Exercise 1
This report attempts to evaluate the factors that determine the level of earnings by an employee. It assesses how factors such as schooling, work experience, and gender impact the personal earnings level. Figure 1 and figure 2 below shows the scatter plot for earnings against experience and years spend in school. It is evident that both years of experience and schooling have a mild positive relationship with the level earnings. Both datasets have clear outliers, and further analysis is required to determine the nature of the relationship between the variables conclusively.
Figure 1: A Scatter Plot of the Earnings and Experience
Figure 2: A Scatter Plot of the Earnings and Years in School
Over the past few years, organizations have sought continuously to retain employees as changes and disruption due to employee turnover are often expensive. The hiring and training of new personnel is both time and resource consuming. New staff must learn and get accustomed to the policies and culture of the organization. It is likely that their effectiveness, efficiency, and productivity is less than that of the employee who left. Therefore, companies seek to retain their employees by rewarding them for staying. However, the dataset does not show a clear relationship between tenure and the level of earning as seen I figure 3 below.
Figure 3: A Scatter Plot of the Earnings and Tenure
Regression Analysis
This report employs regression analysis to determine the extent to which earnings by individuals is explained by working experience, years of education and gender. The results from regressing the corresponding variables are shown in table 1 below. The dataset consists of 200 observations. The estimated model can be summarized as follows:
Earnings = (2.6389 × Schooling) + (0.33737 × Experience) + (8.2732 × Gender) – 26.4886 (1)
Equation 1 above shows is a regression model capturing the relationship between earnings and the working experience, years of education and gender of an employee. It suggests that the experience, years of education, and gender variables have a positive relationship with the earnings. The estimated coefficients suggest the longer an employee spend in education, the higher the earnings. Also, employees with more job experience are likely to earn a higher salary than their counterparts with less experience. The estimated coefficient of the role of gender in earrings suggests that male employees earn more than their female counterparts.
Table 1: Summary of the Regression Analysis Results
Source  SS  df  MS  Number of obs = 200  
F( 3, 196) = 25.78  
Model  13467.66  3  4489.22  Prob > F = 0.0000  
Residual  34135.93  196  174.16  Rsquared = 0.2829  
Adj Rsquared = 0.2719  
Total  47603.60  199  239.21  Root MSE = 13.197  
EARNINGS  Coef.  Std. Err.  t  P>t  [95% Conf. Interval]  
S  2.6389  0.3710  7.11  0.0000  1.907279 3.370423  
EXP  0.3737  0.2474  1.51  0.1320  .1141602 .8615998  
MALE  8.2732  1.9492  4.24  0.0000  4.429143 12.11731  
_cons  26.4886  7.2427  3.66  0.0000  52.9772 
The results of the F Test show the overall significance of the model for further extrapolation. It points out if the independent variable is likely to have occurred by chance with a given degree of freedom and particular observations. The pvalue of the Fstatistic is 0.00 implying that at a 5% significance level, we would reject the null hypothesis that the interceptonly estimated model fits the data as well as the proposed model. Hence, the data provides sufficient evidence to conclude that the proposed regression model is significant as it gives a better fit than the model with no independent variables. The model has an Adjusted R^{2} of 0.2719 implying that the model explains only 27.19% of the variations of the dependent variable. Hence, it can accurately forecast the earnings of an employee using working experience, years of education and gender 27.19% of the times.
The tstatistic is used to examine the significance of the coefficients of the estimated model. It is employed in testing the following hypothesis:
H_{0}: Estimated coefficient = 0
H_{1}: Estimated coefficient ≠ 0
The pvalues of the tstatic for the estimated coefficients of working experience, years of education and gender are 0.13, 0.00, and 0.00 as seen in table 1. Therefore, at a 0.05 significance level, we reject the null hypothesis that the coefficients for the variables of years in school and gender are equal to zero. However, we fail to reject the null hypothesis that the estimated coefficient of the experience variable is zero as the pvalue is greater than 0.05. Hence, at 5% significance level the variables for years in school and gender are significant in the model while experience is not.
Residuals Analysis
Following the regression analysis and estimation of the model describing the nature of the relationship between earnings and, working experience, years of education and gender the residuals are tested for correlation with the independent variables. Residuals are required to be independent and random to minimize type 1 error. Figure 4 it is evident that the residuals are centered on zero seem randomly distributed as they have no distinct pattern.
The classical assumptions of ordinary least squares linear regression advance the following assumptions about the residuals:
 Residuals have a population mean of zero.
 The independent variables are uncorrelated with the error term.
 The residuals have a constant variance.
 The residuals are normally distributed.
The residuals of the model are distributed around zero as seen in figure 4. Hence the assumption that the residuals have a population mean zero is true. The assumption on the variance of errors implies that the variance should be consistent for the range of observations. The plot of residuals versus fitted value is applied to check for the assumption of homoscedasticity. In figure 4 it is evident that the spread of the residuals increases with the increase in fitted values implying that variance changes with values of the fitted model. The coneshaped spread of distribution implies heteroscedasticity of residuals.
Figure 4: A scatter Plot of the Residuals Versus Fitted Values
Table 2 below shows the results of the ShapiroWilk test for Normality. The pvalue is less than 0.05. Hence, we reject the null hypothesis that the residuals are normally distributed.
Table 2: Results of the ShapiroWilk Test for Normality
Variable  Observations  W  V  z  Prob>z 
residual  200  0.98605  2.081  1.686  0.04589 
Testing for Heteroscedasticity
For a linear regression model to be reliable, the variance of the errors should be consistent for all observations. This report employs the BreuschPagan test to test for heteroscedasticity of the independent variables. The analysis produces a chisquared statistics with 3 degrees of freedom when the null hypothesis of no heteroscedasticity is satisfied which is 46.09 in this case as seen in table 3 below. The estimated pvalue of the chisquared statistics is 0.00. Therefore, the null hypothesis of no heteroscedasticity is rejected. The results suggest that the variance of the residuals of the model increases or decrease as a function of one of the independent variables. The presence of heteroscedasticity negatively affects the reliability of the results of the regression analysis as the tvalues of the estimated coefficients are affected by the biased standard error terms.
Table 3: Results of the BreuschPagan Test
BreuschPagan / CookWeisberg test for heteroscedasticity Ho: Constant variance Variables: S EXP MALE chi2(3) = 46.09 Prob > chi2 = 0.0000 
Remedy for Heteroscedasticity
Various approaches can be employed to correct the heteroscedasticity problem. This report will apply use the logarithm approach. It involves generating the logarithm of the dependent variable and using the same as the dependent variable in the regression model. Table 4 below shows the results of regression analysis with the Log of earnings as the dependent variable. Based on the Fstatistic the estimated model is significant at a 0.05 significance level. The estimated coefficients for working experience, years of education and gender are all significant at a 0.05 significance level.
Table 4: Summary of the Regression Analysis Results with Log of Earnings as Dependent Variable
. regress LogEarnings S EXP MALE  
Source   SS  df  MS  Number of obs = 200  
F( 3, 196) = 51.07  
Model  35.39846  3  11.7994871  Prob > F = 0.0000  
Residual   45.28231  196  0.23103217  Rsquared = 0.4387  
Adj Rsquared = 0.4302  
Total   80.68077  199  0.405430989  Root MSE = .48066  
LogEarnings   Coef.  Std. Err.  t  P>t  [95% Conf. Interval]  
S   0.1403  0.0135  10.38  0.0000  0.1136  0.1669 
EXP   0.0350  0.0090  3.88  0.0000  0.0172  0.0528 
MALE   0.3419  0.0710  4.82  0.0000  0.2019  0.4819 
_cons   0.1064  0.2638  0.4  0.6870  0.4138  0.6266 
Table 5 shows the results of the BreuschPagan test for the new model. The results show that the new model is homoscedastic since we fail to reject the null hypothesis for no heteroscedasticity as the pvalue of the chisquared statistic is greater than 0.05. Therefore, the logarithm approach solves the heteroscedasticity problem of the model.
Table 5: Results of the BreuschPagan Test for the New Model
BreuschPagan / CookWeisberg test for heteroskedasticity Ho: Constant variance Variables: fitted values of LogEarnings chi2(1) = 0.08 Prob > chi2 = 0.7811 
Improving the Model
The model does not allow for testing whether the marginal effect of schooling on earnings depends on gender. It would require a change of the model’s gender variable to examine if the coefficient of schooling on earnings depends on gender. The variable female is added to the model to replace the male variable. Table 6 below shows the results of the regression model with the variable Female. The estimated coefficient of the number of years is the same for both models in table 4 and table 6 suggesting that the effect of earnings depends on gender. Further correlation analysis shows that the gender variable does not have a relationship with the number of years in school.
Table 6: Summary of the Regression Analysis Results After Replacing the Variable Male With Female
regress LogEarnings S EXP FEMALE  
Source  SS  df  MS  Number of obs = 200  
F( 3, 196) = 51.07  
Model  35.3985  3  11.7995  Prob > F = 0.0  
Residual  45.2823  196  0.2310  Rsquared = 0.4387  
Adj Rsquared = 0.4302  
Total  80.6808  199  0.4054  Root MSE = .48066  
LogEarnings  Coef.  Std. Err.  t  P>t  [95% Conf. Interval]  
S  0.1403  0.0135  10.38  0.0000  0.1136  0.1669 
EXP  0.0350  0.0090  3.88  0.0000  0.0172  0.0528 
FEMALE  0.3419  0.0710  4.82  0.0000  0.4819  0.2019 
_cons  0.4483  0.2804  1.60  0.1110  0.1047  1.0012 
Table 7: Correlation Matrix
LogEarnings  MALE  S  EXP  
LogEarnings  1  
MALE  0.3523  1  
S  0.5348  0.0405  1  
EXP  0.1692  0.2697  0.2196  1 
Adding Variables
The tenure in current employment should be added to the model. Over the past few years, organizations have sought continuously to retain employees as changes and disruption due to employee turnover are often expensive. The hiring and training of new personnel is both time and resource consuming. New personnel must learn and get accustomed to the policies and culture of the organization. It is likely that their effectiveness, efficiency, and productivity is less than that of the employee who left. Therefore, companies seek to retain their employees by rewarding them with higher salaries for staying in the company.
Conclusion
The study established that working experience, years of education and gender are significant factors that impact the level of income. The results suggest that the more years in work experience and schooling yield higher income levels. The finding highlighted a key issue of gender inequality as the regression model showed that male employees earn more than their female counterparts. Organizations should aim at eliminating gender biases by ensuring all employees are compensated by merit.
Exercise 2
Figure 5 below shows the time series plot of the variable of the study. It is evident that the variable follows a seasonal pattern of an increasing trend followed by a sharp decline. The correlogram for the dataset is shown in figure 6 below. The characterbased plot of the autocorrelations shows that the dataset has a seasonal component.
Figure 5: Time Series Plot
Figure 6: Correlogram
Time series analysis seeks to provide a reliable description of the data and to understand unique features of trends, randomness, and seasonality. As observed in figure 5 and figure 6 the dataset has a significant seasonality component. A model can be derived from the time series data to predict future values based on the identified features of historical data. Before building any forecasting model from a time series, it is necessary that the series is stationary. Where the stationarity condition is breached, the first solution involves making the time series stationary by differencing or detrending and using stochastic models to make predictions in the time series. A timeseries is said to be stationary if the mean of the series is constant such that it is not a function of time and homoscedastic such that the variance does not change over time. Hence, stationarity implies that the probabilistic character of a series does not vary with time.
From figure 5 it is evident that the dataset has a seasonality component but no visible trend component. Based on the time series plot, the mean and the variance of the dataset do not seem to be constant over time hence the dataset is not stationary. In this case, we solve the stationarity problem by creating a new variable, z, derived from differencing the time series data. Figure 7 and figure 8 below show that the time series plot of the new variable and the characterbased plot of the autocorrelations of the new variable. It is evident that the first difference of the variable is not stationary as the new variable shows changes in mean and variance as seen in figure 7 and figure 8.
Figure 7: Time Series Plot of Z
Figure 8: A Correlogram of Z
The second difference is obtained since the first difference does not yield stationarity of the dataset. From figure 9 and figure 10 below it is evident that the stationarity is achieved at the second difference.
Figure 9: A Time Series Plot of the Second Difference
Figure 10: A Correlogram of the Second Difference
Once stationary is achieved for the dataset, we fit the autoregressive model. The study runs an ARIMA (3,2,1) model, and the results are shown in table. The choice of the autoregressive order (3), integrated difference order (2), and moving average order (1) were derived from different approaches. The choice of the autoregressive order was derived from the number of peak points in the partial correlogram in figure 11 below.
Table 8: Summary of the Estimated ARIMA Model
arima z, arima(3,2,1) ARIMA regression Sample: 4 – 111 Number of obs = 108 Wald chi2(4) = 112.06 Log likelihood = 431.3875 Prob > chi2 = 0.0000 ——————————————————————————  OPG D2.z  Coef. Std. Err. z P>z [95% Conf. Interval] ————+————————————————————— z  _cons  .000151 .0236076 0.01 0.995 .046421 .046119 ————+————————————————————— ARMA  ar  L1.  .7428243 .0830595 8.94 0.000 .9056179 .5800307 L2.  .412752 .1040706 3.97 0.000 .6167266 .2087774 L3.  .2386445 .0937128 2.55 0.011 .4223182 .0549708  ma  L1.  1.000014 169.442 0.01 0.995 333.1002 331.1002 ————+————————————————————— /sigma  12.71143 1077.085 0.01 0.495 0 2123.759 —————————————————————————— 
Figure 11: A Partial Correlogram
Stata Log File
___ ____ ____ ____ ____ (R)
/__ / ____/ / ____/
___/ / /___/ / /___/ 12.0 Copyright 19852011 StataCorp LP
Statistics/Data Analysis StataCorp
4905 Lakeway Drive
Special Edition College Station, Texas 77845 USA
800STATAPC http://www.stata.com
9796964600 stata@stata.com
9796964601 (fax)
Singleuser Stata network perpetual license:
Serial number: 93611859953
Licensed to: —–
——
Notes:
1. (/v# option or set maxvar) 5000 maximum variables
. use “C:\Users\dell\Dropbox\jobs\Experiment5\Logfile.dta”, clear
. twoway (scatter EARNINGS EXP)
. twoway (scatter EARNINGS S)
. twoway (scatter EARNINGS TENURE)
. regress EARNINGS S EXP MALE
Source  SS df MS Number of obs = 200
————+—————————— F( 3, 196) = 25.78
Model  13467.6617 3 4489.22056 Prob > F = 0.0000
Residual  34135.9337 196 174.162927 Rsquared = 0.2829
————+—————————— Adj Rsquared = 0.2719
Total  47603.5953 199 239.214047 Root MSE = 13.197
——————————————————————————
EARNINGS  Coef. Std. Err. t P>t [95% Conf. Interval]
————+—————————————————————
S  2.638851 .3709533 7.11 0.000 1.907279 3.370423
EXP  .3737198 .2473859 1.51 0.132 .1141602 .8615998
MALE  8.273226 1.949192 4.24 0.000 4.429143 12.11731
_cons  26.48859 7.242666 3.66 0.000 40.77215 12.20503
——————————————————————————
. log using “C:\Users\dell\Dropbox\jobs\Experiment5\Econometrics.smcl”
———————————————————————————————————————————————
name: <unnamed>
log: C:\Users\dell\Dropbox\Experiment5\Econometrics.smcl
log type: smcl
opened on: 29 Jan 2019, 15:44:31
. twoway (scatter EARNINGS EXP)
. twoway (scatter EARNINGS S)
. twoway (scatter EARNINGS TENURE)
. regress EARNINGS S EXP MALE
Source  SS df MS Number of obs = 200
————+—————————— F( 3, 196) = 25.78
Model  13467.6617 3 4489.22056 Prob > F = 0.0000
Residual  34135.9337 196 174.162927 Rsquared = 0.2829
————+—————————— Adj Rsquared = 0.2719
Total  47603.5953 199 239.214047 Root MSE = 13.197
——————————————————————————
EARNINGS  Coef. Std. Err. t P>t [95% Conf. Interval]
————+—————————————————————
S  2.638851 .3709533 7.11 0.000 1.907279 3.370423
EXP  .3737198 .2473859 1.51 0.132 .1141602 .8615998
MALE  8.273226 1.949192 4.24 0.000 4.429143 12.11731
_cons  26.48859 7.242666 3.66 0.000 40.77215 12.20503
——————————————————————————
. rvfplot
. swilk REGS
ShapiroWilk W test for normal data
Variable  Obs W V z Prob>z
————+————————————————–
REGS  200 0.99205 1.186 0.393 0.34707
. predict resid, Residuals
option Residuals not allowed
r(198);
. predict resid, Resid
option Resid not allowed
r(198);
. predict resid, Res
option Res not allowed
r(198);
. hettest S EXP MALE
BreuschPagan / CookWeisberg test for heteroskedasticity
Ho: Constant variance
Variables: S EXP MALE
chi2(3) = 46.09
Prob > chi2 = 0.0000
. regress LogEarnings S EXP MALE
Source  SS df MS Number of obs = 200
————+—————————— F( 3, 196) = 51.07
Model  35.3984614 3 11.7994871 Prob > F = 0.0000
Residual  45.2823054 196 .231032171 Rsquared = 0.4387
————+—————————— Adj Rsquared = 0.4302
Total  80.6807668 199 .405430989 Root MSE = .48066
——————————————————————————
LogEarnings  Coef. Std. Err. t P>t [95% Conf. Interval]
————+—————————————————————
S  .1402813 .0135107 10.38 0.000 .1136363 .1669263
EXP  .0350012 .0090102 3.88 0.000 .0172319 .0527706
MALE  .341875 .0709926 4.82 0.000 .2018675 .4818824
_cons  .1064149 .2637891 0.40 0.687 .4138144 .6266443
——————————————————————————
. regress LogEarnings S EXP FEMALE
Source  SS df MS Number of obs = 200
————+—————————— F( 3, 196) = 51.07
Model  35.3984614 3 11.7994871 Prob > F = 0.0000
Residual  45.2823054 196 .231032171 Rsquared = 0.4387
————+—————————— Adj Rsquared = 0.4302
Total  80.6807668 199 .405430989 Root MSE = .48066
——————————————————————————
LogEarnings  Coef. Std. Err. t P>t [95% Conf. Interval]
————+—————————————————————
S  .1402813 .0135107 10.38 0.000 .1136363 .1669263
EXP  .0350012 .0090102 3.88 0.000 .0172319 .0527706
FEMALE  .341875 .0709926 4.82 0.000 .4818824 .2018675
_cons  .4482899 .280383 1.60 0.111 .1046649 1.001245
——————————————————————————
. ASVAB03
unrecognized command: ASVAB03
r(199);
. use “C:\Users\dell\Desktop\CHRISTINE\Exercise2.dta”, clear
. twoway (tsline variable)
. ac variable
. twoway (tsline z)
. ac z
. tsset time
time variable: time, 1 to 111
delta: 1 unit
. twoway (tsline z)
. dvariable = d.variable
unrecognized command: dvariable
r(199);
. gen dvariable = d.variable
(1 missing value generated)
. twoway (tsline dvariable)
. ac dvariable
. twoway (tsline d.dvariable)
. ac d.dvariable
. arima d.dvariable time, arima(3,2,0)
note: D2.time dropped because of collinearity
(setting optimization to BHHH)
Iteration 0: log likelihood = 604.36417
Iteration 1: log likelihood = 576.33622
Iteration 2: log likelihood = 545.47559
Iteration 3: log likelihood = 534.24595
Iteration 4: log likelihood = 517.03404
(switching optimization to BFGS)
Iteration 5: log likelihood = 508.97783
Iteration 6: log likelihood = 495.75782
Iteration 7: log likelihood = 488.90125
Iteration 8: log likelihood = 487.12916
Iteration 9: log likelihood = 485.15753
Iteration 10: log likelihood = 483.30611
Iteration 11: log likelihood = 483.18608
Iteration 12: log likelihood = 483.18251
Iteration 13: log likelihood = 483.18244
Iteration 14: log likelihood = 483.18243
ARIMA regression
Sample: 5 – 111 Number of obs = 107
Wald chi2(3) = 578.14
Log likelihood = 483.1824 Prob > chi2 = 0.0000
——————————————————————————
 OPG
D3.dvariable  Coef. Std. Err. z P>z [95% Conf. Interval]
————+—————————————————————
dvariable 
_cons  .0803223 .5956257 0.13 0.893 1.087083 1.247727
————+—————————————————————
ARMA 
ar 
L1.  1.573478 .0720765 21.83 0.000 1.714745 1.43221
L2.  1.292952 .1015338 12.73 0.000 1.491955 1.09395
L3.  .5498229 .0730351 7.53 0.000 .692969 .4066768
————+—————————————————————
/sigma  21.82779 1.038335 21.02 0.000 19.79269 23.86288
——————————————————————————
Note: The test of the variance against zero is one sided, and the twosided confidence interval is truncated at zero.
. arima dvariable time, arima(3,2,0)
note: D2.time dropped because of collinearity
(setting optimization to BHHH)
Iteration 0: log likelihood = 494.4952
Iteration 1: log likelihood = 479.18552
Iteration 2: log likelihood = 473.62105
Iteration 3: log likelihood = 471.01932
Iteration 4: log likelihood = 468.10056
(switching optimization to BFGS)
Iteration 5: log likelihood = 467.94661
Iteration 6: log likelihood = 463.02373
Iteration 7: log likelihood = 457.17288
Iteration 8: log likelihood = 455.27644
Iteration 9: log likelihood = 454.45738
Iteration 10: log likelihood = 454.28194
Iteration 11: log likelihood = 454.27183
Iteration 12: log likelihood = 454.2716
Iteration 13: log likelihood = 454.2716
ARIMA regression
Sample: 4 – 111 Number of obs = 108
Wald chi2(3) = 931.84
Log likelihood = 454.2716 Prob > chi2 = 0.0000
——————————————————————————
 OPG
D2.dvariable  Coef. Std. Err. z P>z [95% Conf. Interval]
————+—————————————————————
dvariable 
_cons  .0798044 .5813969 0.14 0.891 1.059713 1.219321
————+—————————————————————
ARMA 
ar 
L1.  1.227382 .0518261 23.68 0.000 1.328959 1.125804
L2.  .8758729 .1036829 8.45 0.000 1.079088 .6726581
L3.  .3894012 .0953119 4.09 0.000 .5762092 .2025933
————+—————————————————————
/sigma  16.10992 .642368 25.08 0.000 14.8509 17.36894
——————————————————————————
Note: The test of the variance against zero is one sided, and the twosided confidence interval is truncated at zero.
. arima d.dvariable time, arima(3,2,1)
note: D2.time dropped because of collinearity
(setting optimization to BHHH)
Iteration 0: log likelihood = 512.10878
Iteration 1: log likelihood = 506.90462
Iteration 2: log likelihood = 482.4457
Iteration 3: log likelihood = 464.29893
Iteration 4: log likelihood = 464.25877 (backed up)
(switching optimization to BFGS)
Iteration 5: log likelihood = 462.13673
Iteration 6: log likelihood = 458.29143
Iteration 7: log likelihood = 456.83832
Iteration 8: log likelihood = 455.46375
Iteration 9: log likelihood = 454.44286
Iteration 10: log likelihood = 454.25128
Iteration 11: log likelihood = 454.18901
Iteration 12: log likelihood = 454.10725
Iteration 13: log likelihood = 454.10334
Iteration 14: log likelihood = 454.10021
(switching optimization to BHHH)
Iteration 15: log likelihood = 454.09988
Iteration 16: log likelihood = 454.09988 (backed up)
Iteration 17: log likelihood = 454.09987 (backed up)
Iteration 18: log likelihood = 454.09987 (backed up)
Iteration 19: log likelihood = 454.09987 (backed up)
(switching optimization to BFGS)
Iteration 20: log likelihood = 454.09987 (backed up)
Iteration 21: log likelihood = 454.0998
Iteration 22: log likelihood = 454.09977
Iteration 23: log likelihood = 454.09976
ARIMA regression
Sample: 5 – 111 Number of obs = 107
Wald chi2(4) = 1335.98
Log likelihood = 454.0998 Prob > chi2 = 0.0000
——————————————————————————
 OPG
D3.dvariable  Coef. Std. Err. z P>z [95% Conf. Interval]
————+—————————————————————
dvariable 
_cons  .0046315 .0140347 0.33 0.741 .0228761 .032139
————+—————————————————————
ARMA 
ar 
L1.  1.221487 .0527039 23.18 0.000 1.324784 1.118189
L2.  .8672662 .1048654 8.27 0.000 1.072799 .6617337
L3.  .3825357 .0942883 4.06 0.000 .5673374 .197734

ma 
L1.  1 .0793408 12.60 0.000 1.155505 .8444951
————+—————————————————————
/sigma  16.17989 . . . . .
——————————————————————————
Note: The test of the variance against zero is one sided, and the twosided confidence interval is truncated at zero.
. arima dvariable time, arima(3,2,1)
note: D2.time dropped because of collinearity
(setting optimization to BHHH)
Iteration 0: log likelihood = 451.7533
Iteration 1: log likelihood = 447.88398
Iteration 2: log likelihood = 446.88608
Iteration 3: log likelihood = 445.10459
Iteration 4: log likelihood = 444.5857
(switching optimization to BFGS)
Iteration 5: log likelihood = 444.41447
Iteration 6: log likelihood = 441.38757 (backed up)
Iteration 7: log likelihood = 436.57411
Iteration 8: log likelihood = 433.19382
Iteration 9: log likelihood = 432.48895
Iteration 10: log likelihood = 431.47139
Iteration 11: log likelihood = 431.42239
Iteration 12: log likelihood = 431.39591
Iteration 13: log likelihood = 431.39124
Iteration 14: log likelihood = 431.38854
(switching optimization to BHHH)
Iteration 15: log likelihood = 431.38777
Iteration 16: log likelihood = 431.38777 (backed up)
Iteration 17: log likelihood = 431.38777 (backed up)
Iteration 18: log likelihood = 431.38777 (backed up)
Iteration 19: log likelihood = 431.38777 (not concave)
(switching optimization to BFGS)
Iteration 20: log likelihood = 431.38764
Iteration 21: log likelihood = 431.38754
Iteration 22: log likelihood = 431.38754
Iteration 23: log likelihood = 431.38751
Iteration 24: log likelihood = 431.3875
Iteration 25: log likelihood = 431.3875
Iteration 26: log likelihood = 431.38749
ARIMA regression
Sample: 4 – 111 Number of obs = 108
Wald chi2(4) = 112.06
Log likelihood = 431.3875 Prob > chi2 = 0.0000
——————————————————————————
 OPG
D2.dvariable  Coef. Std. Err. z P>z [95% Conf. Interval]
————+—————————————————————
dvariable 
_cons  .000151 .0236076 0.01 0.995 .046421 .046119
————+—————————————————————
ARMA 
ar 
L1.  .7428243 .0830595 8.94 0.000 .9056179 .5800307
L2.  .412752 .1040706 3.97 0.000 .6167266 .2087774
L3.  .2386445 .0937128 2.55 0.011 .4223182 .0549708

ma 
L1.  1.000014 169.442 0.01 0.995 333.1002 331.1002
————+—————————————————————
/sigma  12.71143 1077.085 0.01 0.495 0 2123.759
——————————————————————————
Note: The test of the variance against zero is one sided, and the twosided confidence interval is truncated at zero.
. pac z
. save “C:\Users\dell\Desktop\Exercise2.dta”, replace
file C:\Users\dell\Desktop\Exercise2.dta saved