Data Analysis: Coefficient Interpretation - Assignment Example

Name of the student Name of the institution Date of submission Question one Dependent Variable: NARR86 Sample: 1 1495 Included observations: 1495 Variable Coefficient Std. Error t-Statistic Prob.   Constant 0.533 0.047 11.275 .00 AVGSEN -0.009828 0.013001 -0.755967 0.4498 ETHASIAN 0.790140 0.060477 13.06519 0.0000 ETHBLACK 0.966599 0.063410 15.24369 0.0000 ETHWHITE 0.532573 0.047233 11.27539 0.0000 PTIME86 -0.056414 0.011281 -5.000610 0.0000 QEMP86 -0.116265 0.014860 -7.824293 0.0000 TOTTIME 0.012242 0.010038 1.219583 0.2228 R-squared 0.088148     Mean dependent var 0.379933 Adjusted R-squared 0.084471     S.D. dependent var 0.911843 S.E. of regression 0.872481     Akaike info criterion 2.569720 Sum squared resid 1132.700     Schwarz criterion 2.594582 Log likelihood -1913.866     Hannan-Quinn criter. 2.578984 Durbin-Watson stat 1.969219 a. Coefficient interpretation From the regression table above, one percentage increase in the average sentence served from past conviction reduces the number of times a man will be arrested by 9.828%. One percentage increase in the Asian population increases the number of times a man will be arrested by 79, black ethnic 96.7% and white 53.25%. One percentage increase in the average sentence served from past conviction reduced the month spent in prison by 5.64%, similarly for number of quarters in 1986 during which the man was employed reduces by 11. 626%. Lastly, one percentage increase in the average sentence served from past conviction reduces total time the man has spent in prison prior to 1986 since reaching the age of 18 (in months) by 1.22%. The intercept shows that if all the variables are zero, the number of times a man would be arrested will be 0.533. b. Perform tests for the statistical significance of the parameters of the independent variables avgsen, tottime, ptime86 and qemp86 using the critical value of the corresponding t-distribution and the test p-value. Interpret the tests results. The statistical significance of the parameters AVGSEN, TOTTIME, PTIME86 and QEMP86 can be tested through the t-distribution and p-value tests’. These tests’ will provide insight as to whether or not the dependent variable and independent variables are relevant to each other. The statistical t-test formula is as follows:  To determine whether or not to accept or reject the  we must calculate the t* and observe if this value falls within our Acceptance Region (A.R.). The formula to determine the Acceptance Region is as follows:  The following values are known to carry out the A.R. region   AR = -1.873, 1.86 If we observe the value of t* within our Acceptance Region [A.R.], then we accept. Accepting  thereby informs us that the variable provides no significance as it relates to our overall model. Conversely, should we accept, proving  to be false, then we will conclude that the variable is in fact significant as it relates to our model. By entering our given information into the t-statistic formula, we can conclude the following values for AVGSEN, TOTTIME, PTIME86, and QEMP86 provided below. Independent Variable t-test Value/Comment Conclusion AVGSEN -0.755967#A.R Accept ; Not Significant TOTTIME 1.219583# A.R Accept ; Not Significant PTIME86 -5.000610 # A.R Accept ; Not Significant QEMP86 -7.824293# A.R Accept ; Not Significant Question two The joint significance test for the independent variables is an applicable test to determine if all the independent variables are relevant enough together to explain the dependent variable, C NARR86 in our model. Moreover, the F-test can be used for overall joint significance of the variables. The null hypothesis of the F-test indicates that the model does not have explanatory power. The null hypothesis suggests that no variable is significant enough to explain the dependent variable, whereas  implies that at least one of our independent variables is significant. The hypothesis for our particular model is as follows:   That we have described the necessary hypothesis for our model, we can carry out the F* value and declare whether or not it falls within our Acceptance Region. The F* test is as follows, using t  F. = 0.088148/ (10-1)/ (1-0.088148)/ (141-10) 0.009794 / 0.006961 = 1.40698 With P-value less than 5% and F = 1.40698, we can reject H0 and determine that all variables are jointly significant. a. Comment on the goodness-of fit of the model. The coefficient of determination of  measures the percentage of the variance of the dependent variable explained by regression. It will always increase when a new variable is added to the model, even when the new variable does not have explanatory power (as long as previous explanatory variables are retained). From the output table provided on the previous page, we can see that R-squared is 0.088148, or 8.8148%. Therefore 8.8148% of NARR86 can be explained by independent variables included in the regression model b. What are the consequences of the results of this F-test together with those of the t-tests R-squared 0.088148     Mean dependent var 0.379933 Adjusted R-squared 0.084471     S.D. dependent var 0.911843 S.E. of regression 0.872481     Akaike info criterion 2.569720 Sum squared resid 1132.700     Schwarz criterion 2.594582 Log likelihood -1913.866     Hannan-Quinn criter. 2.578984 F-statistics 4.87321 Pr(F-statistics 0.00000 In question 1, the t-test showed that AVGSEN, TOTTIME and QEMP86 were the only significant variables out of the four (including PTIME86, which was not significant) to explain narr86. When we performed the f-test above, we also produced results that signaled all variables were jointly significant to explain narr86. Observationally, it appears the t-test could be more precise as all the variables are not valid to significantly explain narr86, unlike the f-test, which provides all variables to significantly explain narr86. Question three: Test the hypothesis of that: an extra quarter employed has tripled the effect of an extra month spent in prison in 1986, on narr86. a. Use the command available in Eviews to test for the corresponding coefficient restriction. C (0.25) = 3*C (1) The above input reflects the coefficient restriction we are seeking to test. C (0.25) PTIME86 while C (6) narr86, with the value of 3 representing triple month spent in prison The output table given on the previous page provides us with a new F-statistic value of -0.051416. We must now test to see whether the new F-statistic value fall in, or outside, the Acceptance Region in an effort to either reject or accept the hypothesis. The Wald Test Acceptance Region formula is as follows  Unrestricted parameters = 4 Parameters in restricted model = 2 N=1495  Therefore, 0.54416, 1.982 The test shows that F-statistics does not fall within the accepted region 1.982 hence the hypothesis rejected. b. Perform the test analytically In order to provide further evidence of the decision to accept H0 we will conduct analytical test using the original unrestricted model, as well as the restricted model. The unrestricted variables are shown below NARR86 = ᵝ1 +ᵝ2 + AVGSEN1 + PTIME862 + QEMP863 + TOTTIME4 + C Dependent variable nrr86 Included observations: 1495 White heteroskedasticity-consistent standard errors & covariance Variable Coefficient Std. Error t-Statistic Prob.   AVGSEN -0.006291 0.014336 -0.438843 0.6608 PTIME86 -0.051416 0.008102 -6.345888 0.0000 QEMP86 -0.130318 0.016712 -7.797768 0.0000 TOTTIME 0.012783 0.013526 0.945102 0.3448 C 0.688819 0.058463 11.78216 0.0000 R-squared 0.054669     Mean dependent var 0.379933 Adjusted R-squared 0.052131     S.D. dependent var 0.911843 S.E. of regression 0.887757     Akaike info criterion 2.603102 Sum squared resid 1174.288     Schwarz criterion 2.620861 Log likelihood -1940.819     Hannan-Quinn criter. 2.609719 F-statistic 21.54197     Durbin-Watson stat 1.949440 Prob(F-statistic) 0.000000 Using the above output tables for our unrestricted and restricted models, we can perform an analytical test to determine similarity in our F*-statistic, thereby confirming our initial conclusion of rejecting the null hypothesis staying in prison two more times has the different effect. For non restriction Dependent Variable: NARR86 Sample: 1 1495 Included observations: 1495 Variable Coefficient Std. Error t-Statistic Prob.   Constant 0.533 0.047 11.275 .00 AVGSEN -0.009828 0.013001 -0.755967 0.4498 ETHASIAN 0.790140 0.060477 13.06519 0.0000 ETHBLACK 0.966599 0.063410 15.24369 0.0000 ETHWHITE 0.532573 0.047233 11.27539 0.0000 PTIME86 -0.056414 0.011281 -5.000610 0.0000 QEMP86 -0.116265 0.014860 -7.824293 0.0000 TOTTIME 0.012242 0.010038 1.219583 0.2228 R-squared 0.088148     Mean dependent var 0.379933 Adjusted R-squared 0.084471     S.D. dependent var 0.911843 S.E. of regression 0.872481     Akaike info criterion 2.569720 Sum squared resid 1132.700     Schwarz criterion 2.594582 Log likelihood -1913.866     Hannan-Quinn criter. 2.578984 Durbin-Watson stat 1.969219 = (1174.288 -1132.700)/ (10-9) / (1132.700 / (1495-10) = 41.588 / 0.763 = 54.51% Using the analytical testing format through the use of the unrestricted and restricted model, we can affirm our conclusions in part (a) that we reject. Our analytical test produces F* = 0.5451  5% (.05) significance level. Moreover, 0.5451  Acceptance Region, confirming that one more time spent in prison reduces the number of times a man is arrested. 4. Answer the sub questions below on multicollinearity. [8 marks] (a) Test for multicollinearity between the independent variables ptime86 and qemp86 in the model. Explain your answer using Eviews outputs. PTIME86 QEMP86 PTIME86  0.9231  0.2568 QEMP86  0.2568 0.9231 The above results indicate a low estimated correlation (0.2568) between QEMP86 and PTIME86, providing no clear indication of multicollinearity between the two variables. Therefore, we must now administer a statistical test by regressing one variable on the other in an attempt to test for individual significance. (b) Assuming that there is multicollinearity between those variables: i. Explain how you would resolve this problem. Explain your answer using Eviews outputs. Multicollinearity was found between independent variables, PTIME86 and QEMP86. Therefore we must resolve this issue. Since we do not need both variables, we can solve the problem of multicollinearity by removing either one of PTIME86 or QEMP86 (other methods include increasing the sample size and applying logarithms to the regression). In our case, we will regress both PTIME86 and QEMP86 against our dependent variable NARR86, comparing the goodness-of-fit. The variable with the higher  will be kept, as it provides greater explanatory value than the other variable, which is to be excluded. Below are two regressed output models, one including PTIME86, the other including QEMP86, with each void of one another. Dependent Variable: PTIME86 Method: Least Squares Date: 03/12/15 Time: 02:52 Sample: 1 1495 Included observations: 1495 Variable Coefficient Std. Error t-Statistic Prob.   QEMP86 -0.402875 0.033715 -11.94951 0.0000 C 1.413496 0.093829 15.06458 0.0000 R-squared 0.087292     Mean dependent var 0.501003 Adjusted R-squared 0.086680     S.D. dependent var 2.205893 S.E. of regression 2.108122     Akaike info criterion 4.330809 Sum squared resid 6635.160     Schwarz criterion 4.337913 Log likelihood -3235.280     Hannan-Quinn criter. 4.333456 F-statistic 142.7908     Durbin-Watson stat 2.023076 Prob(F-statistic) 0.000000 ii. What the consequences of multicollinearity are for the OLS estimator? Pertaining to the OLS estimator, multicollinearity presents some noticeable consequences. Perfect multicollinearity has an exact linear relationship between two or more variables, which makes it impossible to calculate the OLS estimators. Standard errors also become larger than normal when multicollinearity in present, meaning the resulting estimates is not reliable for analysis. Lastly, as the degree of multicollinearity increases, correlation between variables cause a larger variance, this leads to inaccurate OLS estimators, and unusable coefficients. Question five Heteroscedasticity exists when there is an unequal presence of residuals in the CLRM, ultimately violating Gauss-Markov’s assumption #5, stating that all disturbance terms are homoscedastic where  is always constant. A line chart, and scatter plot, will provide a graphical mechanism for Heteroscedasticity in the model by analysing the square residuals The above line &symbol chart provides an illustration showing different values of  [square residual is an approximation for variance], signaling Heteroscedasticity. The scatter plots of the independent variables illustrate the residuals of the regression model compared to the independent variables. The second chart, the line & symbol, provides us with the movements of the square residuals. The apparent Heteroscedasticity signals that the OLS estimators are inefficient. Therefore, the standard errors are likely to be underestimated, which will inevitably produce t-statistics and f-statistics with higher values than desired. 6. Perform a White test for Heteroscedasticity I. Provide the auxiliary regression and explain the meaning of the null hypothesis for this test. Dependent Variable: NARR86 Method: Least Squares Date: 03/12/15 Time: 04:38 Sample: 1 1495 Included observations: 1495 White heteroskedasticity-consistent standard errors & covariance Variable Coefficient Std. Error t-Statistic Prob.   AVGSEN -0.009828 0.014188 -0.692698 0.4886 ETHASIAN 0.790140 0.066224 11.93137 0.0000 ETHBLACK 0.966599 0.100894 9.580320 0.0000 ETHWHITE 0.532573 0.054724 9.731984 0.0000 PTIME86 -0.056414 0.008259 -6.830250 0.0000 QEMP86 -0.116265 0.015424 -7.537972 0.0000 TOTTIME 0.012242 0.013146 0.931225 0.3519 R-squared 0.088148     Mean dependent var 0.379933 Adjusted R-squared 0.084471     S.D. dependent var 0.911843 S.E. of regression 0.872481     Akaike info criterion 2.569720 Sum squared resid 1132.700     Schwarz criterion 2.594582 Log likelihood -1913.866     Hannan-Quinn criter. 2.578984 Durbin-Watson stat 1.969219 An auxiliary regression model provides more possible variables. To provide an auxiliary regression, the estimated residuals  must be obtained from our original regression model. Our subsequent auxiliary regression model is as follows: NARR86= ᶏ0 + ᶏ1AVGSEN + ᶏ2ETHASIAN + ᶏ3 TOTTIME + ᶏ4QEMP86 + ᶏ5PTIME86 + ᶏ6ETHWHITE + ᶏ7 ETHBLACK To carry out the test for Heteroscedasticity we must define our null hypothesis (which indicates homoscedasticity) and our alternative, , which signals Heteroscedasticity H0 ᶏ0 = ᶏ2 = ….. ᶏ7 = 0 (HOMODCEDASTICITY) H1 at least 1 of the coefficients ≠ 0 (HETEROSCEDASTICITY) For the White Test to work adequately,  must distribute under the null. To prove this, we need the residuals to be normally distributed. The formula for the test is as follows: II. Why is the White test preferred to the Breusch-Pagan and Goldfeld-Quandt tests for Heteroscedasticity? Explain your answer. In relation to the Breusch-Pagan and Goldfeld-Quandt tests, the White Test is more favorable. The White Test is preferred as a test for Heteroscedasticity because the auxiliary regression is more general. This generality is because we must consider that variance may change with original variables, but also with squares and cross products, providing us more probabilities. Second, the White Test proposes a specific question for the auxiliary regression. The result of this is that there are no possibilities left out. The last preference in favor of the White Test is that the White Test doesn’t assume normality for, the residuals from the original equation. Question 7 If there is Heteroscedasticity, the model can be re-estimated using GLS (Generalized List Squares). The primary component of GLS is that it attempts to reduce our error term variance to a constant. Because we know the form of Heteroscedasticity, provided in, we can use the GLS regression. In our case,  is the factor that produces Heteroscedasticity? Instead of being constant, the variance of our error term is heteroskedasticity. We will carry out the question by dividing the entire regression by the factor we are seeking to delete. In our GLS Regression, we will divide each term in our original model by  (which is equal to). NARR86= ᶏ0 + (ß1+ AVGSEN)/z1 + (ß 2 + ETHASIAN)/z2 + (ß3 + TOTTIME)/z3 + (ß 4 + QEMP86)/ z4 + (ß 5 + PTIME86)/z5 + (ß 6 + ETHWHITE)/z6 + (ß 7 + ETHBLACK)/z8 Now, the variance of our error term is constant due to the GLS Regression Model. A homoscedastic regression model is now present because the factor causing variability on the error term has been eliminated. By regressing NARR86 we have acquired estimates for AVGSEN, ETHASIAN, TOTTIME, QEMP86, PTIME86, ETHWHITE and ETHBLACK E. Furthermore through the regression of NARR86 our estimates are ‘best linear unbiased estimator To elaborate further, if her have Heteroscedasticity, then our initial equation 01 in Eviews is not accurate Therefore: Standard errors aren’t correct F-stat is not correct Coefficients aren’t correct Our  is unbiased and consistent, however, it is inefficient, meaning that the standard errors aren’t valid. In fact, they are underestimated (lower) than they should be. Moreover, the T-test and F-test aren’t valid, so we cannot do tests from the standard errors table. Questions 8: Estimate the model using Dependent Variable: NARR86 Method: Least Squares Date: 03/12/15 Time: 04:38 Sample: 1 1495 Included observations: 1495 White heteroskedasticity-consistent standard errors & covariance Variable Coefficient Std. Error t-Statistic Prob.   AVGSEN -0.009828 0.014188 -0.692698 0.4886 ETHASIAN 0.790140 0.066224 11.93137 0.0000 ETHBLACK 0.966599 0.100894 9.580320 0.0000 ETHWHITE 0.532573 0.054724 9.731984 0.0000 PTIME86 -0.056414 0.008259 -6.830250 0.0000 QEMP86 -0.116265 0.015424 -7.537972 0.0000 TOTTIME 0.012242 0.013146 0.931225 0.3519 R-squared 0.088148     Mean dependent var 0.379933 Adjusted R-squared 0.084471     S.D. dependent var 0.911843 S.E. of regression 0.872481     Akaike info criterion 2.569720 Sum squared resid 1132.700     Schwarz criterion 2.594582 Log likelihood -1913.866     Hannan-Quinn criter. 2.578984 Durbin-Watson stat 1.969219 Dependent Variable: NARR86 Method: Least Squares Date: 03/12/15 Time: 04:38 Sample: 1 1495 Included observations: 1495 White heteroskedasticity-consistent standard errors & covariance Variable Coefficient Std. Error t-Statistic Prob.   AVGSEN -0.009828 0.014188 -0.692698 0.4886 ETHASIAN 0.790140 0.066224 11.93137 0.0000 ETHBLACK 0.966599 0.100894 9.580320 0.0000 ETHWHITE 0.532573 0.054724 9.731984 0.0000 PTIME86 -0.056414 0.008259 -6.830250 0.0000 QEMP86 -0.116265 0.015424 -7.537972 0.0000 TOTTIME 0.012242 0.013146 0.931225 0.3519 R-squared 0.088148     Mean dependent var 0.379933 Adjusted R-squared 0.084471     S.D. dependent var 0.911843 S.E. of regression 0.872481     Akaike info criterion 2.569720 Sum squared resid 1132.700     Schwarz criterion 2.594582 Log likelihood -1913.866     Hannan-Quinn criter. 2.578984 Durbin-Watson stat 1.969219 When we compare the results from the outputs provided above, the original (theoretical) compared to the practical White Test results will show underestimated standard errors for the variables. The coefficients (estimators) do not change in either model; mean the OLS estimators are unbiased and consistent. Conversely, the standard errors illustrate a higher value in the White Heteroscedasticity-consistent standard errors and covariance, meaning the variances and standard errors appear underestimated in the OLS. Hence, this causes an increase in the t-statistic value that was not previously anticipated. Therefore, the White Heteroscedasticity-consistent standard errors and covariance provides a more accurate estimate regarding an unknown variance Question nine Smaller scatter of residuals, becoming larger in the middle before decreasing in size at the end of the plot. This is an informal graphical analysis. The second plot shows a mild trend, which may signal some autocorrelation. However, to be more precise and accurate, a statistical analysis is required to formally address the issue of autocorrelation. There are two primary consequences of autocorrelation on the OLS estimator. Question ten: The most common way to test for autocorrelation is through a formal, statistical test known as the Durbin-Watson Test. There are three assumptions that characterize the Durbin-Watson Test: The model includes a constant The test detects autocorrelation for AR(1) only The model does not include lagged dependent variables as explanatory variables NARR86 AVGSEN ETHASIAN ETHBLACK ETHWHITE PTIME86 QEMP86 TOTTIME NARR86 1 0.04601561958768719 0.06387108274358491 0.1781420983224773 -0.1903019570307919 -0.03679440723268456 -0.2036451068000658 0.05234005227684724 AVGSEN 0.04601561958768719 1 0.008319567966003483 0.1528193910365567 -0.1231284365783148 0.2680107520234372 -0.1384889932596224 0.9261920510707011 ETHASIAN 0.06387108274358491 0.008319567966003483 1 -0.2450468346602019 -0.6774397637411685 0.04712266392389303 0.0335688113947389 0.01905214882209259 ETHBLACK 0.1781420983224773 0.1528193910365567 -0.2450468346602019 1 -0.5471469161926825 0.112104840679903 -0.2011915732066689 0.1464093873025102 ETHWHITE -0.1903019570307919 -0.1231284365783148 -0.6774397637411685 -0.5471469161926825 1 -0.1257388996260479 0.1236641901009991 -0.127531174102759 PTIME86 -0.03679440723268456 0.2680107520234372 0.04712266392389303 0.112104840679903 -0.1257388996260479 1 -0.2954515533133766 0.3252736839946073 QEMP86 -0.2036451068000658 -0.1384889932596224 0.0335688113947389 -0.2011915732066689 0.1236641901009991 -0.2954515533133766 1 -0.1591199536527264 TOTTIME 0.05234005227684724 0.9261920510707011 0.01905214882209259 0.1464093873025102 -0.127531174102759 0.3252736839946073 -0.1591199536527264 1 Question 11 The number of times a man was arrested, is effected by the average sentence served from past conviction, the total time the man has spent in prison prior to 1986 since reaching the age of 18, the months spent in prison in 1986 and the number of quarters in 1986 during which the man was employed. The Chow Test will break the samples into subsamples and estimate the model for each of the subsamples, which can be two or more. The hypothesis of this test will indicate whether there is stability in the parameters or there are structural breaks. Question 12 The number of quarters employed (qemp86) have a significant on narr86 for individuals of different ethnicity. Dependent Variable: NARR86 Method: Least Squares Date: 03/12/15 Time: 05:50 Sample: 1 1495 Included observations: 1495 Variable Coefficient Std. Error t-Statistic Prob.   QEMP86 -0.097545 0.014352 -6.796666 0.0000 ETHASIAN 0.716180 0.058056 12.33607 0.0000 ETHBLACK 0.894535 0.060100 14.88409 0.0000 ETHWHITE 0.475685 0.045521 10.44977 0.0000 R-squared 0.072763     Mean dependent var 0.379933 Adjusted R-squared 0.070897     S.D. dependent var 0.911843 S.E. of regression 0.878925     Akaike info criterion 2.582439 Sum squared resid 1151.812     Schwarz criterion 2.596646 Log likelihood -1926.373     Hannan-Quinn criter. 2.587732 Durbin-Watson stat 1.965669 Question 13 To determine the best functional form for our original model, we must consider three functional forms: linear functional form, semi-logarithmic functional form, and logarithmic functional form. NARR86 = ᵝ1 +ᵝ2 + AVGSEN1 + PTIME862 + QEMP863 + TOTTIME4 + C Semi a logarithm Log (NARR86) = ᵝ1 +ᵝ2 + AVGSEN1 + PTIME862 + QEMP863 + TOTTIME4 + C LOGARITHMIC FUNCTIONAL FORM [log taken on both sides] Log (NARR86) = ᵝ1 +ᵝ2 + log (AVGSEN1) + log (PTIME862) + log (QEMP863) + log (TOTTIME4) + C To select the most functional form of the model, we will utilize the Box-Cox Test to select between the ‘winner’ of the log models (semi-logarithmic and logarithmic) and the linear functional form The Box-Cox steps are as follows: Calculate the geometric mean of the dependent variable: NARR86 in our case (NARR86) = exp 1/n Ʃ log (NARR86)) Using the white command window in Eviews, perform the following command Value NARR86  4.03612 Transform the dependent variable Estimate the linear functional form, and the semi-logarithmic form using the transformed variable, . We obtain the RSS from the linear and semi-logarithmic models. Test the statistic. Then, the model with the lower RSS is not superior.   Performing the test statistic we will reject or accept the null hypothesis Question 14 In order to test for normality in the residuals, we have to conduct the Jarque-Bera Test. To test for the Jarque-Bera test, we perform the following formula: Our degree of freedom is constant at a value of 2, and our significance level is 5%. In the Jarque-Bera test, our objective is to comprehend the skewness and kurtosis of the residuals. Previously in Question 5, we generated a residual series on our original model to produce resid01. To administer the Jarque-Bera test, we must perform the following: Q15. For what purpose can your analysis above be used by the government? Explain your answer. The government can use the statistics in rehabilitation issues to help and prevent the crime rate which is taking place within the country. The employment history and the number of times one has taken in prison among others factors influences the rate of doing crime and possibilities of being taken to jail again hence the government will follow this statistics in following the rehabilitation process. Read More