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Statistics - Lecture 05Nicodème Paul Faculté de médecine, Université de Strasbourg
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Descriptive statistics and probability
Data description and graphical representation
Mean, median, quartiles, standard deviation, interquartile range (IQR)
Barplot, histogram and boxplot
Decisions are based on probability calculation
Notion of random variables and distributions
Binomial and normal distributions
·
·
·
·
·
·
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Estimation
Notion of parameters ( , )
Di�erence between estimate and estimator
The sample mean and the sample variance are estimators
The Central Limit Theorem and sampling distribution
Interval estimate or con�dence interval
· μ σ2
·
· X̄ S2
·
·
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Hypothesis testing
Parametric tests and test procedure
Notion of null and alternative hypotheses
Test statistic and its sampling distribution
Critical values and critical regions
P-values
·
·
·
·
·
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Relation between variables and goodness of �t
Notion of correlation
Non parametric tests and goodness of �t
Notion of association between categorical variables
The test
The Fisher exact test
·
·
·
· χ2
·
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Examples
You have been assigned 12 consenting experimental subjects, each of whom has a brain tumour of the same sizeand type. Four are allocated at random to an untreated control group, four are treated with the drug Tumostat andfour more with the drug Inhibin 4. After two months of treatment, the diameter of each tumour is remeasured.
Survival times in �ve human cancer (stomach, bronchus, colon, ovary, breast). Cameron, E. and Pauling, L. (1978)Supplemental ascorbate in the supportive treatment of cancer: re-evaluation of prolongation of survival times interminal human cancer. Proceedings of the National Academy of Science USA, 75, 4538-4542
Comparison of 5 pretreated patches to reduce mosquito human contact. Bhatnagar, A and Mehta, VK (2007)E�cacy of Deltamethrin and Cy�uthrin Impregnated Cloth over Uniform against Mosquito Bites. Medical JournalArmed Forces India, 63, 120-122
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QuestionYou have been assigned 12 consenting experimental subjects, each of whom has a brain tumour of the samesize and type. Four are allocated at random to an untreated control group, four are treated with the drugTumostat and four more with the drug Inhibin 4. After two months of treatment, the diameter of each tumour isremeasured. What would be the appropriate test here?
Parametric test
Non parametric test
Submit Show Hint Show Answer Clear
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ANOVA: Analysis Of Variance
Population 1: Sample 1:
Population 2: Sample 2:
Population k: Sample k:
Objective: Comparing the means of multiple populations
( , )μ1 σ21 → ( , , . . . , )X11 X12 X1,n1 → ( , )X̄1 S2
1
( , )μ2 σ22 → ( , , . . . , )X21 X22 X2,n2 → ( , )X̄2 S2
2
. . . . . . . . . . . . . . . . . . . . . . . . . .
( , )μk σ2k
→ ( , , . . . , )Xk1 Xk2 Xk,nk → ( , )X̄k S2k
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ANOVA assumption and objectivesEach of the k population or treatment response distributions is normal
(The k normal distributions have identical standard deviations)
The observations in the sample from any particular one of the k populations or treatmentsare independent of one another
When comparing population means, the k random samples are selected independently ofone another
:
: At least two of the ’s are di�erent
Estimation of simultaneous con�dence intervals for the mean di�erences for
and
·
· = =. . . =σ1 σ2 σk
·
·
· H0 = =. . . =μ1 μ2 μk
· H1 μ
· −μi μji, j = 1, . . . ,k i ≠ j
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ExampleYou have been assigned 12 consenting experimental subjects, each of whom has a brain tumour of the same sizeand type. Four are allocated at random to an untreated control group, four are treated with the drug Tumostat andfour more with the drug Inhibin 4. After two months of treatment, the diameter of each tumour is remeasured.
Population parameters:
: There is no di�erence in mean tumour diameter among the treatments
: There is a di�erence in mean tumour diameter among the treatments. At least two ofthe ’s are di�erent
·
: population mean of the control group
: population mean of the Neurohib group
: population mean of the Tumostop group.
- μ1
- μ2
- μ3
· H0
= =μ1 μ2 μ3
· H1
μ
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Data
is the th observation resulting from th treatment
: total of th treatment. : mean of the th treatment
: total of all observations
: the grand mean where
· xij i j
· =T.j ∑nji=1 xij j =x̄.j
T.j
njj
· = =T.. ∑kj=1 T.j ∑k
j=1 ∑nji=1 xij
· =x̄..T..
NN = ∑k
j=1 nj
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Example
· = (7 + 8 + 10 + 11) = 9x̄.114
= (4 + 5 + 7 + 8) = 6x̄.214
· = (4 + 5 + 1 + 2) = 3x̄.314
= (7 + 8 + 10 + 11 + 4 + 5 + 7 + 8 + 4 + 5 + 2 + 1) = 6x̄..112
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Within groups sum of squares
Variation within group: · ssw = ( −∑3j=1 ∑
4i=1 xij x̄.j)
2
ssw = (4 + 1 + 1 + 4) + (4 + 1 + 1 + 4) + (4 + 1 + 1 + 4) = 30
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Between groups sum of squares
Variation between groups: · ssb = 4( −∑3j=1 x̄.j x̄..)
2
ssb = 4 × 9 + 0 + 4 × 9 = 72
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Total sum of squares
Variation between groups: · sst = ( −∑3i=1 ∑
4i=1 xij x̄..)
2
sst = (1 + 4 + 16 + 25) + (1 + 1 + 4 + 4) + (1 + 4 + 16 + 25) = 102
· sst = ssw + ssb = 30 + 72 = 102
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Check yourselfUnder the null hypothesis, the ratio should be close to 1.
True
False
Submit Show Hint Show Answer Clear
ssbssw
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Test for equal meansThe hypotheses are:
If the null hypothesis is true, we combine k samples to estimate overall mean and the samplemean for the group as:
Total sum of squares,
sum of squares within groups and sum of squares between
groups
We can show that:
·
H0
H1
:
:
= =. . . =μ1 μ2 μksome means are different
·i
= =X̄..1
N∑j=1
k
∑i=1
nj
Xij X̄.j1
nj∑i=1
nj
Xi,j
· SST = ( −∑kj=1 ∑
nji=1 Xij X̄..)
2SSW = ( −∑k
j=1 ∑nji=1 Xij X̄.j)
2
SSB = ( −∑kj=1 nj X̄.j X̄..)
2
· SST = SSW + SSB
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Check yourselfLet with and unknown. Let .
What is the distribution of ?
Submit Show Hint Show Answer Clear
, , . . . , ∼ N (μ, )X1 X2 Xn σ2 μ σ = ( −S2 1n−1
∑ni=1 Xi X̄)2
n−1σ2 S2
N (μ; )σ2
n
tn−1
χ21
χ2n−1
χ2n
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Check yourselfIn the anova framework, we de�ned as the sum of squares within group.
Let , what is the distribution of ?
Submit Show Hint Show Answer Clear
SSW = ( −∑kj=1 ∑
nji=1 Xij X̄.j)
2
N = + +. . . +n1 n2 nkSSW
σ2
χ21
χ2k
χ2k−1
χ2N
χ2N−1
χ2N−k
χ2N−k−1
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Parameter estimationComparing means from multiple populations assuming the variances are the same andequal to
Pooled variance estimator:
where , and
Notice that:
As for ,
·σ2
·
= =S2pool
( − 1)∑kj=1 nj S2
j
( − 1)∑kj=1 nj
( −∑kj=1 ∑
nji=1 Xij X̄.j)
2
N − k
N = + +. . . +n1 n2 nk =X̄.j1nj∑
nji=1 Xi,j = ( −S2
j1−1nj
∑nji=1 Xij X̄.j)
2
·
= + +. . . + ∼S2pool
σ2
( − 1)n1 S21
σ2
( − 1)n2 S22
σ2
( − 1)nk S2k
σ2χ2N−k
j = 1, 2, . . . ,k ∼( −1)nj S 2
j
σ2 χ2−1nj
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Parameter estimation
is an estimator of
As , then
con�dence intervals for population means are:
· =X̄.j1nj∑
nji=1 Xij μj
· , , . . . , ∼ N ( , )X1j X2j X jnj μj σ2 ∼ N ( , )X̄.j μjσ2
nj
· ∼−X̄.j μj
Spool1nj
√tN−k
· (1 − α)
= [ − , + ]Iα x̄.j tN−k1− α
2
spool
nj−−
√x̄.j tN−k
1− α
2
spool
nj−−
√
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Example
A 95% con�dence interval for respectively , and is:· μcontrol μneurohib μmitostop
- [9 − 2.262 × × 1/2; 9 − 2.262 × × 1/2] = [6.935; 11.065]30/9− −−−
√ 30/9− −−−
√
- [6 − 2.262 × × 1/2; 6 − 2.262 × × 1/2] = [3.935; 8.065]30/9− −−−
√ 30/9− −−−
√
- [3 − 2.262 × × 1/2; 3 − 2.262 × × 1/2] = [0.935; 5.065]30/9− −−−
√ 30/9− −−−
√
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Check yourselfIn the anova framework, we de�ned as the sum of total squares. What is the
distribution of ?
Submit Show Hint Show Answer Clear
SST = ( −∑kj=1 ∑
nji=1 Xij X̄..)
2
SST
σ2
χ21
χ2k
χ2N−1
χ2N
χ2N−k
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Check yourselfIn the anova framework, we de�ned as the sum of squares between groups.
What is the distribution of ?
Submit Show Hint Show Answer Clear
SSB = ( − . .∑kj=1 nj X̄.j X̄ )2
SSB
σ2
χ21
χ2k
χ2N−1
χ2N−k
χ2k−1
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Test for equal means· = ( ∼SST
σ2 ∑kj=1 ∑
nji=1
−Xij X̄..
σ)2
χ2N−1
· = ( ∼SSW
σ2 ∑kj=1 ∑
nji=1
−Xij X̄.j
σ)2
χ2N−k
· = ( ∼SSB
σ2 ∑kj=1
−X̄.j X̄..
σ
nj√
)2
χ2k−1
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Test for equal means
has a distribution with and degrees of freedom.
De�nition: Let and be independent random variables such that has the distribution with m degrees of freedom and has the distribution with n degrees offreedom, where m and n are given positive integers. The random variable T de�ned asfollows:
Then the distribution of is , the F distribution with m and n degrees of freedom.
Under the null hypothesis, the random variable:
· Y W Y χ2m
W χ2n
T =Y /m
W/n
T Fm,n
·
T =SSB/(k − 1)
SSW/(N − k)
Fk−1,N−k k − 1 N − k
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Test for equal means
To test with a signi�cant level , we calculate the value of the test statistic from thesamples
Reject the null distribution if where is the critical value.
The for the test is:
If the null hypothesis is rejected, what next?
· α f
· H0 f > f 1−αk−1,N−k f 1−α
k−1,N−k
· p − value
p − value = P(T > f) where T ∼ Fk−1,N−k
·
Tests for contrasts
Pairwise comparison
-
-
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Example
We have: , , , and .
We reject the null hypothesis of equal mean of tumour diameters
· k − 1 = 2 N − k = 9 f = = 10.872/2
30/9= 4.256f 0.95
2,9 p − value = 0.004
·
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Example: Comparison of 5 pretreated patches toreduce mosquito human contact
Reference: Bhatnagar, A and Mehta, VK (2007) E�cacy of Deltamethrin and Cy�uthrin Impregnated Cloth overUniform against Mosquito Bites. Medical Journal Armed Forces India, 63, 120-122
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Example: parameter estimationmodel01 = aov(Measure~Treatment, data=bites)
model.tables(model01, type="means")
Tables of means
Grand mean
7.153333
Treatment
Treatment
C+O Cyfluthrin Deltamethrin D+O Odomos
5.367 8.033 8.133 6.333 7.901
= 7.153x̄
= 8.033 = 8.133 = 7.901 = 5.367 = 6.333x̄C x̄D x̄O x̄C+O x̄D+O
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Example: testsummary(model01)
Df Sum Sq Mean Sq F value Pr(>F)
Treatment 4 184.6 46.16 4.48 0.00192 **
Residuals 145 1494.1 10.30
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We have:
If then , we reject the null hypothesis of equal mean mosquito bite
rates.
·
k − 1 = 4 SSB = 184.6N − k = 145 SSW = 1494.1f = 4.48 p − value = 0.00192
· α = 0.05 = 2.434f 1−α4,145
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Example: Survival times in terminal human cancerReference: Cameron, E. and Pauling, L. (1978) Supplemental ascorbate in the supportive treatment of cancer: re-evaluation of prolongation of survival times in terminal human cancer. Proceedings of the National Academy ofScience USA, 75, 4538-4542
STOMACH BRONCHUS COLON OVARY BREAST124 81 248 1234 123542 461 377 89 2425 20 189 201 158145 450 1843 356 1166412 246 180 2970 4051 166 537 456 7271112 63 519 380846 64 455 791103 155 406 1804876 859 365 3460146 151 942 719340 166 776396 37 372
223 163138 10172 20245 283
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Example: Survival times in terminal human cancer
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Example: Survival times in terminal human cancer
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Example: parameter estimationmodel02 = aov(Survival~Type, data=cdat)
model.tables(model02, type="means")
Tables of means
Grand mean
5.555785
Type
stomach bronchus colon ovary breast
4.968 4.953 5.749 6.151 6.559
rep 13.000 17.000 17.000 6.000 11.000
= 5.555785x̄
= 4.968 = 4.953 = 5.749 = 6.151 = 6.559x̄s x̄b x̄c x̄o x̄b
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Example: testsummary(model02)
Df Sum Sq Mean Sq F value Pr(>F)
Type 4 24.49 6.122 4.286 0.00412 **
Residuals 59 84.27 1.428
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We have:
If then , we reject the null hypothesis of equal mean survival days.
·
k − 1 = 4 SSB = 24.49N − k = 59 SSW = 84.27
f = 4.286 p − value = 0.00412
· α = 0.05 = 2.434f 1−α4,59
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Contrasts
A contrast is any linear combination of the population means
such that and integer.
If are means of populations, some examples of contrasts are:
·
C = + +. . . +c1μ1 c2μ2 ckμk
= 0∑ki=1 ci ci
· , , . . . ,μ1 μ2 μ5 k = 5
- −μ1 μ2
- 2 − −μ1 μ3 μ4
- + + − − 2μ1 μ2 μ3 μ4 μ5
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Check yourselfYou want to test:
What would be the contrast:
Submit Show Hint Show Answer Clear
: − ( + + + ) = 0 : + ( + + + ) ≠ 0H0 μ11
4μ2 μ3 μ4 μ5 H1 μ1
1
4μ2 μ3 μ4 μ5
− ( + + + )μ114
μ2 μ3 μ4 μ5
4 − − − −μ1 μ2 μ3 μ4 μ5
5 − − − −μ1 μ2 μ3 μ4 μ5
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Test for a contrast
has a t-distribution with degrees of freedom.
The hypotheses:
Note , the test statistic:
·
: = 0 versus : ≠ 0H0 ∑j=1
k
cjμj H1 ∑j=1
k
cjμj
· = SSW/(N − k)Sw
T =∑k
j=1 cjX̄.j
Sw ∑kj=1
c2j
nj
− −−−−−−√
N − k
The 100% con�dence interval for the contrast is:· (1 − α)
[ − ; + ]∑j=1
k
cjx̄.j tN−k1− α
2
sw ∑j=1
k c2j
nj
− −−−−−
⎷
∑
j=1
k
cjx̄.j tN−k1− α
2
sw ∑j=1
k c2j
nj
− −−−−−
⎷
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Check yourselfIn the anova framework with k conditions or treatments, when you reject the null hypothesis how manycomparisons would you do to compare the means?
Submit Show Hint Show Answer Clear
k − 1
k
k2
k(k+1)
2
k(k−1)
2
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Pairwise comparisonsThere are pairwise comparisons or tests
Recall that when testing a single hypothesis , a type I error is made if it is rejected, even ifit is actually true.
The probability of making a type I error in a test is usually controlled to be smaller than acertain level of , typically equal to 0.05
When there are several null hypotheses, , and all of them are testedsimultaneously, one may want to control the type I error at some level .
A type I error is then made if at least one true hypothesis in the family of hypotheses beingtested is rejected. This signi�cance level is called the familywise error rate (FWER). If thehypotheses in the family are independent, then:
where for are individual signi�cance levels.
· k(k−1)
2
· H0
·α
· , , . . . ,H01 H02 H0m
α
·
FWER = 1 − (1 − αi)m
αi i = 1, 2, . . . ,m
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Pairwise comparisons
is the upper-tail critical value of the Studentized range for comparing di�erentpopulations.
Bonferroni: To control , reject all among for which thep-value is less than .
Studentized range distribution (Tukey) procedure:
· FWER ≤ α H0i , , . . . ,H01 H02 H0m
α/m
·
Rank the k sample means
Two population means and are declared signi�cantly di�erent if the
con�dence interval of :
-
- μi μj (1 − α)100
−μi μj
− ±x̄i x̄j qN−k,k,1−αsw ( + )1
2
1
ni
1
nj
− −−−−−−−−−−
√
qN−k,k,1−α k
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Comparison of 5 pretreated patches Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = Measure ~ Treatment, data = bites)
$Treatment
diff lwr upr p adj
D+O-C+O 0.9663333 -1.3232391 3.255906 0.7707275
Odomos-C+O 2.5336667 0.2440942 4.823239 0.0220410
Cyfluthrin-C+O 2.6656667 0.3760942 4.955239 0.0136686
Deltamethrin-C+O 2.7660000 0.4764276 5.055572 0.0093589
Odomos-D+O 1.5673333 -0.7222391 3.856906 0.3268078
Cyfluthrin-D+O 1.6993333 -0.5902391 3.988906 0.2476696
Deltamethrin-D+O 1.7996667 -0.4899058 4.089239 0.1965293
Cyfluthrin-Odomos 0.1320000 -2.1575724 2.421572 0.9998540
Deltamethrin-Odomos 0.2323333 -2.0572391 2.521906 0.9986342
Deltamethrin-Cyfluthrin 0.1003333 -2.1892391 2.389906 0.9999510
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Comparison of 5 pretreated patches
Cy�uthrin patches when applied in presence of odomos were found to have much morerepellent action as compared to only odomos. The di�erence in the repellent action was veryhighly signi�cant (p < 0.01). Thus it can be inferred that signi�cant bene�t is achieved inreducing man-mosquito contact when cy�uthrin patches are applied over the uniform by thetroops in addition to using odomos as compared to those using odomos only
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Survival times in terminal human cancer Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = Survival ~ Type, data = cdat)
$Type
diff lwr upr p adj
stomach-bronchus 0.01474955 -1.2242933 1.253792 0.9999997
colon-bronchus 0.79595210 -0.3575340 1.949438 0.3072938
ovary-bronchus 1.19744617 -0.3994830 2.794375 0.2296079
breast-bronchus 1.60543320 0.3041254 2.906741 0.0083352
colon-stomach 0.78120255 -0.4578403 2.020245 0.3981146
ovary-stomach 1.18269662 -0.4770864 2.842480 0.2763506
breast-stomach 1.59068365 0.2129685 2.968399 0.0158132
ovary-colon 0.40149407 -1.1954351 1.998423 0.9540004
breast-colon 0.80948110 -0.4918267 2.110789 0.4119156
breast-ovary 0.40798703 -1.2987803 2.114754 0.9615409
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10/8/2018 Statistics - Lecture 05
http://statnipa.appspot.com/cours/05/index.html#47 46/47
Survival times in terminal human cancer
is signi�cantly di�erent to 0. In fact, ascorbate, when used withthe treatment, seems to improve survival times better in breast cancer than in bronchuscancer.
is also signi�cantly di�erent to 0, showing a signi�cantimprovement of survival in breast cancer compared to stomach cancer when ascorbatesupplement is used in the treatment.
· log( ) − log( )μbreast μbronchus
· log( ) − log( )μbreast μstomach
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10/8/2018 Statistics - Lecture 05
http://statnipa.appspot.com/cours/05/index.html#47 47/47
See you next time
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