287
Chapter 14: Analysis of Categorical Data
14.1
a.
H
0
:
p
1
= .41,
p
2
= .10,
p
3
= .04,
p
4
= .45 vs.
H
a
: not
H
0
.
The observed and expected
counts are:
A
B
AB
O
observed
89
18
12
81
expected
200(.41) = 82 200(.10) = 20
200(.04) = 8 200(.45) = 90
The chi–square statistic is
90
)
90
81
(
8
)
8
12
(
20
)
20
18
(
82
)
82
89
(
2
2
2
2
2
−
−
−
−
+
+
+
=
X
= 3.696 with 4 –1 = 3
degrees of freedom.
Since
2
05
.
χ
= 7.81473, we fail to reject
H
0
; there is not enough
evidence to conclude the proportions differ.
b.
Using the Applet,
p
–value =
P
(
χ
2
> 3.696) = .29622.
14.2
a.
H
0
:
p
1
= .60,
p
2
= .05,
p
3
= .35 vs.
H
a
: not
H
0
.
The observed and expected counts are:
admitted unconditionally admitted conditionally
refused
observed
329
43
128
expected
500(.60) = 300
500(.05) = 25
500(.35) = 175
The chi–square test statistic is
175
)
175
128
(
25
)
25
43
(
300
)
300
329
(
2
2
2
2
−
−
−
+
+
=
X
= 28.386 with 3 – 1 = 2
degrees of freedom.
Since
2
05
.
χ
= 7.37776, we can reject
H
0
and conclude that the current
admission rates differ from the previous records.
b.
Using the Applet,
p
–value =
P
(
χ
2
> 28.386) = .00010.
14.3
The null hypothesis is
H
0
:
p
1
=
p
2
=
p
3
=
p
4
=
4
1
vs.
H
a
: not
H
0
.
The observed and
expected counts are:
lane
1
2
3
4
observed 294 276 238 192
expected
250 250 250 250
The chi–square statistic is
250
)
250
192
(
)
250
238
(
)
250
276
(
)
250
294
(
2
2
2
2
2
−
+
−
+
−
+
−
=
X
= 24.48 with 4 –1 = 3
degrees of freedom.
Since
2
05
.
χ
= 7.81473, we reject
H
0
and conclude that the lanes are
not preferred equally.
From Table 6,
p
–value < .005.
Note that
R
can be used by:
> lanes < c(294,276,238,192)
> chisq.test(lanes,p = c(.25,.25,.25,.25))
# p is not necessary here
Chisquared test for given probabilities
data:
lanes
Xsquared = 24.48, df = 3, pvalue = 1.983e05
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Chapter 14: Analysis of Categorical Data
Instructor’s Solutions Manual
14.4
The null hypothesis is
H
0
:
p
1
=
p
2
= … =
p
7
=
7
1
vs.
H
a
: not
H
0
.
The observed and
expected counts are:
SU
M
T
W
R
F
SA
observed
24
36
27
26
32
26
29
expected 28.571 28.571 28.571 28.571 28.571 28.571 28.571
The chi–square statistic is
571
.
28
)
571
.
28
29
(
)
571
.
28
36
(
)
571
.
28
24
(
2
2
2
2
−
+
+
−
+
−
=
…
X
= 24.48 with 7 –1 = 6
degrees of freedom.
Since
2
05
.
χ
= 12.5916, we can reject the null hypothesis and conclude
that there is evidence of a difference in percentages of heart attacks for the days of the
week
14.5
a.
Let
p
= proportion of heart attacks on Mondays.
Then,
H
0
:
p
=
7
1
vs.
H
a
:
p
>
7
1
.
Then,
p
ˆ = 36/200 = .18 and from Section 8.3, the test statistic is
200
)
7
/
6
)(
7
/
1
(
7
/
1
18
.
−
=
z
= 1.50.
Since
z
.05
= 1.645, we fail to reject
H
0
.
b.
The test was suggested by the data, and this is known as “data snooping” or “data
dredging.”
We should always apply the scientific method: first form a hypothesis and
then collect data to test the hypothesis.
c.
Monday has often been referred to as the most stressful workday of the week: it is the
day that is farthest from the weekend, and this realization gets to some people.
14.6
a.
j
i
j
i
j
i
np
np
n
E
n
E
n
n
E
−
=
−
=
−
)
(
)
(
)
(.
b.
Define the sample proportions
n
n
p
i
i
/
ˆ
=
and
n
n
p
j
j
/
ˆ
=
.
Then,
j
i
p
p
ˆ
ˆ
−
is unbiased
for
p
i
–
p
j
from part
a
above.
c.
j
i
j
j
i
i
j
i
j
i
j
i
p
np
p
np
p
np
n
n
n
V
n
V
n
n
V
2
)
1
(
)
1
(
)
,
(
Cov
2
)
(
)
(
)
(
+
−
+
−
=
−
+
=
−
.
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 Spring '08
 Staff
 Degrees Of Freedom, Probability, Chisquare distribution, Pearson's chisquare test, INSTRUCTOR’S SOLUTIONS

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