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CERAM
February-March-April 2008
Class 2
Statistical Inference
Lionel Nesta
Observatoire Français des Conjonctures Economiques
[email protected]
Hypothesis Testing
The Notion of Hypothesis in Statistics
 Expectation
 An hypothesis is a conjecture, an expected explanation of why a given
phenomenon is occurring
 Operational -ity
 An hypothesis must be precise, univocal and quantifiable
 Refutability
 Le result of a given experiment must give rise to either the refutation or the
corroboration of the tested hypothesis
 Replicability
 Exclude ad hoc, local arrangements from experiment, and seek universality
Examples of Good and Bad Hypotheses

« The stakes Peugeot and Citroen have the same variance »

« God exists! »

« In general, the closure of a given production site in Europe is
positively associated with the share price of a given company
on financial markets. »

« Knowledge has a positive impact on economic growth »
Hypothesis Testing
 In statistics, hypothesis testing aims at accepting or rejecting a
hypothesis
 The statistical hypothesis is called the “null hypothesis” H0
 The null hypothesis proposes something initially presumed true.
 It is rejected only when it becomes evidently false, that is, when the
researcher has a certain degree of confidence, usually 95% to 99%,
that the data do not support the null hypothesis.
 The alternative hypothesis (or research hypothesis) H1 is the
complement of H0.
Hypothesis Testing
 There are two kinds of hypothesis testing:
 Homogeneity test compares the means of two samples.
 H0 : Mean(x) = Mean(y) ; Mean(x) = 0
 H1 : Mean(x) ≠ Mean(y) ; Mean(x) ≠ 0
 Conformity test looks at whether the distribution of a given sample
follows the properties of a distribution law (normal, Gaussian, Poisson,
binomial).
 H0 : ℓ(x) = ℓ*(x)
 H1 : ℓ(x) ≠ ℓ*(x)
The Four Steps of Hypothesis Testing
1.
Spelling out the null hypothesis H0 et and the alternative
hypothesis H1.
2.
Computation of a statistics corresponding to the distance
between two sample means (homogeneity test) or between the
sample and the distribution law (conformity test).
3.
Computation of the (critical) probability to observe what one
observes.
4.
Conclusion of the test according to an agreed threshold around
which one arbitrates between H0 and H1 .
The Logic of Hypothesis Testing

We need to say something about the reliability (or
representativeness) of a mean

Large number theory; Central limit theorem

The notion of confidence interval

Once done, we can whether two mean are alike

If so (not), their confidence intervals are (not) overlapping
Statistical Inference

In real life calculating parameters of populations is
prohibitive because populations are very large.

Rather than investigating the whole population, we take
a sample, calculate a statistic related to the parameter of
interest, and make an inference.

The sampling distribution of the statistic is the tool that
tells us how close is the statistic to the parameter.
Prerequisite
Standard Normal Distribution
Two Prerequisites

Large number theory


Large number theory tells us that the sample mean will
converge to the population (true) mean as the sample size
increases.
Central Limit Theorem

Central Limit Theorem tells us that for many samples of like
and sufficiently large size, the histogram of these sample
means will appear to be a normal distribution.
The Dice Experiment
Value
P(X = x)
1
1/6
2
1/6
1 x6
21
E  X   X   x 
 3.5
6 x 1
6
0.20
3
1/6
4
1/6
0.12
5
1/6
0.08
1/6
0.04
6
0.16
0.00
1
2
3
x
4
5
6
The Dice Experiment (n = 2)
Sample
1
2
3
4
5
6
7
8
9
10
11
12
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
 
Mean Sample
Mean
1
13
3,1
2
1.5
14
3,2
2.5
2
15
3,3
3
2.5
16
3,4
3.5
3
17
3,5
4
3.5
18
3,6
4.5
1.5
19
4,1
2.5
2
20
4,2
3
2.5
21
4,3
3.5
3
22
4,4
4
3.5
23
4,5
4.5
4
24
4,6
5
E X  X 
1
1
X1 
X2 
36
36

Sample
25
26
27
28
29
30
31
32
33
34
35
36
Mean
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
1
X 36  3.5   X
36
3
3.5
4
4.5
5
5.5
3.5
4
4.5
5
5.5
6
Sample
1
2
3
4
5
6
7
8
9
10
11
12
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
Mean Sample
Mean
1
13
3,1
2
1.5
14
3,2
2.5
2
15
3,3
3
2.5
16
3,4
3.5
3
17
3,5
4
3.5
18
3,6
4.5
1.5
19
4,1
2.5
2
20
4,2
3
2.5
21
4,3
3.5
3
22
4,4
4
3.5
23
4,5
4.5
4
24
4,6
5
Sample
25
26
27
28
29
30
31
32
33
34
35
36
Mean
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
3
3.5
4
4.5
5
5.5
3.5
4
4.5
5
5.5
6
6/36
5/36
4/36
3/36
2/36
1/36
1
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5 6.0
x
The Normal Distribution
In probability, a random variable follows a normal distribution
law (also called Gaussian, Laplace-Gauss distribution law) of
expectation μ and standard deviation σ if its probability
density function (pdf) is such that
f ( x) 
1
 2
e
1  x 
 

2  
2
This law is written N (μ,σ ²). The density function of a normal
distribution is symmetrical.
Normal Distributions For Different values
of μ and σ
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-5
-4
-3
-2
(μ=0;σ=1)
-1
0
(μ=0.5;σ=1.1)
1
2
3
(μ=-2;σ=0.5)
4
5
The Standard Normal Distribution
The standard normal distribution, also called Z distribution,
represents a probability density function with mean μ = 0 and
standard deviation σ = 1. It is written as N (0,1).
All random variable following a normal law can be standardized via
the following transformation
z
x

The Standard Normal Distribution
(μ=0; σ=1)
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-5.0 -4.5 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
The Standard Normal Distribution
0.45
68% of
observations
0.4
0.35
0.3
95% of
observations
0.25
0.2
99.7% of
observations
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
The Standard Normal Distribution
0.45
0.4
0.35
0.3
0.25
95% of observations
0.2
0.15
0.1
2.5%
2.5%
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
The Standard Normal Distribution (z scores)
0.45
0.4
0.35
0.3
0.25
P(Z ≥ 0)
P(Z < 0)
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Probability of an event (z = 0.51)
0.45
0.4
0.35
0.3
0.25
P(Z ≥ 0.51)
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Probability of an event (z = 0.51)

The z-score is used to compute the probability of
obtaining an observed score.

Example
 Let z = 0.51. What is the probability of observing
z=0.51?
 It is the probability of observing z ≥ 0.51: P(z ≥ 0.51)
= ??
Standard Normal Distribution Table
z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.500
0.496
0.492
0.488
0.484
0.480
0.476
0.472
0.468
0.464
0.1
0.460
0.456
0.452
0.448
0.444
0.440
0.436
0.433
0.429
0.425
0.2
0.421
0.417
0.413
0.409
0.405
0.401
0.397
0.394
0.390
0.386
0.3
0.382
0.378
0.375
0.371
0.367
0.363
0.359
0.356
0.352
0.348
0.4
0.345
0.341
0.337
0.334
0.330
0.326
0.323
0.319
0.316
0.312
0.5
0.309
0.305
0.302
0.298
0.295
0.291
0.288
0.284
0.281
0.278
0.6
0.274
0.271
0.268
0.264
0.261
0.258
0.255
0.251
0.248
0.245
0.7
0.242
0.239
0.236
0.233
0.230
0.227
0.224
0.221
0.218
0.215
0.8
0.212
0.209
0.206
0.203
0.201
0.198
0.195
0.192
0.189
0.187
0.9
0.184
0.181
0.179
0.176
0.174
0.171
0.169
0.166
0.164
0.161
1.0
0.159
0.156
0.154
0.152
0.149
0.147
0.145
0.142
0.140
0.138
1.6
0.055
0.054
0.053
0.052
0.050
0.050
0.049
0.048
0.047
0.046
1.9
0.029
0.028
0.027
0.027
0.026
0.026
0.025
0.024
0.024
0.023
2.0
0.023
0.022
0.022
0.021
0.021
0.020
0.020
0.019
0.019
0.018
2.5
0.006
0.006
0.006
0.006
0.006
0.005
0.005
0.005
0.005
0.005
2.9
0.002
0.002
0.002
0.002
0.002
0.002
0.002
0.002
0.001
0.001
Probability of an event (Z = 0.51)

The Z-score is used to compute the probability of
obtaining an observed score.

Example
 Let z = 0.51. What is the probability of observing
z=0.51?
 It is the probability of observing z ≥ 0.51: P(z ≥ 0.51)
 P(z ≥ 0.51) = 0.3050
Example
 Suppose that for a population students of a famous business school in
Sophia-Antipolis, grades are distributed normal with an average of 10
and a standard deviation of 3. What proportion of them
 Exceeds 12 ; Exceeds 15
 Does not exceed 8 ; Does not exceed 12
 Let the mean μ = 10 and standard deviation σ = 3:
12  10
3
15  10
z
3
8  10
z
3
12  10
z
3
z
 0.66. P( z  0.66)  0.255  25.5%
 1.66. P( z  1.66)  0.049  4.9%
 0.66. P( z  0.66)  P( z  0.66)  0.255  25.5%
 0.66. P( z  0.66) 1 - P( z  0.66)  1  0.255  74.5%
Confidence Interval
Inverting the way of thinking

Until now, we have thought in terms of observations x
and sample values μ and σ to produce the z score.

Let us now imagine that we do not know x, we know μ
and σ. If we consider any interval, we can write:
z
x-

   z   x
  z   x    z 
?
?
Inverting the way of thinking

If z∈[-2.55;+2.55] we know that 99% of z-scores will fall
within the range

If z∈[-1.64;+1.64] we know that 90% of z-scores will fall
within the range

Let us now consider an interval which comprises 95%
of observations. Looking at the z table, we know that
z=1.96
Pr    1.96    x    1.96     0.95
Confidence Interval

In statistics, a confidence interval is an interval within
which the value of a parameter is likely to be (the
mean). Instead of estimating the parameter by a single
value, an interval of likely estimates is given.

Confidence intervals are used to indicate the reliability
of an estimate.

A1. The sample mean is a random variable following a normal
distribution

A2.The sample values μ and σ are good approximation of the
population values.
The Central Limit Theorem



If a random sample is drawn from any
population,
the sampling distribution of the sample mean is
approximately normal for a sufficiently large
sample size.
The larger the sample size, the more closely the
sampling distribution of x will resemble a
normal distribution.
Moments of Sample Mean: The Mean
1
 X 1  X 2  ...  X n 
n
1
E X   E  X 1   E  X 2   ...  E  X n  
n
1
E X       ...   
n
1
E X  n
n
X
 
 
 
 
E X 
On average, the sample mean will be on target,
that is, equal to the population mean.
Moments of Sample Mean: The Variance
1
1
1

var X   var X 1  var X 2  ...  var X n 
n
n
n

1
var X  2  var X 1  var X 2  ...  var X n 
n
1
var X  2  2   2  ...   2 
n
n 2  2
var X  2 
n
n
Standard error of X 

n
The standard deviation of the sample means
represents the estimation error of the sample
mean, and therefore it is called the standard
error.
The Sampling Distribution of the
Sample Mean
1.  x  X
2

2
x
2.  x 
n
3. If x is normal, x is normal. If x is nonnormal
x is approximately normally distributed for
sufficiently large sample size.
Confidence Interval
X  z pc 
X  1.96 
X  1.64 

N

N

N
   X  z pc 
   X  1.96 
   X  1.64 

N

General definition
Definition for 95% CI
N

N
Definition for 90% CI
Standard Normal Distribution and CI
0.45
90% of
observations
0.4
0.35
0.3
95% of
observations
0.25
0.2
99.7% of
observations
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Application of Confidence Interval
 Let us draw a sample of 25 students from CERAM (n = 25), with X = 10
and σ = 3. Let us build the 95% CI
10  1.96 
3
3
   10  1.96 
25
25
 8.8    11.2
CERAM Average grades
0.14
0.12
95% of chances that the mean is
indeed located within this interval
0.10
0.08
0.06
0.04
0.02
0.00
0
5
8.8
10
11.2
15
20
Application of Confidence Interval
 Let us draw a sample of 25 students from CERAM (n = 25), with X = 10
and σ = 3. Let us build the 95% CI
10  1.96 
3
3
   10  1.96 
25
25
 8.8    11.2
 Let us draw a sample of 25 students from HEC (n = 30), with X = 11.5
and σ = 4.7. Let us build the 95% CI
11.5  1.96 
4.7
4.7
   11.5  1.96 
30
30
 9.8    13.2
HEC Average grades
0.09
0.08
95% of chances that the mean is
indeed located within this interval
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00
0
5
10
9.8
13.2
15
20
Hypothesis Testing
 Hypothesis 1 : Students from CERAM have an average grade which is
not significantly different from 11
 H0 : μ(CERAM) = 11
 H1 : μ(CERAM) ≠ 11
I Accept H0 and reject H1
 Hypothesis 2 : Students from CERAM have similar grades as students
from HEC
 H0 : μ(CERAM) = μ(HEC)
 H1 : μ(CERAM) ≠ μ(HEC)
I Accept H0 and reject H1
Comparing the Means Using CI’s
0.25
HEC
0.20
0.15
(μ=11.5;σ=4.7)
(μ=10;σ=3)
CERAM
0.10
0.05
The Overlap of the two CIs means that at 95% level, the two
means do not differ significantly.
0.00
0
5
10
15
20
The Student Test
 Thus far, we have assumed that we know both the mean
and the standard deviation of the population. But in fact,
we do not know them: both μ and σ are unknown.
 The Student t statistics is then preferred to the z statistics.
Its distribution is similar (identical to z as n → +∞). The CI
becomes
s
  X t 
N
df
cp
Application of Student t to CI’s
 Let us draw a sample of 25 students from CERAM (n = 25), with μ = 10
and σ = 3. Let us build the 95% CI
3
3
24
10  t 
   10  t2.5 
25
25
3
3
10  2.06 
   10  2.06 
8.76    11.23
25
25
24
2.5
 Let us draw a sample of 25 students from HEC (n = 30), with μ = 11.5
and σ = 4.7. Let us build the 95% CI
11.5  2.06 
4.7
4.7
   11.5  2.06 
30
30
 9.73    13.26
SPSS Application: Student t
 Import CERAM_LMC into SPSS
 Produce descriptive statistics for sales; labour, and R&D expenses
 Analyse  Statistiques descriptives  Descriptive
 Options: choose the statistics you may wish
 A newspaper writes that by and large, LMCs have 95,000 employees.
 Test statistically whether this is true at 1% level
 Test statistically whether this is true at 5% level
 Test statistically whether this is true at 10% and 20% level
 Write out H0 and H1
 Analyse  Comparer les moyennes  Test t pour échantillon unique
 Options: 99; 95, 90%
SPSS Application: t test at 99% level
Statistiques sur échantillon unique
labour
N
1634
Moyenne
Ecart-type
91298.87 96400.957
Erreur
standard
moyenne
2384.818
Test sur échantillon unique
Valeur du test = 95000
labour
t
-1.552
ddl
1633
Sig.
(bilatérale)
.121
Différence
moyenne
-3701.130
Intervalle de confiance
99% de la différence
Inférieure
Supérieure
-9851.20
2448.94
SPSS Application: t test at 95% level
Statistiques sur échantillon unique
labour
N
1634
Moyenne
Ecart-type
91298.87 96400.957
Erreur
standard
moyenne
2384.818
Test sur échantillon unique
Valeur du test = 95000
labour
t
-1.552
ddl
1633
Sig.
(bilatérale)
.121
Différence
moyenne
-3701.130
Intervalle de confiance
95% de la différence
Inférieure
Supérieure
-8378.75
976.50
SPSS Application: t test at 80% level
Statistiques sur échantillon unique
labour
N
1634
Moyenne
Ecart-type
91298.87 96400.957
Erreur
standard
moyenne
2384.818
Test sur échantillon unique
Valeur du test = 95000
labour
t
-1.552
ddl
1633
Sig.
(bilatérale)
.121
Différence
moyenne
-3701.130
Intervalle de confiance
80% de la différence
Inférieure
Supérieure
-6758.63
-643.63
SPSS Results (at 1% level)
 X  95000  t


0.01

X  95000  2.573 




s2
s2
0.01

    95000   X  95000  t 
N
N
96400
96400
    95000   X  95000  2.573 
1634
1634
9851.20     95000   2448.94
85148.8    97448.94
Pr  85148.8    97448.94   0.99
Critical probability
 The confidence interval is designed in such a way that for each t
statistics chosen, we define a share of observations which this CI is
comprising.
 For large n, when t = 1.96, we have 95% CI
 For large n, when t = 2.55, we have 99% CI
 Actually, for each t, there corresponds a share of observations
 One can compute directly the t value from our observations as follows:
t
X 
s2 n
Critical probability
 The confidence interval is designed in such a way that for each t
statistics chosen, we define a share of observations which this CI is
comprising.
 For large n, when t = 1.96, we have 95% CI
 For large n, when t = 2.55, we have 99% CI
 Actually, for each t, there corresponds a share of observation
 http://www.socr.ucla.edu/Applets.dir/T-table.html
 One can compute directly the t value from our observations as follows:
X  95000   91298  95000  3702

t


 1.552
s2
N
96400
1634
2384
Critical probability
 With t = 1.552, I can conclude the following:
 12% probability that μ belongs to the distribution
where the population mean = 95,000
 I have 12% chances to wrongly reject H0
 88% probability that μ belongs to another
distribution where the population mean ≠ 95,000
 I have 88% chances to rightly reject H0
Shall I the accept or reject H0?
88.0%
6.1%
6.1%
Critical probability
 With t = 1.552, I can conclude the following:
 12% probability that μ belongs to the distribution
where the population mean = 95,000
 I have 12% chances to wrongly reject H0
 88% probability that μ belongs to another
distribution where the population mean ≠ 95,000
 I have 88% chances to rightly reject H0
I accept H0 !!!
Critical probability
 The practice is to reject H0 only when the
critical probability is lower than 0.1, or 10%
 Some are even more cautious and prefer to
reject H0 at a critical probability level of 0.05,
or 5%.
 In any case, the philosophy of the statistician
is to be conservative.
A Direct Comparison of Means Using
Student t
 Another way to compare two sample means is to calculate the CI of
the mean difference. If 0 does not belong to CI, then the two
sample have significantly different means.
 1  2    X 1  X 2   t pc 
s
2
p
X



1
 X1
   X
2
2
(n1  1)  (n2  1)
sp
n1  n2
 X2

2
Standard error, also called pooled
variance
SPSS Application: t test comparing means
 Another newspaper argues that US companies are much larger than
those from the rest of the world. Is this true?
 Produce descriptive statistics labour comparing the two groups
 Produce a group variables which equals 1 for US firms, 0 otherwise
 This is called a dummy variable
 Write out H0 and H1
 Analyse  Comparer les moyennes  Test t pour échantillon
indépendants
 What do you conclude at 5% level?
 What do you conclude at 1% level?
SPSS Application: t test comparing means
Statistiques de groupe
labour
AM
1
0
N
628
1006
Moyenne Ec art-type
97808.99
112765.1
87234.90 84403.469
Erreur
standard
moyenne
4499.817
2661.101
Te st d'échantillons indépendants
Test de Levene sur
l'égalit é des variances
F
labour
Hy pothèse de
varianc es égales
Hy pothèse de
varianc es inégales
.024
Sig.
.877
Test-t pour égalité des moyennes
t
ddl
Sig.
(bilatérale)
Différence
moyenne
Différence
éc art-t ype
Int ervalle de confiance
95% de la différenc e
Inférieure
Supérieure
2.159
1632
.031
10574.084
4897.135
968.751
20179.417
2.023
1061.268
.043
10574.084
5227.792
316.102
20832.067
SPSS Application: t test comparing means
Statistiques de groupe
labour
AM
1
0
N
628
1006
Moyenne Ec art-type
97808.99
112765.1
87234.90 84403.469
Erreur
standard
moyenne
4499.817
2661.101
Te st d'échantillons indépendants
Test de Levene sur
l'égalit é des variances
F
labour
Hy pothèse de
varianc es égales
Hy pothèse de
varianc es inégales
.024
Sig.
.877
Test-t pour égalité des moyennes
t
ddl
Sig.
(bilatérale)
Différence
moyenne
Différence
éc art-t ype
Int ervalle de confiance
99% de la différenc e
Inférieure
Supérieure
2.159
1632
.031
10574.084
4897.135
-2054. 870
23203.038
2.023
1061.268
.043
10574.084
5227.792
-2916. 075
24064.243