statistics applied to biomedical sciences
DESCRIPTION
Seminar entitled "Statistics Applied to Biomedical Sciences" presented at UCLA on April 10, 2014.TRANSCRIPT
Statistics Applied to Biomedical Sciences
Luca Massarelli
@ UCLA - Anesthesiology Department Division of Molecular Medicine
April 10, 2014
Outline
Random variables: definition & properties
Estimators: sample mean & sample variance
Distributions of estimators
Confidence interval
Hypothesis test
Application to biomedical experiments: z-Test, t-Test & ANOVA
Classification:- Discrete- Continuous
Random Variable: Definition
- A random variable is a real-valued function defined on a set of possible outcomes, the sample space Ω.
-Probability distribution is a function which maps each value of the random variable to a probability
-Cumulative Distribution is a function that, given the probability distribution, determines the probability at a value less than or equal to x
We will extract 2 persons from West LA population and we will consider the number of persons affected by a certain allergy in spring.
Probability of people with allergy = 20%random variable
Probability distribution function
Cumulative distribution function
(D1,D2) ---- > 0(D1,D2) ---- > 1(D1,D2) ---- > 1(D1,D2) ---- > 2
X f(x) Φ(x)0 0.64 0.64 1 0.32 0.96 2 0.04 1.00
Probability distribution function: f(x) = P(X=x)
xxi
x)P(X (x) xifCumulative distribution function
X: Ω ---- > R
Ω = (D1,D2), (D1,D2), (D1,D2), (D1,D2)
Random Variable: Discrete
Probability density function:
Cumulative distribution function
f(x) is related to the following eq.
x
-
dx f(x) x)P(X (x)
Random Variable: Continuous
Normal Probability Density Function
Prob
abili
ty D
ensi
ty
x
22 2/)(
2
1),;()(
x
x exfxf
The random variable X is identified by its probability density function.
),( NX
Normal Probability Density Function
Prob
abili
ty D
ensi
ty
a
-
b
-
b
dx f(x)dx f(x)dx f(x) (a)-(b))(a
bxaP
Continuous Random Variable: Normal Distribution
T-student Probability Density FunctionPr
obab
ility
Den
sity
Continuous Random Variable: Distributions
Moment is a quantitative measure of the shape of a set of points.
dxxfcxcxE nn
n)()()(Moment
The nth moment of a random variable about a value c
dxxxfxE )()(Mean: Central tendency.
dxxfxExVAR x )()()( )(222 Variance Dispersion around the mean.
Information about the shape of a distribution
dxxfxE x )()( )(33 Skewness Asymmetry around the mean.
dxxfxE x )()( )(44 Kurtosis Measure of flatness
Moments: Tendency & Shape
dispersion
asymmetry
flatness
Moments: Tendency & Shape
It is a function of random variables whose values are used to estimate a certain parameter.
T = t(Xi) where Xi has a given distribution with unknown θ (parameter), which is the target of our statistic.
Xi is the i-th observable random variable
X1, X2, X3, … Xn is a sample of random variables extracted from a population;
Population has a certain probability density function f(x)
Estimators: Definition
The Transformation of Random variable is still a random variable: Addition, Subtraction, Multiplication and Divisions results in another random variable
Estimator has some desirable properties:
UNBIASED The estimator is an unbiased estimator of θ if and only if E(T) = θ --- > (E(T) – θ) = 0 In other words the distance between the average of the collection of estimates and the single parameter being estimated is null.Bias is a property of the estimator, not of the estimate.
1||lim
TPn
EFFICIENCYThe estimator has minimal mean squared error (MSE) or variance of the estimatorMSE(T) = E[T- θ]2
If I have an estimator with smaller dispersion, I will have more probability to find an estimation which is closer to the TRUE Parameter.
CONSISTENCY
Increasing the sample size increases the probability of the estimator being close to the population parameter.
Estimators: Properties
SAMPLE MEAN estimator for μ
Sample mean is an unbiased estimator of μ INDIPENDENTLY of the distribution of the random variable X.
Sample mean is an efficient estimator of μ in case of Normal, Poisson, Exponential, Bernoulli distribution.
n
iiXn
T1
1
Estimators: Sample Mean
Exp1
X1
RV
x1 = 55%
Observed Value
Exp2
X2
RV
x2 = 75%
Observed Value
Exp3
X3
RV
x3 = 95%
Observed Value
Transfection efficiency after 24h
Transfection efficiency after 24h
Transfection efficiency after 24h
Estimators: Sample Mean
)()1
()(1
i
n
ii XEX
nETE
nXnVAR
nX
nVARTVAR i
n
ii
2
21
)(1
)1
()(
It is unbiased and (under certain conditions) efficient for μ!
Using the sample mean I can take some conclusions:
• the E(T) on average tends to μ (Unbiased Property)• my estimation is reasonable in a certain level of uncertainty (Std Error)• by increasing n, I can decrease the error of my estimation
nStdError
n
iiXn
T1
1 Why Sample Mean?
According to the observed values the estimation of μ = 75%
(x1, x2, x3) is just one vector of the potential one. I could have had any other value.
Are we close or far from the real value of μ?
Estimators: Sample Mean
SAMPLE VARIANCE estimator for δ22
1
)(1
TXn
Sn
ii
n
nSE
1)( 2
2
1
' )(1
1
1TX
nS
n
nS
n
ii
2)'( SE
According to the observed values the estimation of σ2 = 4%(variance has been corrected by n/(n-1))
Assuming that Xi are INDIPENDENT it can be demonstrated that:
--- > distorted estimator
(sample variance as defined above would be unbiased only if μ of population is known)
SAMPLE VARIANCE
Estimators: Sample Variance
This estimator depends on n random variables (Xi) and the random variable sample mean.
SAMPLE MEAN
n
iiXn
T1
1
Assuming that ),( 2NX
),(2
nNT
N(μ, δ2/n)
n
2n
2
then
Distribution of Estimators
SAMPLE VARIANCE 2
1
' )(1
1TX
nS
n
ii
Assuming that ),( 2NX
12
2'
1
nn
S then
Distribution of Estimators
n
iiXn
T1
1
),( 2NX
2
1
' )(1
1TX
nS
n
ii
12
2'
1
nn
S
),(2
nNT
11
22'
1
1
nn
Tstudent
nn
T
nS
TD
with μ and σ2 are unknownAssuming that
Normal Distr.
χ2 Distr.
Distribution of Estimators: Distance
A confidence interval (CI) is a type of interval which estimates a certain parameter of a population.
Confidence interval (which is calculated from the observations), is that interval that frequently includes the parameter of interest if the experiment is repeated.
The probability that the observed interval contains the parameter is determined by the confidence level or confidence coefficient.
CI95% for μ means that I want to determine an interval being sure that 95% of the time the TRUE MEAN of the population lies somewhere within my interval.
Confidence Interval: Definition
),( 2NX
Assuming my population has a normal distribution
Set of n experiments
Set of n experiments
Set of n experiments
Set of n experiments
Set of n experiments
Set of n experiments
Set of n experiments
...
We will never determine the TRUE VALUE of μ
Confidence Interval: Definition
)(TEn
TVAR2
)(
n
StdError
n
iiXn
T1
1
If ),( 2NX ),(2
nNT
Define an interval around our estimate of μ with confidence coeff. = 95%
1][Pr bTaobWith the transformation to a Standard Normal Distribution
1][Pr
222
n
b
n
T
n
aob
1][Pr
22
n
bZ
n
aob
1][Pr
212ZZZob
Confidence Interval of the Mean
Normal Probability Density Function
Prob
abili
ty D
ensi
ty
x
Here we are assuming σ2 is known. What if the variance is unknown?
1][Pr2
12ZZZob
nZb2
21
1][Pr bTaob
Zα/2 0 Z1-α/2 a μ b
nZa2
2
nZCI2
21
Confidence Interval of the Mean
Zα/2 Z1- α /2
tα/2 t1- α /2
nStb
21
nSta
2
nStCI
21
Lack of information brings about higher uncertainty
Confidence Interval of the Mean
• Coefficient 1-α. Increasing the probability that the interpretation of an experiment is correct requires to make the interval larger.
• Number of Experiments. SE of Estimator can be reduced increasing n.
• Available information about the population. The lack of information about the population brings about a bigger uncertainty which is reflected in a larger interval.
These considerations will be similarly apply to the Hypothesis Test.
Confidence Interval: Considerations
The test is based on the following MODEL:a) Assume that the treatment has NO effect on the underlying population (H0)b) Set a variable which measures the DISTANCE of the meansc) Distance is associated with a probability under the assumption that the treatment
has no effect (H0 is TRUE)
Hypothesis Test: Definition
It is a statistical tool used to determine what results would lead to accept or reject a certain hypothesis for a pre-specified level of significance (α).
H0 - Null Hypothesis: μ=μ0
H1 - Alternative Hypothesis μ ≠ μ0
The hypothesis test here assumes the following statements:a) we have 1 population where we know the distribution and sometimes its parameters (i.e. mean and variance)
b) from the underlying population we have extracted one or more groups of subjects (sample or set of the experiment)
c) we have applied a certain treatment to our samples
n
T
n
D2
0
2
0
DISTANCE(Critical Ratio)
If the distance is too large (observed value is too far from my value μ0) it is likely my null hypothesis is NOT true (H0 is REJECTED)
The probability to reject the null hypothesis, when H0 is true, is α
Notice that D is a random variable, simple transformation of T
ACCEPT REJECTREJECT
Criticalvalue
Criticalvalue
α/2α/2
0
Notice that this is D distribution under the assumption H0 is TRUE
0
Hypothesis Test: Example
REJECT H0 if |D|≥dα
ACCEPT H0 if |D|≤dα
p-valueGiven the DISTANCE, what is the probability that the sample differs from the underlying population, when the NULL HYPOTHESIS is TRUE?
p-value is the probability of observing a distance that is as extreme or more extreme than currently observed, assuming that the NULL HYPOTHESIS is TRUE
It is a measure of making a mistake. It is the risk that you reject NULL HYPOTESIS given the fact that it is true.
ACCEPT REJECTREJECT
Criticalvalue
Criticalvalue
α/2α/2
0-dα dα
n
T
n
D2
0
2
0
d
.p-value
p-value > αAccept H0
d
.p-value
p-value < αReject H0
Hypothesis Test: p-value
Scientific Assumptions:- the true mean of underling population is known- the true SD of underling population is known
Statistical Assumptions:- the underlying distribution is NORMAL- the sample is chosen randomly from the underlying population
Hypothesis definitionH0 - Null Hypothesis: μ=μ0 or μ-μ0 = 0H1 - Alternative Hypothesis μ ≠ μ0
Hypothesis Test: one Sample z-Test
QuestionBased on Mean comparison, on average, did TRTM treatment really change the level of survival of the general population?
HEK cells [H2O2] = 200 μM
We assume that the survival probability of HEK cells follows the normal distribution with the following known parameters.
μ0 = 62%σ = 15.5%
Assume that one sample of HEK population is extracted and submitted to a certain treatment (TRTM)
Observed value:
Hypothesis Test: one Sample z-Test
Exp. Observed Values1 0.605 2 0.592 3 0.661 4 0.367 5 0.323
6 0.307
Sample Mean 0.476 SD 0.160 SE 0.065
n
T
n
D2
0
2
0
Critical Ratio Level of Significance α = 0.05
ACCEPT REJECTREJECT
The distance between sample and known mean is 2.278 units.We can conclude that the treatment significantly decreased the percentage of survival. The chance of wrongly reject the null hypothesis is much less than 5%.
Hypothesis Test: one Sample z-Test
Description Observed Values
Mean (Population) 0.6200 Mean (observed) 0.4758 SD known 0.1550 Observations 6 Hypothesized Mean Difference
-
D -2.2783
p value - 1 tail 0.0116 p value - 2 tails 0.0233
Scientific Assumptions:- the true mean of underling population is known- the true SD of underling population is NOT known
Hypothesis definitionH0 - Null Hypothesis: μ=μ0 or μ-μ0 = 0H1 - Alternative Hypothesis μ ≠ μ0
Hypothesis Test: one Sample t-Test
Statistical Assumptions:- the underlying distribution is NORMAL- the sample is chosen randomly from the underlying population
QuestionBased on Mean comparison, on average, did TRTM treatment really change the level of survival of the general population?
Let’s assume that the survival probability of HEK cells follow the normal distribution with the following known parameters.
μ0 = 62%σ = unknown
Assume that one sample of HEK population is extracted and submitted to treatment TRTM
Observed value:
HEK cells [H2O2] = 200 μM
Hypothesis Test: one Sample t-Test
Exp. Observed Values1 0.6052 0.5923 0.6614 0.3675 0.3236 0.307
Sample Mean 0.476 SD 0.160 SE 0.065
nS
T
nS
D2
0
2
0
Critical Ratio Level of Significance α = 0.05
The distance between sample and known mean is 2.206 units.We can conclude that the treatment did NOT significantly decreased the percentage of cell survival. The chance of wrongly reject the null hypothesis is grater than 5%.
ACCEPT REJECTREJECT
2.447-2.447
Hypothesis Test: one Sample t-Test
0
Description Observed Values
Mean (Population) 0.6200 Mean (observed) 0.4758 Variance estimated 0.1601 Observations 6 Hypothesized Mean Difference
- Degree of Freedom (n-1) 5
D -2.2060
p value - 1 tail 0.0392 p value - 2 tails 0.0784
Scientific Assumptions:- we estimate the MEAN DIFFERENCE between pairs for an underlying population- we estimate the SD of the distribution of differences
Hypothesis definitionH0 - Null Hypothesis: μDiff = 0H1 - Alternative Hypothesis μDiff > 0 (one tail)
Hypothesis Test: 2-Sample Paired t-Test
Statistical Assumptions:- the underlying distribution is NORMAL- the sample is chosen randomly from the underlying population
Hypothesis Test: 2-Sample Paired t-Test
dataMembrane Potential (mV)
Membrane Potential (mV) Membrane Potential (mV)
Nor
mal
ized
Con
duct
ance
Nor
mal
ized
Con
duct
ance
Nor
mal
ized
Con
duct
ance
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-100 -50 0 50 100
G(V) (before)G(V) CARDAMONIN (after)
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-100 -50 0 50 100
G(V) (before)
G(V) CARDAMONIN (after)
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-100 -50 0 50 100
G(V) (before)
G(V) CARDAMONIN (after)
QuestionOn average is the observed difference significantly more than 0? Is the distance big enough to conclude that the treatment/drug had effect?
Hypothesis Test: 2-Sample Paired t-Test
nS
TD Diff
2
0
Critical Ratio
Vm (mV)
Diff. Exp1
Diff. Exp2
Diff. Exp3
Mean Diff.
(Tdiff)
SD Diff.
D - Critcal Ratio
-100 0.005 -0.005 -0.002 -0.001 0.005 -0.211-90 0.003 -0.007 0.010 0.002 0.008 0.445-80 0.003 -0.002 0.010 0.004 0.006 1.072-70 -0.002 -0.003 0.005 0.000 0.005 -0.010-60 0.000 0.000 -0.004 -0.001 0.003 -0.970-50 -0.001 -0.004 -0.009 -0.005 0.004 -1.904-40 0.009 -0.018 -0.017 -0.009 0.016 -0.957-30 0.002 -0.042 0.014 -0.009 0.029 -0.530-20 0.003 -0.061 0.073 0.005 0.067 0.126-10 0.018 -0.025 0.076 0.023 0.051 0.7900 0.042 0.022 0.065 0.043 0.022 3.42710 0.067 0.069 0.053 0.063 0.009 12.43620 0.096 0.104 0.044 0.082 0.033 4.32330 0.117 0.116 0.047 0.093 0.040 4.04640 0.123 0.119 0.040 0.094 0.047 3.48650 0.115 0.099 0.041 0.085 0.039 3.78560 0.104 0.076 0.017 0.066 0.045 2.54670 0.087 0.052 0.027 0.055 0.030 3.16580 0.064 0.039 0.013 0.038 0.025 2.61190 0.042 0.026 0.005 0.025 0.019 2.259100 0.038 0.021 0.020 0.026 0.010 4.442
nS
TD Diff
2
0Critical Ratio
Level of Significance α = 0.05
Case -10 mV: After cardamonin treatment, the increase of mean conductance is indicated by a distance of 0.79 units away from 0. Thus, the NULL Hypothesis is ACCEPTED.
Hypothesis Test: 2-Sample Paired t-Test
Hypothesis definitionH0 - Null Hypothesis: μDiff = 0H1 - Alternative Hypothesis μDiff > 0
Case 30 mV : After cardamonin treatment, the increase of mean conductance is indicated by a distance of 4.05 units away from zero. Thus, the NULL Hypothesis is REJECTED.
With α set at 5%, then the critical value for this study is 2.92.
ACCEPT REJECT
2.92 00 2.92
Vm (mV)
Diff. Exp1
Diff. Exp2
Diff. Exp3
Mean Diff.
(Tdiff)
SD Diff.
D - Critcal Ratio
p-value
-20 0.003 -0.061 0.073 0.005 0.067 0.126 0.4556-10 0.018 -0.025 0.076 0.023 0.051 0.790 0.25620 0.042 0.022 0.065 0.043 0.022 3.427 0.037810 0.067 0.069 0.053 0.063 0.009 12.436 0.003220 0.096 0.104 0.044 0.082 0.033 4.323 0.024830 0.117 0.116 0.047 0.093 0.040 4.046 0.028040 0.123 0.119 0.040 0.094 0.047 3.486 0.0367
Scientific Assumptions:- we compare only TWO groups- both the 2 samples have comparable experimental conditions
Statistical Assumptions:- the underlying distribution is NORMAL- both the 2 samples have equal variance
Hypothesis definitionH0 - Null Hypothesis: μ1 =μ2 or μ1 - μ2 = 0H1 - Alternative Hypothesis μ ≠ μ0
Hypothesis Test: 2-Sample Unpaired t-Test
Observed value:
Hypothesis Test: 2-Sample Unpaired t-Test
H2O2 Concentration (μM)
Cell
Surv
ival
(%)
0%
20%
40%
60%
80%
100%
0 100 200 300 400 500
HEK + BK
HEK
Conc Mean SD N
100 0.936 0.854 0.774 0.887 0.921 0.897 0.958 0.892 0.892 0.890 0.053 9200 0.741 0.803 0.697 0.549 0.674 0.67 0.665 0.629 0.712 0.682 0.071 9300 0.662 0.757 0.305 0.366 0.305 0.362 0.32 0.334 0.426 0.178 8500 0.424 0.388 0.398 0.205 0.174 0.176 0.245 0.254 0.283 0.104 8
Conc Mean SD N100 0.64 0.714 0.717 0.895 0.937 0.956 0.711 0.698 0.673 0.771 0.122 9200 0.605 0.592 0.661 0.576 0.558 0.489 0.367 0.323 0.307 0.498 0.133 9300 0.562 0.551 0.349 0.348 0.33 0.168 0.154 0.172 0.329 0.163 8500 0.33 0.257 0.409 0.231 0.25 0.229 0.165 0.136 0.134 0.238 0.090 9
X1: HEK+BK
X2: HEK
QuestionIs the distance between the 2 means “big enough” to conclude that the 2 samples are significantly different?
Critical Ratio Level of Significance α = 0.05
21
21
21
21
1111NN
S
XX
NNS
D
pp
where)2(
)1()1(
21
222
211
NN
SNSNS p Weighted average of the 2 sample variances
ACCEPT REJECTREJECT
-2.120 2.120
Hypothesis Test: 2-Sample Unpaired t-Test
0
ACCEPT REJECTREJECT
-2.120 2.120
t-Test: Two-Sample Assuming Equal Variancesconcentration = 100
Variable 1 Variable 2Mean 0.890 0.771 Variance 0.003 0.015 Observations 9 9Pooled Variance 0.009 Hypothesized Mean Difference 0.0
df 16t Stat 2.681 P(T<=t) one-tail 0.008 t Critical one-tail 1.746 P(T<=t) two-tail 0.016 t Critical two-tail 2.120
Conc S2p X1-X2 SE Critical Ratio - D DF p-value100 0.00885 0.119 0.044 2.681 16 0.016397200 0.01132 0.185 0.050 3.682 16 0.002018300 0.02910 0.097 0.085 1.139 14 0.273818500 0.00938 0.045 0.047 0.959 15 0.352762
Hypothesis Test: 2-Sample Unpaired t-Test
When more than 2 means within one analysis need to be compared simultaneously, pairwise t-Test would be less appropriate.
The composite chance of making mistake increases with the number of pairwise tests.
# Parwise Test Variables Confidence Int. Error Type I1 2 95.0% 5.0%2 3 90.3% 9.8%3 4 85.7% 14.3%4 5 81.5% 18.5%5 6 77.4% 22.6%
Hypothesis Test: ANOVA
Scientific Assumptions:-You are comparing two or more groups-The true mean of the underlying population is UNKNOWN-The true SD of the underlying population is UNKNOWN
Hypothesis definitionH0 - Null Hypothesis: μ1 = μ2 = μ3 = μ4 … = μk
H1 - Alternative Hypothesis μi ≠ μj at least for 1 pairwise
Hypothesis Test: ANOVA
Statistical Assumptions:- the underlying distribution is NORMAL- samples have equal variance
How big this variance (VB) must be to indicate that the samples probably did not come from the same underlying population?
Hypothesis Test: ANOVA
If the NULL hypothesis is true:
All the sample means should be valid estimators of μ: the sample means should be fairly close to each other.
VB = variance between the sample meansVW=variance of each sample
VarianceDecomposition.xlsx
The variance of the underlying population σ2 could be estimated by 2 different methods:
a) The variance of the sample meansb) The average of the sample variances
1
])(...)()([ 2222
211
K
XxnXxnXxnVB kk
Assuming that each group has the same n, VB can be simplified as follows:
1
)(1
2
K
XxnVB
K
kk
K
k
n
i
kikK
kk n
Xx
KS
KVW
1 1
2
1
2
1
)(11
Variance Between:We know that the variance of the sample means is equal to the SE or δ2 /n
Variance Within:
Hypothesis Test: ANOVA
Under these assumptions, we expect that:
Critical RatioVW
VB
Hypothesis Test: ANOVA
OR
at least the VB tend to be as small as possible.
VB and VW should converge to the same value
Critical Ratio)(),1(2
12
1KNK
KN
K
F
KN
KVW
VB
NULL Hypothesis will be acceptable when VB tends to be very little (the sample means lay on same line) or at most when the VB is close enough to the VW (the sample means differ from each other because of the internal variation of the population)
Reject H0
Hypothesis Test: ANOVA
0 ……………………1 …….………………2 ……….……………3 …………………4………………… 5………………… 6
ACCEPT REJECT
1FVW
VB
Prob
abili
ty D
ensi
ty
F(1-α)
H2O2 Concentration (μM)
Cell
Surv
ival
(%)
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0 100 200 300 400 500
HEK
HEK+BK
HEK+BK Mut
0 ……………………1 …….………………2 ……….……………3 …………………4………………… 5………………… 6
ACCEPT REJECT
Prob
abili
ty D
ensi
ty
Anova: Single FactorGroups Count Sum Average Variance
Row 1 6 4.1530 0.6922 0.0009 Row 2 6 5.3060 0.8843 0.0043 Row 3 6 5.3360 0.8893 0.0098
ANOVASource of Variation SS df MS F P-value F crit
Between Groups 0.1517 2 0.0758 15.2225 0.0002 3.6823 Within Groups 0.0747 15 0.0050
Total 0.2264 17
data
Exp1 Exp2 Exp3 Exp4 Exp5 Exp6 Conc 100 Mean VAR SS - VB SS - VW0.6400 0.7140 0.7170 0.7110 0.6980 0.6730 HEK 0.6922 0.0009 0.0046 0.9360 0.8540 0.7740 0.9580 0.8920 0.8920 HEK+BK 0.8843 0.0043 0.0213 0.9450 0.9840 0.9920 0.8430 0.7490 0.8230 HEK+BK Mut 0.8893 0.0098 0.0488
0.2264 SS 0.1517 0.0747
VB 0.0758 0.0758 VW 0.0050 0.0050
H2O2 Concentration (μM)
Cell
Surv
ival
(%)
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0 100 200 300 400 500
HEK
HEK+BK
HEK+BK Mut
*
**