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On February 13th, 2017

Tests for Autocorrelated Errors

In ordinary least square regression model, we specify the equation as
y = b0 + b1 x1 + b2 x2 + b3 x3 + b4 x4 + ut
and we can test the assumption of autocorrelation or we can test whether the disturbances are autocorrelated.
To test the autocorrelation, we can follow the steps below:
(i) Estimate the regression model above using ordinary least square approach/OLS:
sysuse auto, clear
gen t=_n
tsset t
reg price rep78 trunk length
. reg price rep78 trunk length
Source |       SS           df       MS      Number of obs   =        69
-------------+----------------------------------   F(3, 65)        =      6.42
Model |   131790806         3  43930268.8   Prob > F        =    0.0007
Residual |   445006152        65   6846248.5   R-squared       =    0.2285
-------------+----------------------------------   Adj R-squared   =    0.1929
Total |   576796959        68  8482308.22   Root MSE        =    2616.5
------------------------------------------------------------------------------
price |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
rep78 |   578.7949   348.6211     1.66   0.102    -117.4495    1275.039
trunk |  -31.78264   108.8869    -0.29   0.771    -249.2447    185.6794
length |   70.17701   22.01136     3.19   0.002     26.21729    114.1367
_cons |  -8596.181   3840.351    -2.24   0.029    -16265.89   -926.4697
------------------------------------------------------------------------------

(ii)  Now calculate the residuals from the above regression:
predict errors, res
(iii)  Run another regression by inserting lagged residuals or the lag values of error terms, (errors) as predicted from above regression model into a regression model of the residuals as a dependent variable. Our regression model will be estimated through the following regression code in Stata: reg errors rep78 trunk length l.errors. We can consider this regression as auxiliary regression. The results from this auxiliary regression is given below
. reg errors rep78 trunk length l.errors
Source |       SS           df       MS      Number of obs   =        63
-------------+----------------------------------   F(4, 58)        =      4.66
Model |   104717963         4  26179490.9   Prob > F        =    0.0025
Residual |   325759819        58   5616548.6   R-squared       =    0.2433
-------------+----------------------------------   Adj R-squared   =    0.1911
Total |   430477782        62  6943190.04   Root MSE        =    2369.9
------------------------------------------------------------------------------
errors |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
rep78 |  -47.23696   332.1243    -0.14   0.887    -712.0559     617.582
trunk |   62.68025   103.8779     0.60   0.549    -145.2539    270.6144
length |   9.373147   20.97416     0.45   0.657     -32.6112    51.35749
|
errors |
L1. |   .5273932   .1225638     4.30   0.000      .282055    .7727313
|
_cons |  -2375.671   3741.339    -0.63   0.528    -9864.774    5113.432
------------------------------------------------------------------------------

(iv) Using the estimated results from auxiliary regression above, note the the R-squared value and multiply it by the number of included observations:
scalar N=_result(1)
scalar R2=_result(7)
scalar NR2= N*R2
scalar list N  R2  NR2
N =         63
R2 =  .24325986
NR2 =  15.325371

(v)  Now, the null hypothesis of BP test is that there is no autocorrelation, we can use the standard Chi-Square distribution to find the tabulated values of the Chi-Square to check if the null hypothesis of no autocorrelation needs to be rejected. According to theory, the Chi-Square statistic calculated using the NRSquare approach above, the test statistic NR2 converges asymptotically where degrees of freedom for the test is s which the number of lags of the residuals included in the auxiliary regression and we have included 1 lagged value of errors/residuals so degrees of freedom in this case is 1. We can use Stata conventional functions for distribution to determine the tabulated values at 5% level of significance using the following code:
scalar chi151=invchi2tail(1, .05)
scalar list chi151
chi15 =  3.8414598
We got from the above tutorial in tests for autocorrelation, NR2 = 15.325371 > 3.84 = Chi-Square (1, 5%). As the calculated value of Chi2 is greater than tabulated values of Chi2, so we reject the null hypothesis of no autocorrelation on the disturbances.

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