#### Description of this paper

##### Stats Final Exam Part 2

Description

solution.

Question

Stats Final Exam Part 2

Description

solution

Question

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

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

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

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

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?

Paper#61787 | Written in 10-Dec-2015

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