HW8
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University of North Carolina, Chapel Hill *
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Course
600
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
docx
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Uploaded by BaronCrownBoar30
BIOS 600 Spring 2019.
Homework 8.
Chapter 18.
Chapter 18. Exercises 18.24.
+
−
Total
Treated patients
26
178
204
Control patients
2
62
64
Total
28
240
268
(a) Will test if chi-square test can be used:
+
−
Total
Treated patients
21.31
182.69
204
Control patients
6.69
57.31
64
Total
28
240
268
We can use chi-square test since all expected values are greater than 5.
(b)
1. H
0
: The proportion of deaths due to cardiovascular event is independent/not associated with
treatment
vs.
H
a
: The proportion of deaths due to cardiovascular event is associated with treatment.
2. χ
2
stat
=
E
i
O
i
−
¿
¿
¿
2
¿
21.31
26
−
¿
¿
¿
2
¿
182.69
178
−
¿
¿
¿
2
¿
2
−
6.69
¿
2
¿
¿
6.69
+
(
62
−
57.31
)
2
57.31
¿
¿
∑
¿
2
df = (2-1)*(2-1)=1
From table E we have p-value between 0.005 and 0.10.
R response:
> a=matrix(c(26,178,2,62), ncol=2, byrow=T)
> chisq.test(a, correct=FALSE)
Pearson's Chi-squared test
data:
a
X-squared = 4.8188, df = 1, p-value = 0.02815
3.
At alpha = 0.05 (p=0.02815) we find that our test statistic is significant.
We reject the null
hypothesis in favor or our alternative hypothesis. We can conclude that there is an association
of deaths due to cardiovascular event with the treatment.
Chapter 18. Exercises 18.26.
Cases
Controls
Total
Infertility
89
283
372
IUD use
640
3833
4473
Total
729
4116
4845
^
O R
=
89
∙
3833
640
∙
283
=
1.88
ln
^
O R
=
ln
(
1.88
)
=
0.6313
SE
ln
^
OR
=
√
1
89
+
1
283
+
1
640
+
1
3833
=
0.1288
95% CI:
e
0.6313
±
1.96
∗
0.1288
=
e
0.6313
±
0.2524
=
(
e
0.3789
,e
0.8837
)
=(
1.46,2.42
)
The 95% confidence interval for the IUD use is (
1.46,2.42
).
As 1 is not in this interval, we
conclude that infertility is associated with IUD use.
Chapter 18. Exercises 18.32.
Case
Exposed
Case
Nonexposed
Total
Control Exposed
5
13
18
Control
Nonexposed
30
107
137
Total
35
120
155
3
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(a)
^
¿
=
30
13
=
2.31
, 2.3-fold increase in the risk of stroke among the female using oral
contraceptives.
(b)The data with the match broken:
Case
Control
Total
Exposed
5+30=35
5+13=18
53
Nonexposed
13+107=120
30+107=137
257
Total
155
155
310
Odds ratio with the match broken is =(35*137)/(130*18)=2.05, we see a decrease in estimation.
4