hypothesis testing previous lecture notes pval confidence interval t-stat
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Hypothesis Testing Previous Lecture Notes Pval Confidence Interval T-statTRANSCRIPT
4.1
Confidence Interval &
Hypothesis Testing
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4.2 How reliable is this OLS estimation?
(Computing Tutorial #2: Application of Phillips Curve Theory for the Case of Hong Kong)
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4.3
^
Properties of OLS estimators–two variable case
β ˆ 1 σ ˆ
2 β ˆ 2
1. Unbiased Ε(β1) = β1 Ε(β2)=β2 ˆ ˆ
efficient
3. Consistent: as n gets larger, estimator is more accurate
∑
σ = = β x
ˆ var(β2) 2
2 2 ˆ 2
σ
σ ⋅ ∑
∑ = β 2 2
ˆ 1
x n
X var( ) 2. Min. Variance
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^
^
σ ̂ ̂ 1 β
2
4.4
( ) x 2
) 2 n ( 2
2
~ 2 n ˆ
- σ
σ -
Properties of OLS estimators (continue)
( ) σβ β β 2 ˆ 1 1 1 , N ~ ˆ
( ) σβ β β 2 ˆ 2 2 2 , N ~ ˆ
6.
ˆ ˆ 4 & 5. β1 and β2 are normally distribution
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4.5 Hypothesis Testing and Confidence Interval
( ) β ˆ f 2
Density
δ - β ˆ 2 δ + β ˆ 2 β 2 β ˆ
2
Random interval (confidence interval)
true
Estimated β2 falls in area
^
How reliable is the OLS estimation ? How “close” is β1 to β1 ? How “close” is β2 to β2 ?
^ ^
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4.6 Hypothesis Testing and Confidence Interval
ˆ ˆ 0.99 0.95 0.90
Pr( β2-δ ≤ β2 ≤ β2+δ ) = (1-α)
where (1-α) is confidence coefficient: (0< α <1)
α is also called the level of significance.
ˆ β2 - δ is called lower confidence bound β2 + δ is called upper confidence bound the interval between (β2 - δ) and (β2 + δ) is called random interval (confidence interval)
ˆ
ˆ ˆ
0.01 0.05 0.10
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4.7 Constructing Confidence Interval for βi
∑
∑ σ = σ ⇒ β x n
x 2 i
2 i
2 2 ˆ
1 ( ) σ β β β 2
1 1 ˆ 1 , N ~ ˆ
( ) σ β β β 2
2 2 ˆ 2 , N ~ ˆ
∑
σ = σ ⇒ β x
2 i
2 2 ˆ
2
( ) σ 2 u i , O N ~ u E(u) = 0
Var(u) =σu2
By assumption:
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4.8 Constructing Confidence Interval for (cont.) β i
( ) β ˆ f 2
β ˆ 2 ( ) β = β ˆ E 2 2
Actual estimated β2 could fall into these regions
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4.9
( ) β2
β - β = ˆ se
ˆ Z 2 2
Constructing Confidence Interval for (cont.) β i
Transform into normal standard distribution
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4.10 Constructing Confidence Interval for βi (cont.)
Use the normal distribution to make probabilistic statements about σ2 provided the true β2 is known
In practice this is unobserved
( ) ( )
( ) σ
β - β =
β
β - β =
∑ x ˆ
1 , 0 N ˆ Se
ˆ Z
2
2 2
2
2 2 ~
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4.11
For example:
Constructing Confidence Interval for βi (cont.)
Accept region
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4.12
( ) 95 . 0 96 . 1 Z 96 . 1 Pr = ≤ ≤ -
95% confidence interval:
96 . 1 ) ˆ ( Se
ˆ 96 . 1
2
2 2 ≤ β
β - β ≤ -
Constructing Confidence Interval for βi (cont.)
ˆ β - β 95 . 0 96 . 1
) ˆ ( Se 96 . 1 Pr
2
2 2 = ≤ β
≤ - ⇒
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4.13
( ) * β ± β ⇒ ˆ se 96 . 1 ˆ 2 2
σ 2 In practice, is unknown, we have to use the unbiased estimator
2 n
2 ← RSS u ˆ i 2 ˆ -
∑ = σ
Then, instead of normal standard distribution, t-distribution is used.
Constructing Confidence Interval for βi (cont.)
( ) * ( ) β * + β ≤ β ≤ β - β ⇒ ˆ se 96 . 1 ˆ ˆ se 96 . 1 ˆ 2 2 2 2 2
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Assumed or test value
4.14
= standard error of estimator estimated - true parameter t
β
β - β = ) ˆ ( se
ˆ t
2
2 2
( )
σ
∑ β - β = ˆ
x ˆ t
2 2 2
Use the t to construct a confidence interval for β 2
Constructing Confidence Interval for βi (cont.)
SEE
Or some specific values that want to compare
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4.15
( )
σ
∑ β - β = ˆ
x ˆ t
2 2 2
Constructing Confidence Interval for βi (cont.)
( ) β ⇒
β
β - β = 2 2
2 2 ˆ se
ˆ t a specified value
( ) ∑
σ = β x
ˆ se where 2
2
2
^
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4.16
Use the tc to construct a confidence interval for β2 as
Constructing Confidence Interval for βi (cont.)
where is the critical t value at two-tailed
level of significance. α is level of significance
and (n-2) is degrees of freedom (in 2-variable case).
t c
2 n , 2 - α ±
2 α
α - = ≤ ≤ - - α
- α 1 t t t Pr c c 2 n ,
2 2 n , 2
*
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4.17 Constructing Confidence Interval for βi (cont.)
( ) 90 . 0 t
ˆ se
ˆ t Pr
c
2
2 2 c 2 n , 05 . 0 2 n , 05 . 0 = ⎜
⎜
⎝
⎛ ≤
β
β - β ≤ -
- -
Therefore (assume α = 0.10 α/2 = 0.05 )
( ) ( ) ( ) ˆ ˆ ˆ ˆ c * + ≤ β ≤ * -
90 . 0
se t se t Pr 2 2
c 2 2 2 2 n , 05 . 0 2 n , 05 . 0
=
β β β β - -
Rearranging, Pr( -tc
0.025, n-2 ≤ (β2- β2)/se(β2) ≤ tc0.025, n-2 ) = 0.95
^ ^
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4.18 Then 90% confidence interval for β2 is:
( ) β * ± β - ˆ se t ˆ 2
c 2 2 n , 05 . 0
( ) β ˆ se 2 - t c 2 n , 05 . 0 Check it from t-table Check it from
estimated result β ˆ 2 &
The 95% confidence interval interval for β2 becomes
( ) β ± ˆ *se t 2
c β ˆ 2 0.025, n-2
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4.19 The t-statistic in computer (EVIEWS) output Example: Gujarati (2003)pp.123
SEE= σ RSS ^
H0: β2= 0 H1: β2≠ 0
t= 0.5091 - 0
0.0357
se(β2) ^
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4.20
δ
Example: Gujarati (2004), p.123
Given β2 = 0.5091, n = 10, se(β2) = 0.0357, ˆ ˆ
95% confidence interval is:
ˆ se( ) t ˆ 2
c 2 2 n ,
2 β * ± β
- α
] 5914 . 0 , 4268 . 0 [
0823 . 0 5091 . 0
) 0357 . 0 ( t 5091 . 0 c
8 , 025 . 0
⇒
± ⇒
± ⇒
2.306 x 0.0357 5091 . 0 ± ⇒
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4.21 90% confidence interval is:
) 0357 . 0 ( t 5091 . 0 c 8 , 05 . 0 ±
δ
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4.22 The t-statistic in computer (EVIEWS) output Example: tutorial #2, unemployment & inflation rate of HK
SEE= σ RSS ^
H0: β2= 0 H1: β2≠ 0
t= -0.395 - 0
0.0332
se(β2) ^
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4.23
δ
Example: tutorial #2, unemployment & inflation of HK
Given β2 = -0.395, n = 84, se(β2) = 0.0332, ˆ ˆ
95% confidence interval is:
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1. Establish hypotheses: state the null and alternative hypotheses.
2. Determine the appropriate statistical test and sampling distribution.
3. Specify the Type I error rate (α). 4. State the decision rule. 5. Gather sample data (estimate the model=. 6. Calculate the value of the test statistic. 7. State the statistical conclusion. 8. Make decision.
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The Null and Alternative Hypotheses are mutually exclusive. Only one of them can be true.
The Null and Alternative Hypotheses are collectively exhaustive. They are stated to include all possibilities. (An abbreviated form of the null hypothesis is often used.)
The Null Hypothesis is assumed to be true. The burden of proof falls on the Alternative
Hypothesis.
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A soft drink company is filling 12 oz. cans with cola.
The company hopes that the cans are averaging 12 ounces.
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µ=12 oz Non Rejection Region
Rejection Region
Critical Value
Rejection Region
Critical Value
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Type I Error ◦ Rejecting a true null hypothesis ◦ The probability of committing a Type I error is
called α, the level of significance.
Type II Error ◦ Failing to reject a false null hypothesis ◦ The probability of committing a Type II error is
called β.
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(
( )
Null True Null False
Fail to reject null
Correct Decision
Type II error β)
Reject null Type I error α
Correct Decision
α
β
Reduce probability of one error and the other one goes up holding everything else unchanged.
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4.31 Test-Significance Approach: One-tailed T-test decision rule
t c 2 n , - α
Step 3: check t-table for
look for critical t value
Step 4: compare tc and t
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( )
( ) β < β β > β
β ≥ β β ≤ β
2 2 1 2 2 1
2 2 0 2 2 0
ˆ : H : H
: H : H Step 1: State the hypothesis
0
0
0
0
Computed value Step 2: ( ) β
β - β =
ˆ se t 2 2
2
0
4.32 One-tailed t-test decision rule
Left-tail (If t < - tc ==> reject H0 ) (If t > - tc ==> not reject H0 )
Decision Rule Step 5: If t > tc ==> reject H0 If t < tc ==> not reject H0
Right-tail
0 tc < t
Right-tail
0 -tc t <
left-tail
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4.33 Two-Tailed T-test
3. Check t-table for critical t value:
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β ≠ β
β = β
2 2 1
2 2 0
: H
: H 1. State the hypothesis 0
0
β - ˆ
( ) β
β = ˆ se
t 2
2 2 2. Compute
0
4.34 Two-Tailed t-test (cont.)
4. Compare t and
β 2
Accept region
reject H0 region
( ) β * + β - α ˆ se t ˆ
2 c
2 2 n , 2
reject H0 region
( ) β * - β - α ˆ se t ˆ
2 c
2 2 n , 2
5. If t > tc or -t < - tc , then reject Ho or | t | > | tc |
Decision Rule:
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4.35 One-Tailed t-test
3 . 0 : H
3 . 0 : H
2 1
2 0
> β
≤ β We also could postulate that:
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( )
5.857 0357 . 0
2091 . 0
0357 . 0
3 . 0 5091 . 0 t
ˆ se
ˆ t
2
2 2
= = - =
β
β - β =
1. Compute: 0
4.36 One-Tailed t-test (cont.)
α = 0.05 2. Check t-table for where =1.860
H reject
860 . 1 t 857 . 5 t
0
c 8 , 05 . 0
∴
= > =
3. Compare t and the critical t
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4.37 One-Tailed t-test (cont.)
β < β 0 2 2 1 : H
β ≥ β 0 2 2 0 : H “ Decision rule for left-tail test”
If t < - tcα, df => reject H0
^ β*
β*- tc se(β)
left-tail
β ^ ^ EMU, Econometrics I, M. Balcılar
4.38
Suppose we postulate that
Is the observed compatible with true ?
=
3 . 0 : H
3 . 0 : H
2 1
2 0
≠ β
β
β 2 β ˆ 2
Two-Tailed t-test
(1) From Confidence-interval approach: 95% confidence-interval is [0.4268, 0.5914] which does not contain the true β2. The estimated β2 is not equal to 0.3
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4.39 (2) From Significance test approach:
Compare t-value and the critical t-value:
5.857 0357 . 0
2091 . 0 = = 0357 . 0
3 . 0 5091 . 0 - = ( ) 2 ˆ se
ˆ t 2 2
β
β - β =
tc0.025, 8 = 2.306
,
==> reject H0
It means the estimated β2 is not equal to 0.3
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0
4.40 Tests about σ2
4.41 Forming the Null and Alternative Hypotheses Given the null and the alternative hypotheses, testing them for
statistical significance should no longer be a mystery. But how does one formulate these hypotheses? There are no hard-and-fast rules. Very often the phenomenon under study will suggest the nature of the null and alternative hypotheses.
For example, consider the capital market line (CML) of portfolio theory, which postulates that Ei = β1 + β2σi , where E = expected return on portfolio and σ = the standard deviation of return, a measure of risk. Since return and risk are expected to be positively related—the higher the risk, the higher the return—the natural alternative hypothesis to the null hypothesis that β2 = 0 would be β2 > 0. That is, one would not choose to consider values of β2 less than zero.
4.42 Forming the Null and Alternative Hypotheses Prior studies of the money demand functions have shown that
the incomeelasticity of demand for money (the percent change in the demand for money for a 1 percent change in income) has typically ranged between 0.7 and 1.3. Therefore, in a new study of demand for money, if one postulates that the income-elasticity coefficient β2 is 1, the alternative hypothesis could be that β2 ≠ 1, a two-sided alternative hypothesis.
Thus, theoretical expectations or prior empirical work or both can be relied upon to formulate hypotheses. But no matter how the hypotheses are formed, it is extremely important that the researcher establish these hypotheses before carrying out the empirical investigation. Otherwise, he or she will be guilty of circular reasoning or selffulfilling prophesies.
4.43 “Accepting” or “Rejecting”
"Accept "the null hypothesis: All we are saying is that on the basis of the sample evidence we have no reason to reject it; We are not saying that the null hypothesis is true beyond any doubt. Therefore, in “accepting” a Ho , we should always be aware that another null hypothesis may be equally compatible with the data. So, the conclusion of a statistical test is “do not reject” rather than “accept”.
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4.44 p-value Approach to Testing Convert Sample Statistic (e.g., β2 ) to Test
Statistic (e.g., Z, t or F –statistic) Obtain the p-value from a table or computer p-value: probability of obtaining a test statistic as
extreme or more extreme ( ≤ or ≥ ) than the observed sample value given H0 is true
Called observed level of significance Smallest value of α that an H0 can be rejected
Compare the p-value with If p-value ≥ α, do not reject H0 If p-value < α, reject H0
^
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4.45
α = 0.05, se(β2)=0.0357 n = 10, df=8 Critical Value: ±2.306
Example: Two-Tail Test
Test Statistic:
Decision:
Conclusion:
Reject at α = 0.05.
t 0 2.306
.025 Reject
-2.306
.025
H0: β2 = 0.3 H1: β2 ≠ 0.3
5.857
Insufficient Evidence that β2 is equal to 0.3.
t=(0.5091-0.3)/0.0357 = 5.857
^
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4.46 p-value solution (get the p-value from my t table at the web site)
(p-value = 0.0005) < (α = 0.05) Reject the Null.
0.3 5.857 t
Reject p-value=0.0005
2.306
α /2= 0.025
Test Statistic 5.857 is in the Reject Region
Reject
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4.47 2-t Rule of Thumb
4.48 REGRESSION ANALYSIS AND ANALYSIS OF VARIANCE
TSS = ESS + RSS Consider variable F:
4.49 REGRESSION ANALYSIS AND ANALYSIS OF VARIANCE Therefore, if β2 is in fact zero the explanatory variable X has no linear inuence on Y whatsoever and the entire variation in Y is explained by the random disturbances ui . If, on the other hand, β2 is not zero, a part of the variation in Y will be ascribable to X. Therefore, the F ratio provides a test of the null hypothesis H0: β2 = 0.
4.50 REGRESSION ANALYSIS AND ANALYSIS OF VARIANCE
F0.05(1,8)=5.32 Reject H0