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PCI_Test<-read_excel("PCI_Test.xlsx")

Using the database of ArcGIS maps and their attribute tables, I have cross-referenced the map of Pavement Condition Index (PCI) scores with the Infrastructure Management Program (IMP) projects to create a filtered dataset displaying streets that were NOT scheduled for a maintenance project.

Projects=PCI_Test
stat.desc(Projects$PCI)
##      nbr.val     nbr.null       nbr.na          min          max        range 
## 2.734000e+03 6.500000e+01 0.000000e+00 0.000000e+00 9.949000e+01 9.949000e+01 
##          sum       median         mean      SE.mean CI.mean.0.95          var 
## 1.549055e+05 5.668500e+01 5.665892e+01 4.993701e-01 9.791810e-01 6.817788e+02 
##      std.dev     coef.var 
## 2.611089e+01 4.608435e-01

According to this, median and mean PCI for the “left-out” projects is relatively-low at 56.68 and 56.65, respectively, meaning the median and mean values would receive a “D” rating from A to F in the grading system.

hist(Projects$PCI)

ks.test(Projects$PCI,pnorm)
## Warning in ks.test.default(Projects$PCI, pnorm): ties should not be present for
## the one-sample Kolmogorov-Smirnov test
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##  Asymptotic one-sample Kolmogorov-Smirnov test
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## data:  Projects$PCI
## D = 0.97416, p-value < 2.2e-16
## alternative hypothesis: two-sided

The previous histogram did not give a clear enough picture for me to understand yet, so I will break it down in more detail.

hist(Projects$PCI,breaks=50)

hist(Projects$PCI,breaks=50,probability = T)
lines(density(Projects$PCI),col="red",lwd=3)

I will use the log function to see its effect.

LogPCI=log(Projects$PCI)

hist(LogPCI,breaks=50,probability = T)
lines(density(LogPCI),col="red",lwd=3)

ks.test(LogPCI,pnorm)
## Warning in ks.test.default(LogPCI, pnorm): ties should not be present for the
## one-sample Kolmogorov-Smirnov test
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##  Asymptotic one-sample Kolmogorov-Smirnov test
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## data:  LogPCI
## D = 0.94673, p-value < 2.2e-16
## alternative hypothesis: two-sided
shapiro.test(Projects$PCI)
## 
##  Shapiro-Wilk normality test
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## data:  Projects$PCI
## W = 0.96697, p-value < 2.2e-16
shapiro.test(LogPCI)
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##  Shapiro-Wilk normality test
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## data:  LogPCI
## W = NaN, p-value = NA