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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.

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Presentation on theme: "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."— Presentation transcript:

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2 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

3 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.

4 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

5 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

6 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 ) ? ^ ^

7 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

8 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:

9 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

10 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

11 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 ~

12 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

13 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   

14 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

15 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

16 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 

17 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 ^ ^

18 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

19 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

20 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)

21 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

22 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.

23 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:

24 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:

25 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

26 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

27 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*

28 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. ^ ^

29 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*

30 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

31 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

32 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

33 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  ^^

34 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”.

35 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 YY X1X1 =  1 = 0? ^ ^ From the printout

36 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 YY X2X2 =  2 = 0? ^ ^ From the printout


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