Statistics for EES 3. From standard error to t-Test

Statistics for EES3. From standard error to t-Test

Dirk Metzler

http://evol.bio.lmu.de/_statgen

May 14, 2012

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http://evol.bio.lmu.de/_statgen

Contents

1 The Normal Distribution

2 Taking standard errors into account

3 The t-Test for paired samplesExample: Orientation of pied flycatchers

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The Normal Distribution

Contents




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Density of the normal distribution

−1 0 1 2 3 4 5

0.0

0.1

0.2

0.3

0.4

Nor

mal

dich

te

µµ µµ ++ σσµµ ++ σσ

The normal distribution is also called Gauß distribution(after Carl Friedrich Gauß, 1777-1855)

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−1 0 1 2 3 4 5

0.0

0.1

0.2

0.3

0.4

Nor

mal

dich

te

µµ µµ ++ σσµµ ++ σσ

The normal distribution is also called Gauß distribution

(after Carl Friedrich Gauß, 1777-1855)

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The normal distribution is also called Gauß distribution(after Carl Friedrich Gauß, 1777-1855)

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400 450 500 550 600

0.00

00.

005

0.01

00.

015

0.02

00.

025

400:600

dbin

om(4

00:6

00, 1

000,

0.5

)

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Density of standard normal distributionA random variable Z with density

f (x) =1√2π· e−

x22

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

is called standard normally distributed.

“Gaussian bell”

for short: Z ∼N (0,1)

EZ = 0

Var Z = 1

6 / 52


If Z is N (0,1) distributed, then X = σ · Z + µ is normallydistributed with mean µ and variance σ2, for short:

X ∼ N (0,1)

X has the density

f (x) =1√2πσ

· e−(x−µ)2

2σ2 .

7 / 52


Keep in mind

If Z ∼ N (µ, σ2), then:Pr(|Z − µ| > σ) < 1

3

Pr(|Z − µ| > 2 · σ) < 0.05Pr(|Z − µ| > 3 · σ) < 0.003

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Keep in mind


3

Pr(|Z − µ| > 2 · σ) < 0.05

Pr(|Z − µ| > 3 · σ) < 0.003

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Keep in mind


3

Pr(|Z − µ| > 2 · σ) < 0.05Pr(|Z − µ| > 3 · σ) < 0.003

8 / 52


f (x) =1√2πσ

· e−(x−µ)2

2σ2

0 2 4 6 8 10

0.0

00.0

50.1

00.1

50.2

00.2

5

µ− σ µ+ σµ

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Densities need integralsIf Z is a random variable with density f (x)

0 2 4 6 8 10

0.0

00

.05

0.1

00

.15

0.2

00

.25

ba ,then

Pr(Z ∈ [a,b]) =

∫ b

af (x)dx .

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Densities need integralsIf Z is a random variable with density f (x)

0 2 4 6 8 10

0.0

00

.05

0.1

00

.15

0.2

00

.25

ba ,then

Pr(Z ∈ [a,b]) =∫ b

af (x)dx .

10 / 52


Example: Let Z ∼ N (µ = 5, σ2 = 2.25).

Pr(Z < a) can be computed with pnorm(a,mean=5,sd=1.5)

Computation of Pr(Z ∈ [3,4]):

Pr(Z ∈ [3,4]) = Pr(Z < 4)− Pr(Z < 3)

> pnorm(4,mean=5,sd=1.5)-pnorm(3,mean=5,sd=1.5)

[1] 0.1612813

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Let Z ∼ N (µ, σ2). How to compute Pr(Z = 5)?

Answer: For each x ∈ R holds Pr(Z = x) = 0

How about EZ =∑

a a · Pr(Z = a) ?

Needs a new definition:

EZ =

∫ ∞−∞

x · f (x)dx

But we already know the result EZ = µ.

12 / 52

Taking standard errors into account

Contents




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Important consequence

Consider the intervalx − s/

√n x + s/

√n

x

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x − s/√

n x + s/√

n

This interval contains µ with probabil-ity of ca. 2/3

x

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x − s/√

n x + s/√

n

This interval contains µ with probabil-ity of ca. 2/3

probability that µ is outside of intervalis ca. 1/3

x

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Thus:

It may happen that xdeviates from µ

by more thans/√

n.

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Application 1:Which values of µ are plausible?

x = 0.12s/√

n = 0.007

Question: Could it be thatµ = 0.115?

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Answer: Yes, not unlikely.

Deviationx − µ = 0.120− 0.115 = 0.005.

Standard Errors/√

n = 0.007

Deviations like thisare not unusual.

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Application 2:Comparison of mean values

Example: Galathea

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Galathea: Carapace lengthsin a sample

Males:x1 = 3.04 mms1 = 0.9 mm

n1 = 25

Females:x2 = 3.23 mms2 = 0.9 mm

n2 = 2919 / 52


The femalesare apparently larger.

Is this significant?

Or could it be just random?

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How precisely do we know the true meanvalue?

Males:x1 = 3.04 mms1 = 0.9 mm

n1 = 25s1/√

n1 = 0.18 [mm]

We have to assume uncertainty in themagnitude of±0.18 (mm) in x1

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n2 = 29s2/√

n2 = 0.17 [mm]

It is not unlikelythat x2 deviates from the true mean by more

than ±0.17 (mm).

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n2 = 29s2/√

n2 = 0.17 [mm]

It is not unlikelythat x2 deviates from the true mean by more

than ±0.17 (mm).22 / 52


The difference of the means

3.23− 3.04 = 0.19 [mm]

is not much larger thanthe expected inaccuracies.

It could also be due to pure randomthat

x2 > x1

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MORE PRECISELY:

If the true meansare actually equalµFemales = µMales

it is still quite likely thatthe sample means

x2 are x1

that different.

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In the language of statistics:

The differenceof the mean values

is(statistically)

not significant.

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not significant

=

can be just random

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Application 3:

If the mean valuesare represented graphically,

you should als showtheir precision

(±s/√

n)

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Not Like this:Carapace lengths:

Mean values for males and females2.

62.

83.

03.

23.

4

Car

apac

e le

ngth

[mm

]

●

●

Males Females

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Like that:Carapace lengths:

Mean values ± standard errors for males and females2.

62.

83.

03.

23.

4

Car

apac

e le

ngth

[mm

]

●

●

Males Females

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Application 4:

Planning an experiment:

How many observationsdo I need?

(How large should n be?)

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If you know which precision you need for(the standard error s/

√n of)

x

and if you already have an idea of sthen you can estimate

the value of nthat is necessary:

s/√

n = g(g = desired standard error)

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√n of)

x

and if you already have an idea of s

then you can estimatethe value of n

that is necessary:s/√


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√n of)

x

and if you already have an idea of sthen you can estimate

the value of nthat is necessary:

s/√


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Example:Stressed transpiration values

in another sorghum subspecies:x = 0.18s = 0.06n = 13

How often do we have to repeat theexperiment to

get a standard error of ≈ 0.01?

Which n do we need?

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Example:Stressed transpiration values

in another sorghum subspecies:x = 0.18s = 0.06n = 13

How often do we have to repeat theexperiment to

get a standard error of ≈ 0.01?

Which n do we need?31 / 52


Solution:

desired:s/√

n ≈ 0.01

From the previous experiment we know:

s ≈ 0.06,so:√

n ≈ 6n ≈ 36

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Solution:

desired:s/√

n ≈ 0.01


s ≈ 0.06,

so:√

n ≈ 6n ≈ 36

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Solution:

desired:s/√

n ≈ 0.01


s ≈ 0.06,so:√

n ≈ 6

n ≈ 36

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Solution:

desired:s/√

n ≈ 0.01


s ≈ 0.06,so:√

n ≈ 6n ≈ 36

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Summary

Assume a population has mean value µ and standarddeviation σ.

We draw a sample of size n from this population withsample mean x .x is a random variablewith mean value µ and standard deviation σ/

√n.

Estimate the standard deviation of x by s/√

n.s/√

n is the Standard Error (of the Mean).Deviations of x of the magnitude of s/

√n are usual.

They are not significant: they can be random.

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Summary

Assume a population has mean value µ and standarddeviation σ.We draw a sample of size n from this population withsample mean x .

x is a random variablewith mean value µ and standard deviation σ/

√n.


n.s/√


√n are usual.


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Summary

Assume a population has mean value µ and standarddeviation σ.We draw a sample of size n from this population withsample mean x .x is a random variable

with mean value µ and standard deviation σ/√

n.Estimate the standard deviation of x by s/

√n.

s/√


√n are usual.


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Summary

Assume a population has mean value µ and standarddeviation σ.We draw a sample of size n from this population withsample mean x .x is a random variablewith mean value µ and standard deviation σ/

√n.


n.s/√


√n are usual.


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Summary


√n.


n.

s/√


√n are usual.


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Summary


√n.


n.s/√

n is the Standard Error (of the Mean).

Deviations of x of the magnitude of s/√

n are usual.They are not significant: they can be random.

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Summary


√n.


n.s/√


√n are usual.


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The t-Test for paired samples

Contents




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sample means ± standard errors

D

C

A

B

Is A significantly different from B?

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D

C

A

B

Is A significantly different from B?No: A is in 2/3 range of µB

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D

C

A

B

Is B significantly different from C?

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D

C

A

B

Is B significantly different from C?No: 2/3 ranges of µB and µC overlap

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D

C

A

B

Is C significantly different from D?

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D

C

A

B

Is C significantly different from D?We do not know yet.

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D

C

A

B

Is C significantly different from D?We do not know yet.(Answer will be “no”.)

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The t-Test for paired samples Example: Orientation of pied flycatchers

Contents




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Pied flycatchers (Ficedula hypoleuca)

http://en.wikipedia.org/wiki/File:Ficedula hypoleuca NRM.jpg

Foto (c) Simon Eugster

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http://en.wikipedia.org/wiki/File:Ficedula_hypoleuca_NRM.jpg

http://de.wikipedia.org/w/index.php?title=Datei:Trauerschnäpper_auf_Esche_cutted.jpg


Wiltschko, W.; Gesson, M.; Stapput, K.; Wiltschko, R.

Light-dependent magnetoreception in birds: interaction of at leasttwo different receptors.

Naturwissenschaften 91.3, pp. 130-4, 2004.

Wiltschko, R.; Ritz, T.; Stapput, K.; Thalau, P.;Wiltschko, W.

Two different types of light-dependentresponses to magnetic fields in birds.

Curr Biol 15.16, pp. 1518-23, 2005.

Wiltschko, R.; Stapput, K.; Bischof, H. J.;Wiltschko, W.

Light-dependent magnetoreception in birds:increasing intensity of monochromatic lightchanges the nature of the response.

Front Zool, 4, 2007.38 / 52


the more variable the directions the

shorter the arrows!

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Direction of flight with blue light.

the more variable the directions the shorter the arrows!

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Direction of flight with blue light.Another flight of the same bird withblue light.


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All flight’s directions of this bird inblue light.


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All flight’s directions of this bird inblue light.corresponding exit points


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Directions of this bird’s flights ingreen light.


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Directions of this bird’s flights ingreen light.corresponding exit points


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corresponding exit pointsarrow heads: barycenter of exitpoints for green light


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arrow heads: barycenter of exitpoints for green light


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arrow heads: barycenter of exitpoints for green lightthe same for “blue” directions


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Question of interest

Does the color of monochromatic lightinfluence the orientation?

Experiment: For 17 birds comparebarycenter vector lengths for blue and green

light.

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Pies flycathchers :barycenter vector lengths for

green light and for blue light b, n=17

●

●

●●

●

●

●●

●

●

●

●

●

●

●

●

●

0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.1

0.2

0.3

0.4

0.5

with blue light

with

gre

en li

ght

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Compute for each bird the distance to thediagonal,

i.e.

x := “green length”− “blue length”

−0.05 0.00 0.05 0.10 0.15 0.20

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−0.05 0.00 0.05 0.10 0.15 0.20

Can the theoretical mean be µ = 0? (i.e. the expectation valuefor a randomly chosen bird)

x = 0.0518s = 0.0912

SEM =s√n

=0.912√

17= 0.022

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−0.05 0.00 0.05 0.10 0.15 0.20


x = 0.0518

s = 0.0912

SEM =s√n

=0.912√

17= 0.022

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−0.05 0.00 0.05 0.10 0.15 0.20


x = 0.0518s = 0.0912

SEM =s√n

=0.912√

17= 0.022

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−0.05 0.00 0.05 0.10 0.15 0.20


x = 0.0518s = 0.0912

SEM =s√n

=0.912√

17

= 0.022

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−0.05 0.00 0.05 0.10 0.15 0.20


x = 0.0518s = 0.0912

SEM =s√n

=0.912√

17= 0.022

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Is |x − µ| ≈ 0.0518 a large deviation?

Large? Large compared to what?

Large compared the the standard error?

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Large?

Large compared to what?


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Large? Large compared to what?


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|x − µ|s/√

n≈ 0.0518

0.022≈ 2.34

So, x is more than 2.3 standard deviations away from µ = 0.

How probable is this when 0 is the true theoretical mean? in

other words:

Is this deviation significant?

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|x − µ|s/√

n≈ 0.0518

0.022≈ 2.34


How probable is this when 0 is the true theoretical mean?

in

other words:


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|x − µ|s/√

n≈ 0.0518

0.022≈ 2.34


How probable is this when 0 is the true theoretical mean? in

other words:


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We know:x − µσ/√

n

is asymptotically (for large n) standard-normally distributed.

σ is the theoretical standard deviation, but we do not know it.Instead of σ we use the sample standard deviation s. We haveapplied the “2/3 rule of thumb” which is based on the normaldistribution.For questions about the significance this approximation is tooimprecise if σ is replaced by s (and n is not extremely large).

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n

is asymptotically (for large n) standard-normally distributed.σ is the theoretical standard deviation, but we do not know it.

Instead of σ we use the sample standard deviation s. We haveapplied the “2/3 rule of thumb” which is based on the normaldistribution.For questions about the significance this approximation is tooimprecise if σ is replaced by s (and n is not extremely large).

46 / 52



n

is asymptotically (for large n) standard-normally distributed.σ is the theoretical standard deviation, but we do not know it.Instead of σ we use the sample standard deviation s.

We haveapplied the “2/3 rule of thumb” which is based on the normaldistribution.For questions about the significance this approximation is tooimprecise if σ is replaced by s (and n is not extremely large).

46 / 52



n

is asymptotically (for large n) standard-normally distributed.σ is the theoretical standard deviation, but we do not know it.Instead of σ we use the sample standard deviation s. We haveapplied the “2/3 rule of thumb” which is based on the normaldistribution.

For questions about the significance this approximation is tooimprecise if σ is replaced by s (and n is not extremely large).

46 / 52



n

is asymptotically (for large n) standard-normally distributed.σ is the theoretical standard deviation, but we do not know it.Instead of σ we use the sample standard deviation s. We haveapplied the “2/3 rule of thumb” which is based on the normaldistribution.For questions about the significance this approximation is tooimprecise if σ is replaced by s (and n is not extremely large).

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General RuleIf X1, . . . ,Xn are independently drawn from a normal distributedwith mean µ and

s2 =1

n − 1

n∑i=1

(Xi − X )2,

thenX − µs/√

n

is t-distributed with n − 1 degrees of freedom (df).

The t-distribution is also called Student-distribution, sinceGosset published it using this synonym.

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Density of the t-distribution

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

dens

ity

Density of t-distribution with 16 df

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−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

dens

ity

Density of t-distribution with 16 dfDensity of standard normal distribution

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−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

dens

ity

Density of t-distribution with 16 dfDensity of standard normal distribution µ = 0 and σ2 = 16

14

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−4 −2 0 2 4

0.0

0.1

0.2

0.3

dens

ity

Density of t-distribution with 5 dfDensity of normal distribution with µ = 0 and σ2 = 5

3

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How (im)probable is a deviation by

at least

2.35 standard errors?

Pr(T = 2.34) =

0 does not help!

Compute Pr(T ≥ 2.34), the so-called p-value.

2.34−2.34

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

de

nsity The total area of the

magenta-colored regionsis the p-value.

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How (im)probable is a deviation by

at least

2.35 standard errors?

Pr(T = 2.34) = 0

does not help!


2.34−2.34

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

de



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How (im)probable is a deviation by at least2.35 standard errors?

Pr(T = 2.34) = 0 does not help!


2.34−2.34

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

de



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2.34−2.34

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

de

nsity

The total area of themagenta-colored regionsis the p-value.

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2.34−2.34

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

de



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R does that for us:

> pt(-2.34,df=16)+pt(2.34,df=16,lower.tail=FALSE)

[1] 0.03257345

Comparison with normal distribution:

> pnorm(-2.34)+pnorm(2.34,lower.tail=FALSE)

[1] 0.01928374

> pnorm(-2.34,sd=16/14)+

+ pnorm(2.34,sd=16/14,lower.tail=FALSE)

[1] 0.04060902

50 / 52


t-Test with R

> x <- length$green-length$blue

> t.test(x)

One Sample t-test

data: x

t = 2.3405, df = 16, p-value = 0.03254

alternative hypothesis: true mean is not equal to 0

95 percent confidence interval:

0.004879627 0.098649784

sample estimates:

mean of x

0.05176471

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We note:p −Wert = 0.03254

i.e.: If the Null hypothesis “just random”, in our case µ = 0holds, any deviation like the observed one or even more is veryimprobable.If we always reject the null hypothesis if and only if the p-value isbelow a level of significance of 0.05, then we can say followingholds:If the null hypothesis holds, then the probability that we reject it,is only 0.05.

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i.e.: If the Null hypothesis “just random”, in our case µ = 0holds, any deviation like the observed one or even more is veryimprobable.

If we always reject the null hypothesis if and only if the p-value isbelow a level of significance of 0.05, then we can say followingholds:If the null hypothesis holds, then the probability that we reject it,is only 0.05.

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i.e.: If the Null hypothesis “just random”, in our case µ = 0holds, any deviation like the observed one or even more is veryimprobable.If we always reject the null hypothesis if and only if the p-value isbelow a level of significance of 0.05, then we can say followingholds:

If the null hypothesis holds, then the probability that we reject it,is only 0.05.

52 / 52



i.e.: If the Null hypothesis “just random”, in our case µ = 0holds, any deviation like the observed one or even more is veryimprobable.If we always reject the null hypothesis if and only if the p-value isbelow a level of significance of 0.05, then we can say followingholds:If the null hypothesis holds, then the probability that we reject it,is only 0.05.

52 / 52

Statistics for EES 3. From standard error to t-Test

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Transcript of Statistics for EES 3. From standard error to t-Test