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2015-11-09
Biostatistics for the
biomedical profession
Lecture 4
BIMM34
Karin Källen & Linda Hartman
November 2015
1
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
2
•
•
•
•
•
•
The mean, median, and mode all have
the same value
The curve is symmetric around the
mean; the skew and kurtosis is 0
The curve approaches the X-axis
asymptotically
Mean ± 1 SD covers 2*34.1% of data
Mean ± 2 SD covers 2*47.5% of data
Mean ± 3 SD covers 99.7% of data
3
The (perfect) normal distribution
2015
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Excercise:
Mean:
SD:
50.5 cm
2.3 cm
• +2 z-scores:
50.5 + 2*2.3 = 55.1cm
4
1. Estimate the value corresponding to a z-score of -2 and +2,
respectively, for birth length, based on the following sample
measurements:
• -2 z-scores:
50.5 – 2*2.3 = 45.9cm
2. Estimate the z-score corresponding to a birth length of 48 cm.
• (48-50.5)cm / 2.3cm = - 1.1
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Birth length example, continued
95% of all Swedish
babies will have a
birth lenght betwen
45.9 and 55.1cm
Exercise:
How large proportion of the
population will have a birth length
above 48 cm?
• Hint: approximate that 70% av
all will be between +/- 1.1 SD
• Approx 85%
5
A birth lenght of 48cm
corresponds to a z-value of -1.1.
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The standard error of the mean is a measurement of the spread of the mean.
.
6
Standard error of the mean
(SEM)
Let
SD= the population standard deviation
n=sample size
Then
SEM=
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Confidence interval
 E.g., a 95% confidence interval tells us between which
limits the ’true’ estimate (with 95% certainty) lies.
7
 A confidence interval tells us within which interval the
’true’ estimate of a parameter probably lies-
 Repetition: 95% of the data will lie between +/- 2 SD
(1.96 exactly).
 A 95% CI could be constructed (assuming large sample):
(mean-1.96*SEM to mean+1.96*SEM)
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Excercise, birth lengths, continued:
Mean: 50.5 cm
• +2 z-scores:
50.5 + 2*2.3 = 55.1cm
SD:
2.3 cm
2. Estimate the z-score corresponding to a birth length of 48 cm.
• (48-50.5)cm / 2.3cm = - 1.1
3. How large proportion (approximately) of the infants will have a birth
below 48 cm?
• Approx 15%
4. The mean (50.5cm) was based on 77 000 births.
A)Construct a 95% CI for mean birth lenght in this sample
• 50.5 +/- 1.96 * SEM where SEM=2.3/√70 000 = 0.0087
• Mean (95%CI)= 50.5 (50.48 – 50.52)
B)Construct a 95% reference interval for birth length
50.5 +/- 1.96*SD: (46.0 – 55.0)cm
8
1. Estimate the value corresponding to a z-score of -2 and +2,
respectively, for birth length, based on the following sample
measurements:
• -2 z-scores:
50.5 – 2*2.3 = 45.9cm
2015
-1109
9
The 95%CI are often represented by
error bars
Statistical inference: Is there a true difference between the means?
2015
-1109
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
10
Hypothesis testing
• In order to get significant findings, we want to reject H0 ex. no effect
If H0 is rejected, the alternative hypothesis is left (H1)
• The p-value is the probability that you get that result you got or even more
extreme if H0 is true.
• P-value is a probability between 0% and 100%
11
• Set up a null hypothesis (H0): No effect
• Set up an alternative hypothesis (H1): Effect
 If the p-value is small enough:  reject H0
 Small enough is something you decide before the analysis, significance level
Ex. 1%, 5% or 10%
 Calculation of the p-value can be done even if the data is not normally
distributed, but in different ways
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Elements of statistical inference
Sample 2: 𝑥12 , …, 𝑥𝑛2
Mean= 𝑥 2
St dev = s2
H0: μ1 = μ2, i.e. μ1 - μ2 = 0
H1: μ1 ≠ μ2, i.e. μ1 - μ2 ≠ 0
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Sample 1: 𝑥11 , …, 𝑥𝑛1
Mean= 𝑥 1
St dev = s1
• The p-value is the probability of obtaining a test statistic at least as
extreme as the one that was actually observed, assuming that the
null hypothesis (H0)is true.
• The test statistic is 𝑥 1-𝑥 2
• If H0 is true 𝑥 1-𝑥 2 is a sample from N(0, SE(𝑥 1-𝑥 2)
12
H0: μ1 = μ2, i.e. μ1 - μ2 = 0
H1: μ1 ≠ μ2, i.e. μ1 - μ2 ≠ 0
_____ Expected distribution of 𝑥 1-𝑥 2
_ _ _ _ Sample mean, 𝑥 1-𝑥 2
• Large p-value -> Data probable if H0
is true
• Thus: H0 is not rejected (= considered
to be true until further evidence)
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Elements of statistical inference,
hypothesis testing, two-sided
The p-value is the
probability that you get that
result you got or even more
extreme if H0 is true.
0
𝑥 1-𝑥 2
13
Elements of statistical inference,
hypothesis testing, two-sided
• Data unlikely if H0 is true.
• Thus: H0 is rejected (=considered to
be false)
_____ Expected distribution of 𝑥 1-𝑥 2
_ _ _ _ Sample mean, 𝑥 1-𝑥 2
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H0: μ1 = μ2, i.e. μ1 - μ2 = 0
H1: μ1 ≠ μ2, i.e. μ1 - μ2 ≠ 0
The p-value is the
probability that you get that
result you got or even more
extreme if H0 is true.
0
𝑥 1-𝑥 2
14
Uses the fact that the difference between two means that both
comes from normally distributed data, will follow a normal
distribution with expected mean=0,and expected standard
deviation SE.
t=
Difference between group means
Standard error of difference in means
SE=s√(1/n0+1/n1); s=
√[
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The t-test
= x1 - x0
SE
(n1-1)s12 +(n0-1)*s02 .
(n1 + n0 – 2)
]
The test statistic (t) could then be compared with the student’s
t- distribution. If the samples are large, the test statistic will
follow a normal distribution (small samples will follow a t(df)
distribution, where df=n1+n2-2).
15
The t-test can be used to determine if two small sets of normal
distributed data are significantly different from each other, and
the spread is not known but instead estimated from the data.
Under certain conditions, the test statistic will follow a
Student's t distribution.
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The t-test
16
Elements of statistical inference,
hypothesis testing, two-sided, large samples
•
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H0: μ1 = μ2, i.e. μ1 - μ2 = 0
H1: μ1 ≠ μ2, i.e. μ1 - μ2 ≠ 0
1.96
𝑥 1−𝑥 2
𝑆𝐸
17
Expected mean difference = μ1 - μ2 =0
Difference between samples 𝑥 1-𝑥 2
-1.96
0
Z-value below -1.96 or above +1.96 
p(𝑥 1-𝑥 2 from distribution 0) =(.0249 + .0249)
= 0.05
• Our data unlikely if H0 is true 
H0 is rejected
Elements of statistical inference,
hypothesis testing, two-sided, large samples
•
2015-11-09
H0: μ1 = μ2, i.e. μ1 - μ2 = 0
H1: μ1 ≠ μ2, i.e. μ1 - μ2 ≠ 0
1.96
𝑥 1−𝑥 2
𝑆𝐸
18
Expected mean difference = μ1 - μ2 =0
Difference between samples 𝑥 1-𝑥 2
-1.96
0
Z-value below -1.96 or above +1.96 
p(𝑥 1-𝑥 2 from distribution 0) =(.0249 + .0249)
= 0.05
• Our data unlikely if H0 is true 
H0 is rejected
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Example:
Do the birth weight differ significantly between girls
and boys? (data from the birth-database)
Discuss: What do you conclude regarding the distributions?
Would a t-test be suitable?
19
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T-test, birth weight boys vs girls, continued
SPSS-output:
Significance of the T-test
birthweight boys vs girls
Difference between the
means, with 95%CI for
the difference
20
Just to decide which row to read from. If Sig. Is large (e.g., >.1): read from the upper row
Statistical significance vs.
clinical relevance
Statistical significance:
”There is a difference”
2015-11-06
• Low p-value
• How large is the difference?
Clinical relevance:
”Is the difference of importance?”
Effect estimation (CI) is needed!
21
• The p-value does not tell us anything about the size of
the effect. Only how probable it is to obtain an effect of
the size in our sample if the null hypothesis is true
• The P-value is a function of both sample size and of true
effect size
2015-11-09
Important notes….
• With large samples, statistically significant results could
be found even if the size of the absolute effects are so
small that they are of no clinical interest.
• The fact that no significant results were found, does not
mean that no difference exists. Perhaps the study had to
low power to detect a true difference/effect/association.
22
1. In a sample with two arms with 200 000
observations each, a statistically significant 25g
weight loss difference between two diet
groups was detected.
A. High risk for a Type II-error
2. In a sample with 300 individuals, the
associations between 50 environmental toxins
and 5 different outcomes (cancer, psychiatric
disorders etc etc) were investigated. Two
strong associations, and 10 associations of
moderate strength were statistically significant.
3. In a sample with 300 individuals, the risk for
ADHD after exposure with a certain pollutant
during pregnancy was investigated. In the
exposed group (n=100) 4 children had ADHD,
compared to 6 in the control group. The
difference was not statistically significant
(p=0.28).
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Exercise: Combine the results
with the comments
B. High risk for a Type I- error
C. Statistical significant results
without clinical significance.
23
Multiple T-tests could result in mass-significance!
Do ANOVA instead of repeated T-tests.
•
•
•
ANOVA:
H0: Mean1=mean2=mean3
H1: At least two of the means are different
•
In short: In an ANOVA, the total variance is devided
into the within-groups, and between-groups
variance.
2015-11-09
Repetition:
More than two groups, t-test assumptions holds,
one way ANOVA (analysis of variance)
24
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
25
ANOVA
• Compare variances
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• Between groups (VB)
• Within groups (VW)
VB
VW
Ratio VB/VW
Large
Small
Large
Small
Large
Small
The quotient (F=VB/VW) is equal to 1 if group means are equal and >1 if they are not.
26
The corresponding test is called an F-test – and is based on the F-distribution
ANOVA example birthweights
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Descriptives:
Created with
Analyze > Tables >
CustomTables in SPSS
Exercise:
Interpret the output in words
27
ANOVA example birthweights
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Descriptives:
“A one-way between subjects ANOVA
was conducted to compare the effect
of smoking mothers (non-smokers,
smokers(≤10 cig/day), smokers (>10
cig/day)) on birthweight . There was
no significant effect of smoking on
birth weight at the p<.05 level for the
three groups [F(2, 246) = 2.73, p =
0.067].”
28
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
29
2015-11-09
Cotinine is the main metabolite from nicotine. The graph shows
the cotinine levels in the umbelical artery.
Mean = 31.2
Median=0.2
30
2015-11-09
Child cotinine levels, continued…
31
Exercise: Normal distribution?
NO!
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Exercise:
1. Would it be possible to nevertheless use a T-test to
compare the cotinine level among children of smoker versus
non-smokers? Under what conditions?
32
Repetition –
Central Limit Theorem
• no. of observations is large (faster if distribution is symmetric)
• Independent observations
• from the same distribution
We could often use
normal distribution to

2015-11-09
• Mean has approximately normal distribution if
test difference in mean
– even if observations are not
normal
10000 samples of mean values from
dice-rolls
Based on
• 10 rolls
• 100 rolls
33
Population distribution
N=10
Distribution of
means from
1000 samples,
by different
sample sizes
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Difference between distribution in the
population, and distribution of means of
samples, by sample sizeifferent sizes
N=50
N=100
34
Repetition:
T-test
1.
The mean is a relevant summary measure
2.
Independent observations
(e.g. no patient contributes more than one observation)
3.
Observations are of Normal distribution
OR
Both groups are large
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Assumptions
Exercise:
Would you perform a T-test if you, e.g., would like to compare the cotinine
levels in children to smokers vs non-smokers?
Doubtful…. (does not meet the first criteria)
Other solutions?
Perform a non-parametric test (e.g., Mann-Whitney)
35
2015-11-09
Box-plot:
(cotinine levels in children in mothers of
smokers vs non-smokers)
36
Exercise: Describe the distributions by studying the box-plot
2015-11-09
The SPSS-output from a T-test (against better
knowledge). Cotinine levels in children of smokers
compared to non-smokers.
Equal variances could not
be assumed. Read from the
lower row.
A significant
difference between
the means…
…but the means
(and thus, the
difference
between the
means) do not
make sense.
37
2015-11-09
So…. How could we investigate
data that is not normally
distributed?
Using non-parametric tests!!!
38
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
39
• Original measurements are
converted to ranks in the analysis
• H0: Distributions are equal in all groups
Median useful marker for differences in distribution
2015-11-09
Non-parametric methods
• Insensitive to skewed distributions, extreme values
• Can be used for ordinal data
E.g. 0 = No response, 1 = Mild response, 2 = Strong response
40
Difference between two independent
groups:
Mann-Whitney’s test
• Rank the observations from the lowest to highest
• Calculate rank sum in group A (WA) and in group B (WB)
 Straightforward generalization to more than two groups (KruskalWallis test)
• The larger the difference is in mean ranks WA/nA and WB/nB ,
the lower p-value will be
• Mann-Whitney
U  WA 
n A  n A  1
2
41
Another name for the same test is ”Wilcoxon Rank sum test” which utilizes WA
2015-11-09
Mann-Whitney…
Sex
Female
Female
Female
Female
Female
Male
Male
Female
Male
Male
Male
Female
Male
Female
Male
Female
Male
Male
Female
Male
Female
Female
Male
Rank
Small Group
Discussion...
• Calculate the rank sum and the mean rank for
males (and females if you have time)
• For the group sizes nA = 11 (males) and nB = 12
(females), p < 0.05 if the rank sum for the
smallest group (males) is below 100 or
above 175
2015-11-09
Creatinine
38
57
64
65
70
79
81
81
82
85
93
105
107
110
113
123
219
232
262
297
313
320
845
• Conclusion?
• How would you summarize the test?
42
Presenting Mann-Whitney results
• Median (possibly mean as well for comparison)
• Variability in each group
• Percentiles or quartiles (in smaller groups)
or min-max (in even smaller groups)
• Standard deviation not relevant
2015-11-09
• Average in each group
• Difference between the groups
• P-value for M-W test
• U-statistic sometimes relevant
• Ideally: Median difference ± 95% CI sometimes relevant (could be
calculated in e.g. SPSS)
43
2015-11-09
Mann-Whitney
Creatinine, cont
44
Extension to more than two groups
Kruskal-Wallis test
• Mann-Whitney U-test (k = 2)
H0: Distribution A = Distribution B
H1: Distribution A  Distribution B
2015-11-09
E.g.
• Kruskal-Wallis test (k > 2 groups)
E.g. k = 3:
H0: Distribution A = Distribution
H1: Distribution A  Distribution
Distribution A  Distribution
Distribution B  Distribution
B = Distribution C
B or
C or
C
• Independent groups, independent observations within each group
• Median useful marker for differences in distribution
• The more the mean ranks differ, the lower the p-value will be
45
Now… back to our cotinine
data…
2015-11-09
• How to compare levels between groups?
46
2015-11-09
The SPSS-output from a non-parametric test MannWhitney U-test: (cotinine levels in children in
mothers of smokers vs non-smokers)
Exercise:
How could you interpret the output? Do you need to do more to show the
results?
Present medians, perhaps percentiles or inter-quartile-range, or boxplot
Non-smokers:
Smokers:
Median
Interquartile Range
.13
.29
Median
Interquartile Range
76.1
86.2
47
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
48
Paired data, cell-count example
ControlsDay2
915600
953300
650000
700000
1050000
984000
772000
920000
1080000
920000
840000
533000
510000
722000
SalDay2
357800
502200
470000
560000
736000
556000
418000
600000
680000
520000
560000
620000
704000
696000
Two ways of comparing means:
1. Calculate the means of the groups, and estimate the difference
2. Estimate the difference for each row. Then calculate the mean of the differences
2015-11-09
Preparation
11
2
3
4
5
6
7
8
9
10
11
12
13
14
Means
Difference between
values
49
Preparation
11
2
3
4
5
6
7
8
9
10
11
12
13
14
Means
ControlsDay2
915600
953300
650000
700000
1050000
984000
772000
920000
1080000
920000
840000
533000
510000
722000
824992,8571
SalDay2
357800
502200
470000
560000
736000
556000
418000
600000
680000
520000
560000
620000
704000
696000
570000
Difference between
values
557800
451100
180000
140000
314000
428000
354000
320000
400000
400000
280000
-87000
-194000
26000
2015-11-09
T-test for paired data
50
Preparation
11
2
3
4
5
6
7
8
9
10
11
12
13
14
Means
ControlsDay2
915600
953300
650000
700000
1050000
984000
772000
920000
1080000
920000
840000
533000
510000
722000
824992,8571
SalDay2
357800
502200
470000
560000
736000
556000
418000
600000
680000
520000
560000
620000
704000
696000
570000
Difference between
values
557800
451100
180000
140000
314000
428000
354000
320000
400000
400000
280000
-87000
-194000
26000
254992,9
Difference between means=
mean of the differences
2015-11-09
T-test for paired data
51
Preparation
11
2
3
4
5
6
7
8
9
10
11
12
13
14
Means
ControlsDay2
915600
953300
650000
700000
1050000
984000
772000
920000
1080000
920000
840000
533000
510000
722000
824992,8571
SalDay2
357800
502200
470000
560000
736000
556000
418000
600000
680000
520000
560000
620000
704000
696000
570000
Difference between
values
557800
451100
180000
140000
314000
428000
354000
320000
400000
400000
280000
-87000
-194000
26000
254992,9
s=
181454,0097
111808
216636,9
s (combined)
SEM
150709,3394
56962,77603
57898,64
2015-11-09
T-test for paired data
52
Preparation
11
2
3
4
5
6
7
8
9
10
11
12
13
14
Means
ControlsDay2
915600
953300
650000
700000
1050000
984000
772000
920000
1080000
920000
840000
533000
510000
722000
824992,8571
SalDay2
357800
502200
470000
560000
736000
556000
418000
600000
680000
520000
560000
620000
704000
696000
570000
Difference between
values
557800
451100
180000
140000
314000
428000
354000
320000
400000
400000
280000
-87000
-194000
26000
254992,9
s=
181454,0097
111808
216636,9
Thus, the mean is not influenced on whether the
data are paired or not, but the estimate of the
standard deviation is likely to differ with method.
2015-11-09
T-test for paired data
Use analyses for paired data when adequate!
s (combined)
SEM
150709,3394
56962,77603
57898,64
53
Paired samples t-test
Sometimes it is more powerful to test for differences within the same patient (or another
paired measurement)
In a study of weight loss from spicy food, 12 subjects were weighed before and after a
month on spicy food diet, see the table
2015-11-09
Previous t-test was made to find differences between
independent groups of observations
Discuss:
How would you test if
the diet gave
weightloss?
54
Paired samples t-test
Calculate the
differences di for each
subject’s weights.
2015-11-09
Do a paired samples
t-test!
Test if mean(d) = 0
95% CI for mean(d):
Mean(d)± t0.025(N-1)∙𝑠
Here:
-2.1 ± 2.2∙3.03 12 =
-2.1 ± 1.9 = (-4.0,-0.17)
𝑁
Discuss:
• How do you interpret the CI?
• Was the treatment effective
55
Paired test:
95% CI for weight loss
(-4.0,-0.17)
P=0.036
If the researchers wouldn’t recognize the paired design, but
did an independent groups’ t-test:
Why so wide?
CI = (-18.6;22.76)
Large variability BETWEEN subjects inflates the
P=0.84
variability of the difference in an independent
groups’ design!
2015-11-09
Paired samples t-test
56
Normally distributed outcomes/’large’ studies
Focus on mean comparisons
• Two independent groups
t-test
• Paired groups (paired measurements)
Paired t-test
• > 2 groups
Analysis of variance (ANOVA)
2015-11-09
Parametric methods for
group comparisons
If assumptions are not met:
Non-parametric tests!
• Two Independent groups
• More than two independent groups
• Paired groups (paired measurements)
Mann-Whitney U-test
Kruskal-Wallis test
Wilcoxons signed rank test
57
Non-parametric test for paired samples:
Wilcoxon signed rank test
Spicy diet continued:
Subject
Subject
11
22
33
44
55
66
77
88
99
10
10
11
11
12
12
Pre
Pre Post
Post Diff
Diff Sign
Sign Rank
Rank
65
65
88
88
125
125
103
103
90
90
76
76
85
85
126
126
97
97
142
142
132
132
110
110
62
62
86
86
118
118
105
105
91
91
72
72
81
81
122
122
95
95
145
145
132
132
105
105
-3
-3
-2
-2
-7
-7
22
11
-4
-4
-4
-4
-4
-4
-2
-2
33
00
-5
-5
Signed
Signed
rank
rank
---++
++
----++
5,5
5,5
33
11
11
33
11
88
88
88
33
5,5
5,5
-5,5
-5,5
-3
-3
-11
-11
33
11
-8
-8
-8
-8
-8
-8
-3
-3
5,5
5,5
--
10
10
-10
-10
(Sum the negative ranks
=56.5 (=11*12/2-9.5) )
…
2015-11-09
Sum the positive ranks:
3+1+5.5=9.5
58
Test-situation
Parametric test
Non-parametric
test
Independent
samples, 2 groups
T-test
Mann-Whitney
Independent
ANOVA
samples, ≥ 2 groups
Kruskal-Wallis
Paired samples, 2
groups
Wilcoxon rank sum
test
Paired t-test
2015-11-09
Comparison of different tests
59
Parametric
methods
Nonparametric
methods
Ex
Positive
Negative
T-test
+
Results in both effectmeasure (w CI) and pvalue.
-
Based on assumptions
about the distribution of
data - typically 𝑁(𝜇, 𝜎)
+
More effective to detect
differences if data is
(close to) normal
-
Test results could be
sensitive to deviation
from Normal distribution,
especially in small studies
+
No assumptions about
the distribution of data
-
+
useful also for data
measured on an ordinal
scale
Less powerful than
parametric methods
(if normal distribution
applies)
-
Typically results only in
p-value (but sometimes
an effect measure with CI
could be computed)
MannWhitney
+
Suitable for small
studies
2015-11-09
Two broad categories of statistical methods
60
Exercise:
Which methods would you use to investigate the relationship between
the mothers and their childrens values?
2015-11-09
Comparison of maternal cotinine levels and
child’s cotinine levels
1. We know that cotinine levels are not suitable for a T-test.
2. This type of analysis should have a paired design
- Wilcoxon signed test
61
2015-11-09
The SPSS-output from an
analyses with paired data
Exercise:
Interpret the output
What more should be shown?
Childs levels
Mothers levels:
Median
Interquartile Range
.22
22.6
Median
.22
Interquartile Range
27.5
The information is not very illustrative. We’ll learn more about
correlations next lession
62
• Repetition
• Normal distribution
• Reference intervals and confidence intervals
• Elements of statistical inference, Hypothesis testing
• T-test, ANOVA
• Central limit theorem
• Criteria to be met for performing a T-test.
2015-11-09
Today
• Lecture 4:
• Non-parametric tests
• Paired samples tests.
Next lecture:
• Linear regression
• Correlation
• R2
63