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4.1 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Lecture #4 Studenmund (2006): Chapter 5 Review of hypothesis testing Confidence Interval and estimation t-test and Partial significance test Objectives
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4.2 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Hypothesis Testing in regression: 1. Testing individual (partial) coefficient 2. Testing the overall significance of all coefficients 3. Testing restriction on variables (add or drop): X k = 0 ? 4. Testing partial coefficient under some restrictions Such as 1 + 2 = 1; or 1 = 2 (or 1 + 2 = 0); etc. 5. Testing the stability of the estimated regression model -- over time -- in different cross-sections 6. Testing the functional form of regression model.
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4.3 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Properties of OLS estimators –two variable case ˆ 0 ˆ 2 ˆ 1 Unbiased 1. Unbiased ˆ ˆ efficiency Consistency 3. Consistency: as n gets larger, estimator is more accurate x ˆ [Se( 2 2 2 ˆ 1 2 2 2 ˆ 0 x n x Se Min. Variance 2. Min. Variance
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4.4 All rights reserved by Dr.Bill Wan Sing Hung - HKBU x 2 )2n( 2 2 ~ 2n ˆ Properties of OLS estimators (continue) 2 ˆ 0 0 0, N~ ˆ 2 ˆ 1 1 1,N~ ˆ 6. ˆ ˆ 4 & 5. 0 and 1 are normal distribution
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4.5 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Hypothesis Testing and Confidence Interval ˆ f 1 Density ˆ 1 ˆ 1 1 ˆ 1 Random interval (confidence interval ) true Estimated falls in area ^ How reliable is the OLS estimation ? How “close” is to (true or theoretical value) ? How “close” is to (true or theoretical value ) ? ^ ^
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4.6 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Hypothesis Testing and Confidence Interval ˆ ˆ 0.99 0.95 0.90 Pr( 1 - < 1 < 1 + ) = (1- ) which (1- ) is confidence coefficient: (0< <1) level of significance is also called the level of significance. ˆ lower confidence bound is called lower confidence bound upper confidence bound is called upper confidence bound random intervalconfidence interval the interval between ( and ( is called random interval (confidence interval) ˆ ˆ ˆ 0.01 0.05 0.10 Select one level to construct the interval
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4.7 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Confidence Interval Constructing Confidence Interval for i x n x 2 i 2 i 2 2 ˆ 0 2 0 0 ˆ 0,N~ ˆ 2 1 1 ˆ 1,N~ ˆ x 2 i 2 2 ˆ 1 2 i,0 N ~ E( ) = 0 Var( ) = By assumptions:
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4.8 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Confidence Interval Constructing Confidence Interval for (cont.) i ˆ f 1 ˆ 1 ˆ E 1 1 Actual estimated 1 could be fallen into these regions ^ True value
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4.9 All rights reserved by Dr.Bill Wan Sing Hung - HKBU ˆ Se ˆ Z 1 1 1 Confidence Interval Constructing Confidence Interval for (cont.) i Transform into Normal standard distribution
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4.10 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Confidence Interval Constructing Confidence Interval for i (cont.) Use the normal distribution to make probabilistic statements about 2 provided the true 1 is known In practice this is unobserved x ˆ 1,0N ˆ Se ˆ Z 2 1 1 1 1 1 ~
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4.11 All rights reserved by Dr.Bill Wan Sing Hung - HKBU For example: Confidence Interval Constructing Confidence Interval for i (cont.) Accept region Critical Values Rejected region
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4.12 All rights reserved by Dr.Bill Wan Sing Hung - HKBU 95.096.1Z.1Pr 95% confidence interval: 96.1 ) ˆ (Se ˆ 96.1 1 1 1 Confidence Interval Constructing Confidence Interval for i (cont.) ˆ 95.096.1 ) ˆ (Se 96.1Pr 1 1 1
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4.13 All rights reserved by Dr.Bill Wan Sing Hung - HKBU ˆ Se96.1 ˆ 1 1 2 In practice, is unknown, we have to use the unbiased estimator 2n 2 RSS ˆ i 2 ˆ Instead of using normal standard distribution, t-distribution is used. Confidence Interval Constructing Confidence Interval for i (cont.) * ˆ Se 96.1 ˆˆ Se96.1 ˆ 11 1 1 1
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4.14 All rights reserved by Dr.Bill Wan Sing Hung - HKBU standard error of estimator estimated - true parameter t ) ˆ (Se ˆ t 1 1 1 ˆ x ˆ t 2 1 1 t Use the t to construct a confidence interval for 1 Confidence Interval Constructing Confidence Interval for i (cont.) SEE a specified value
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4.15 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Use the t c to construct a confidence interval for 1 as Confidence Interval Constructing Confidence Interval for i (cont.) where is critical t value at two-tailed level of significance. is level of significance and (n-2) is degree of freedom (in 2-variable case). t c 2n, 2 2
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4.16 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Confidence Interval Constructing Confidence Interval for i (cont.) ˆˆˆˆ c * 90.0 Se t t Pr 11 c 1 1 1 2n,05.0 2n,.0 Rearranging, 90.0 t ˆ Se ˆ t Pr c 1 1 1 c 2n,05.02n,.0 Therefore Pr( -t c 0.025, n-2 ( 1 - 1 )/Se( 1 ) t c 0.025, n-2 ) = 0.95 ^ ^
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4.17 All rights reserved by Dr.Bill Wan Sing Hung - HKBU confidence interval Then 90% confidence interval for 1 is : ˆ Se t ˆ 1 c 1 2n,05.0 ˆ Se 1 t c 2n,05.0 Check it from t-table Check it from estimated result ˆ 1 & The 95% confidence interval interval for becomes ˆ * Se t 1 c ˆ 1 0.025, n-2
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4.18 All rights reserved by Dr.Bill Wan Sing Hung - HKBU t c 2N, 2 Critical Critical t-value: α= 0.1 two-tailed = 1.860 t c 8,05.0 = 2.306 t c 8,025.0 α= 0.05 two-tailed
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4.19 All rights reserved by Dr.Bill Wan Sing Hung - HKBU ˆ Se( ) t ˆ 1 c 1 2n, 2 For example For example: Given = 0.5091, n = 10, Se ( = 0.0357, ˆ ˆ 95% confidence interval is: α= 0.05 )5914.0,4268.0( 0823.05091.0 )0357.0( t 5091.0 c 8,025.0 2.306 0.5091 ± 2.306(0.0357)
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4.20 All rights reserved by Dr.Bill Wan Sing Hung - HKBU 90% confidence interval is 90% confidence interval is: ) 0357.0( t 5091. 0 c 8,05.0 α= 0.1
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4.21 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Procedures for Hypothesis Testing 1. Determine null (H 0 ) and alternative (H 1 ) hypotheses 2. Specify the test statistic and its distribution as if the null hypothesis were true. 3. Select and determine the rejection region. 4. Calculate the sample value of test statistic (t*). 5. State your conclusion.
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4.22 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Test-Significance Approach: Two-Tailed T-test ˆ ˆ Se t*t* 1 1 1 2. Compute 1 1 1 1 1 0 ˆ : H ˆ : H 1. State the hypothesis 3. Check t-table for critical t value:
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4.23 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Two-Tailed t-test decision rule 4. Compare t* and 1 Accept region reject H 0 region ˆ Se t ˆ 1 c 1 2n, 2 reject H 0 region ˆ Se t ˆ 1 c 1 2n, 2 5. If t* > t c or –t* < - t c, then reject H o or | t* | > | t c | Decision Rule Decision Rule:
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4.24 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Two-Tailed t-test Confidence-interval (1) From Confidence-interval approach: 95% confidence-interval is (0.4268, 0.5914) which does not contain the true 1. 0.3 The estimated 1 is not equal to 0.3 3.0: H 3.0: H 1 1 1 0 1 ˆ 1 postulate Suppose we postulate that Is the observed compatible with true ? ˆ ˆ
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4.25 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Significance test (2) From Significance test approach: t*-value”critical t-value” Compare “t*-value” and the “critical t-value”: 5.857 0357.0 2091.0 0357.0 3.05091.0 1 ˆ Se ˆ t*t* 1 1 t c 0.025, 8 = 2.306 check from the t-table, ==> reject H 0 It means the estimated 1 is not equal 0.3
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4.26 All rights reserved by Dr.Bill Wan Sing Hung - HKBU computer output The t-statistic in computer output SEE= RSS ^ H 0 : 1 =0 H 1 : 1 0 ^ ^ Var(β i ) ^ 1 ˆ Se ˆ t* 1 1 = 0.5091 - 0 0.0357 t*
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4.27 All rights reserved by Dr.Bill Wan Sing Hung - HKBU p-values The p-values Reported by Regression Software p-value Ho When it (p-value) is the lowest level of significance at which we could reject Ho. 2. Testing p-value probability, or marginal significant level, The p-value of a coefficient reported by the computer is the probability, or marginal significant level, of obtaining that estimated coefficient (β i ) if the null hypothesis Hoβ i =0 (Ho: β i =0) were true. ^ ^
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4.28 All rights reserved by Dr.Bill Wan Sing Hung - HKBU One-tailed T-test t c 2n, Step 3: check t-table for look for critical t value Computed value Step 2: ˆ Se ˆ t*t* 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 ˆ : H ˆ : H ˆ : H ˆ : H Step 1 : State the hypothesis Step 4: compare t c and t*
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4.29 All rights reserved by Dr.Bill Wan Sing Hung - HKBU One-tailed t-test decision rule Left-tail (If t reject H 0 ) (If t > - t c ==> not reject H 0 ) Decision Rule Step 5: If t > t c ==> reject H 0 If t not reject H 0 Right-tail 0 tctc < t Right-tail 0 -t c t < left-tail
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4.30 All rights reserved by Dr.Bill Wan Sing Hung - HKBU Example: One-Tailed t-test 5.857 0357.0 2091.0 0357.0 3.05091.0 t*t* ˆ Se ˆ t*t* 1 1 1 1. Compute: 3.0: H 3.0: H 1 1 1 0 We also could postulate that: ˆ ˆ
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4.31 All rights reserved by Dr.Bill Wan Sing Hung - HKBU One-Tailed t-test (cont.) = 0.05 2. Check t-table for where =1.860 H reject 860.1 t 857.5t 0 c 8,05.0 3. Compare t and the critical t
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4.32 All rights reserved by Dr.Bill Wan Sing Hung - HKBU One-Tailed t-test (cont.) * 1 1 1 : H * 1 1 0 : H ˆ ˆ “ Decision rule for left-tail test” If t reject H 0 ^ ** *- t c Se( ) left-tail ^^
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4.33 All rights reserved by Dr.Bill Wan Sing Hung - HKBU “Accepting” or “Rejecting” H 0 "Accept "the null hypothesis (H 0 ): sample 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. H o Therefore, in “accepting” a H o, 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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4.34 All rights reserved by Dr.Bill Wan Sing Hung - HKBU 1. Individual partial coefficient test t* = 1 - 0 ^ Se ( 1 ) ^ = 0.726 0.048 = 14.906 Compare with the critical value t c 0.025, 12 = 2.179 Since t* > t c reject H o Answer : Yes, 1 is statistically significant and is significantly different from zero. ^ Y = 0 + 1 X 1 + 2 X 2 ^^ ^ ^ H 0 : 1 = 0 H 1 : 1 0 X 1 effectY Holding X 2 constant: Whether X 1 has the effect on Y ? 1 YY X1X1 = 1 = 0? ^ ^ From the printout
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4.35 All rights reserved by Dr.Bill Wan Sing Hung - HKBU 1. Individual partial coefficient test (cont.) t = 2 - 0 ^ Se ( 2 ) ^ = 2.736-0 0.848 = 3.226 Critical value: t c 0.025, 12 = 2.179 Since | t | > | t c | ==> reject H o Answer: Yes, 2 is statistically significant and is significantly different from zero. ^ X 2 effectY holding X 1 constant: Whether X 2 has the effect on Y? 2 H 0 : 2 = 0 H 1 : 2 0 YY X2X2 = 2 = 0? ^ ^ From the printout
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