Econometrics – lecture 4 – statistical inference

KTEE 310 FINANCIAL ECONOMETRICS STATISTICAL INFERENCE Chap & – S & W Dr TU Thuy Anh Faculty of International Economics TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT Model: Null hypothesis: Y =  +  2X + u H :    20 Alternative hypothesis: H :    20 We will suppose that we have the standard simple regression model and that we wish to test the hypothesis H0 that the slope coefficient is equal to some value 20 We test it against the alternative hypothesis H1, which is simply that 2 is not equal to 20 TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT Model: Y =  +  2X + u Null hypothesis: H :    20 Alternative hypothesis:H :    20 Example model: p =  +  2w + u Null hypothesis: H :   1.0 Alternative hypothesis: H :   1.0 As an illustration, we will consider a model relating p, the rate of growth of prices and w, the rate of growth of wages We will test the hypothesis that the rate of price inflation is equal to the rate of wage inflation The null hypothesis is therefore H0: 2 = 1.0 TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT probability density function of b2 Distribution of b2 under the null hypothesis H0: 2 =1.0 is true (standard deviation equals 0.1 taken as given) 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 b2 We will assume that we know the standard deviation and that it is equal to 0.1 This is a very unrealistic assumption In practice you have to estimate it TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT probability density function of b2 Distribution of b2 under the null hypothesis H0: 2 =1.0 is true (standard deviation equals 0.1 taken as given) 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 b2 Suppose that we have a sample of data for the price inflation/wage inflation model and the estimate of the slope coefficient, b2, is 0.9 Would this be evidence against the null hypothesis 2 = 1.0? And what if b2 =1.4? TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT probability density function of b2 2-4sd Distribution of b2 under the null hypothesis H0: 2 =2 is true (standard deviation taken as given)  0-3sd0-2sd 2 0-sd 20 0 0+4sd 2+sd 2+2sd2+3sd b2 The usual procedure for making decisions is to reject the null hypothesis if it implies that the probability of getting such an extreme estimate is less than some (small) probability p TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT probability density function of b2 Distribution of b2 under the null hypothesis H0: 2 =2 is true (standard deviation taken as given) 2.5% 2-4sd  0-3sd0-2sd 2 2.5% 0-sd 20 0 0+4sd 2+sd 2+2sd2+3sd b2 For example, we might choose to reject the null hypothesis if it implies that the probability of getting such an extreme estimate is less than 0.05 (5%) According to this decision rule, we would reject the null hypothesis if the estimate fell in the upper or lower 2.5% tails TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT Decision rule (5% significance level): probability reject H0 :    20 density (1) if b2   20  1.96 s.d (2) if b2   20  1.96 s.d function of b2 (b2   20 ) / s.d  1.96 (b2   20 ) / s.d  1.96 2.5% 2.5% 2-1.96sd2-sd 02 20+sd20+1.96sd b2 The 2.5% tails of a normal distribution always begin 1.96 standard deviations from its mean Thus we would reject H0 if the estimate were 1.96 standard deviations (or more) above or below the hypothetical mean Or if the difference, expressed in terms of standard deviations, were more than 1.96 in absolute terms (positive or negative) TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT Decision rule (5% significance level): probability reject H0 :    20 density (1) if b2   20  1.96 s.d (2) if b2   20  1.96 s.d function of b2 (1) if z > 1.96 (2) if z < -1.96 acceptance region for b2:  20  1.96 s.d  b2   20  1.96 s.d b2   20 z s.d  1.96  z  1.96 2.5% 20-1.96sd 2-sd 2.5% 20 20+sd  20+1.96sd b2 The range of values of b2 that not lead to the rejection of the null hypothesis is known as the acceptance region Type II error (the probability of accepting the false hypothesis) The limiting values of z for the acceptance region are 1.96 and -1.96 (for a 5% significance test) TESTING A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT probability density Type I error: rejection of H0 when it is in fact true function of b2 Probability of Type I error: in this case, 5% Significance level of the test is 5% reject H0 :    20 reject H0 :    20 acceptance region for b2 2.5% 20-1.96sd 2-sd 2.5% 20 20+sd  20+1.96sd b2 Rejection of the null hypothesis when it is in fact true is described as a Type I error With the present test, if the null hypothesis is true, a Type I error will occur 5% of the time because 5% of the time we will get estimates in the upper or lower 2.5% tails The significance level of a test is defined to be the probability of making a Type I error if the null hypothesis is true EXAMPLE Model 4: OLS, using observations 1899-1922 (T = 24) Dependent variable: q coefficient std error t-ratio p-value const -10,7774 16,4164 -0,6565 0,5190 l 0,822744 0,190860 4,311 0,0003 *** k 0,312205 0,195927 1,593 0,1267 sq_k -0,000249224 0,000310481 -0,8027 0,4316 Mean dependent var 165,9167 Sum squared resid 2455,130 R-squared 0,944239 F(3, 20) 112,8920 Log-likelihood -89,58910 Schwarz criterion 191,8904 rho -0,083426 19 S.D dependent var 43,75318 S.E of regression 11,07955 Adjusted R-squared 0,935875 P-value(F) 1,05e-12 Akaike criterion 187,1782 Hannan-Quinn 188,4284 Durbin-Watson 1,737618 Hypothesis testing using p-value  Step 1: Calculate tob = ˆi   i se( ˆi )  Step 2: Calculate p-value = P (|t| > |tob|)  Step 3: Gor a given α: • Two-tail test: p-value < α  reject H0 • One-tail test: p-value/2 < α:  reject H0 20 EXAMPLE Model 3: OLS, using observations 1899-1922 (T = 24) Dependent variable: q coefficient std error t-ratio p-value -const -4,85518 14,5403 -0,3339 0,7418 l 0,916609 0,149560 6,129 4,42e-06 *** k 0,158596 0,0416823 3,805 0,0010 *** Mean dependent var 165,9167 Sum squared resid 2534,226 R-squared 0,942443 F(2, 21) 171,9278 Log-likelihood -89,96960 Schwarz criterion 189,4734 rho 0,098491 21 S.D dependent var 43,75318 S.E of regression 10,98533 Adjusted R-squared 0,936961 P-value(F) 9,57e-14 Akaike criterion 185,9392 Hannan-Quinn 186,8768 Durbin-Watson 1,535082 EXAMPLE Model 4: OLS, using observations 1899-1922 (T = 24) Dependent variable: q coefficient std error t-ratio p-value const -10,7774 16,4164 -0,6565 0,5190 l 0,822744 0,190860 4,311 0,0003 *** k 0,312205 0,195927 1,593 0,1267 sq_k -0,000249224 0,000310481 -0,8027 0,4316 Mean dependent var 165,9167 Sum squared resid 2455,130 R-squared 0,944239 F(3, 20) 112,8920 Log-likelihood -89,58910 Schwarz criterion 191,8904 rho -0,083426 22 S.D dependent var 43,75318 S.E of regression 11,07955 Adjusted R-squared 0,935875 P-value(F) 1,05e-12 Akaike criterion 187,1782 Hannan-Quinn 188,4284 Durbin-Watson 1,737618 CONFIDENCE INTERVALS probability density function of b2 max (1) conditional on 2 = 2 being true (2) conditional on 2 = 2 being true (2) min 2 -1.96sd 2 -sd 2 (1) 2 +sd max max b2 2 - sd 2 max max 2 + 2 + 1.96sd sd The diagram shows the limiting values of the hypothetical values of their associated probability distributions for b2 2, together with CONFIDENCE INTERVALS reject any 2 > max = b2 + 1.96 sd reject any 2 < min = b2 - 1.96 sd 95% confidence interval: b2 - 1.96 sd < 2 < b2 + 1.96 sd (2) min 2 -1.96sd 2 -sd 2 (1) 2 +sd b2 max 2 - sd max max 2+sd 2 max 2 + 1.96sd Any hypothesis lying in the interval from 2min to 2max would be compatible with the sample estimate (not be rejected by it) We call this interval the 95% confidence interval CONFIDENCE INTERVALS Standard deviation known 95% confidence interval b2 - 1.96 sd < 2 < b2 + 1.96 sd 99% confidence interval b2 - 2.58 sd < 2 < b2 + 2.58 sd Standard deviation estimated by standard error 95% confidence interval b2 - tcrit (5%) se < 2 < b2 + tcrit (5%) se 99% confidence interval b2 - tcrit (1%) se < 2 < b2 + tcrit (1%) se TESTING FOR A SINGLE RESTRICTION H0 H1 j = * j ≠  * tob|>tn-k; α/2 j ≥ * j <  * tob  * tob>tn-k; α tob  26 Reject H0 if b j  * se(b j ) TESTING BETWEEN COEFFCIENTs H0 Reject H0 if j = i j ≠  i tob|>tn-k; α/2 j ≥ i j <  i tob  i tob>tn-k; α tob  27 H1 b j  bi se(b j  bi )  b j  bi bj bi ˆ  ˆ  2ˆ b j ,i MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE Model 3: OLS, using observations 1899-1922 (T = 24) Dependent variable: q coefficient std error t-ratio p-value -const -4,85518 14,5403 -0,3339 0,7418 l 0,916609 0,149560 6,129 4,42e-06 *** k 0,158596 0,0416823 3,805 0,0010 *** Covariance matrix of regression coefficients: const 211,422 28 k 0,371062 0,00173742 l -2,01166 -0,00533574 0,0223683 const k l TESTING FOR MORE THAN ONE RESTRICTIONS - F TEST  meaning: all variables in the 29  If 2 = 3= = K= 0? model not affect y  model is insignificant  R2 =0  if 2 = 4=0  Meaning: X2 and X4 should not be included SIGNIFICANCE OF MODEL – F TEST  If the model is significant?  H0: R2 =0; H1: R2>0 R / (k  1) Fob  (1  R ) / (n  k )  If Fob > Fα(k-1,n-k)  reject H0 30 EXAMPLE Model 1: OLS, using observations 1899-1922 (T = 24) Dependent variable: q coefficient std error t-ratio p-value const -38,7267 14,5994 -2,653 0,0145 ** l 1,40367 0,0982155 14,29 1,29e-012 *** Mean dependent var 165,9167 Sum squared resid 4281,287 R-squared 0,902764 F(1, 22) 204,2536 Log-likelihood -96,26199 Schwarz criterion 198,8801 rho 0,836471 31 S.D dependent var 43,75318 S.E of regression 13,95005 Adjusted R-squared 0,898344 P-value(F) 1,29e-12 Akaike criterion 196,5240 Hannan-Quinn 197,1490 Durbin-Watson 0,763565 EXAMPLE Model 3: OLS, using observations 1899-1922 (T = 24) Dependent variable: q coefficient std error t-ratio p-value -const -4,85518 14,5403 -0,3339 0,7418 l 0,916609 0,149560 6,129 4,42e-06 *** k 0,158596 0,0416823 3,805 0,0010 *** Mean dependent var 165,9167 Sum squared resid 2534,226 R-squared 0,942443 F(2, 21) 171,9278 Log-likelihood -89,96960 Schwarz criterion 189,4734 rho 0,098491 32 S.D dependent var 43,75318 S.E of regression 10,98533 Adjusted R-squared 0,936961 P-value(F) 9,57e-14 Akaike criterion 185,9392 Hannan-Quinn 186,8768 Durbin-Watson 1,535082 F-TEST  Y = 1 + 2X2+ + 5X5 + u (1)  if 2 = 4=0? H0: 2 = 4=0; H1: at least one of them is nonzero Step1: run unrestricted model (1) => R2(1) Step 2: run: restricted model: Y = 1 + 3X3+ 5X5 + u (2) => R2(2) ( R (1)  R (2)) / m Fob  (1  R (1)) / (n  k ) If Fob > F α(m,n-k)=> reject H0  Example: H0: 2 =0 :  Fob = 14,477 > F 5%(1,21)=4.32 => reject H0 33 ... -const -4, 85518 14, 540 3 -0,3339 0, 741 8 l 0,916609 0, 149 560 6,129 4, 42e-06 *** k 0,158596 0, 041 6823 3,805 0,0010 *** Mean dependent var 165,9167 Sum squared resid 25 34, 226 R-squared 0, 942 443 F(2,... -10,77 74 16 ,41 64 -0,6565 0,5190 l 0,822 744 0,190860 4, 311 0,0003 *** k 0,312205 0,195927 1,593 0,1267 sq_k -0,000 249 2 24 0,00031 048 1 -0,8027 0 ,43 16 Mean dependent var 165,9167 Sum squared resid 245 5,130... -10,77 74 16 ,41 64 -0,6565 0,5190 l 0,822 744 0,190860 4, 311 0,0003 *** k 0,312205 0,195927 1,593 0,1267 sq_k -0,000 249 2 24 0,00031 048 1 -0,8027 0 ,43 16 Mean dependent var 165,9167 Sum squared resid 245 5,130