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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_155_библиотеки_им_акад_М_И_Перельмана
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380 Answer to Exercises
(continued)
Chapter 1
Chapter 9
Q1 OR = 1.3125 (a = 42, b = 496, c = 28, and d = 434)
Q3 Sen =
11
81
= 0.1375, Spe =
2037
2069
= 0.984, PPV =
11
43
= 0.256, NPV =
2037
2107
= 0.967
Q5 OR = 11.65
Q7 P
e
= 0.577, P
0
= 0.94 and K = 0.858
Se = 0.02981 and CI = 0.80–0.916
Chapter 11
Q1 Because Sd is known (18 min), z criteria should be used. Calculated Z = 1.23 < 1.96
and we can’t reject the factory claim
Q2 x = 23.7, Sd = 6.7364,μ
0
= 20, n = 20 and t
19
= 2.093
Calculated t = 2.456 > t
19
= 2.093 and the researcher claim is rejected
Q5 A: Se
(
p
1
− p
2
)
= 0.1637
CI95% =
(
0.4000 −
0.3333
)
± 1.96 × 0.1637 =−0.254 to 0.3877
B: Calculated Z = 0.407 < 1.96
C: Using Chi-square χ
2
= 0.1624 < 3.841
Q7 Using paired t-test, t = 1.990 < t
15
= 2.131 indicates that there is no difference
between 2 methods
Q14
Sp =
(
37−1
)
×1.59
2
+
(
36−1
)
2.36
2
37+36−2
= 2.01
Se = 2.01 ×
1
37
+
1
36
= 0.47
t =
7.79−7.63
0.47
= 0.34 < t
71
= 1.994
Chapter 12
Q1 A: Correlation is significant r = 0.765
B:Pl = 0.077 + 0.044W
C:If w = 60, Pl = 2.72
If w = 65, Pl = 2.94
If w = 70, Pl = 3.16
D: We are unable to estimate plasma for those individuals as this equation is only
applicable
to weights
that fall between the minimum and maximum values
Q3 Correlation coefficient (r = 0.689) is strong and significant
Q5 Correlation coefficient for group 1 is (r1 =−0.18) which is not significant and for
group 2 is (r2 = –0.239) which is significant
(continued)
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Answer to Exercises 381
(continued)
Chapter 1
Q8 A: Correlation between height of children and father’s height = 0.604
Correlation between height of children and mother’s height = 0.262
To compare this correlation, we use Z criteria
Z
1
=
1
2
ln
1+0.604
1−0.604
= 0.699, Z
2
=
1
2
ln
1+0.262
1−0.262
= 0.268
Z
(
r
1
− r
2
)
=
0.699−0.268
1
40−3
+
1
40−3
= 1.854 < 1.96
The difference in correlation coefficient between the fathers and mothers is not
significant
B: Correlation between height of girls and mother’s height = 0.795
Correlation between height of Boys and father’s height = 0.241
To compare this correlation, we use Z criteria
Z
1
=
1
2
ln
1+0.795
1−0.795
= 1.085, Z
2
=
1
2
ln
1+0.241
1−0.241
= 0.246
Z
(
r
1
− r
2
)
=
1.085−0.246
1
20−3
+
1
20−3
= 2.446 > 1.96
The difference in correlation coefficient between the boys with fathers and girls with
mothers is
significant
Chapter 13
Q1 There are more than two groups to compare, so the ANOVA method is appropriate. In
this instance, the average pulse rate of these groups is the same (F = 0.079, p = 0.924)
Q3 Since there are more than two groups to compare, the ANOVA technique is appropriate,
as evidenced by the F = 2.823 with p = 0.082 > 0.05. So these teaching strategies have
thesameaveragescore
Q5 The ANOVA technique is appropriate for comparing four groups. In this case, F =
3.795, p = 0.015 < 0.05. Then, there’s a difference in the average scores for these types
of insurance
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