Statistics for Molecular Medicine -- Probability ... · PDF fileStatistics for Molecular Medicine -- Probability & Diagnostic Testing -- ... Probability: Definition and Calculation

Statistics for Molecular Medicine -- Probability & Diagnostic Testing --

Barbara Kollerits & Claudia LaminaMedical University of Innsbruck, Division of Genetic Epidemiology

Molekulare Medizin, SS 2016

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Probability:

Definition and Calculation rules

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3

Descriptive Statistics: Pure description of your observed data

But: You cannot conclude on the underlying population

Example: It is not enough to know, that a medication is effective in the

specific study is it effective in all patients for whom the medication

was meant for?

Was the result a chance finding?

With which certainty can the result be transferred to the population?

Introduction

Random experiment • A situation involving chance leading to mutually excluding events.

• The result of the experiment is not known beforehand.

• The experiment can be uncontrolled (observational studies) or controlled, which can be repeated under the same conditions.

Outcome • All possible realizations of an experiment

• Example: The set of all possible realizations of the experiment “Rolling a dice” is 1,2,3,4,5,6

• Lets denote the set of all possible events as Ω

Event / Realization • One specific realization of a single trial of an experiment

• Example: At one single throw, you get “2”

Probability • is the measure of how likely an event is.

• The probability of landing “2” in one dice throw is 1/6.

Definitions

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Sets A and B are sets for possible realizations:

0 ≤ P(A) ≤ 1 (for each A Ω)

P(Ω) = 1

P(Ø) = 0

P(A)=1-P(A) with A = Ω\A = „Not-A“

“B is part of A”: B A

P(B) ≤ P(A), if B A

Complementary set A\B

Calculation rules for probabilities

AB

Ω

∩

BA

∩_ _

„A without B“A B

∩

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Intersecting set of A and B: A ∩ B

Set union A B: „A or B“


A BA ∩ B

∩

A B

Common saying: „A and B“

A B

A and B not distinct A and B distinct

P(A B) = P(A) + P(B) – P(A ∩ B)

∩

P(A B) = P(A) + P(B)

∩

More general: If all Ai are pairwise distinct:

P(A1 A2… Ak) = P(A1) + P(A2) + ….+ P(Ak)

∩

∩

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“Counting rule” for Probability:

“Counting rule” for Odds:

Definition of Odds:

Example: Experiment = Tossing a coin once

P(„Head“) = 0.5

Odds („Head“) = 1:1

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The number of ways event A can occurThe total number of possible outcomes

P(A)=

7The number of ways event A can occur P(A)=

The number of ways event A can not occur

P(A) Odds(A)=1-P(A)

Example: The probability to get an even number (2,4,6), when rolling a

dice once (6 possible outcomes): P(even number) =

Probabilities and relative frequencies: The relative frequency of an

event approximates the probability, if the number of trials is high enough


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Number of trials (rolling the dice)

Rel

ativ

e fr

eque

ncy

ofth

eev

ent

„eve

nnu

mbe

r“

A simulated „experiment“: The approximation ofa theoretical probability by a relative frequency

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Example: In a practical seminar, students (n=250) determine their blood

types and get the following results:


Blood type Absolute frequencies Relative frequencies

A 106 0.424

B 34 0.136

AB 15 0.06

0 95 0.38

Rhesus factor

+ 210 0.84

- 40 0.16

Relative frequencies asapproximationof probabilities

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Relative frequencies as approximation of the probabilities:

Blood groups: P(A) = 0.424, P(B) = 0.136, P(AB) = 0.06, P(0) = 0.38

Rhesus Factor: P(R+) = 0.84, P(R-) = 0.16

Applying the rules for calculation:

P(A or B) =

(it can only apply either A or B A and B are distinct)

P(not AB) =

Calculation rules for probabilities: Exercise

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Conditional probability = P(A|B) = „Probability of A given B“

The probability of the event A, if it is known, that B already occurred

P(A|B) = or equivalently: P(A ∩ B) = P(A|B)*P(B)

P(A ∩ B) = P(A|B)*P(B) = P(B|A)*P(A)

If the events A and B are independent P(A|B) = P(A) and P(B|A) = P(B)

For the occurrence of event A it is irrelevant , if B occurred or not

A and B independent P(A ∩ B) = P(A)*P(B)

Example: Two subsequent dice rolls do not depend on each other

P(getting a 6 in two subsequent trials) = 1/6 * 1/6 = 1/36

Conditional probabilities

P(A ∩ B)P(B)

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Example for independence:

Relative frequencies as approximation of the probabilities:

Blood groups: P(A) = 0.424, P(B) = 0.136, P(AB) = 0.06, P(0) = 0.38

Rhesus Factor: P(R+) = 0.84, P(R-) = 0.16

Applying the rules for calculation:

P(A and R+) =


6

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Example for dependence:

P(Diabetes|male) = 0.07; P(Diabetes|female) = 0.02

The probability to get Type 2 diabetes depends on the gender

How high is the probability that one randomly chosen person (gender

unknown) has diabetes?

P(Diabetes) = ?


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Joint event Probability

Probability tree

male

female

P(m)=0.5

P(f)=0.5

Diabetes P(D|m) = 0.07

NoDiabetes P(not D|m) = 0.93

Diabetes

NoDiabetes P(not D|f) = 0.98

P(D|f) = 0.02

P(male and Diabetes) = 0.5*0.07

P(female and Diabetes) = 0.5*0.02

P(Diabetes) = 0.07*0.5 + 0.02*0.5 Total probability rule

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Rule for the total probability:

P(B) = P(B|A1)*P(A1) +…+ P(B|Ak)*P(Ak)

Example:

P(Diabetes) = 0.07*0.5 + 0.02*0.5

Bayes Theorem:

P(A|B) = =

Important for evaluation of the validity of diagnostic tests


ΩA1 A2 A3 …. Ak

B

P(A ∩ B)P(B)

P(A ∩ B) Σi P(B|Ai)*P(Ai)

P(m) P(f)

P(D|f)P(D|m)


On a cruise ship with 70% female and 30% male passengers, a mysteriousinfectious disease D has broken out. The infection rate is 5 % in womenand 10% in men.

How high is the probability that a randomly selected passenger is infected?

Solution with total probability rule:

Solution with probability tree:

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Diagnostic Testing

Intention:

To find a marker / specific test, that discriminates patients into diseased or

non-diseased

How accurate is a test? Can it detect all currently diseased and exclude all non-diseased (e.g. Elisa test, Westernblot, PCR etc.). How likely is it, that a person is actually diseased/non-diseased, if the test is positive/negative?

Make predictions for individuals, who will get the disease in the future

Example 1: Predict the progression of a chronic disease using a biomarker

Example 2: Predict the probability to get a disease in the future based on genetic variants

Example: Ideal situation – perfect test

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Example: In a study, there are 200 individuals, 100 of them are diseased and 100 are healthy. A scientist claims that he has found a new test which perfectly detects those who are ill and those who are healthy.

The “perfect” test would yield such a table:


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Test result „Truth“

Disease No Disease

Positive 100 (a) 0 (b) 100 (a + b)

Negative 0 (c) 100 (d) 100 (c + d)

100 (a+c) 100 (b + d) 200

Sensitivity and Specificity

Generally used with diagnostic tests

Sensitivity: Percentage of persons with positive test results ( ) of all diseased persons ( ) =

Specificity : Percentage of persons with negative test results ( ) of all non-diseased persons ( ) =

What would be the “perfect” test?

→ one with no false negative results (sensitivity would be 1) or false positive results (specificity would be 1)

When the sensitivity increases, the specificity decreases (or vice versa)

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Sensitivity and Specificity

Sensitivity = True positive / (True positive + False negative) x 100 =

a / (a+c) x 100

Specificity = True negative / (True negative + False positive) x 100 =

d / (b+d) x 100


Disease No disease

Positiv diseased and positive test result = true positive (a)

not diseased but positive test result = false positive (b)

Negativ diseased, but negative test result = false negative (c)

not diseased, and negative test result = true

negative (d)


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Disease No Disease

Positive 100 (a) 0 (b) 100 (a + b)

Negative 0 (c) 100 (d) 100 (c + d)

100 (a+c) 100 (b + d) 200

The “perfect test” revisited:

In a study, there are 200 individuals, 100 of them are diseased and 100 are healthy. A scientist claims that he has found a new test which perfectly detects those who are ill and those who are healthy.

Sensitivity = True positive / (True positive + False negative) x 100 = a / (a+c) x 100 = (100 / 100) x 100 = 100%

Specificity = True negative / (True negative + False positive) x 100 = d / (b+d) x 100 = (100 / 100) x 100 = 100%

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Sensitivity and Specificity: Example 2

Real example: Intention: Predict the progression of renal disease in patients with chronic kidney disease (CKD)

Cohort study of 227 nondiabetic patients with CKD

Fibroblast growth factor 23 (FGF23) plasma concentrations (involved in calcium-phosphate metabolism)

177 of the patients were followed prospectively for up to 7 years to assess progression of renal disease

Sensitivity = a / (a+c) x 100 = 52 / (52 + 11) x 100 = 82.5 %

Specificity = d / (b+d) x 100 = 76 / (76+34) x 100 = 69.1 %

23Kollerits et al. Fibroblast Growth Factor 23 (FGF23) Predicts Progression of Chronic Kidney Disease: The Mild to Moderate Kidney Disease (MMKD) Study. J Am Soc Nephrol 18: 2601–2608, 2007


Progression No Progression

FGF 23 above median

52 (a) 34 (b)

FGF 23 below median

11 (c) 76 (d)

63 (a+c) 110 (b + d)

Positive and negative predictive value

Sensitivity and specificity describe the accuracy of a test

Estimation of the probability of the presence or absence of disease: positive predictive value (PV+) and the negative predictive value (PV-)

Now we are interested in how likely it is for one patient to have the disease if a test is positive:

PV+ = percentage of persons who actually have the disease given a positive test result =

PV- = percentage of persons who don’t have the disease given a negative test result =

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Probability tree (conditional probabilities of composed events)

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Sick and Test positive

Sick and Test negative = false negative

Not sick and Test positive = false positive

Not sick and Test negative

State: : healthy

: sick

Test: : Test neg.

: Test pos.Joint event Probability

= Sensitivity

= 1- Sensitivity

= Specificity

= 1- Specificity

Example for calculating positive and negative predictive value

Example: “Elisa Aids-Test”

1989: screening of all civil service applicants of the land Bavaria was planned

The result of the Elisa Aids-Test gives the information, whether HIV antibodies are in the blood or not

The Elisa Aids-Test has a very high sensitivity and specificity (99,9% and 99,5%, respectively)

There are the following assumptions:

: there are antibodies in the blood

: there are no antibodies in the blood

: the test is positive on antibodies

: the test is negative on antibodies

Sensitivity: = 0.999

Specificity: = 0.995

Prevalence of HIV in the whole population = 0.001

Question: If the Elisa Aids-Test is positive, how high is the probability to really have HIV antibodies in the blood?

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0.001

No HIV

HIV

Test positive

Test positive

0.999

Test negative

Test negative

0.999 x 0.001

0.999

0.0050.999 x 0.005

Positive predictive value = 17% of persons with a positive Elisa Aids-Test result actually

have HIV antibodies in the blood

83% are falsely diagnosed

Formula based on Bayes Theorem:

Sensitivity

1- Specificity: false positive

Test positive


0.001

No HIV

HIV

Test positive

Test positive

Test negative

Test negative

0.999

0.995

0.001

Negative predictive value = 99% of persons with a negative Elisa Aids-Test result actually

have no HIV antibodies in the blood

= 1- Sensitivity: false negative

= Specificity

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Relation between positive and negative predictive value

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Dependency of PV- / PV+ on the prevalence / incidence of a disease:

Here, Sensitivity = 99,9% and Specificity = 99,5%

Even if a test has very high sensitivity and specificity, a screening in the whole population is not appropriate for rare diseases

define high-risk populations first

Example: screening for prostate cancer only in men above a certain age (in Austria >45 years)

You want to validate a risk marker for the progression of a degenerative disease and get the following results in your study after several years of follow-up:

A clinician claims that values greater than 100 can predict the progression ofdisease. If you use this value as a clinical test for defining progression, how high is the sensitivity and specificity of this test?

Sensitivity =

Specificity=

What can you do to increase the Sensitivity?

What happens then with the Specificity then?

Categorized risk marker Number of patients withprogression of disease

Number of stable patients

Minimum-60 0 1061-80 5 2581-100 25 50101-120 20 10121-Maximum 10 5Sum Σ 60 100

Sensitivity / Specificity / PPW: Exercise

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Sensitivity =

Specificity =

It is known, that the incidence of the disease progression (D) is 1%.

Fill in all the parameters that you already know into this probability tree:

Based on this tree, how high is the PPW of this test?

Sensitivity / Specificity / PPW: Exercise

D

not D

Test positive

Test negative

Test positive

Test negative

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Probability distributions

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Random variable X: The values of a random variable X are the

outcomes of a random experiment. A number is assigned to all possible

realizations of a random experiment.

Realization of a random variable X in an experiment: xi

Discrete random variable: Qualitative variables, like gender, disease

status etc.

Continuous random variable: Qualitative variable like age, cholesterol

levels etc.

Definitions

Why are gender, disease status, age etc. „random“?

They are outcomes of the random experiment„Drawing one person from the population“

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A random variable is discrete, if it can only take a finite number of

realizations

To each realization, a specific probability can be assigned:

f(x) = P(X = xi) = pi Probability function

Distribution function :

F(x) = P(X ≤ x) Probability that X takes the values of x or smaller than x

A distribution function is increasing monotonically and is the sum of the

probabilities of all realizations ≤ x

Discrete random variables

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Example: „Rolling a dice“ once

The random variable is X = “The sum of the pips when rolling a dice once”

Each realization has the probability 1/6: f(x1)=…f(x6) = 1/6

F(2) = F(X ≤ 2) = f(x1) + f(x2) = 2/6

F(3) = 3/6 etc…


Realisation xi Probabilityf(xi) = pi

Distribution function F(xi)

1 1/6 1/6

2 1/6 2/6

3 1/6 3/6

4 1/6 4/6

5 1/6 5/6

6 1/6 6/6

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Example for a uniform distribution

Histogram of the probability

= Density function

Distribution function

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Characteristics of probability distributions

Expectation E(X):

Measures of location for a sample: Mean or Median

Measure of location for the underlying population: E(X) or μ

For a discrete random variable with k realizations: E(X) = x1p1 + …+xkpk

Example: X = “The sum of the pips when rolling a dice once”

E(X) = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 3.5

Expected average number of pips when rolling the dice several times

Expectation of a sum: E(X1+X2+…+Xn) = E(X1) + E(X2) + … + E(Xn)

Example: Z = “The sum of the pips when rolling two dices”

E(Z) = E(Xdice1) + E(Xdice2) = 3.5 + 3.5 = 7

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Characteristics of probability distributions

Variance Var(X):

Measure of dispersion for a sample: Sample variance

Measure of dispersion for the underlying population: Var(X) or σ2

Var(X) = E(X - μ)2 = E(X2) - μ2

For a discrete random variable : Var(X) = (x1 – μ)2p1 + …+ (xk – μ)2pk

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The Binomial distribution

Bernoulli-Experiment: The realizations of a Bernoulli-Experiment can

take exactly two different values

The probability is described as follows: P(X=1) = p; P(X=0) = 1-p = q

Examples:

A Binomial distribution describes a process, where a Bernoulli-

Experiment is repeated several times and independently from each

other Binomial distributed random variable: X = X1+X2 + …. Xn

(n independent Bernoulli-Experiments)

Experiment Possible realisations Probability P(X)

Tossing a coin HeadTail

0.50.5

Birth of a child FemaleMale

0.50.5

Getting a diseaseup to a specific age

DiseaseNo disease

0.1 (for example)0.9

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Binomial distributed random variable X is described unambiguously by:

n: number of times the Bernoulli-Experiment is repeated

p = P(X=1)

X ~ B(n,p) : X is Binomial distributed with its parameters n and p

E(X) = np

Var(X) = np(1-p)

P(X=k) = ( )pk(1-p)n-k ( )= nk

nk

1*2*…*n (1*…*k)*(1*…*(n-k))

Binomial-coefficient

Example: ( )=103

10*9*8 1*2*3

simplifiedversion

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Example: Passing a MC-exam just by guessing

5 answers per question p = 0.2

10 questions

E(X) = n*p = 2; Var(X) = n*p*(1-p) = 1.6

The exam is passed, if 6 or more answers are correct P(X ≥ 6) ?

~B(10,0.2)

you can expect to have 2 answers right

k P(X = k) P(X ≤ k)

0 1*0.20*0.810 = 0.1074 0.1074

1 10*0.21*0.89 = 0.2684 0.3758

2 45*0.22*0.88 = 0.3020 0.6778

3 120*0.23*0.87 = 0.2013 0.8791

4 210*0.24*0.86 = 0.0881 0.9672

5 252*0.25*0.85 = 0.0264 0.9936

6 210*0.26*0.84 = 0.0055 0.9991

7 120*0.27*0.83 = 0.0008 0.99992

8 45*0.28*0.82 = 0.00007 0.999999

9 10*0.29*0.81 = 0.000004 ~ 1

10 1*0.210*0.80 = 0.0000001 ~1

P(X ≥ 6) =

= 1- P(X ≤ 5) =

= 1- 0.9936 =

= 0.0064

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Histogram ofthe probability

= Density


k P(X = k) P(X ≤ k)

0 0.1074 0.1074

1 0.2684 0.3758

2 0.3020 0.6778

3 0.2013 0.8791

4 0.0881 0.9672

5 0.0264 0.9936

6 0.0055 0.9991

7 0.0008 0.99992

8 0.00007 0.999999

9 0.000004 ~ 1

10 0.0000001 ~1

Distribution functionP(X ≤ 5) = ?

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Histogram of theprobability


P(X ≤ 5) = F(5) = 0.9936

Area under thecurve = 0.9936

The Binomial distribution: Exercise

In a laboratory experiment with 30 students, a virus is set free. The infectionprobability is 20% for each student Binomial distributed

How many infections are expected (Expectation E(X)) ?

Determine the following probabilities using the formula

P(X=k) = ( )pk(1-p)n-k

a) no student will be infected

b) exactly one student will be infected

nk

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The Binomial distribution: Exercise

Now, the distribution function of this probability distribution is given. Determine the following probabilities using this figure:

a) at most 3 students will be infected

b) more than 8 students will be infected

Number of students

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Continuous random variables

A variable is continuous, if each value between two values a and b (with a

< b) can possibly be realized.

How can a probability be assigned to P(X ≤ x) ?

For discrete random variables sum of the probabilities of all values ≤ x

For continuous random variables “area under the curve”

f(x) F(x)

x0 x0

F(x0)

Density


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Continuous random variables

Distribution function of a continuous variable:

F(x) = P(X ≤ x) = f(t) dt

f(x) F(x)

x0 x0

F(x0)

Density

x

∫-∞ The total area between the x-Axis and the density function f(x) is 1

P(a ≤ X ≤ b) = F(b) – F(a) and P(X ≥ a) = 1- F(a)


1

0

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The normal distribution

Normal distribution = the most important distribution function in statistics

It can be described unambiguously by μ and σ2 : X ~ N(μ, σ2)

The densities of the following normal distributions are given:

N(0, 1) standard normal distribution

N(0,4) higher variance

N(1,1.44) slightly higher variance and “shifted” to the right

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Normal distribution

μ = 4

σ2 = 4

Standard Normal distribution

μ = 0

σ2 = 1

Standardization: Z=


N(4, 4)

N(0, 1)

X - μσ

In this example: Z = X - 42

Distribution of X:

Distribution of Z:

An illustration of the standard normal distribution with quantiles:

20% of the data

50% of the data

5% of the data

p 0.2 0.5 0.8 0.95 0.975

Quantile zp -0.85 0 0.85 1.64 1.96

zp = -z1-p


Symmetry

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Idea of a Quantile-Quantile-Plot (QQ-Plot):

Plot the quantiles of two distributions against each other:

20%-Quantile

50%-Quantile

97.5%-Quantile


If the two distributions are the same, all points lie on one line !

If QQ-plots are used to be compared with the normal distribution, they areoften also called Normal-Quantile-Plots (NQ-Plot).

50%-Quantile


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Do our observed data follow a normal distribution?

Histogram of the observed data with the estimated density function:


Do our observed data follow a normal distribution?

Compare with normal distribution:Blue line:

Normal distribution withmean = 53.25 and stdev = 9.19

(as estimated from the data)


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NQ-Plot Histogram with normal curve

Example:

Test the distribution of the variable age in observed data with the normal distribution:

The tails of the distribution do not fit !


Quantiles of Normal distribution: Exercise

Fasting blood glucose is a normally distributed variable with =90 and =10. What is the probability for a randomly drawn person to have a blood glucoselevel of

a) ≤ 75 mg/dl b) >100 mg/dl c) >85 & ≤95

Indicate the probabilities as area under the curve:

N(90,100)) Density function

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Quantiles of Normal distribution: Exercise

Calculate the probabilities using quantiles of the standard normal distribution:

a) ≤ 75 mg/dl

b) >100 mg/dl

c) >85 & ≤95

Distribution function of a standard normal distribution

Steps:

1. Standardize to getquantiles (z) of a standard normal distribution

2. Derive theprobability(graphically) fromthe distributionfunction

Other continuous distributions

t-distribution: X ~ t(df)

The t-distribution only depends on the degrees of freedom df=(n-1)

The t-distribution approximates N(0,1) for increasing n

f(x)

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F-distribution: X ~ F(df1, df2) : it depends on two „degrees of freedom“ df

f(x)


Chi-Square-distribution: X ~ 2 (df) :

it depends on one „degree of freedom“ dff(x)

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Point and Confidence Estimates

Point and confidence estimates

Intention: Conclude from sample on underlying population

Complete population of interest (e.g. all Austrians, all patients with previous myocardial infarctions etc…) cannot be observed samples drawn from the population

Samples should be chosen to be representative for the population.

Sample1

Sample 3

Sample 2Population

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Arithmetic Mean in the study sample:

Descriptive measures in the study sample

Conclude on

unknown parameter / characteristic of the underlying population

For example:

is the estimate of

Expectation value of the underlying population


Example:

Population of interest = All patients with previous MI

„Parameter“ of interest: blood pressure

Study sample: Representative sample of all patients with previous MI

Expected value of the blood pressure in the underlying population, that cannot be observed

Arithmetic Mean of blood pressure in the study sample:

is the estimate of

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The precision of the estimate depends on the quality of the sample (representativity) and the sample size

Properties of a good estimator:

Unbiasedness: There is no systematic under- or overestimation

Consistency: increasing the sample size increases the probability of the estimator being close to the population parameter small variance

Bias-Variance-Tradeoff: find an estimating function T, which reduces both:

Reducing Mean Squared Error:

Bias large Bias large

Variation small

Bias small

Variation small Variation large Variation large

Bias small


The precision of the estimate depends on the quality of the sample (representativity) and the sample size

Properties of a good estimator:

Unbiasedness: There is no systematic under- or overestimation

Consistency: increasing the sample size increases the probability of the estimator being close to the population parameter small variance

Bias-Variance-Tradeoff: find an estimating function T, which reduces both:

Reducing Mean Squared Error:

Bias large Bias large

Variation small

Bias small

Variation small Variation large Variation large

Bias small

Most efficient

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Examples for unbiased, consistent and most efficient estimators:

Arithmetic Mean X

Sample Variance S2

Relative frequency

expected value

variance

proportion p of a

dichotomous trait

_


There is uncertainty in parameter estimation because it is based on a random sample of finite size from the population of interest

Construct an interval, that includes the population parameter with given certainty: Confidence Interval CI

The measure of certainty is given by the error probability

= 5% : 95% CI; = 1% : 99% CI etc.

Assumptions on the underlying distribution have to be made, e.g. it is assumed, that the values X1, X2, …, Xn are the measurements of a normally distributed variable X (the variable is N(,2)-distributed)

Confidence intervals for all parametric models can be described in this way:

Parameter estimate +/-(1-/2)-Quantile of the respective probability

distribution

* Standard error of the estimate

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Standard error of the mean

The standard error is the standard deviation of the sampling distribution of a statistical estimate (e.g. the mean)

Standard error of the mean:

How accurate is the mean?

How much would the mean change when different samples are taken?

Gets smaller, the bigger sample size ( estimate is more accurate)

Difference to standard deviation: Standard deviation describes the variability of the single observations in the sample you cannot conclude on the population

Confidence Interval

0.025-Quantile

= -1.96

0.975-Quantile:

= + 1.96

Parameter estimate +/-(1-/2)-Quantile of the respective probability

distribution

* Standard error of the estimate

Mean +/-Quantiles of the Standard

Normal Distribution* Standarderror of the Mean

Example: CI for the Mean:

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Confidence Interval

and = variance of the standard normal distribution

= (1-/2)-Quantile of the standard normal distribution

Interpretation of the confidence interval in general:

If the study is repeated 100 times on 100 different samples, the point estimate (here: mean) will be within this range in 1-percent of cases

For a 95% confidence interval:

= 5% z0.975 ~ 1.96

If is not known & the sample size is large (>30, better >120): take S (sample variance)

Mean +/-Quantiles of the

Standard Normal Distribution

* Standarderror of the Mean

Example for 95% confidence interval for the mean of total cholesterol within a sample of n=1475 individuals

If this study is repeated 100 times, it is expected that the mean of total cholesterol falls into this range in 95 times

Confidence Interval

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Confidence Interval

Simulation Example: 95% CI for mean of body height in mean born 1993

Assumption: We know the „truth“: mean = 182 cm; σ = 7

A different sample is drawn 100 times (either with n=100 or n=20)

in 5 out of 100 experiments, the true expected mean value is not included

If you only conduct an experiement once (as usual), you do not know, ifthe true value is included (green bars) or not (red bars)

n=100 n=20

p 0.95 0.975 0.995

zp 1.64 1.96 2.58

Determine the 95%-CI for the following parameters:

Mean = 100

S = 15

n=25

95 % CI =

If you repeat an experiment 100 times with the assumption, that these parameters are true, how many times can you expect to observe a mean that is lower or higher than these confidence limits?

Quantiles of the standard normal distribution:

Confidence Interval: Exercise

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Confidence Interval: Exercise

p 0.95 0.975 0.995

zp 1.64 1.96 2.58

The manufacturer of a laboratory measurement device claims that one measurement will take 5 minutes on average. You want to test that statement with 10 measurements.

You get the following estimates: Mean = 5.3, Standard deviation = 0.3

Determine a 99% Confidence Interval (CI):

99 % CI =

Interpretation:

Quantiles of the standard normal distribution:

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