Stats Final Exam Complete Solution Part 2 Complete Solution

Question: #7726

Stats Final Exam Part 2

Given the data set called County Demographic Information, construct a predictive model for the variable “Total Serious Crime” using some or all of the other variables in the set of data.
The model should be mathematically valid, accurate and reliable.
Total Serious Crime is Variable #8
Other Variables:
#2   Land Area
#3   Total Population
#4   Percent of Population aged 18-34
#5   Percent of Population 65 or over
#6   Number of Active Physicians
#7   Number of Hospital Beds
#9   Percent of High School Graduates
#10   Percent of Population with College Degrees
#11   Percent of Population below poverty level
#12   Unemployment Percent
#13   Per Capita Income
#14   Total Personal Income
#15   Geographic Region
Note: I am omitting the data set to simplify this problem; the following analyses use the data set described above, and you can assume the math is calculated correctly. I am testing to see if you can identify what analytical techniques may be validly employed and how effective are they building a model.
Variables 2 to 14 are numeric variables and variable 15 is categoric.
Analysis #1
In the given data set, we were asked to determine if an accurate predictive model for Variable #8, Serious Crime could be found using the attached data.
   Since Variable 15 was determined to be categoric, regression was not appropriate to use; so I used Analysis of Variance (ANOVA) to examine if there was a significant relationship between Variable 8 and 15. The results (using Systat 13.0) are printed above.

Variables   Levels
VAR(15) (4 levels)   1.000   2.000   3.000   4.000

Dependent Variable   VAR(8)
N   440
Multiple R   0.110
Squared Multiple R   0.012

Estimates of Effects B = (X'X)-1X'Y
Factor   Level   VAR(8)
CONSTANT       28,017.368
VAR(15)   1   -4,931.339
VAR(15)   2   -6,236.627
VAR(15)   3   -1,026.394

Analysis of Variance
Source   Type III SS   df   Mean Squares   F-Ratio   p-Value
VAR(15)   1.795E+010   3   5.985E+009   1.774   0.151
Error   1.471E+012   436   3.374E+009

ANOVA results suggest that Variable 15 is significantly related to Variable 8, but Variable 15 can only explain approximately 15.1% of the variation in Variable 8.
Therefore, I conclude that variable 15 is significantly related to variable 8 although variable 15 is only a minor factor in predicting variable 8.

Analysis #2
Using Systat, I employed Multiple Linear Regression to attempt to create a predictive model, using all of the available variables as independent variables.
The results are shown below.

Dependent Variable   VAR(8)
N   440
Multiple R   0.919
Squared Multiple R   0.844
Adjusted Squared Multiple R   0.839
Standard Error of Estimate   23,367.069

Regression Coefficients B = (X'X)-1X'Y
Effect   Coefficient   Standard Error   Std.
Coefficient   Tolerance   t   p-Value
CONSTANT   -50,925.731   35,344.226   0.000   .   -1.441   0.150
VAR(2)   -3.054   0.849   -0.081   0.719   -3.599   0.000
VAR(3)   0.234   0.020   2.422   0.008   11.560   0.000
VAR(4)   221.063   424.685   0.016   0.393   0.521   0.603
VAR(5)   32.120   380.640   0.002   0.539   0.084   0.933
VAR(6)   -5.189   3.150   -0.159   0.039   -1.647   0.100
VAR(7)   3.404   2.280   0.134   0.046   1.493   0.136
VAR(9)   -265.566   321.799   -0.032   0.244   -0.825   0.410
VAR(10)   140.915   373.505   0.019   0.152   0.377   0.706
VAR(11)   1,142.711   488.132   0.091   0.241   2.341   0.020
VAR(12)   -159.661   658.025   -0.006   0.526   -0.243   0.808
VAR(13)   2.335   0.699   0.163   0.154   3.339   0.001
VAR(14)   -7.070   0.946   -1.564   0.008   -7.475   0.000
VAR(15)   1,456.610   1,319.387   0.026   0.668   1.104   0.270

Analysis of Variance
Source   SS   df   Mean Squares   F-Ratio   p-Value
Regression   1.256E+012   13   9.664E+010   176.989   0.000
Residual   2.326E+011   426   5.460E+008

Since the combined model had a p-value of 0.000, I concluded that this model could accurately predict variable 8, Total Serious Crime. The R-Squared value of approximately .84 suggests that the model explains about 84% of the variation in Serious Crime. Therefore, I conclude that this is a fairly accurate, valid, predictive model of Total Serious Crime.

Analysis #3
Since many individual, independent variables of the previous regression model had p-values above .05, they were not significant factors. I discarded them, redid the regression analysis, and got the results listed below.

Dependent Variable   VAR(8)
N   440
Multiple R   0.918
Squared Multiple R   0.842
Adjusted Squared Multiple R   0.840
Standard Error of Estimate   23,274.901

Regression Coefficients B = (X'X)-1X'Y
Effect   Coefficient   Standard Error   Std.
Coefficient   Tolerance   t   p-Value
CONSTANT   -63,890.789   10,233.100   0.000   .   -6.244   0.000
VAR(2)   -3.109   0.758   -0.083   0.894   -4.101   0.000
VAR(3)   0.250   0.016   2.580   0.013   15.282   0.000
VAR(11)   1,449.915   307.144   0.116   0.603   4.721   0.000
VAR(13)   2.460   0.469   0.171   0.341   5.250   0.000
VAR(14)   -7.899   0.787   -1.748   0.012   -10.037   0.000

Analysis of Variance
Source   SS   df   Mean Squares   F-Ratio   p-Value
Regression   1.254E+012   5   2.508E+011   462.898   0.000
Residual   2.351E+011   434   5.417E+008

This model is a better predictive model than analysis #2 since it has a higher F-value, and therefore a smaller p-value. Also, each factor of the model has a p-value smaller than .05; this indicates that each component is significant in itself. The R-Squared value of .84 indicates that I can predict Variable 8 with approximately 84% accuracy, using only five variables and a constant.

Analysis #4
Repeating the previous analysis, but deleting the constant allowed me to raise the R-Squared value to almost .87.
Dependent Variable   VAR(8)
N   440
Multiple R   0.932
Squared Multiple R   0.869
Adjusted Squared Multiple R   0.868
Standard Error of Estimate   23,381.775

Regression Coefficients B = (X'X)-1X'Y
Effect   Coefficient   Standard Error   Std.
Coefficient   Tolerance   t   p-Value
VAR(2)   -3.010   0.763   -0.088   0.612   -3.942   0.000
VAR(3)   0.245   0.016   2.739   0.009   15.107   0.000
VAR(9)   -697.218   118.026   -0.846   0.015   -5.907   0.000
VAR(10)   496.913   209.212   0.174   0.056   2.375   0.018
VAR(11)   683.363   248.743   0.105   0.206   2.747   0.006
VAR(13)   1.727   0.472   0.511   0.015   3.657   0.000
VAR(14)   -7.658   0.780   -1.800   0.009   -9.818   0.000

Analysis of Variance
Source   SS   df   Mean Squares   F-Ratio   p-Value
Regression   1.576E+012   7   2.251E+011   411.714   0.000
Residual   2.367E+011   433   5.467E+008

Using seven variables and no constant, I found a model that had each component with a low p-value (under .05) and an overall p-value of 0.000. I would conclude similar to what I did in analysis #3, but I would prefer this model because of its higher R-Squared value.

Analysis #5
Trying to optimize the model, I repeated the earlier analytical methods. I discarded the constant and tried to lower the number of variables. I was able to find a model (see results listed below, and compare to analyses #3 and #4 ) that used only four variables. Each variable had a p-value under .05, the F-value was higher than earlier models (therefore, the overall p-value was lower for the overall model) and the R-Squared value was still approximately .84.

Dependent Variable   VAR(8)
N   440
Multiple R   0.916
Squared Multiple R   0.840
Adjusted Squared Multiple R   0.839
Standard Error of Estimate   25,805.795

Regression Coefficients B = (X'X)-1X'Y
Effect   Coefficient   Standard Error   Std.
Coefficient   Tolerance   t   p-Value
VAR(2)   -2.141   0.814   -0.062   0.656   -2.629   0.009
VAR(3)   0.088   0.002   0.979   0.644   41.013   0.000
VAR(11)   1,240.562   217.578   0.191   0.327   5.702   0.000
VAR(13)   -0.846   0.116   -0.251   0.314   -7.328   0.000

Analysis of Variance
Source   SS   df   Mean Squares   F-Ratio   p-Value
Regression   1.522E+012   4   3.805E+011   571.367   0.000
Residual   2.903E+011   436   6.659E+008

Therefore, I concluded that Model #5 was the preferred model since it only had four input variables and achieved approximately the same predictive accuracy. Thus I needed only four independent variables to predict variable #8 with accuracy of approximately 84%.

A)   Are each of the five analyses valid? (if not, why not?)
B)   Are each of the five analyses significant? (why?)
C)   Are each of the five analyses accurate? (why?)
D)   Which is the best predictive model and why?

Solution: #7736

Stats Final Exam Complete Solution Part 2 Complete Solution

Yes, all the five analyses are accurate, since all the analyses...