Transcript PowerPoint Sunusu

Example: In a heart study the systolic blood pressure
was measured for 24 men aged 25 and for 30 men
aged 40. Do these data show sufficient evidence to
conclude that the older men have a higher systolic
blood pressure, at the 0.05 level of significance?
Since
The variable concerning systolic blood
pressure is continuous
The sample size of each group is greater
than 10
Systolic blood pressure values in each
group is normally distributed
There are two groups and they are
independent
Independent
samples t-test
is used
Subject
1
2
3
4
5
6
7
8
9
10
11
12
20- year-old
Sbp
Subject
95
13
122
14
130
15
148
16
130
17
150
18
105
19
110
20
130
21
156
22
108
23
124
24
Sbp
132
100
120
125
115
138
100
118
136
110
140
106
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
40- year-old
Sbp
Subject
150
16
152
17
154
18
160
19
164
20
176
21
108
22
126
23
132
24
142
25
136
26
146
27
114
28
118
29
130
30
Sbp
148
116
128
136
110
126
130
122
140
110
124
136
120
142
114
GROUP
20- year-old
40- year-old
N
24
30
Mean
Std. Deviation
122,8333
16,7790
133,6667
17,3013
24
30
20- year-old
40- year-old
160
Mean  1 SD SBP
150
140
130
120
110
100
N=
GROUP
(1) H0:1=2
Ha: 1<2
(2) Testing the equality of variances
H0:21= 22
Ha: 21 22
S2max 299.33
F 2 
 1.06  F(30, 24,0.05)  1.94
Smin 281.54
Accept H0. Variances are equal.
(3)
(n1  1) s  (n2  1) s
s 
n1  n2  2
2
1
2
p
2
2
(24  1)281.54 2  (30  1)299.332

 291.46
24  30  2
t
( x1  x2 )  (1   2 )
s 2p
n1
(4)

s 2p
n2
t(52,0.05)=1.65<

(122 . 83  133 . 67)  0
291 . 46 291 . 46

24
30
t cal  2.31
 2.31
p<0.05, Reject H0.
(5) The older men have higher systolic blood pressure
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 Moderate Score High Score
22,20
15,10
10,20
57,80
23,20
11,30
29,10
10,50
11,40
37,00
13,90
5,30
35,80
9,70
14,50
44,20
19,00
11,00
52,00
19,80
13,60
56,00
9,10
33,40
39,30
30,10
25,00
19,90
15,50
27,00
39,50
10,30
36,30
22,80
11,00
17,70
37,40
13
12
12
37,92
15,60
18,06
12,41
6,42
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
Groups
Low score
Moderate
score
High score
Total
60
13
Mean
37,92
Std.
Deviation
12,41
12
15,60
6,42
11,52
19,68
12
37
18,06
24,24
9,97
14,12
11,72
19,53
24,39
28,95
N
Mean  1 SD NKA
50
40
30
20
10
0
N=
13
Low
95% CI for Mean
Lower
Upper
Bound
Bound
30,42
45,42
12
Moderate
12
High
H0: 1=  2= 3
Ha: Not all the i are equal.
2
2
T
896
.
9
SST    xij2  ..  22.2 2  57.8 2    17.7 2 
N
37
j 1 i 1
 28923.65  21741.34
 7182.31
k nj
k nj
k
T. 2j
j 1 i 1
j 1
nj
SSW    xij2  
2
2
2


493
187
.
2
216
.
7
2
2
2
 22.2  57.8    17.7  



13
12
12


 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
Between Groups
Within Groups
Total
Sum of
Squares
3788,297
3394,012
7182,309
df
2
34
36
Mean Square
1894,148
99,824
F
18,975
Sig .
,000
We conclude that not all population means are equal.
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 score
3,00 High score
(J) GROUP
2,00 Moderate score
3,00 High score
1,00 Low score
3,00 High score
1,00 Low score
2,00 Moderate score
Mean
Difference
(I-J)
Std. Error
22,3231*
3,9997
19,8647*
3,9997
-22,3231*
3,9997
-2,4583
4,0789
-19,8647*
3,9997
2,4583
4,0789
*. The mean difference is significant at the .05 level.
Sig .
,000
,000
,000
,551
,000
,551
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.
Since
The variable concerning digital dexterity
test scores is continuous
The sample size is greater than 10
digital dexterity test score is normally
distributed
There are two groups and they are
dependent
Paired
sample
t-test
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 Difference
54
-5
42
14
63
7
77
6
83
0
51
17
82
2
54
9
62
5
71
8
82
6
50
-2
41
11
67
6
57
-5
70
3
72
6
62
2
64
7
44
-2
44
7
42
14
35
5
73
8
Total
129
Mean 65,46
60,08
5,38
SD
14,38
14,46
5,65
H0: d = 0
Ha: d > 0
di

d
129

 5.38
n
24
2
(
di

d
)
sd2  
 31.90
n 1
d  d
5.38  0
t

sd / n
31.90 / 24
 4.66
t(23,0.05)=1.7139
t
t
Since, calculated
table
reject H0.
We conclude that the new
treatment is effective.
Example: We want to know if children in two
geographic areas differ with respect to the proportion
who are anemic. A sample of one-year-old children
seen in a certain group of county health departments
during a year was selected from each of the
geographic areas composing the departments’
clientele. The followig information regarding anemia
was revealed.
Geographic
Area
1
Number in
sample
450
Number
anemic
105
Proportion
2
375
120
0.32
0.23
H 0 : P2  P1  0
p1  105 / 450  0.23
H a : P2  P1  0
p2  120 / 375  0.32
(450 )(0.23)  (375)(0.32)
p
 0.27
450  375
z
(0.23 - 0.32)  0
 2.78 p  0.0027  0.025
(0.27)(0.73) (0.27)(0.73)

450
375
Reject H0.
We concluded that the proportion of anemia is different
in two geographic areas.
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
34,4
34,2
34,0
Means
33,8
33,6
33,4
33,2
33,0
32,8
1
2
Time
3
Sources of variation
SS
df
Times
16.79 2
Subjects
567.72 19
Error
3.33
38
MS
8.39
29.89
0.09
F
Sig.
95.92 0.000
Diet program is effective on obese subjects’ BMI.
Example: To test the median level of energy intake of
2 year old children as 1280 kcal reported in another
study, energy intakes of 10 children are calculated.
Energy intakes of 10 children are as follows:
Child
Energy
Intake
1
2
3
4
5
6
7
8
9
10
1500
825
1300
1700
970
1200
1110
1270
1460
1090
Since
The variable concerning energy intake is
continuous
The sample size is not greater than 10
Energy intake is not normally distributed
There is only one group
Sign
test
H0: The population median is 1280.
HA: The population median is not 1280.
Child
Energy
İntake
1
2
3
4
5
6
7
8
9
10
1500
825
1300
1700
970
1200
1110
1270
1460
1090
+
-
+
+
-
-
-
-
+
-
Number of (-) signs = 6
and number of (+) signs = 4
For k=4 and n=10
From the sign test table p=0.377
Since p > 0.05 we accept H0
We conclude that the median energy intake level
in 2 year old children is 1280 kcal.
Example: Cryosurgery is a commonly used therapy
for treatment of cervical intraepithelial neoplasia
(CIN). The procedure is associated with pain and
uterine cramping. Within 10 min of completing the
cryosurgical procedure, the intensity of pain and
cramping were assessed on a 100-mm visual analog
scale (VAS), in which 0 represent no pain or
cramping and 100 represent the most severe pain
and cramping. The purpose of study was to compare
the perceptions of both pain and cramping in women
undergoing the procedure with and without
paracervical block.
5 women were selected randomly in each
groups and their scores are as follows:
Group
Women without
a block
Score
14
88
37
27
0
50
Women with a
paracervical
block
70
37
66
75
Since
The variable concerning pain/cramping
score is continuous
The sample size is less than 10
There are two groups and they are
independent
Mann
Whitney
U test
H 0 : M I  M II
H A : M I  M II
Group
I
I
I
I
II
II
II
II
II
I
Score
0
14
27
37
37
50
66
70
75
88
R1= 1+2+3+4.5+10 = 20.5
Rank
1
2
3
4.5
4.5
6
7
8
9
10
n1 (n1  1)
U1  n1n2 
 R1
2
5(5  1)
 55 
 20.5  19.5
2
U 2  n1n2  U1  5  5 19.5  5.5
U  19.5
From the table, critical value is 21
19.5 < 21 accept H0
We conclude that the median pain/
cramping scores are same in two groups.
Example: A study was conducted to analyze the
relation between coronary heart disease (CHD) and
cigarette smoking. 40 patients with CHD and 50
control subjects were randomly selected from the
records and smoking habits of these subjects were
examined. Observed values are as follows:
Smoking
CHD
Total
+
-
10
30
40
No
4
46
50
Total
14
76
90
Yes
Observed and expected frequencies
Cigarette
Smoking
+
Total
-
10
6.2
30 33.8
40
No
4
7.8
46 42.2
50
Total
14
76
90
Yes
2
CHD
2
χ 2  
(O ij  E ij ) 2
E ij
i 1 j1

10  6.2

2
6.2

30  33.8

2
33.8

4  7.8

2
7.8

46  42.2

2
42.2
 4.95
df = (r-1)(c-1)=(2-1)(2-1)=1
2
2 =4. 95 >  (1,0.05)=3.845
reject H0
Conclusion: There is a relation between CHD and
cigarette smoking.
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
Subject
Since
Before After
1
85
82
The variable concerning
weight is continous.
2
91
92
3
68
62
The sample size is less than 10
4
5
6
76
82
87
73
81
83
7
8
105
93
85
88
9
98
90
There are two groups and they
are dependent
Wicoxon signed
ranks test
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
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% cinfident that diet is effective.
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
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
R 2j
12
KW 
 3( n  1)

n( n  1) j 1 n j
 54.52 128.52 117 2 
12




  3( 24  1)
24( 24  1)  8
8
8 
 7.93
KW=7.93 > 2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  R j
n(n  1) n  1  KW  1 1 
t


12
n  k  ni n j 
Statistical
Decision
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
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.
Subject
1
I
8.4
Treatments
II
III
9.4
9.8
2
3
4
5
12.8
9.6
9.8
8.4
15.2
9.1
8.8
8.2
12.9
11.2
9.9
8.5
14.4
9.8
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
IV
12.1
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
 k 2
12
Fr 
 R j   3N (k  1)
Nk (k  1)  j 1 


12

112  16 2  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.
Example
Subject Lipid Energy
10,00 2,10
Measures of lipid content and total energy content 1
of the stools were recorded.
Data for the cystic fibrosis children are given
in the table.
Find and interpret the correlation
between stool lipid and stool energy
2
3
4
2
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
11,00
9,90
9,80
15,50
5,00
10,70
13,00
13,80
16,70
3,20
4,00
6,00
8,90
9,10
4,10
17,00
22,20
2,90
5,00
1,10
1,10
0,90
0,70
0,40
1,00
1,50
1,20
1,40
1,00
0,50
0,90
0,80
0,60
0,50
1,20
1,10
0,90
1,00
2,2
2,0
1,8
1,6
ENERGY
1,4
1,2
1,0
0,8
0,6
0,4
0,2
2
4
6
8
10
12
14
LIPID
16
18
20
22
24
n
n
r
r
x y
i 1
i
i

n
x y
i 1
i
i 1
i
n
n
n


2 
( xi )  n
(  yi ) 2 
 n
 x 2  i 1
 y 2  i 1



i
i
 i 1
 i 1

n
n






Correlation between
stool lipid and stool
energy is 0.42. Is it
significant?
213.46  (197.8)(19.90)
2
2



197.8 
19.90 
 2489.44 
 22.71 





20
20



 0.4227
n2
20  2
tr
 0.4227
 1.98  p=0.06251
2
2
1 r
1  0.4227
There is no significant correlation between stool lipid and stool
energy.
Example
Intravenous glucose tolerance tests were performed on 14
hyperthyroid women, 6 of whom were overweight, and in
19 volunteers with normal thyroid levels matched for age
and weight. Insulin sensitivity and BMI of these women are
given in the table.
•Find the correlation coefficients between BMI and insulin
sensitivity in the control group and hyperthyroid group.
•In which group, bmi is more effective?
•Find the regression equation for each group?
CONTROL GROUP
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
BMI
18,10
23,60
24,00
20,40
21,50
22,20
20,60
23,90
17,90
21,90
20,20
30,70
28,30
31,60
28,80
25,50
30,10
26,00
31,00
IS
0,97
0,88
0,66
0,52
0,38
0,71
0,46
0,29
0,68
0,96
0,97
0,76
0,44
0,48
0,39
1,10
0,19
0,19
0,19
HYPERTHYROID GROUP
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
BMI
23,50
21,40
21,40
24,80
24,90
25,00
25,00
22,00
27,30
31,40
26,00
28,10
27,70
27,60
IS
0,56
0,89
0,55
0,66
0,11
0,27
0,56
0,80
0,19
0,11
0,13
0,21
0,32
0,01
BMI_0:INSULIN_0: r2 = 0,2083; r = -0,4564, p = 0,0495; y =
1,32639193 - 0,0299838014*x
BMI_0:INSULIN_0: r2 = 0,6006; r = -0,7750, p = 0,0011; y =
2,33571397 - 0,0767480919*x
1,2
Normal Thyroid
Hyperthyroid
1,0
Insulin Sensitivity
0,8
0,6
0,4
0,2
0,0
-0,2
16
18
20
22
24
26
BMI
28
30
32
34
Model Summaryb
GRUP
,00 CONTROL
1,00 HYPERTHYROID
Model
1
1
R
,456a
,775a
R Square
,208
,601
Adjusted
R Square
,162
,567
Std. Error of
the Estimate
,26751
,18553
a. Predictors: (Constant), BMI
b. Dependent Variable: INS_SENS
In the hyperthyroid group, 60.1% of the variation in
insulin sensitivity is explained by BMI. Rest of the
variation is explained by another independent
variable(s).
ANOVAb
GRUP
,00 CONTROL
Model
1
1,00 HYPERTHYROID
1
Reg ression
Residual
Total
Reg ression
Residual
Total
a. Predictors: (Constant), BMI
b. Dependent Variable: INS_SENS
Sum of
Squares
,320
1,217
1,537
,621
,413
1,034
df
1
17
18
1
12
13
Mean Square
,320
,072
F
4,474
Sig .
,049a
,621
,034
18,048
,001a
Coefficientsa
GRUP
,00 CONTROL
Model
1
1,00 HYPERTHYROID
1
(Constant)
BMI
(Constant)
BMI
Unstandardized
Coefficients
B
Std. Error
1,326
,353
-,030
,014
2,336
,462
-,077
,018
Standardized
Coefficients
Beta
-,456
-,775
t
3,755
-2,115
5,054
-4,248
Sig .
,002
,049
,000
,001
a. Dependent Variable: INS_SENS
In hyperthyroid group, each time a woman’s BMI increases
by 1 unit, her predicted insulin sensitivity decreases by
0.077. For example, as the BMI increases from 25 to 30,
insulin sensitivity decreases from about 0.41 to about 0.03.
Predictive values of insulin sensitivity in hyperthyroid group.
BMI I NS_ SENS
24
0, 56
21
0, 89
21
0, 55
25
0, 66
25
0, 11
25
0, 27
25
0, 56
22
0, 80
27
0, 19
31
0, 11
26
0, 13
28
0, 21
28
0, 32
28
0, 01
25
30
PRE_ 2
0, 53
0, 69
0, 69
0, 43
0, 42
0, 42
0, 42
0, 65
0, 24
- 0, 07
0, 34
0, 18
0, 21
0, 22
0, 42
0, 03
RES_ 2
0, 03
0, 20
- 0, 14
0, 23
- 0, 31
- 0, 15
0, 14
0, 15
- 0, 05
0, 18
- 0, 21
0, 03
0, 11
- 0, 21