資工系 Lecture#11-2 p.1
生物統計學 96-1 2008/01/02
生物統計學 11-2：SPSS for Descriptive Statistics & Testing
01/02/2008
O. Descriptive Statistics
Exercise 3.12 (a), (b), (c)。（p. 63）- data set – unicef.dat
Percentages of low birth weight infants – weighing less than 2500 gram.
Variables – nation, lowbwt, life60, life92
(a) Compute the mean and the median. (b) Compute the 5% trimmed mean.
(c) For this data set, which number would you prefer as a measure of central tendency?
Result of SPSS:
Descriptive Statistics
N
Minimum
percent low birth weight
111
Valid N (listwise)
111
Maximum
4
50
Mean
Std. Deviation
11.96
6.524
Case Processing Summary
Valid
N
percent low birth weight
111
Percent
77.1%
Cases
Missing
N
Percent
33
22.9%
Total
N
144
Percent
100.0%
De scri ptives
percent low birt h weight Mean
95% Confidenc e
Int erval for Mean
5% Trimmed Mean
Median
Variance
St d. Deviat ion
Minimum
Maximum
Range
Int erquartile Range
Sk ewness
Kurtos is
Lower Bound
Upper Bound
St atist ic
11.96
10.74
St d. E rror
.619
13.19
11.38
10.00
42.562
6.524
4
50
46
8
2.283
9.941
.229
.455
資工系 Lecture#11-2 p.2
生物統計學 96-1 2008/01/02
His to g ra m
50
Fre q u e n c y
40
30
20
10
Me a n = 11.96
S td. De v. = 6.524
N = 111
0
0
10
20
30
40
50
p e rc e n t lo w b irth we ig h t
percent low birth weight Stem-and-Leaf Plot
Frequency
Stem & Leaf
.00
0 .
9.00
0 . 444455555
21.00
0 . 666666666677777777777
17.00
0 . 88888888999999999
18.00
1 . 000000000111111111
6.00
1 . 222333
13.00
1 . 4444455555555
12.00
1 . 666666677777
3.00
1 . 899
7.00
2 . 0000011
1.00
2 . 3
2.00
2 . 55
2.00 Extremes
(>=33)
Stem width: 10
50
Each leaf: 1 case(s)
10
40
56
30
20
10
0
pe rce nt low birth we ight
資工系 Lecture#11-2 p.3
生物統計學 96-1 2008/01/02
Exercise 3.15 (a), (b), (c)。（p. 64）- data set – lowbwt.dat
100 low birth weight infants in two hospitals in Boston, MA.
Variables – sbp (systolic blood pressure), sex (gender)
(a) Construct a pair of box plots for the SBP – one for boys and one for girls.
(b) Compute the mean and s.d. of SBP for males and for females. Compare them.
(c) Compute the C.V. of SBP for each gender. Compare them. (CVf – 24%, CVm – 25%)
Case Processing Summary
Valid
sys tolic blood press ure
gender
Female
Male
N
Percent
100.0%
100.0%
56
44
Cases
Mis sing
N
Percent
0
.0%
0
.0%
16
80
s ys to lic b lo o d p re s s u re
41
60
40
46
20
Fe ma le
Ma le
gender
His to g ra m
for s e x= Fe ma le
30
25
Fre q u e n c y
20
15
10
5
Me a n = 46.46
S td. De v. = 11.145
N = 56
0
10
20
30
40
50
s ys to lic b lo o d p re s s u re
60
70
Total
N
56
44
Percent
100.0%
100.0%
資工系 Lecture#11-2 p.4
生物統計學 96-1 2008/01/02
His to g ra m
for s e x= Ma le
20
Fre q u e n c y
15
10
5
Me a n = 47.86
S td. De v. = 11.806
N = 44
0
20
30
40
50
60
70
80
90
s ys to lic b lo o d p re s s u re
Report
sys tolic blood press ure
gender
Female
Male
Total
Mean
46.46
47.86
47.08
N
56
44
100
Median
47.00
46.00
47.00
Minimum
19
27
19
Maximum
67
87
87
Std. Deviation
11.145
11.806
11.403
資工系 Lecture#11-2 p.5
生物統計學 96-1 2008/01/02
I. Hypothesis Testing for One Sample
Exercise 10.16 (a), (b) , (c)。（p. 282）- data set – heart.dat
Infant development study on two scores of index - PDI (Psychomotor Development
Index), MDI (Mental Development Index); variables: trtment (0 - ca; 1 - lf), pdi, mdi.
(a) At α = 0.05, is the mean PDI score equal to 100, the mean score for healthy infants?
(b) Test that the mean MDI score for the analogous test of hypothesis.
(a) H0 :μ= 100 vs H1 :μ≠ 100, the p-value = 0.000. We reject H0 :μ= 100 at α = 0.05
level. In fact, the mean PDI is less than 100, and the 95% CI is [100 – 7.84, 100 – 2.60].
Result of SPSS:
One-Sample Statistics
N
ps ychomotor
development index
Mean
143
Std. Deviation
Std. Error
Mean
15.851
1.326
94.78
One-Sample Test
Test Value = 100
t
ps ychomotor
development index
df
-3.936
Sig. (2-tailed)
Mean
Difference
.000
-5.217
142
95% Confidence
Interval of the
Difference
Lower
Upper
-7.84
-2.60
(b) H0 :μ= 100 vs H1 :μ≠ 100, the p-value = 0.000. We reject H0 :μ= 100 at α = 0.05
level. The mean MDI is higher than 100, and the 95% CI is [100 + 2.17, 100 + 7.31].
One-Sample Statistics
N
mental
development index
Mean
144
104.74
Std. Deviation
Std. Error
Mean
15.604
1.300
One-Sample Test
Test Value = 100
t
mental
development index
3.642
df
143
Sig. (2-tailed)
Mean
Difference
.000
4.736
95% Confidence
Interval of the
Difference
Lower
Upper
2.17
7.31
資工系 Lecture#11-2 p.6
生物統計學 96-1 2008/01/02
II. Hypothesis Testing for Matched or Paired Samples
Example 1: Starting salaries for environmental jobs, in dollars, for 5
positions requiring no degree are as follows:
Position
Salary
1
2
3
4
5
32,000
25,000
54,000
40,000
45,000
The starting salaries for the same position for persons holding a degree in
geography or computer science are:
Position
Salary
1
2
3
4
5
32,000
26,000
55,000
43,000
49,000
Result of SPSS:
Paired Samples Statistics
Pair
1
Mean
39200.00
41000.00
salary1
salary2
N
5
5
Std. Deviation
11256.109
11937.336
Std. Error
Mean
5033.885
5338.539
Pa ired Sa mples Correlations
N
Pair 1
salary1 & s alary2
5
Correlation
.992
Sig.
.001
Pa ired Sa mpl es Test
Paired Differenc es
Pair 1
Mean
salary1 - s alary 2 -1800.000
test statistic td =
St d. Deviat ion
1643.168
St d. Error
Mean
734.847
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
-3840.262
240.262
xd  d
= 1800/(1643/ 5 ) = 2.45,
sd / n
The p-value = 0.0352 = 0.070/2
t
-2. 449
df = n – 1 = 4
(one-sided p-value = two-sided /2)
df
4
Sig. (2-tailed)
.070
資工系 Lecture#11-2 p.7
生物統計學 96-1 2008/01/02
Example 2: The study of males with coronary artery disease.
For each of the 63 men, calculate the percent decrease in time to angina
when he is exposed to carbon monoxide minus the percent decrease when
he is exposed to unadulterated air. (exposure to air vs exposure to co4%)
Result of SPSS:
Paired Samples Statistics
Pair
1
100*(air1-air2)/air1
(@4co1-@4co2)/@4co1
Mean
.957895
7.586281
N
Std. Deviation
17.4402749
16.8956695
63
63
Std. Error
Mean
2.1972681
2.1286543
Paired Samples Correlations
N
Pair
1
100*(air1-air2)/air1 &
(@4co1-@4co2)/@4co1
Correlation
63
Sig.
.302
.016
Paired Samples Test
Paired Differences
Mean
100*(air1-air2)/air1 (@4co1-@4co2)/@4co1
test stat. td =
Std. Error
Mean
20.2911395
2.5564433
-6.62839
-11.7386
-1.51813
xd  d
= (-6.63 – 0)/(20.29/ 63 ) = -2.59,
sd / n
The p-value = 0.006 = 0.012/2
t
-2.593
df
Sig. (2-tailed)
62
df = n – 1 = 62
(one-sided p-value = two-sided /2)
(The t62, 0.0125 = -2.299, t62, 0.005 = -2.660 for one-sided t-test, so 0.005 < p-value < 0.0125)
40
35
30
Fre q u e n c y
Pair
1
Std. Deviation
95% Confidence
Interval of the
Difference
Lower
Upper
25
20
15
10
5
Me a n = -6 .6 2 8 4
S td . De v. = 2 0 .2 9 1 1 4
N = 63
0
-8 0 .0 0
-6 0 .0 0
-4 0 .0 0
-2 0 .0 0
0 .0 0
d iff1 -d iff2
2 0 .0 0
4 0 .0 0
6 0 .0 0
.012
資工系 Lecture#11-2 p.8
生物統計學 96-1 2008/01/02
Further Application
Example 2-1: The study of males with coronary artery disease.
(exposure to air vs exposure to co2%; there was one missing data, so n = 62)
Result of SPSS:
Paired Samples Statistics
Pair
1
Mean
.9254
5.8738
100*(air1-air2)/air1
(@2co1-@2co2)/@2co1
N
62
62
Std. Deviation
17.58073
14.18603
Std. Error
Mean
2.23275
1.80163
Paired Samples Correlations
N
Pair
1
100*(air1-air2)/air1 &
(@2co1-@2co2)/@2co1
Correlation
62
.296
Sig.
.020
Paired Samples Test
Paired Differences
Mean
Pair
1
100*(air1-air2)/air1 (@2co1-@2co2)/@2co1
test stat. td =
-4.94833
Std. Deviation
Std. Error
Mean
19.04708
2.41898
95% Confidence
Interval of the
Difference
Lower
Upper
-9.78538
-.11128
xd  d
= (-4.95 – 0)/(19.05/ 62 ) = -2.59,
sd / n
The p-value = 0.0225 = 0.045/2
t
-2.046
df = n – 1 = 61
(one-sided p-value = two-sided /2)
df
Sig. (2-tailed)
61
.045
資工系 Lecture#11-2 p.9
生物統計學 96-1 2008/01/02
III. Hypo. Testing for Independent Groups: Two Population Means
Exercise 11.14 (a), (b)。（p. 281）- data set – lowbwt.dat
100 low birth weight infants in two hospitals in Boston, MA.
Variables – sbp (systolic blood pressure), sex (gender)
(a) A histogram of SBP for this sample. Is it approximately normally distributed?
(b) At α = 0.05, is the mean SBP for boys equal to the mean for girls?
(a) Yes, the values appear to be fairly symmetric, and could be considered to be
approximately normally distributed.
50
Frequency
40
30
20
10
Me a n = 4 7 .0 8
S td . De v. = 1 1 .4 0 3
N = 100
0
0
20
40
60
80
100
s y s t o lic b lo o d p r e s s u r e
(b) H0 :μf =μm vs H1 :μf ≠μm , the p-value = 0.545 for equal variance assumed. We
are unable to reject H0 :μf =μm at the 0.05 level of significance.
Result of SPSS:
Group Statistics
sys tolic blood press ure
gender
Female
Male
N
56
44
Mean
46.46
47.86
Std. Deviation
11.145
11.806
Std. Error
Mean
1.489
1.780
Independent Samples Test
Levene's Test for
Equality of Variances
F
systolic blood press ure
Equal variances
as sumed
Equal variances
not ass umed
.079
Sig.
.779
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-.607
98
.545
-1.399
2.305
-5.973
3.174
-.603
89.858
.548
-1.399
2.321
-6.010
3.211
資工系 Lecture#11-2 p.10
生物統計學 96-1 2008/01/02
Exercise 11.15 (a), (b) , (c)。（p. 282）- data set – heart.dat
Infant development study on two scores of index - PDI (Psychomotor Development
Index), MDI (Mental Development Index); variables: trtment (0 - ca; 1 - lf), pdi, mdi.
(a) At α = 0.05, is the mean PDI score for circulatory arrest group (ca) equal to the
mean for low-flow bypass group (lf)? What is the p-value?
(b) Test that the mean MDI scores are identical for the two treatment groups.
(a) H0 :μca =μlf vs H1 :μca ≠μlf , the p-value = 0.027 for equal variance assumed. We
reject H0 :μca =μlf at α = 0.05 level; mean PDI is higher for the low-flow group.
Result of SPSS:
Group Statistics
ps ychomotor
development index
treatment group
CA
LF
N
Mean
91.92
97.77
73
70
Std. Deviation
16.488
14.685
Std. Error
Mean
1.930
1.755
Independent Samples Test
Levene's Test for
Equality of Variances
F
ps ychomotor
development index
Equal variances
as sumed
Equal variances
not ass umed
2.815
t-test for Equality of Means
Sig.
t
.096
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-2.239
141
.027
-5.854
2.615
-11.023
-.684
-2.244
140.247
.026
-5.854
2.609
-11.011
-.696
(b) H0 :μca =μlf vs H1 :μca ≠μlf , the p-value = 0.214 for equal variance assumed.
Group Statistics
mental
development index
treatment group
CA
LF
N
Mean
103.16
106.40
74
70
Std. Deviation
16.465
14.573
Std. Error
Mean
1.914
1.742
Independent Samples Test
Levene's Test for
Equality of Variances
F
mental
development index
Equal variances
as sumed
Equal variances
not ass umed
.404
Sig.
.526
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-1.247
142
.214
-3.238
2.597
-8.371
1.895
-1.251
141.386
.213
-3.238
2.588
-8.354
1.878
資工系 Lecture#11-2 p.11
生物統計學 96-1 2008/01/02
Example 3 in Lecture#11
Source: http://archpedi.ama-assn.org/cgi/content/abstract/143/8/969
American J. of Diseases of Children Vol. 143 No. 8, August 1989
or
Archives of Pediatrics and Adolescent Medicine Vol. 143 No. 8, August 1989
Effect of pancreatic enzyme supplements on iron absorption
W. T. Zempsky, B. J. Rosenstein, J. A. Carroll and F. A. Oski
Department of Pediatrics, Johns Hopkins University School of Medicine, Baltimore, Md.
Iron deficiency has been reported in one third of patients with cystic
fibrosis. There are data that suggest that iron absorption is increased
with exocrine pancreatic deficiency and that administration of
pancreatic enzymes may impair oral iron absorption. We compared oral
iron absorption over a 3-hour period in the presence and absence of
exogenous pancreatic enzymes in 13 stable young-adult patients with
cystic fibrosis and 9 age-matched control patients. Although none of
the patients with cystic fibrosis had a hemoglobin level less than 119
g/L, serum ferritin levels were less than 25 micrograms/L in 5 of the 13
patients, and the mean corpuscular volume was significantly lower in
the patient group (86.1 +/- 2.7 vs 90.9 +/- 5 fL). Baseline mean serum
iron levels were higher in controls (18.9 +/- 5.9 mumol/L) than in
patients (11.9 +/- 6.3 mumol/L). There was no difference in iron
absorption in the absence of exogenous pancreatic enzymes.
Significant impairment of iron absorption was detected in both patients
with cystic fibrosis and controls after administration of a preparation of
pancreatic enzymes. There was an inverse relationship between iron
stores, as measured by serum ferritin, and iron absorption. These
findings suggest that long-term consumption of pancreatic enzymes by
patients with cystic fibrosis may contribute to iron deficiency. (no data)
資工系 Lecture#11-2 p.12
生物統計學 96-1 2008/01/02
Example 4 in Lecture#11: http://www.nlm.nih.gov/databases/alerts/hypertension.html
Results of the NHLBI and NIA clinical trial of the efficacy of
antihypertensive treatment for isolated systolic hypertension in those
aged 60 years and older
National Heart, Lung, and Blood Institute (NHLBI)
National Institute on Aging (NIA) June 26, 1991
Abstract:
NIH's National Heart, Lung, and Blood Institute and the National
Institute on Aging have announced a treatment that reduces the
complications of isolated systolic hypertension. This form of high blood
pressure, which affects more than 3 million people aged 60 years and
above, can be treated with a stepped-care regimen of low dose
chlorthalidone and atenolol with a resultant reduction of stroke (fatal
plus non-fatal).
The Systolic Hypertension in the Elderly Program (SHEP) was a 5-year
double-blind randomized trial that assessed the ability of
antihypertensive drug treatment to reduce risk of nonfatal plus fatal
(total) stroke in men and women age 60 and above with isolated
systolic hypertension. Persons with average systolic blood pressure
160-219 mm Hg and average diastolic blood pressure 90 mmHg were
eligible. SHEP randomized 4,736 people to either active treatment
(2,365 persons) or placebo (2,371). The two treatment groups were
comparable at baseline. Mean baseline age was 71.6 years; 47.5%
were white women, 38.6% white men, 9.3% black women, and 4.6%
black men. Mean systolic blood pressure was 170 mmHg: mean
diastolic blood pressure, 77 mm Hg. For the active treatment group, a
stepped-care regimen was used which included chlorthalidone 12.5 or
25 mg/day, and as needed, addition of atenolol 25 or 50 mg/day or
reserpine, 0.05 or 0.10 mg/day. Treatment goal was to reduce systolic
blood pressure by at least 20 mm Hg from baseline and to below 160
mm Hg with minimal amounts of study medication. Five-year average
systolic blood pressures was 155 mmHg for the placebo group and 143
mm Hg for the active treatment group; diastolic blood pressure values
were 72 mmHg and 68 mm Hg respectively. Goal blood pressure was
achieved by about 70% of the actively treated group and about 35% of
the placebo group during the trial. Average follow-up was 4.5 years.
Five-year incidence of total stroke was 5.2/100 for active treatment
and 8.2/100 for placebo. The relative risk as computed by proportional
hazards regression was 0.64, 95% confidence interval 0.51 to 0.82, p
= 0.0003. The results were similar across age and race-sex subgroups.
資工系 Lecture#11-2 p.13
生物統計學 96-1 2008/01/02
For SHEP secondary end points, rates were again lower for active
treatment than placebo, e.g., clinical nonfatal myocardial infarction
plus coronary death relative risk was 0.73 (95% confidence interval
0.57 - 0.94); combined major cardiovascular events, relative risk is
0.68 (95% confidence interval 0.58 - 0.79); deaths from all causes,
relative risk 0.87 (95% confidence interval 0.73 -1.05). Occurrence of
adverse effects was of low order, and findings from behavioral
evaluation (including dementia and depression) were similar
throughout the trial for the two treatment groups.
Two additional secondary questions were stated at the time of trial
planning as subgroup hypotheses. First, would treatment of ISH reduce
the frequency of total stroke (fatal and nonfatal) similarly in those
receiving and not receiving antihypertensive medication at initial
contact? Second, would treatment of ISH reduce the incidence of
sudden cardiac death or of coronary plus nonfatal myocardial infarction
similarly in those free of baseline ECG abnormalities and in those with
such abnormalities? Reduction in stroke incidence for those on and not
on medication at initial contact was similar during the trial. The
reduction in nonfatal myocardial infarction plus coronary death was
greater in those with ECG abnormalities at baseline than in those
without such abnormalities. The results for sudden death and for
nonfatal myocardial infarction plus coronary death were similar in the
two groups.
Conclusion: In persons aged 60 years and over with isolated systolic
hypertension antihypertensive stepped-care drug treatment with low
dose chlorthalidone as Step 1 medication reduces incidence of total
stroke by 36% with 5-year absolute benefit estimated at 30/1,000
participants. Major cardiovascular events are also reduced, with 5-year
absolute benefit estimated as 55/1000 participants.
Implications: The SHEP trial conclusively demonstrates that ISH can
be effectively and safely treated with a stepped-care regimen including
low dose diuretic, chlorthalidone, and low dose beta-blocker, atenolol.
These are low cost medications which should be considered as the first
choice of treatment for ISH. Other medication regimens have not been
tested for efficacy in the treatment of ISH. (no data)