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KPSS Test Output Interpretation

Posted: Wed Jun 09, 2010 3:28 am
by cointthesis
Hi!

I am still a beginner with Unit root tests in EViews, and I need urgent help concerning the following output:

KPSS Test Level Intercept
Null Hypothesis: BN_LOG is stationary
Exogenous: Constant
Bandwidth: 18 (Newey-West using Bartlett kernel)

Kwiatkowski-Phillips-Schmidt-Shin test statistic 1.905745
Asymptotic critical values*: 1% level 0.739000
5% level 0.463000
10% level 0.347000

Residual variance (no correction) 0.087861
HAC corrected variance (Bartlett kernel) 1.581625

KPSS Test Equation
Dependent Variable: BN_LOG
Method: Least Squares
Date: 06/09/10 Time: 12:00
Sample: 2/26/2008 3/31/2010
Included observations: 534

Coefficient Std. Error t-Statistic Prob.
C 7.383234 0.012839 575.0586 0.0000

R-squared 0.000000 Mean dependent var 7.383234
Adjusted R-squared 0.000000 S.D. dependent var 0.296692
S.E. of regression 0.296692 Akaike info criterion 0.409624
Sum squared resid 46.91779 Schwarz criterion 0.417639
Log likelihood -108.3695 Hannan-Quinn criter. 0.412760
Durbin-Watson stat 0.008087

This is the output I get for testing in level and with intercept...does this mean that my series are stationary???

Thx in advance for your help!!!

Re: KPSS Test Output Interpretation

Posted: Thu Jul 08, 2010 2:12 am
by Pantera
Hi - According to the test result, the time series has no unit root (you cannot reject the null hypothesis).

Re: KPSS Test Output Interpretation

Posted: Thu Jul 08, 2010 3:54 am
by Pantera
Hi again - I must have mistakenly looked on a different number because the test result shows that the null hypothesis is rejected: So the time series has proably a unit root and is NOT stationary - sorry about the confusion.

Re: KPSS Test Output Interpretation

Posted: Fri Aug 24, 2012 5:39 am
by gobbble
To refer back to the interpretation question.
The important value is the Kwiatkowski-Phillips-Schmidt-Shin test statistic, in this case 1.905745 to decide wether to reject or not. AM I right here in this assumption?

Re: KPSS Test Output Interpretation

Posted: Wed Sep 12, 2012 3:52 am
by Pantera
To refer back to the interpretation question.
The important value is the Kwiatkowski-Phillips-Schmidt-Shin test statistic, in this case 1.905745 to decide wether to reject or not. AM I right here in this assumption?
Hi, yes, that's correct.

Re: KPSS Test Output Interpretation

Posted: Sat May 17, 2014 10:25 am
by zahideconomist
Hi
I need urgent help concerning the following output is Variable Y is stationary or not?

Null Hypothesis: Y is stationary
Exogenous: Constant, Linear Trend
Bandwidth: 4 (Newey-West automatic) using Bartlett kernel

LM-Stat.

Kwiatkowski-Phillips-Schmidt-Shin test statistic 0.211095
Asymptotic critical values*: 1% level 0.216000
5% level 0.146000
10% level 0.119000

*Kwiatkowski-Phillips-Schmidt-Shin (1992, Table 1)


Residual variance (no correction) 36006.27
HAC corrected variance (Bartlett kernel) 114183.4



KPSS Test Equation
Dependent Variable: Y
Method: Least Squares
Date: 05/17/14 Time: 22:12
Sample: 1971 2011
Included observations: 41

Variable Coefficient Std. Error t-Statistic Prob.

C 876.0881 59.67462 14.68108 0.0000
@TREND("1971") 9.145404 2.567987 3.561313 0.0010

R-squared 0.245399 Mean dependent var 1058.996
Adjusted R-squared 0.226050 S.D. dependent var 221.1528
S.E. of regression 194.5578 Akaike info criterion 13.42689
Sum squared resid 1476257. Schwarz criterion 13.51048
Log likelihood -273.2512 Hannan-Quinn criter. 13.45732
F-statistic 12.68295 Durbin-Watson stat 0.271097
Prob(F-statistic) 0.000991

Re: KPSS Test Output Interpretation

Posted: Sat May 17, 2014 10:43 am
by Pantera
According to the results the null-hypothesis is rejected, which indicates non-stationarity. The data seem to have a deterministic trend as well. Eliminate the deterministic trend and test the null-hypothesis again.

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Re: KPSS Test Output Interpretation

Posted: Sat May 31, 2014 1:52 am
by zahideconomist
hi
thanks for comments now my results are:

Null Hypothesis: D(LY) is stationary
Exogenous: Constant, Linear Trend
Bandwidth: 7 (Newey-West automatic) using Bartlett kernel

LM-Stat.

Kwiatkowski-Phillips-Schmidt-Shin test statistic 0.092292
Asymptotic critical values*: 1% level 0.216000
5% level 0.146000
10% level 0.119000

*Kwiatkowski-Phillips-Schmidt-Shin (1992, Table 1)


Residual variance (no correction) 0.007112
HAC corrected variance (Bartlett kernel) 0.002123



KPSS Test Equation
Dependent Variable: D(LY)
Method: Least Squares
Date: 05/31/14 Time: 10:12
Sample (adjusted): 1972 2011
Included observations: 40 after adjustments

Variable Coefficient Std. Error t-Statistic Prob.

C -0.044376 0.027883 -1.591500 0.1198
@TREND("1971") 0.002582 0.001185 2.178300 0.0357

R-squared 0.111007 Mean dependent var 0.008548
Adjusted R-squared 0.087612 S.D. dependent var 0.090585
S.E. of regression 0.086526 Akaike info criterion -2.008038
Sum squared resid 0.284496 Schwarz criterion -1.923594
Log likelihood 42.16076 Hannan-Quinn criter. -1.977506
F-statistic 4.744992 Durbin-Watson stat 2.332809
Prob(F-statistic) 0.035657

Please comment

Re: KPSS Test Output Interpretation

Posted: Sat May 31, 2014 8:33 am
by Pantera
According to your results the null hypothesis cannot be rejected. The transformed variable is level stationary.

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Re: KPSS Test Output Interpretation

Posted: Thu Nov 19, 2015 2:55 pm
by Hasna
So in summary we can say if

Kwiatkowski-Phillips-Schmidt-Shin test statistic Value is less than (<) any of the below than we would accept the null i.e the variable is stationary.
Asymptotic critical values*: 1% level 0.216000
5% level 0.146000
10% level 0.119000

Please answer.

Re: KPSS Test Output Interpretation

Posted: Fri Nov 20, 2015 3:10 pm
by londonphd
So in summary we can say if

Kwiatkowski-Phillips-Schmidt-Shin test statistic Value is less than (<) any of the below than we would accept the null i.e the variable is stationary.
Asymptotic critical values*: 1% level 0.216000
5% level 0.146000
10% level 0.119000

Please answer.
if the estimated test statistic is less than the critical values tabulated above, yes, you fail to reject the null of stationarity.