Solutions to Practice Final; Statistics 371;
Fall 2011; Professor Wardrop
1. First we note that in the collapsed table
ˆ
p
1
=0
.
34
is smaller than
ˆ
p
2
=0
.
356
.Fo
r
Simpson’s Paradox to occur, this inequality
must be reversed in both component tables.
In the frst component table, we need
c/
100
<
60
/
200
or
c<
30
or
c
≤
29
.
In the second component table, we need
d/
150
<
42
/
100
or
d<
63
or
d
≤
62
.
But we also need the component tables to
be consistent with the collapsed table; i.e.
we need
c
+
d
=89
.
So, we have three conditions:
c
≤
29
,
d
≤
62
and
c
+
d
=89
.Cana
l
lthreecond
i
t
ions
be satisfed? Yes; there are three possible
answers:
(
c
=29
;
d
=60
)
;
(
c
=28
;
d
=
61)
;and
(
c
=27;
d
=62)
.
2. We note that in the collapsed table
ˆ
p
1
=
0
.
362
is smaller than
ˆ
p
2
=0
.
42
.ForS
imp
son’s Paradox to occur, this inequality must
be reversed in both component tables.
In the frst component table, we need
c/
35
<
21
/
75
or
c<
9
.
8
or
c
≤
9
.
In the second component table, we need
d/
65
<
26
/
55
or
d<
30
.
7
or
d
≤
30
.
But we also need the component tables to
be consistent with the collapsed table; i.e.
we need
c
+
d
=42
.
So, we have three conditions:
c
≤
9
,
d
≤
30
and
c
+
d
=42
.C
a
na
l
lt
h
r
e
ec
o
n
d
i

tions be satisfed? No, because the largest
c
+
d
can be and still keep both reversals is
39. On the exam it would su±fce to say:
It is impossible ±or
c
≤
9
,
d
≤
30
and
c
+
d
=42
;Iw
i
l
lassumeyouknowwhy
.
(You won’t have time ±or long essay an
swers on the exam.)
3. I will defne population 1 to be ±emales and
population 2 to be males. Assume inde
pendent random samples ±rom these pop
ulations. Based on the data given, the 95%
CI ±or
p
1

p
2
is
(0
.
407

0
.
364)
±
1
.
96
±
0
.
407(0
.
593)
450
+
0
.
364(0
.
636)
538
=
0
.
043
±
1
.
96(0
.
0311) = 0
.
043
±
0
.
061 =
[

0
.
018
,
0
.
104]
.
4. I will defne population 1 to be males and
population 2 to be ±emales. Assume inde
pendent random samples ±rom these popu
lations. Based on the data given, the 95%
CI ±or
p
1

p
2
is
(0
.
257

0
.
173)
±
1
.
96
±
0
.
257(0
.
743)
537
+
0
.
173(0
.
827)
446
=
0
.
084
±
1
.
96(0
.
0260) = 0
.
084
±
0
.
051 =
[0
.
033
,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document6.
(a) The frequency of the interval is 5.
(b) It cannot be a relative frequency his
togram because a relative frequency
cannot exceed one.
(c) The relative frequency of the interval
is its area:
5(0
.
10) = 0
.
50
.Thu
s
,i
t
s
frequency is
0
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 hanlon
 Statistics, The Table, Histogram

Click to edit the document details