Statistical Significance of the Parameters - Statistics Project Example

?Question Dependent Variable: Y Method: Least Squares 03/15/12 Time: 22:12 Sample 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C 148.2206 221.6269 0.668784 0.5126 X1 -1.287395 0.805735 -1.597788 0.1285 X2 1.809622 0.515248 3.512139 0.0027 X3 0.590396 1.800093 0.327981 0.7469 X4 -21.48169 10.22264 -2.101384 0.0508 X5 5.619403 14.75619 0.380817 0.7081 X6 -14.51467 4.226150 -3.434490 0.0032 X7 29.36026 6.370371 4.608877 0.0003 R-squared 0.961206     Mean dependent var 2109.386 Adjusted R-squared 0.945233     S.D. dependent var 1946.249 S.E. of regression 455.4699     Akaike info criterion 15.33487 Sum squared resid 3526698.     Schwarz criterion 15.72491 Log likelihood -183.6859     Hannan-Quinn criter. 15.44305 F-statistic 60.17375     Durbin-Watson stat 1.916498 Prob(F-statistic) 0.000000 Y: Monthly manhours needed to operate an establishment X1: Average daily occupation X2: Monthly average number of check-ins X3: Weekly hours of service desk operation X4: Common use area (in square feet) X5: Number of building wings X6: Operational berthing capacity X7: Number of rooms ? 0= -148.2206 ? 1= -1.287395 ? 2= 1.809622 ? 3= 0.590396 ? 4= -21.48169 ? 5= 5.619403 ? 6= -14.51467 ? 7= 29.36026 From the above table, it can be noticed that the constant is -148.22 which means these manhours are available to operate an establishment if there is no factor available. Average daily occupation has the negative relationship with the manhours such that to add 1 extra manhour the average daily occupation will reduce by -1.28. Monthly average number of check-ins has the positive impact on the manhours such that for 1 extra man hour, 1.80 check-in are required. The coefficient for weekly hours of service desk opreration is 0.59 which causes an additional manhour. The common use area has a negative impact on the manhours as every 21.48 square feet will reduce one manhour. Number of building wings has a positive influence upon manhours such that around 5.6 building wings can cause additional one manhour. Operational berthing capacity has a negative impact upon manhours needed such that 14.5 units of operational berthing capacity reduce the 1 manhour required. Number of rooms, are in a direct relationship with manhours such that around 29 rooms create a need of extra 1 manhour. Since the probability (F-test) is less than the benchmark cutoff point of 0.05 which constitutes that overall model is good enough. R-squared value is around 0.96 which means that all the independent variables predict the manhours needed by around 96%. In other words, 96% variation in the manhours needed is explained by these seven variables included in the model. Statistical Significance of the Parameters If the p-values of each parameter are considered, it can be observed that for variables X1, X3, X4 and X5, the p-values are greater than the cutoff significant level of 0.05, therefore, these parameters are not considered as significant in predicting then manhours needed on individual basis. However, by staying in the overall model, they jointly predict the dependent variable of manhours needed. The other three independent parameters i.e. X2, X6 and X7 are statistically significant to predict the manhours needed as their p-values are less 0.05/ Question 2 H0: ?1 = ?2 = ?3 = ?4 = ?5 = ?6 = ?7 = 0 H1: At least one of the coefficients is not equal to 0 which would suggest that the model has explanatory power. F-statistics = 60.173 Therefore the hypothesis acceptance region is [0,F 8-1,25-8 ]=[0, F 7,17] From the F-statistics table the acceptance region is [0 and 2.61] Since F-statistics computed lies outside this region which is 60.173, therefore the hypothesis can be rejected as the all the parameters can jointly predict the manhours needed. This thing can also be proved by p-values of F-statistics which is less than the threshold of 0.05. R-squared tells about the goodness-of-fit of the model which is around 0.96. Therefore, 96% of the variation in the manhours, can be explained by all seven independent parameters jointly. The consequence of the results of F-statistics is that F-statistics describes that all the parameters jointly have the explanatory power of predicting the dependent variable. However, the t-statistics check the explanatory power each parameter in individual basis where some of the parameters do explain the dependent variable and some of them simply do not. But on overall basis, F-statistics provides a better yardstick for the overall explanatory power of all parameters on joint basis. Question 3 Dependent Variable: Y Method: Least Squares Date: 03/15/12 Time: 23:46 Sample: 1 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C 787.1661 418.5982 1.880481 0.0728 COMB 3.597651 0.814733 4.415741 0.0002 R-squared 0.458808     Mean dependent var 2109.386 Adjusted R-squared 0.435278     S.D. dependent var 1946.249 S.E. of regression 1462.567     Akaike info criterion 17.49039 Sum squared resid 49199351     Schwarz criterion 17.58790 Log likelihood -216.6299     Hannan-Quinn criter. 17.51744 F-statistic 19.49876     Durbin-Watson stat 1.488417 Prob(F-statistic) 0.000200 Wald Test: Equation: Untitled Test Statistic Value   df     Probability F-statistic 0.438964 (1, 17)   0.5165 Chi-square 0.438964 1   0.5076 Null Hypothesis Summary: Normalized Restriction (= 0) Value   Std. Err. C(2) - 2*C(4) -2.468186 3.725320 Restrictions are linear in coefficients. H0: ?2 = 2*?4 (The restriction on the coefficient is true) H1: ?2 ? 2*?4 (The restriction on the coefficient is not true) The acceptance region is [0, F 8-7, 25-8] = [0, F 1, 17]. From the table of F-statistics, the acceptance region is [0 and 4.45] From the results of Wald Test, the computed value of F-statistics is 0.438964 which lies under the acceptance region. Therefore the null hypothesis can be accepted which means that the restriction on the coefficient is true such that X1 is twice as X3. The above conclusion can be inferred from the p-value of F-statistics i.e. 0.5165 which is greater than the threshold of 0.05. Hence, it can be claimed that X3 is half of the X1. Question 4 Y X5 X7 Y  1.000000  0.735451  0.943140 X5  0.735451  1.000000  0.758939 X7  0.943140  0.758939  1.000000 From the above table it can be seen that variables X5 and X7 are correlated. But we need to test them once more by regressing X5 on X7 so that the value of R-squared can confirm the effect of multicollinearity. Here are the results of regression. Dependent Variable: X5 Method: Least Squares Date: 03/16/12 Time: 01:09 Sample: 1 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C 1.212531 2.389214 0.507502 0.6166 X7 0.078456 0.014036 5.589614 0.0000 R-squared 0.575988     Mean dependent var 11.12000 Adjusted R-squared 0.557553     S.D. dependent var 12.04270 S.E. of regression 8.010407     Akaike info criterion 7.075979 Sum squared resid 1475.832     Schwarz criterion 7.173489 Log likelihood -86.44973     Hannan-Quinn criter. 7.103024 F-statistic 31.24378     Durbin-Watson stat 2.141441 Prob(F-statistic) 0.000011 From the above table, it can be observed that the value of R-squared in 0.5759 which shows that there is high multicollinearity among the independent variables X5 and X7. In order to resolve the issue of multicollinearity from the model, there are few strategies which include the omission of any variable, obtaining more data, standardization of the independent variables, mean-center approach, Shapely value approach and do nothing approach. Consequences of multicollinearity for OLS estimators The consequences of multicollinearity for OLS estimators are as follows: The inclusion or deletion of some observations can influence the model substantially on statistical basis in the estimated coefficients. Opposite signs of the estimated coefficients can be observed as compared to those of expected. There is a likelihood that a wrong influential variable can be dropped out from the model because of the impacted coefficients as they cannot be statistically significant due to the lower values of t-statistics. OLS coefficient estimates may not be precise as the large standard errors cause in widening of the confidence intervals. Question 5 From the above three graphs, it is evident that there is not any dispersion found in any of the three graphs which reflects that hetroskedasticity is not there. As the data is more concentrated towards the below left corner, therefore the signs of dispersion are not visible. If the error term of the OLS model is found to be hetroskedastic, then following are consequences on OLS estimators. There is a possibility that the OLS coefficients still remain impartial and persistent as there is no correlation among the parameters and the error terms. There is likelihood that the OLS estimators are not efficient as other neutral estimators can also be found with smaller variances. The standard error estimators seems to be underestimated as they cause substantially higher values of t-statistics and F-statistics so these estimators can be incorrect. Question 6 There are reasons which lead to the statement that White test should be preferred over Breush-Pagan and Goldfeld-Quandt test because: White Test takes into more number of variables as compared Breush-Pagan. The prior knowledge of hetroskdeacity is not assumed by the White test. Normality assumption of Breush-Pagan is not required for White test. In the auxiliary regression, it proposed the choices of specific Zs. Goldfeld-Quandt provides more accurate results because it works on the basis of variances of the residuals and assumes the variances of all the residuals are same. Question 7 Y = ?1 + ?2X1 + ?3X2 + ?4X3 + ?5X4 + ?6X5 + ?7X6 + ?8X7 + µt … (1) Given = ?2µt = ?2µ. X41/4t Variance = ?2t = ?2.Z2t Let Z2t be X41/4t So Zt = X41/8t Dividing (1) by X41/8t We have Y/ X41/8t = ?1/X41/8t + ?2X1/X41/8t + ?3X2/ X41/8t + ?4X3/X41/8t + ?5X4/X41/8t + ?6X5/X41/8t + ?7X6/X41/8t + ?8X7/X41/8t + µt/X41/8t µt = µt/X41/8t So Var (µt) = Var (µt/X41/8t) = ?2 So the hetroskedacity is resolved by re-estimating the initial model on the basis of generalized least squares method. Question 8 Dependent Variable: Y Method: Least Squares Date: 03/19/12 Time: 14:46 Sample: 1 25 Included observations: 25 White Heteroskedasticity-Consistent Standard Errors & Covariance Coefficient Std. Error t-Statistic Prob.   C 148.2206 102.7056 1.443160 0.1671 X1 -1.287395 0.537525 -2.395042 0.0284 X2 1.809622 0.626183 2.889925 0.0102 X3 0.590396 1.101279 0.536100 0.5988 X4 -21.48169 9.726582 -2.208554 0.0412 X5 5.619403 14.15687 0.396938 0.6964 X6 -14.51467 5.813918 -2.496539 0.0231 X7 29.36026 7.714837 3.805687 0.0014 R-squared 0.961206     Mean dependent var 2109.386 Adjusted R-squared 0.945233     S.D. dependent var 1946.249 S.E. of regression 455.4699     Akaike info criterion 15.33487 Sum squared resid 3526698.     Schwarz criterion 15.72491 Log likelihood -183.6859     Hannan-Quinn criter. 15.44305 F-statistic 60.17375     Durbin-Watson stat 1.916498 Prob(F-statistic) 0.000000 Dependent Variable: Y Method: Least Squares Date: 03/15/12 Time: 22:12 Sample: 1 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C 148.2206 221.6269 0.668784 0.5126 X1 -1.287395 0.805735 -1.597788 0.1285 X2 1.809622 0.515248 3.512139 0.0027 X3 0.590396 1.800093 0.327981 0.7469 X4 -21.48169 10.22264 -2.101384 0.0508 X5 5.619403 14.75619 0.380817 0.7081 X6 -14.51467 4.226150 -3.434490 0.0032 X7 29.36026 6.370371 4.608877 0.0003 R-squared 0.961206     Mean dependent var 2109.386 Adjusted R-squared 0.945233     S.D. dependent var 1946.249 S.E. of regression 455.4699     Akaike info criterion 15.33487 Sum squared resid 3526698.     Schwarz criterion 15.72491 Log likelihood -183.6859     Hannan-Quinn criter. 15.44305 F-statistic 60.17375     Durbin-Watson stat 1.916498 Prob(F-statistic) 0.000000 From the above tables, it can be observed that when White Hetroskedasticity Coefficient is applied, then there is not change found in the coefficient estimates however, the standard errors have increased and t-statistics have reduced. We use White standard errors because of the various kinds of Hetroskedasticity available. Therefore HC estimators are suggested to tackle this issue. The foremost advantage of this technique is that it does not affect the OLS coefficient, what actually it does, is to adjust the standard errors only. Question 9 From the above charts, no pattern or trend can be particularly found which is evident of the fact that there is no autocorrelation found in the data. The consequences of autocorrelation on OLS estimators are as follows such that: The coefficients remain consistent and impartial but they become inefficient. Estimation of Standard errors provides wrong results due to biasness towards R-squared. t-statistics also provide higher figures which also lead towards erroneous results. Question 10 Dependent Variable: Y Method: Least Squares Date: 03/15/12 Time: 22:12 Sample: 1 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C 148.2206 221.6269 0.668784 0.5126 X1 -1.287395 0.805735 -1.597788 0.1285 X2 1.809622 0.515248 3.512139 0.0027 X3 0.590396 1.800093 0.327981 0.7469 X4 -21.48169 10.22264 -2.101384 0.0508 X5 5.619403 14.75619 0.380817 0.7081 X6 -14.51467 4.226150 -3.434490 0.0032 X7 29.36026 6.370371 4.608877 0.0003 R-squared 0.961206     Mean dependent var 2109.386 Adjusted R-squared 0.945233     S.D. dependent var 1946.249 S.E. of regression 455.4699     Akaike info criterion 15.33487 Sum squared resid 3526698.     Schwarz criterion 15.72491 Log likelihood -183.6859     Hannan-Quinn criter. 15.44305 F-statistic 60.17375     Durbin-Watson stat 1.916498 Prob(F-statistic) 0.000000 d=1.916 as DW is always between (0,4) n =25 number of observations k=7 number of independent variables From the table du = 1.915 dl = 0.610 Since du ? d ? 4 – du Therefore 1.915 ? 1.916 ? 4 - 1.915 1.915 ? 1.916 ? 2.085 So on the basis of above condition, at 1% significant level, H0 cannot be rejected such that no autocorrelation is found. Question 12 (a, b) Consider the linear model y = ?0 + ?1(x1) + ?2 (x2) + ?3(x3) + ?4(x4) + ?5(x5) + ?6 (x6) + ?7 (x7) The semi logarithmic model is Log(y) = ?0 + ?1(x1) + ?2 (x2) + ?3(x3) + ?4(x4) + ?5(x5) + ?6 (x6) + ?7 (x7) As the two models have different dependent variables we use BOX-COX Dependent Variable: Y_STAR Method: Least Squares Date: 03/24/12 Time: 21:20 Sample: 1 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C 0.116913 0.174814 0.668784 0.5126 X1 -0.001015 0.000636 -1.597788 0.1285 X2 0.001427 0.000406 3.512139 0.0027 X3 0.000466 0.001420 0.327981 0.7469 X4 -0.016944 0.008063 -2.101384 0.0508 X5 0.004432 0.011639 0.380817 0.7081 X6 -0.011449 0.003333 -3.434490 0.0032 X7 0.023159 0.005025 4.608877 0.0003 R-squared 0.961206     Mean dependent var 1.663836 Adjusted R-squared 0.945233     S.D. dependent var 1.535157 S.E. of regression 0.359264     Akaike info criterion 1.044821 Sum squared resid 2.194205     Schwarz criterion 1.434861 Log likelihood -5.060260     Hannan-Quinn criter. 1.153001 F-statistic 60.17375     Durbin-Watson stat 1.916498 Prob(F-statistic) 0.000000 The value of RSS1= 2.194 Dependent Variable: LY_STAR Method: Least Squares Date: 03/24/12 Time: 21:22 Sample: 1 25 Included observations: 25 Coefficient Std. Error t-Statistic Prob.   C -1.650635 0.295274 -5.590183 0.0000 X1 -0.000119 0.001073 -0.110494 0.9133 X2 0.000649 0.000686 0.945786 0.3575 X3 0.005056 0.002398 2.108387 0.0501 X4 0.018638 0.013620 1.368431 0.1890 X5 -0.008332 0.019660 -0.423823 0.6770 X6 -0.005223 0.005631 -0.927649 0.3666 X7 0.010342 0.008487 1.218515 0.2397 R-squared 0.809668     Mean dependent var -3.73E-16 Adjusted R-squared 0.731296     S.D. dependent var 1.170645 S.E. of regression 0.606823     Akaike info criterion 2.093179 Sum squared resid 6.259985     Schwarz criterion 2.483219 Log likelihood -18.16474     Hannan-Quinn criter. 2.201360 F-statistic 10.33110     Durbin-Watson stat 0.935327 Prob(F-statistic) 0.000047 The value of RSS2= 6.259 Test: H0: The model with the lower RSS is not superior H1: H0 false (n/2)*Log(RSS2/RSS1) ~ x2 1 Where, RSS2 is the RSS of the equation with the higher RSS. In EViews Statistic is Scalar sat= scalar stat = (25/2)*Log (6.259/2.194) The value of Stat is 13.10. Acceptance Region = [0, 3.84] So Stat=13.10 is not in the region. So reject H0. Therefore the model with the lower RSS is superior. So, linear function form is selected. Question 12 (c) The model is Log (y) = ?0 + ?1Log (x1) + ?2 Log (x2) + ?3 Log (x3) + ?4 Log (x4) + ?4 Log (x5) + ?6 Log (x6) + ?7 Log (x7) Regression: Quick/ estimate equation/ enter L y C Lx1 Lx2 Lx3 Lx4 Lx5 Lx6 Lx7 Dependent Variable: LY Method: Least Squares Date: 03/25/12 Time: 14:34 Sample: 1 25 Included observations: 24 Coefficient Std. Error t-Statistic Prob.   C 3.801110 0.537933 7.066140 0.0000 LX1 -0.070055 0.192293 -0.364314 0.7204 LX2 0.278102 0.089824 3.096075 0.0069 LX3 0.096532 0.104394 0.924684 0.3689 LX4 0.050855 0.101003 0.503498 0.6215 LX5 0.151730 0.157679 0.962271 0.3502 LX6 -1.053468 0.666722 -1.580070 0.1337 LX7 1.408493 0.863207 1.631699 0.1223 R-squared 0.951481     Mean dependent var 7.154951 Adjusted R-squared 0.930255     S.D. dependent var 1.194748 S.E. of regression 0.315526     Akaike info criterion 0.792048 Sum squared resid 1.592902     Schwarz criterion 1.184732 Log likelihood -1.504570     Hannan-Quinn criter. 0.896227 F-statistic 44.82435     Durbin-Watson stat 1.766617 Prob(F-statistic) 0.000000 Interpretation to the parameters estimates of that regression: ?0= 3.801 ?1 =-0.07 ?2 = 0.278 ?3 = 0.096 ?4 =0.050 ?5 =0.151 ?6 =-1.053 ?7 =1.408 ?0 = 3.801, If the value of Log (x1), Log (x2), Log (x3), Log (x4), Log (x5), Log (x6) and Log (x7) are are 0, so Log(y) will be 3.801. ?1 = -0.07, If x1 increases by 1% then y decreases by 7% If x1decreases by 1% then y increases by 7% ?2 = 0.278 If x2 increases by 1% then y increases by 27.8% If x2 decreases by 1% then y decreases by 27.8% ?3 = 0.096 If x3 increases by 1% then y increases by 9.6% If x3 decreases by 1% then y decreases by 9.6% ?4 = 0.050 If x4 increases by 1% then y increases by 5% If x4 decreases by 1% then y decreases by 5% ?5 = 0.151If x5 increases by 1% then y increases by 15.1% If x5 decreases by 1% then y decreases by 15.1% ?6 = -1.053 If x6increases by 1% then y decreases by 105.3% If x6 decreases by 1% then y increases by 105.3% ?7 = 1.408 If x7 increases by 1% then y increases by 140.8% If x7 decreases by 1% then y decreases by 140.8% The model is good because the value of R 2 =0.95 is between 0 and 1. Looking at the p-value for Log (x1), Log (x3), Log (x4), Log (x5), Log (x6) and Log (x7) are bigger then 0.05 we will accept H0 at the 5% level. Question 13 Use the residuals histogram and descriptive statistics to comment on your results from the JB test. To check for normality of residuals in a regression model we need to check on the histogram and the J-B statistic. To do this, we first need to estimate the desired equation by clicking on Quick/Estimate Equation Log (y) C Log (x1) Log (x2) Log (x3) Log (x4) Log (x5) Log (x6) Log (x7) and obtain the estimated residuals( using Eviews: Proc/Make Residual series...) then in Residuals windows, click on View/ Descriptive Statistics/ Histogram and Stats. The Hypothesis: H0= Residuals are normally distributed such that Skewness S=0 and Kurtosis K=3 H1= Residuals are not normally distributed We use this formula to calculate The JB statistical test Where  = is the number of observations = 24  is the number of parameters in the model The critical value is 5.9915 % (from chi-square table) The JB table is 0.129138 From the graph we can see that the skewness is -0.050360 which means that the distribution of the residuals, are slightly skewed to the left. The kurtosis is 3.344955 which is slightly greater than 3 and that indicate that the distribution is almost concentrated around the mean and that we have low chances to observe extreme observations. We accept the null hypothesis and accept that the residuals are normally distributed. Because JB critical > JB table From the JB test it can be stated that none of the assumptions of Markov-Gauss are violated such that residuals are not much varied due which the series is normally distributed. OLS estimator from this kind of model will have variance which is not bigger, therefore implying efficient results. In a nutshell, the desired characteristics of OLS estimators to be consistent, impartial and efficient are met in this model. Question 14 This model highlights the impact of different factors on the need of manpower in the Navy bachelor officer’s requirement such that different factors play an important role in explaining the manpower needed. Some of those factors have a direct impact on the manpower needed and very few factors behave in an inverse relation with the requirement of manpower. Thus, in Ministry of Defense, this model can describe the patterns for the requirement of the manpower. Read More