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| ECN 3972 | | | | | | Name
| Fall 2020 |
| Final Exam |
              READ THIS SECTION CAREFULLY. 
 
This exam is for your work ONLY. You may use: 
1.   Your notes
2.   Your book
3.   The class videos (made by me)
4.   A calculator

You may not use: 
1.   Your computer (other than downloading, printing, or reading the exam). Note: If you really want, you can type your answers in Word, but I don’t really think you will want to with all of
the equations.
2.   Excel or any other spreadsheet program
3.   The internet except D2L notes and videos
4.   Other individuals

Failure to follow these guidelines will result in: 
1.   A grade of zero on the exam
2.   A report of academic misconduct to the Office of Student Standards

Please believe me when I say that I take this seriously. I need to be able to trust that this is your work. Just because this is an online class does not change anything – you should treat this as you would an in-class exam where you are allowed to use your book and notes. 
 
Completion of this exam means you have acknowledged the rules listed above and will 
act accordingly. 
 
**You will upload your answers either in Word or a pdf the same way as you submit your 
homework assignments in ONE document.** NO PICTURES! Make sure it is in order!! 
 
Due date: Tuesday, December 15, 2020 at 11:00 pm. No late exams will be accepted!  

Please make sure your answers are written clearly and IN ORDER. Make sure every question is labeled with the proper number. Read each question carefully. For each hypothesis test, you must write both the null and alternative hypotheses, the decision rule, the calculated test statistic, the critical value of the test statistic and then (finally) the decision. You must have ALL parts! (Don’t assume that I will find it – make it clear.) 

Show all work if you want any partial credit. Point values are in parentheses. 

1. Suppose you are interested in determining the factors that affect the violent crime rate in the 100 largest metropolitan areas (MSAs) in the U.S. You gather information and estimate the following model: 

_VCi _= _β0 _+ _β1 _MedInci _+ _β2 _Povi _+ _β3 _UnEmpi _+ _β4 _AvAgei _+ _β5 _Policei _+ _εi 

              where: 
              ^VCi ^= ^violent ^crime ^rate ^(crimes ^per ^1,000 ^people) ^in ^MSA ^i 
              MedInc_i = median income in MSA i (in thousands of dollars) 
              Pov_i = poverty rate in MSA i 
              UnEmp_i = unemployment rate in MSA i 
              AvAge_i = average age in MSA i  
              Police_i = number of police officers per 1,000 residents in MSA i 
 
              The results from the estimation are as follows: 
 
              Coefficients: 
                                                       Estimate          Std. Error  
              (Intercept)                    12.322               2.566    
              MedInc                           -1.455               0.045   
              Pov                                  0.745                 0.552  
              UnEmp                           0.983                 0.794  
              AvAge                             -1.566               0.567  
              Police                              0.549                 0.223   
              --- 
              R-squared:  0.6402,   Adjusted R-squared:  0.5922  
              F-statistic: 78.98 on 5 and 846 DF 

(6)        a. Test the overall significance of the model at the 5% significance level. 
 
 Ho: 𝛽1= 𝛽2= 𝛽3…..= 𝛽5=0
H1: At least one 𝛽≠0                         
F-statistic: 78.98   numerator(m)=5, denominator=846             and  846= DF
Reject Ho if F-calculated > F-tabulated
F-tabulated(alpha=0.05,5,846) =2.2246
F-calculated=78.98   78.98>2.382
Reject the null hypothesis and conclude that at least one of the  𝛽 is not equal to zero.

 

 

^(6)        ^b. ^Provide ^an ^interpretation ^of ^the ^coefficient ^on ^AvAgei^. 

The coefficient of years of ^AvAgei^.=(-1.566),s.e=0.567.It is negative.
An increase in average age in MSA i  by one year, the ^violent ^crime ^rate ^(crimes ^per ^1,000 ^people) ^in ^MSA ^i I decreases by  1.566 holding median income , poverty rate in MSA , unemployment rate and number of police officers per 1,000 residents  constant.
(6)        c. It is believed that the impact of median income on the violent crime rate is negative. Test 
_this _hypothesis _at _the _5% _level. 
Alpha=0.05  d.f==846           t(0.05,)=1.647
Ho: 𝛽 1 =0                           
H1: 𝛽 1>0
Decision rule; Reject   Ho if |t-calculated|>t-critical    t calculated = 𝛽1/SE(𝛽1)=(-1.455)/ 0.045 )=32.33

t-critical=1.647
t-calculated= -2.814
32.33>1.647, we reject the null hypothesis because the t-calculated is greater than the  t-critical. The coefficient of median income is greater than 0. And sign of the co-efficient is positive

 

 

 
(4)        d. The R^2 of the model estimated is 0.6402. What does this number tell us? 
         R^2    Is a measure of goodness-fit of linear regression .It shows the total variation for the dependent variable that could be explained by the independent variables.
R^2 =0.6402 =62%. The model explains 62%  of the fitted data in the regression model.

 
(6)       e. Both the poverty rate and the unemployment rate are expected to be positively related to ^the ^violent ^crime ^rates. ^Unfortunately, ^the ^t-statistic ^for ^the ^coefficient ^on ^Povi ^is ^1.35 ^and ^the ^t-statistic ^for ^the ^coefficient ^on ^Unempi ^is ^1.24. ^Do ^you ^have ^any ^suggestions ^of ^why ^these ^two coefficients may not be significantly different from zero? 

 There might be heteroskedasticity, multicollinearity .

 
 
(6)        f. Name three ways you could check your answer for (b). 
 
Durbin-Watson test  
Breusch-Pagan test 
Newey-west

 

(6)        g. How would you go about calculating a variance inflation factor for Unemp_i? Be specific 
about the steps involved. Also, how big does the VIF have to be before we would get 
_concerned? 
Conduct multiple  regression with Unemp_i   as the dependent variable and median income , poverty rate in MSA , unemployment rate and ^violent ^crime ^rate ^(crimes ^per ^1,000 ^people) ^as ^the ^independent ^variables.
We consider the Adjusted R-squared
Calculate the VIF f Unemp_i
The variance inflation factors of Unemp_i =1/(1-Adjusted R-squared)
Obtain the VIF, If VIF<5, we don’t have a problem
it suggests we possibly dropping one of the variables out of the model   or finding some way to consolidate them to capture their joint effect.

_2. _Let’s _talk _about _correlation.  

(4)        a. What is the range of possible values that a correlation coefficient can take on? 
 
 The correlation coefficient can take values range between +1 and -1

 
 
(6)        b. Below are two scatterplots showing the relationship between variables x and y. For each, 
provide an estimate of the correlation coefficient. Explain why you chose these values. 

The first scatter plot estimate to be -0.98 which shows a strong negative correlation 
The first scatter plot estimate to ba slope of almost Zeroe -0.18 close to zero which shows a weak negative correlation and the line will have 
3. The following model was estimated using OLS and a time series data set with 86 
              _observations: 

              _where: 

^RConst ^= ^β0 ^+ ^β1 ^RGDPt ^+ ^εt
         ^RConst ^= ^Consumption ^expenditures ^(in ^constant ^dollars) ^in ^year ^t 
    
          RGDP_t = Real GDP in. year t 
 
              The output from the regression model is as follows: 
 
              Coefficients: 
                                         Estimate          Std. Error  
              (Intercept)      -0.01879          32.03 
              RGDP                0.6752              0.004082 
              --- 
              R-squared:  0.9969,   Adjusted R-squared:  0.9969  
              F-statistic: 2.735e+04 on 1 and 84 DF 
 ^(6)        ^a. ^Test ^whether ^the ^coefficient ^on ^RGDPt ^is ^positive ^at ^the ^5% ^level ^of ^significance. 
 Alpha=0.05 N=50 d.f==84
t(0.05,)=1.6787
t calculated = 𝛽1/SE(𝛽1)=(0.6752/0.004082)=165.40
Decision rule; Reject   Ho if |t-calculated|>t-critical and the sign positive
t-critical= 1.66
Ho:  𝛽 1 =0                           H1: 𝛽 2>0
165.40>1.66. We reject the null hypothesis because the t-calculated is greater than the  t-critical and that the coefficient of ^RGDPis positive. It is greater than Zero.

 

 ^(6)        ^b. ^Provide ^an ^interpretation ^of ^the ^coefficient ^on ^RGDPt^. 
 The coefficient of years of ^RGDPt^.=( 0.6752),s.e=0.004082.It is positive.
An increase in average age in Real GDP in. year by one unit, the ^Consumption ^expenditures increases by 67 cents.

 

 
(8)        c. A Durbin-Watson test was performed for this model. The results are: 
 
              Durbin-Watson test 
              data:  model1 
              DW = 0.12803 
 
              Test this model for positive serial correlation at the 5% level of significance. 
 N= 86    k=1   df =N-1-1=84 Alpha=0.05
  Ho: ρ =0
H1: ρ >0
alternative hypothesis: true autocorrelation is greater than 0
Decision rule; If Durbin-Watson (d)<dl  we reject Ho , If d>du do not reject Ho
From the table we get dl=(1.481+1.496)/2=1.4885
                                          Du=(1.529+1.541)/2= 1.535
0.12803<1.4885 we reject Ho and  conclude that there is a positive correlation

 

(6)        d. Explain what it means for a regression to have positive serial correlation. 

                      A positive serial correlation means that the values are likely to change in future time periods in the same way, or direction, that they have in recent past time periods.           

              4. Economists are interested in determining the sales price of used tractors sold at an auction. 
              The following model was estimated using OLS and a cross section data set with 276 
              observations: 
 
_Pricei _= _β0 _+ _β1 _Horsepoweri _– _β2 _Agei _– _β3 _EngHoursi _+ _β4 _Cabi _+ _β5 _JohnDeerei _+ _εi 

              where: 
              ^Pricei ^= ^price ^paid ^for ^tractor ^i ^(in ^dollars) 
              Horsepower_i = horsepower of tractor i 
              Age_i = the number of years since tractor i was manufactured 
              EngHours_i = number of hours of use recorded on tractor i 
              Cab_i = dummy variable that = 1 if tractor i has an enclosed cab; = 0 otherwise 
              JohnDeere_i = dummy variable that =1 if tractor i is a John Deere; = 0 otherwise 
 
              The results from the estimation are as follows: 
 
              Coefficients: 
                                                       Estimate          Std. Error  
              (Intercept)                    11363.5419     2308.3146    
              Horsepower                 223.0509         14.8812   
              Age                                  -790.4896        126.7347   
              EngHours                        -1.9608             0.3926   
              Cab                                  8089.1044       2483.6226  
              JohnDeere                     13731.7988    2899.2501   
              --- 
              R-squared:  0.6512,   Adjusted R-squared:  0.6447  
              F-statistic: 100.8 on 5 and 270 DF 
 
(6)        a. Provide an interpretation of the coefficient on JohnDeere_i.  
 
 The coefficient of years of JohnDeere =(13731.7988),s.e=2899.2501.It is positive.
An increase in average age in  JohnDeere by one unit, the ^price ^paid ^for ^tractor ^i ^(in ^dollars) increases by  13731 dollars,80 cents horsepower of tractor I, Age i ,number of hours of use recorded on tractor I, dummy variable that = 1 if tractor i has an enclosed cab; = 0 otherwise constant.
(6)        b. To test for heteroskedasticity, a White test was performed. Here are the results of the 

        estimation: 

                             

| | | Estimate | | Std. Error
| (Intercept) | | 5.726e+07 | | 1.795e+08
| Horsepower | | 8.909e+05 | | 2.843e+06
| Age | | -7.604e+06 | | 9.317e+06
| EngHours | | -9.565e+04 | | 5.610e+04
| Cab | | 1.362e+08 | | 2.322e+08

              Coefficients: 

| JohnDeere | | 6.678e+08 | | 2.751e+08
| Horsepower2 | | -2.788e-02 | | 1.922e-02
              Age2                                            3.877e-02                     2.196e-02                                                              ```
              EngHours2                                 2.876e-02                      1.786e-02 
              Cab2                                              -3.227e-02                    2.665e-02 
              JohnDeere2                                -5.337e-02                    3.652e-02          
              Horsepower:Age                      -1.741e+04                   7.740e+04    
              Horsepower:EngHours          -4.384e+02                   1.777e+02  
              Horsepower:Cab                       5.984e+06                     2.743e+06   
              Horsepower:JohnDeere         3.346e+06                     1.551e+06   
              Age:EngEours                            8.212e+03                     1.785e+03  
              Age:Cab                                       -4.132e+07                   1.298e+07  
              Age:JohnDeere                          -2.237e+07                   1.467e+07  
              EngHours:Cab                            -6.960e+03                  4.216e+04   
              EngHours:JohnDeere              -1.430e+05                   4.509e+04  
              Cab:JohnDeere                          7.921e+08                     3.040e+08   
              --- 
              R-squared:  0.5113,   Adjusted R-squared:  0.4831  
              F-statistic: 18.14 on 20 and 255 DF 
 
              Calculate the White test statistic and test for heteroskedasticity at the 5% level of 
              significance. 
H0: Ho: α1= α2= α3………………… α9=0(Homoskedasticity)
H1: Heteroskedasticity
Decision rule: Reject Ho χ2-calculated > χ2 critical        df=N-k-1=N-20-1,N=df+20+1                           
χ2-calculated =N* Multiple R-squared                          N=255+20+1=276
                      =276*0.5113=140.12
χ2 critical (  χ2,d.f=20  α=0.05)=31.41,    df 18=28.86 which are lower
140.12>31.41 we reject the null hypothesis and conclude that there is heteroskedasticity

                  
^(6)        ^c. ^The ^researcher ^is ^considering ^adding ^another ^independent ^variable ^to ^the ^analysis: ^Dieseli^, 
              which is a dummy variable =1 if tractor i runs on diesel fuel and =0 otherwise. What variables 
              will have to be added to the White regression testing for heteroskedasticity? (List them all.)
Horsepower
Age
JohnDeere
^Diesel
Horsepower2
Age2
EngHours2                                 
Cab2
JohnDeere2
^Diesel2
Horsepower:Age                      
Horsepower:EngHours          
Horsepower:Cab                       
Horsepower:JohnDeere         
Age:EngEours                            
Age:Cab     
Age:JohnDeere                          
EngHours:Cab                            
EngHours:JohnDeere  
^Diesel:Age    
Diesel: EngHours
Diesel: Cab
Diesel: JohnDeere
Cab:JohnDeere                                                            

 

                                        
(6)        d. Explain why it is important to test for heteroskedasticity. If a researcher fails to correct for 
              heteroskedasticity, what may happen? 

A researcher should correct for heteroskedasticity, corrected standard error so to obtain a less biased.
Presence of residual variance in the model shows that the scattering of the model is dependent on at least one independent one independent variable.

Consequences of heteroskedasticity is that the estimators and the regression predictions based on them remains unbiased and constant.
The estimators are no longer best linear unbiased estimator because they are no longer efficient so the predictions would result to insignificant.
Because of the inconsistency of covariance matrix of the estimated regression coefficients, the tests hypotheses are no longer valid.