
.
97
Since the odds ratio is interpretable in either the row conditional on column, or column
conditional on row direction, I interpret the odds ratio using the explanatory/response
variables from part (a). The estimated odds of cancer conditional on smoking are 2.97
times larger than the odds of cancer conditional on not smoking. Thus, smoking is quite
dangerous in terms of increasing the risk of cancer incidence.
2. (8 points)
Effect Size Measures and Tests of Independence
This question relates to data reported in:
Nasar J, Hecht P and Wener R. ‘Call if You Have Trouble’: Mobile Phones and Safety among College Students.
International Journal of Urban and Regional Research 2007; 31(4): 863-73.
For this research, a survey center at The Ohio State University randomly sampled 264 students
who carry cell phones to answer a survey (consider this number to be fixed). Of these, 45.9%
happened to be female. The results section reported:
A higher percentage of females (42.1%) than males (27.9%) reported walking [with
their cell phone] somewhere after dark they would not normally go [if they did not
have their cell phone with them].
1
This
preview
has intentionally blurred sections.
Sign up to view the full version.
In this question, I would like you to analyze the data to determine if there is a relationship
between gender and whether or not the students would walk somewhere they would not normally
go.
(a)
What kind of sampling distribution does this study’s data approximatly follow?
This study is approximately a multinomial study, since only the total sample size is fixed.
(b)
From the reported conditional and marginal sample proportions, reconstruct the approxi-
mate 2x2 table of counts for gender and walking. (Note that due to rounding error, this
will not be exact, but your table should only contain whole numbers!)
I can estimate the following proportions using the given data:
π
f
+
=
0
.
459
a marginal probability for gender
π
f
=
0
.
421
the conditional probability of walking given female
π
m
=
0
.
279
the conditional probability of walking given male
Using these, I can also derive the probabilities:
π
m
+
=
1
-
π
f
+
= 0
.
541
a marginal probability for gender
π
fw
=
P
(
W
|
F
)
P
(
F
) =
π
f
π
f
+
= 0
.
193
the joint probability of walking and female
π
fn
=
P
(
F
)
-
P
(
W, F
) =
π
f
+
-
π
fw
= 0
.
266
the joint probability of not walking and female
π
mw
=
P
(
W
|
M
)
P
(
M
) =
π
m
π
m
+
= 0
.
151
the joint probability of walking and male
π
mn
=
P
(
M
)
-
P
(
W, M
) =
π
m
+
-
π
mw
= 0
.
390
the joint probability of not walking and male

This is the end of the preview.
Sign up
to
access the rest of the document.
- Spring '10
- Staff
- Conditional Probability, Probability, Probability theory, γ, Πf
-
Click to edit the document details