STAT 360: Computational Statistics and Data Analysis
Load R Libraries, Import and Attach Relevant Data, and Specify Seed
library(rmarkdown); library(knitr); library(readxl)
set.seed(37)EXERCISE 01
Part (a)
poundToNewton <- function(pounds)
{
newt <- pounds*4.44822
return(newt)
}Part (b)
poundToNewton(70.59)## [1] 313.9998poundToNewton(35.75)## [1] 159.0239EXERCISE 02
MooseWolvesData <- read.csv("C:/Users/Sarah Chock/OneDrive - University of St. Thomas/Senior Year/STAT 360 Comp Stat and Data Analysis/Exploratory Data Analysis/Wolfies.csv")Part (b)
nums <- which(sapply(MooseWolvesData, is.numeric))
numwolves <- as.matrix(MooseWolvesData[,nums])
numwolves## ï..year wolves moose Apr.May..temp..F. Jan.Feb..temp..F.
## [1,] 1959 20 538 43.90 1.40
## [2,] 1960 22 564 43.40 8.45
## [3,] 1961 22 572 41.40 9.75
## [4,] 1962 23 579 42.60 2.15
## [5,] 1963 20 596 43.50 -0.35
## [6,] 1964 26 620 45.75 12.40
## [7,] 1965 28 634 43.65 1.25
## [8,] 1966 26 661 39.20 1.70
## [9,] 1967 22 766 39.90 2.75
## [10,] 1968 22 848 43.00 5.85
## [11,] 1969 17 1041 44.95 7.80
## [12,] 1970 18 1045 41.15 3.35
## [13,] 1971 20 1183 42.45 3.20
## [14,] 1972 23 1243 44.10 -0.05
## [15,] 1973 24 1215 42.70 10.85
## [16,] 1974 31 1203 41.55 5.65
## [17,] 1975 41 1139 42.80 9.05
## [18,] 1976 44 1070 45.95 10.45
## [19,] 1977 34 949 50.55 4.70
## [20,] 1978 40 845 44.90 3.25
## [21,] 1979 43 857 39.50 -1.15
## [22,] 1980 50 788 47.95 7.15
## [23,] 1981 30 767 44.50 11.70
## [24,] 1982 14 780 43.75 0.40
## [25,] 1983 23 830 41.10 14.45
## [26,] 1984 24 927 44.85 12.50
## [27,] 1985 22 976 47.05 5.45
## [28,] 1986 20 1014 47.45 9.80
## [29,] 1987 16 1046 49.20 17.95
## [30,] 1988 12 1116 47.30 3.10
## [31,] 1989 12 1260 43.35 6.60
## [32,] 1990 15 1315 43.15 13.05
## [33,] 1991 12 1496 48.35 9.10
## [34,] 1992 12 1697 44.15 15.20
## [35,] 1993 13 1784 42.60 9.10
## [36,] 1994 17 2017 44.60 -0.40
## [37,] 1995 16 2117 41.65 9.50
## [38,] 1996 22 2398 40.30 2.80
## [39,] 1997 24 900 40.70 8.35
## [40,] 1998 14 925 49.40 20.75
## [41,] 1999 25 997 47.35 12.40
## [42,] 2000 29 1031 45.10 12.60
## [43,] 2001 19 1120 46.00 10.05
## [44,] 2002 17 1100 41.00 16.65
## [45,] 2003 19 900 44.20 5.80
## [46,] 2004 29 750 42.05 7.60
## [47,] 2005 30 540 45.85 10.15
## [48,] 2006 30 450 48.80 15.20
## [49,] 2007 21 385 45.40 7.95
## [50,] 2008 23 650 41.55 7.75
## [51,] 2009 24 530 43.55 6.00
## [52,] 2010 19 510 49.20 10.65
## [53,] 2011 16 515 43.85 6.85
## [54,] 2012 9 750 46.50 17.15
## [55,] 2013 8 975 39.80 8.95
## [56,] 2014 9 1050 42.05 -1.05
## [57,] 2015 3 1250 44.65 3.85
## [58,] 2016 2 1300 44.20 11.95
## [59,] 2017 2 1600 43.85 15.40
## [60,] 2018 2 1475 42.70 6.05
## [61,] 2019 15 2060 41.65 3.50
## ice.bridges..0.none..1...present.
## [1,] 0
## [2,] 0
## [3,] 1
## [4,] 1
## [5,] 1
## [6,] 0
## [7,] 1
## [8,] 1
## [9,] 1
## [10,] 1
## [11,] 1
## [12,] 1
## [13,] 1
## [14,] 1
## [15,] 0
## [16,] 1
## [17,] 0
## [18,] 0
## [19,] 1
## [20,] 0
## [21,] 1
## [22,] 0
## [23,] 0
## [24,] 1
## [25,] 0
## [26,] 0
## [27,] 1
## [28,] 0
## [29,] 0
## [30,] 1
## [31,] 0
## [32,] 0
## [33,] 1
## [34,] 0
## [35,] 0
## [36,] 1
## [37,] 0
## [38,] 1
## [39,] 1
## [40,] 0
## [41,] 0
## [42,] 0
## [43,] 0
## [44,] 0
## [45,] 0
## [46,] 0
## [47,] 0
## [48,] 0
## [49,] 0
## [50,] 1
## [51,] 0
## [52,] 0
## [53,] 0
## [54,] 0
## [55,] 0
## [56,] 1
## [57,] 1
## [58,] 0
## [59,] 0
## [60,] 1
## [61,] 1Part (c)
sumsquares <- function(param = c(0,0,0))
{
pred <- param[1] + param[2]*numwolves[,2] + param[3]*numwolves[,4]
res <- numwolves[,3] - pred
resss <- (res)^2
return(sum(resss))
}Part (d)
sumsquares(c(2000,-10,-20))## [1] 10650919sumsquares(c(3000,-15,-25))## [1] 29011491Part (e)
The first estimation with (2000,-10,-20) is better because it had a lower residual sum of squares. This means that the actual points tend to fall closer to the prediction line.Part (f)
optim(par = c(0,0,0), fn = sumsquares, method = "Nelder-Mead")## $par
## [1] 2383.65262 -14.44774 -24.02888
##
## $value
## [1] 9749433
##
## $counts
## function gradient
## 370 NA
##
## $convergence
## [1] 0
##
## $message
## NULLPart (g)
From this model, the number of wolves and average temperature both have a negative impact on the number of moose. For every additional wolf, mooses go down by 14ish, and for every additional degree in temp, mooses go down by 24ish. The mooses just want to live alone in the arctic.library(cats)
here_kitty()
## meow