Upload
kennedy-pickering
View
215
Download
3
Embed Size (px)
Citation preview
1
RESITATION- ONE SAMPLE HYPOTHESIS TESTS
- TWO SAMPLE HYPOTHESIS TESTS (INDEPENDENT AND DEPENDENT SAMPLES)
June 7, 2012
2
Example: A researcher believes that mean hemoglobin value is 12 in the population. He selected 64 adults from population randomly to verify this idea and identified hemoglobin mean as 11.2. According to these findings, is population hemoglobin mean different from 12?
Since
The variable hemoglobin is continuous
The sample size is 64
Hemoglobin is normally distributed
There is only one group
One sample t test is used
3
N Mean Std. Dev.
Hemoglobin 64 11.2 2.8
Hypothesis: H0: = 12
Ha: 12
Sample results:
Population mean: 12
Test statistic: Since the population variance is unknown, our test statistic is t
Level of significance: =0.05
4
29.2=648,212–1
=nsμ–x
=t1.2
t(/2,n-1)= t(0.025,64-1)≈2.00
tcal > ttable Reject H0
Population mean is different from 12.
5
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 1 2 3 4 5 6 7 8 9 10
EnergyIntake
1500 825 1300 1700 970 1200 1110 1270 1460 1090
6
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
7
H0: The population median is 1280.
HA: The population median is not 1280.
Child 1 2 3 4 5 6 7 8 9 10
Energy intakeSign
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
8
Since p > 0.05 we accept H0
We conclude that the median energy intake level in 2 year old children is 1280 kcal.
9
Example: The dean of the faculty wants to know whether the smoking ratio among the Phase I students is 0.25 or not. For this purpose, 50 student is selected and 18 of them said they were smoking. According to this results, can we say that this ratio is different from 0.25 at the 0.05 level of significance?
H0: P=0.25
HA: P0.25
.801=50/)75.0)(25.0(
25.0–36.0=
n/PQP–p
=z
Critical z values are 1.96
1.80<1.96 Accept H0. Smoking ratio among the Phase I student is 0.25.
p=18/50=0.36
One sample z test for proportion / one sample chi-square test
10
We can solve this problem with one sample chi square test at the same time.
Smoking Observed Expected (O-E) (O-E)2 (O-E)2/E
Yes 18 12.5 (50*0.25)
5.5 30.25 2.42
No 32 37.5 (50*0.75)
-5.5 30.25 0.81
Total 50 50 0 3.23
23.3=E
)E–O(=χ
2
1=i i
2ii2 ∑ 0
2cal
2)05.0,1( HAccept ⇒ 84.3
11
18 12,5 5,5
32 37,5 -5,5
50
Yes
No
Total
Observed N Expected N Residual
Test Statistics
3,227
1
,072
Chi-Square a
df
Asymp. Sig.
smoking
0 cells (,0%) have expected frequencies less than5. The minimum expected cell frequency is 12,5.
a.
SPSS Output
Example: In a heart study the systolic blood pressure was measured for 24 men aged 20 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
12
13
Subject Sbp Subject Sbp Subject Sbp Subject Sbp1 95 13 132 1 150 16 1482 122 14 100 2 152 17 1163 130 15 120 3 154 18 1284 148 16 125 4 160 19 1365 130 17 115 5 164 20 1106 150 18 138 6 176 21 1267 105 19 100 7 108 22 1308 110 20 118 8 126 23 1229 130 21 136 9 132 24 140
10 156 22 110 10 142 25 11011 108 23 140 11 136 26 12412 124 24 106 12 146 27 136
13 114 28 12014 118 29 14215 130 30 114
20- year-old 40- year-old
14
24 122,8333 16,7790
30 133,6667 17,3013
GROUP 20- year-old
40- year-old
N Mean Std. Deviation
3024N =
GROUP
40- year-old20- year-old
Mea
n
1 S
D S
BP
160
150
140
130
120
110
100
15
(1) H0:1=2
Ha: 1<2
94.1=F<06.1=54.28133.299
=SS
=F )05.0,24,30(2min
2max
(2) Testing the equality of variances
Accept H0. Variances are equal.
H0:21= 2
2
Ha: 21 2
2
16
2–n+ns)1–n(+s)1–n(
=s21
222
2112
p
46.291=2–30+24
33.299)1–30(+54.281)1–24(=
22
31.2=
3046 . 291
+24
46 . 2910–)67 . 133–83 . 122(
=
ns
+ns
)μ–μ(–)x–x(=t
2
2p
1
2p
2121
(3)
(4) t(52,0.05)=1.675 < p<0.05, Reject H0. 31.2calt
(5) The older men have higher systolic blood pressure
17
Group Statistics
24 122,83 16,779
30 133,67 17,301
Group
20 year old
40 year old
SBP
N Mean Std. Deviation
Independent Samples Test
,013 ,910 -2,317 52 ,024 -10,833 4,675 -20,215 -1,451
-2,325 50,049 ,024 -10,833 4,659 -20,191 -1,475
Equal variances assumed
Equal variances notassumed
SBP
F Sig.
Levene's Test forEquality of Variances
t df Sig. (2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Difference
t-test for Equality of Means
SPSS Output
18
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.
19
5 women were selected randomly in each groups and their scores are as follows:
Group Score
Women without a block
148837270
Women with a paracervical block
5070376675
20
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
21
IIIA
III
H
H
≠:
:0
Group Score RankI 0 1I 14 2I 27 3I 37 4.5II 37 4.5II 50 6II 66 7II 70 8II 75 9I 88 10
R1= 1+2+3+4.5+10 = 20.5
5.19=5.20–2
)1+5(5+5×5=
R–2
)1+n(n+nn=U 1
11211
From the table, critical value is 21
19.5 < 21 accept H0
5.5=5.19–5×5=U–nn=U 1212
5.19=U
We conclude that the median pain/ cramping scores are same in two groups.
22
Ranks
5 4,10 20,50
5 6,90 34,50
10
Group2
1
2
Total
VAS
N Mean Rank Sum of Ranks
Test Statisticsb
5,500
20,500
-1,467
,142
,151a
Mann-Whitney U
Wilcoxon W
Z
Asymp. Sig. (2-tailed)
Exact Sig. [2*(1-tailed Sig.)]
VAS
Not corrected for ties.a.
Grouping Variable: Group2b.
Descriptive Statistics
VAS
5 0 33,20 27,00 33,641 0 88 7,00 27,00 62,50
5 0 59,60 66,00 15,726 37 75 43,50 66,00 72,50
Group
1
2
Valid Missing
N
Mean Median Std. Deviation Minimum Maximum 25 50 75
Percentiles
SPSS Output
23
GroupCervical blockWithout block
VAS
100
80
60
40
20
0
2
24
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
Number in sample
Number anemic
Proportion
1 450 105 0.23
2 375 120 0.32
The difference between two population proportion
25
0P–P:H0=P–P:H
12a
120
≠
27.0=375+450
)32.0)(375(+)23.0)(450(=p
32.0=375/120=p23.0=450/105=p
2
1
0.025<0.0027=p 78.2=
375)73.0)(27.0(
+450
)73.0)(27.0(0–0.32)–.230(
=z
Reject H0
We concluded that the proportion of anemia is different in two geographic areas.
26
Example: A study was conducted to analyze the relation between coronary heart disease (CHD) and 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:
+ -
Yes
No
Total 90
SmokingTotalCHD
30
4 46
14 76
40
50
10
27
Observed and expected frequencies
+ -
Yes
No
Total 90
SmokingTotal
CHD
30
4 46
14 76
40
50
10 6.2 33.8
7.8 42.2
( ) ( ) ( ) ( )95.4=
2.422.42–46
+8.7
8.7–4+
8.338.33–30
+2.6
2.6–10=
E)E–(O
=χ
2222
2
1=i
2
1=j ij
2ijij2 ∑∑
28
df = (r-1)(c-1)=(2-1)(2-1)=1
2(1,0.05)=3.841
Conclusion: There is a relation between CHD and smoking.
2 =4. 95 > reject H0
29
SPSS Output
CHD * Smoking Crosstabulation
10 30 40
6,2 33,8 40,0
4 46 50
7,8 42,2 50,0
14 76 90
14,0 76,0 90,0
Count
Expected Count
Count
Expected Count
Count
Expected Count
Yes
No
CHD
Total
Yes No
Smoking
Total
Chi-Square Tests
4,889b 1 ,027
3,681 1 ,055
4,937 1 ,026
,040 ,027
4,835 1 ,028
90
Pearson Chi-Square
Continuity Correctiona
Likelihood Ratio
Fisher's Exact Test
Linear-by-LinearAssociation
N of Valid Cases
Value dfAsymp. Sig.
(2-sided)Exact Sig.(2-sided)
Exact Sig.(1-sided)
Computed only for a 2x2 tablea.
0 cells (,0%) have expected count less than 5. The minimum expected count is6,22.
b.
30
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.
31
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
32
New Standard Difference49 54 -556 42 1470 63 783 77 683 83 068 51 1784 82 263 54 967 62 579 71 888 82 648 50 -252 41 1173 67 652 57 -573 70 378 72 664 62 271 64 742 44 -251 44 756 42 1440 35 581 73 8
Total 129Mean 65,46 60,08 5,38SD 14,38 14,46 5,65
H0: d = 0
Ha: d > 0
38.5=24
129=
n
d=d i∑
90.31=1–n
)d–di(=s
22d
∑
66.4= 24/90.310–38.5
=n/s
μ–d=t
d
d
t(23,0.05)=1.714
We conclude that the new treatment is effective.
Since, reject H0.tablecalculated t>t
33
Paired Samples Statistics
65,46 24 14,380 2,935
60,08 24 14,461 2,952
New
Standard
Pair 1
Mean N Std. DeviationStd. Error
Mean
Paired Samples Test
5,375 5,648 1,153 2,990 7,760 4,662 23 ,000New - StandardPair 1
Mean Std. DeviationStd. Error
Mean Lower Upper
95% ConfidenceInterval of the Difference
Paired Differences
t df Sig. (2-tailed)
SPSS Output
34
StandardNew
Dig
ital D
exte
rity
Test
Sco
res
(Mea
n±SD
)
80
70
60
50
40
35
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:
SubjectWeights
Before After
1 85 822 91 923 68 624 76 735 82 816 87 837 105 858 93 889 98 90
Since
The variable concerning weight is continous.
The sample size is less than 10
There are two groups and they are dependent
Wilcoxon signed ranks test
36
SubjectWeights Difference
Di
SortedDi Rank
SignedRankBefore After
1 85 82 3 -1 1.5 -1.52 91 92 -1 1 1.5 1.53 68 62 6 3 3.5 3.54 76 73 3 3 3.5 3.55 82 81 1 4 5 56 87 83 4 5 6 67 105 85 20 6 7 78 93 88 5 8 8 89 98 90 8 20 9 9
37
T = 1.5
reject H0 , p<0.05T = 1.5 < T(n=9,a =0.05) = 6
We conclude that the diet is effective.
38
AfterBefore
110
100
90
80
70
60
3
39
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 IDevice II
Total
Arrhythmia (+) Arrhythmia (-)
Arrhythmia (+) 10 3 13
Arrhythmia (-) 13 9 22
Total 23 12 35
The significance test for the difference between two dependent population / McNemar test
40
H0: P1=P2
Ha: P1 P2
25.2=13+3
1––3=
c+b1–c–b
=z
13
Critical z value is ±1.96 Reject H0
41
McNemar test approach:
c+b)c–b(
=χ2
2
c+b)1–c–b(
=χ2
2
1.5=13+3
)1–13–3(=
c+b)1–c–b(
=χ22
2
2(1,0.05)=3.841<5.1 p<0.05; reject H0.
42
Device Arrhythmia Crosstabulation
Count
10 3 13
13 9 22
23 12 35
Yes
No
DeviceI
Total
Yes No
DeviceII
Total
Chi-Square Tests
,021a
35
McNemar Test
N of Valid Cases
Value dfAsymp. Sig.
(2-sided)Exact Sig.(2-sided)
Binomial distribution used.a.
SPSS Output