statistics applied to biomedical sciences

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


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




)()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?


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







...

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


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


Zα/2 Z1- α /2

tα/2 t1- α /2

nStb

21

nSta

2

nStCI

21

Lack of information brings about higher uncertainty


• 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:


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%.


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


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


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


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


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



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?


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 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:


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?


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


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


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


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?


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:


Under these assumptions, we expect that:

Critical RatioVW

VB


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


0 ……………………1 …….………………2 ……….……………3 …………………4………………… 5………………… 6

ACCEPT REJECT

1FVW

VB

Prob

abili

ty D

ensi

ty

F(1-α)


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


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

*

**

statistics applied to biomedical sciences

Presentations & Public Speaking