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Use R software to show parallel results using both the logistic and Gompertz functions ('SSlogis' and 'SSgompertz') , use the self-starting and report: the data

Use R software

to show parallel results using both the logistic and Gompertz functions('SSlogis' and 'SSgompertz'), use the self-starting and report: the data and the two fitted lines all on the same plot for comparisonreasons , the confidence intervals of parameters and tests indicating if any of them is zero, residual plots and diagnostics, methods to identify which of the twomodelsis bettere.g. based on RMSE

#Years

year<-c(1800,1810,1820,1830,1840,1850,1860,1870,1880

,1890,1900,1910,1920,1930,1940,1950,1960,1970

,1980,1990,2000,2010,2020)

#The population

pop<-c(3800,4800,7500,28000,212300,397700,749100,

1200000,1600000,2100000,2400000,2810173,

3668412,4842325,5256106,6371766,7823194,

8875083,9262078,9295297,9938444,9883640,10077331)

#Adding a column x with ranges from 0 to 220

#1810 - 1800 = 10 This means there is a 10 years interval between 1800 and 1810.

#Similarly 2020 - 1800 = 220. This means that there is an interval of 220 years between 1800 and 2020.

use this column for 'SSgompertz'

> data

x

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

180

190

200

210

220

i want to combine those functions and compare which one is better

hint:

# "SSgompertz"

?SSgompertz

#Asym, b2, and b3 are names of objects the gradient matrix with respect to these names is attached as an attribute named gradient.

#Adding a column x with ranges from 0 to 220

#1810 - 1800 = 10 This means there is a 10 years interval between 1800 and 1810.

#Similarly 2020 - 1800 = 220. This means that there is an interval of 220 years between 1800 and 2020.

data<- read.csv(file.choose())

SSgompertz(seq(10,20,by=2)/100, Asym=2, b2=5, b3=1.3)

with(list(Asymptote=2, b2=5, b3=1.3),SSgompertz(seq(10,20,by=2)/100, Asymptote, b2, b3))

output<-nls(Pop ~ SSgompertz(x, Asymptote, b2, b3),data=data.frame(x=(11:20), Pop=sort(log(11:20)+1)))

plot(pop ~ x, data = data)

curve(predict(output, newdata = data.frame(x = x)), add = TRUE)

hint:

SSlogis

model2 <- nls(pop~SSlogis(year,a,b,c))

new.data <- data.frame(year = seq(min(year),max(year),len = 100))

plot(year,pop, col= "Red", main = "Logistic Graph", xlab = "Year", ylab = "Population")

lines(new.data$year,predict(model2,newdata = new.data), col= "Blue")

summary(model2

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