Transcript 1 - Biyoistatistik

June 13, 2012
Example: A study was conducted to see if a new
therapeutic procedure is more effective than the standard
treatment in improving the digital dexterity of certain
handicapped persons.
Twenty-four pairs of twins were used in the study, one of
the twins was randomly assigned to receive the new
treatment, while the other received the standard therapy.
At the end of the experimental period each individual was
given a digital dexterity test with scores as follows.
2
Since
The variable concerning digital dexterity
test scores is continuous
The sample size is greater than 10
digital dexterity test score is normally
distributed
Paired
sample
t-test
There are two groups and they are
dependent
3
Total
Mean
SD
New
49
56
70
83
83
68
84
63
67
79
88
48
52
73
52
73
78
64
71
42
51
56
40
81
Standard
54
42
63
77
83
51
82
54
62
71
82
50
41
67
57
70
72
62
64
44
44
42
35
73
65,46
14,38
60,08
14,46
Difference
-5
14
7
6
0
17
2
9
5
8
6
-2
11
6
-5
3
6
2
7
-2
7
14
5
8
129
5,38
5,65
H0: d = 0
Ha: d > 0
d
∑
d=
i
n
=
129
= 5.38
24
(di – d)
∑
=
2
s2d
n–1
= 31.90
d – μd
5.38 – 0
t=
=
sd / n
31.90 / 24
= 4.66
t(23,0.05)=1.7139
Since,
tcalculated > t table
reject H0.
We conclude that the new treatment
is effective.
4
SPSS Output
Paired Samples Statistics
Mean
Pair 1
N
Std. Deviation
Std. Error
Mean
New
65,46
24
14,380
2,935
Standard
60,08
24
14,461
2,952
Paired Samples Test
Paired Differences
Mean
Pair 1
New - Standard
5,375
Std. Deviation
5,648
Std. Error
Mean
1,153
95% Confidence
Interval of the Difference
Lower
2,990
Upper
7,760
t
df
4,662
Sig. (2-tailed)
23
,000
5
Digital Dexterity Test Scores (Mean±SD)
80
70
60
50
40
New
Standard
6
Example: To test whether the weight-reducing diet is effective, 9
persons were selected. These persons stayed on a diet for two
months and their weights were measured before and after diet. The
following are the weights in kg:
Weights
Since
Subject
Before
After
The variable concerning
1
85
82
weight is continous.
2
91
92
3
4
5
68
76
82
62
73
81
6
7
8
87
105
93
83
85
88
9
98
90
The sample size is less than 10
There are two groups and they
are dependent
Wilcoxon signed
ranks test
7
Subject
1
2
3
4
5
6
7
8
9
Weights
Before
After
85
82
91
92
68
62
76
73
82
81
87
83
105
85
93
88
98
90
Difference
Di
Sorted
Di
Rank
Signed
Rank
3
-1
6
3
1
4
20
5
8
-1
1
3
3
4
5
6
8
20
1.5
1.5
3.5
3.5
5
6
7
8
9
-1.5
1.5
3.5
3.5
5
6
7
8
9
8
T = 1.5
T = 1.5 < T(n=9,a =0.05) = 6
reject H0 , p<0.05
T = 1.5 < T(n=9,a =0.01) = 2
reject H0 , p<0.01
We conclude 99% confident that diet is effective.
9
110
100
90
80
70
3
60
Before
After
10
Example: 35 patients were evaluated for arrhythmia with
two different medical devices. Is there any statistically
significant difference between the diagnose of two devices?
Device II
Device I
Total
Arrhythmia (+)
Arrhythmia (-)
Arrhythmia (+)
10
3
13
Arrhythmia (-)
13
9
22
Total
23
12
35
11
H0: P1=P2
Ha: P1 P2
b–c –1
z=
b+c
3 – 13 – 1
=
= 2.25
3 + 13
Critical z value is ±1.96
Reject H0
12
McNemar test approach:
2
2
( b – c – 1)
χ =
b+c
(b – c)
χ =
b+c
2
2
2
2
( b – c – 1) ( 3 – 13 – 1)
χ =
=
= 5.1
b+c
3 + 13
2
2(1,0.05)=3.841<5.1
p<0.05; reject H0.
13
SPSS Output
Device Arrhythmia Crosstabulation
Count
DeviceII
Yes
DeviceI
No
Total
Yes
10
3
13
No
13
9
22
23
12
35
Total
Chi-Square Tests
Value
df
Exact Sig.
(2-sided)
a
McNemar Test
N of Valid Cases
Asymp. Sig.
(2-sided)
,021
35
a. Binomial distribution used.
14
Example: NK cell activity was measured for three groups of
subjects: those who had low, medium, and high scores on
Social Readjustment Rating Scale. The original
observations, sample sizes, means and standard deviations
are given in table. Is the mean NK cell activity different in
three groups?
n
Mean
SD
Low Score
22,20
57,80
29,10
37,00
35,80
44,20
52,00
56,00
39,30
19,90
39,50
22,80
37,40
13
37,92
12,41
Moderate Score
15,10
23,20
10,50
13,90
9,70
19,00
19,80
9,10
30,10
15,50
10,30
11,00
High Score
10,20
11,30
11,40
5,30
14,50
11,00
13,60
33,40
25,00
27,00
36,30
17,70
12
15,60
6,42
12
18,06
9,97
The variable concerning NK cell activity
measures is continuous.
There are three groups.
Variances are equal in three groups.
NK cell activity values in each group are
normally distributed.
The sample size of each group is greater than
10
One-Way
ANOVA
Natural killer activity
95% CI for Mean
13
Mean
37,92
Std.
Deviation
12,41
Lower
Bound
30,42
Upper
Bound
45,42
Moderate
score
12
15,60
6,42
11,52
19,68
High score
12
18,06
9,97
11,72
24,39
Total
37
24,24
14,12
19,53
28,95
Groups
Low score
N
60
Mean  1 SD NKA
50
40
30
20
10
0
N=
13
Low
12
Moderate
GROUP
12
High
H0: 1=  2= 3
Ha: Not all the i are equal.
k
nj
2
2
T
896
.
9
SST = ∑∑x 2ij – .. = 22.22 + 57.82 +  + 17.72 –
N
37
j=1 i=1
= 28923.65 – 21741.34
= 7182.31
k
nj
k
T.2j
SSW = ∑∑x 2ij – ∑
j=1 i=1
j=1 n j
4932 187.22 216.72
= 22.2 + 57.8 +  + 17.7 –
+
+
13
12
12
2
2
2
= 3394.01
SSA=SST-SSW=7182.31-3394.01=3788.30
MSW=SSW/34=3394.01/27=99.82
MSA=SSA/(3-1)=3788.30/2=1894.15
F=MSA/MSW=1894.15/99.82=18.98
ANOVA
NKA Natural killer activity
Sum of
Squares
df
Mean Square
Betw een Groups
3788,297
2
1894,148
Within Groups
3394,012
34
99,824
Total
7182,309
36
F
Sig.
18,975
We conclude that not all population means are equal.
,000
Since n1  n2, reject H0 if
1 1
xi – x j ≥t MSW( + )
ni n j
37.92 – 15.6
Hypothesis
LSD
Statistical
Decision
H0: 1= 2
1 1
2.03 99.824( + ) = 8.12
13 12
22.32>8.12,
reject H0.
H0: 1= 3
1 1
2.03 99.824( + ) = 8.12
13 12
19.87>8.12,
reject H0.
H0: 2= 3
1 1
2.03 99.824( + ) = 8.28
12 12
2.46<8.28,
accept H0.
Multiple Comparisons
Dependent Variable: NKA Natural killer activity
LSD
(I) GROUP
1,00 Low score
2,00 Moderate s core
3,00 High score
(J) GROUP
2,00 Moderate s core
Mean
Difference
(I-J)
Std. Error
Sig.
22,3231*
3,9997
,000
3,00 High score
19,8647*
3,9997
,000
1,00 Low score
-22,3231*
3,9997
,000
3,00 High score
-2,4583
4,0789
,551
1,00 Low score
-19,8647*
3,9997
,000
2,4583
4,0789
,551
2,00 Moderate s core
*. The mean difference is s ignificant at the .05 level.
Example: Hamilton depression scores was measured for
three groups of subjects and shown in table. Is the
Hamilton depression scores different in three groups?
Subject
1
2
3
4
5
6
7
8
Group1
1
11
12
13
9
1
2
1
Group2
20
19
0
17
14
21
20
15
Group3
15
15
19
4
1
19
21
18
Since
The variable concerning Hamilton
depression score is continuous
The sample size is less than 10
Kruskal
Wallis Test
There are three groups and they are
independent
H0: The population distributions are all identical.
Ha: At least one of the populations tends to exhibit larger values
than at least one of the other populations.
Group1
Score Rank
1
3.5
11
9
12
10
13
11
9
8
1
3.5
2
6
1
3.5
Ri
54.5
Group2
Score Rank
20
21.5
19
19
0
1
17
16
14
12
21
23.5
20
21.5
15
14
128.5
Group3
Score Rank
15
14
15
14
19
19
4
7
1
3.5
19
19
21
23.5
18
17
117
k
R2j
12
KW =
– 3(n + 1)
∑
n(n + 1) j=1 n j
12
54.52 128.52 1172
=
+
+
24(24 + 1)
8
8
8
– 3(24 + 1)
= 7.93
KW=7.93 > χ2(2,a=0.05) = 5.99
Reject H0
We conclude that at least one of the populations tends to
exhibit larger values than at least one of the other
populations.
Multiple Comparisons Table
Groups
Ri – Rj
Statistical
Decision
n(n + 1) n – 1 – KW 1 1
t
+
12
n–k
ni n j
1-2
9.25
5.87
p<0.05
1-3
7.82
5.87
p<0.05
2-3
1.43
5.87
p>0.05
25
Hamilton Depression Score
20
15
10
5
11
0
Group I
Group II
Group III
Example: A researcher wants to know whether the mothers
age is affecting the probability of having congenital
abnormality of neonatals or not. The collected data is given
in the table:
Congenital abnormality
Age groups
Present
Absent
Total
≤25
3
22
25
26-35
8
34
42
>35
18
16
34
Total
29
72
101
H0: There is no relation between the age of mother and presence
of congenital abnormality.
Under the assumption that null hypothesis is true:
(Expected count)
Congenital abnormality
Age groups
Present
Absent
≤25
3 (7.2)
22 (17.8)
26-35
8 (12.1)
34 (29.9)
>35
18 (9.8)
16 (24.2)
Reject H0
Congenital abnormality
Age
groups
χ2
Present
Absent
≤25
3 (7.2)
22 (17.8)
3.44
26-35
8 (12.1)
34 (29.9)
1,95
>35
18 (9.8)
16 (24.2)
9,64
Omit the
>35 age
group
Congenital abnormality
Age
groups
Present
Absent
≤25
3
22
26-35
8
34
H0 is accepted
In older age groups, risk of having a baby with congenital
abnormality is higher than the other age groups.
100.0%
90.0%
80.0%
70.0%
Percent
60.0%
50.0%
Present
Absent
40.0%
30.0%
20.0%
10.0%
0.0%
≤25
26-35
Age groups
>35
Example: A special diet program was given to 20 clinically obese
people. Subjects’ BMI were measured before the diet and they
have been followed for two months. BMI measures before the
diet and after the end of each month following the diet are
given in the table. Is the diet program effective?
Before
32,0
35,1
30,8
39,5
36,0
40,0
31,3
33,7
32,8
31,1
34,3
32,6
30,0
31,5
38,1
34,9
35,0
37,0
30,5
38,9
Month1
31,8
34,9
30,0
39,1
35,0
39,8
30,4
33,3
32,1
31,0
34,0
31,5
30,0
31,0
38,0
33,8
34,0
36,6
30,0
38,0
Month2
30,0
34,1
29,2
38,5
34,3
38,7
29,5
32,8
31,7
30,6
33,2
31,2
29,4
30,3
37,6
33,2
33,0
35,0
29,6
37,4
The variable concerning BMI is
continuous
The sample size is greater than
10
BMI values are normally
distributed
There are three groups and
they are dependent
Repeated measures ANOVA
Sources of variation
Times
Subjects
Error
SS
16.79
567.72
3.33
df
2
19
38
MS
8.39
29.89
0.09
Diet program is effective on obese subjects’ BMI.
F
95.92
Sig.
0.000
34,4
34,2
34,0
Means
33,8
33,6
33,4
33,2
33,0
32,8
1
2
Time
3
Example: To compare the effects on the clotting time of plasma
of four different methods of treatment of the plasma. Samples
of plasma from 8 subjects were assigned to the four
treatments.
Treatments
Subject
1
2
3
I
8.4
12.8
9.6
II
9.4
15.2
9.1
III
9.8
12.9
11.2
IV
12.1
14.4
9.8
4
5
9.8
8.4
8.8
8.2
9.9
8.5
12.0
8.5
6
7
8
8.6
8.9
7.9
9.9
9.0
8.1
9.8
9.2
8.2
10.9
10.4
10
Since
The variable concerning clotting time is
continuous
The sample size is less than 10
There are four groups and they are
dependent
Friedman
Test
Treatments
I
time
8.4
12.8
9.6
9.8
8.4
8.6
8.9
7.9
Sum Ri
II
rank
1
1
2
2
2
1
1
1
11
time
9.4
15.2
9.1
8.8
8.2
9.9
9.0
8.1
III
rank
2
4
1
1
1
3
2
2
16
time
9.8
12.9
11.2
9.9
8.5
9.8
9.2
8.2
IV
rank
3
2
4
3
3.5
2
3
3
23.5
time
12.1
14.4
9.8
12.0
8.5
10.9
10.4
10
rank
4
3
3
4
3.5
4
4
4
29.5
12
Fr =
Nk(k + 1)
k
2
R
∑ j – [3N(k + 1)]
j 1
12
=
112 + 162 + 23.52 + 29.52 – [3(8)(4 + 1)]
8(4)(4 + 1)
= 14.96
[
2(3,0.05)= 7.815 < Fr=14.96, p<0.05
]
Reject H0.
We conclude that at least one of the treatments were
different from the other treatments.
16,00
10
14,00
Clotting Time
2
12,00
10,00
8,00
6,00
Treatment I
Treatment II
Treatment III
Treatment IV
Example: Four radiologists have evalueted 12 patient
whether they have lesion or not;
H0: There is no difference between the decision of
radiologists
Since
The variable is categorical (0 and 1)
There are four groups and they are
dependent
Cochran Q
Test
Subject
Radiologist I
Radiologist II
Radiologist III
Radiologist IV
Ri
1
1
0
0
1
2
2
1
0
0
1
2
3
1
1
1
0
3
4
0
0
0
1
1
5
1
1
1
1
4
6
1
0
1
1
3
7
1
0
1
1
3
8
1
0
0
1
2
9
1
1
0
1
3
10
1
0
1
0
2
11
0
0
1
1
2
12
0
9
0
3
0
6
1
10
1
Cj
18
k
(k – 1 ) k ∑C j –
∑C
2
j= 1
Q=
n
k ∑R i –
i= 1
=
2
j
j= 1
n
[
2
k
∑R
2
i
i= 1
2
2
2
( 4 – 1 ) 4 (9 + 3 + 6 + 10 – 28
2
2
2
]
2
4 (2 + 2 +  + 1 ) – (2 + 2 +  + 1 )
= 9 . 474
QCalculated= 9.474 > 2(3,0.05) =7.815 p<0.05
Reject H0
Type of
variables
Continuous
Continuous
Categorical
Categorical
Continuous
Continuous
Categorical
Categorical
Number of
groups
1
1
1
1
2
2
2
2
Independent
/Dependent
–
–
–
–
Independent
Independent
Independent
Independent
Name of tests
One sample t test
Sign test
One sample proportion test
One sample chi square test
Independent samples t test
Mann Whitney U Test
Difference between two population proportions
2*2 chi square test
2
2
2
Parametric test
assumptions
Satisfied
Not satisfied
Sample size is large
Sample size is small
Satisfied
Not satisfied
Sample size is large
Expected count > 5
Expected count < 5 at least in
Independent
one cell
Dependent
Satisfied
Dependent
Not satisfied
Categorical
Continuous
Continuous
2
2
k
k
k
k
k
k
Dependent
Dependent
Independent
Independent
Independent
Dependent
Dependent
Dependent
Fisher's exact test
Paired samples t test
Wilcoxon signed rank test
Difference between two dependent population
proportions
McNemar Test
One way ANOVA
Krıskal Wallis Test
r*c Chi square test
Repeated measures ANOVA
Friedman test
Cochran Q test
Categorical
Categorical
Continuous
Continuous
Categorical
Continuous
Continuous
Categorical
Sample size is large
Sample size is small
Satisfied
Not satisfied
Expected count > 5
Satisfied
Not satisfied
–