In agracy2246/MATH4753grac0009: Adam Gracy MATH4753

knitr::opts_chunk$set(echo = TRUE)

Task 1

getwd()

Task 2

d=c(5.0581, 4.9707, 5.0893, 4.9334, 4.9777, 5.0285, 4.8555, 4.9565, 
4.9769, 4.9722, 4.999, 4.9925, 4.9686, 5.0662, 4.9239, 4.9781, 
5.0485, 5.0014, 4.9957, 5.0195, 5.0118, 4.9928, 5.0361, 5.0185, 
4.9879)

mp = c(-1,1)

# for 95% ci
ci95= mean(d) + mp*qt(0.975,24)*sd(d)/sqrt(25) 
# for 90% ci
ci90= mean(d) + mp*qt(0.95,24)*sd(d)/sqrt(25) 
# for 80% ci
ci80= mean(d) + mp*qt(0.9,24)*sd(d)/sqrt(25) 
# for 50% ci
ci50= mean(d) + mp*qt(0.75,24)*sd(d)/sqrt(25) 

95% ci using R as calculator

ci95

90% ci using R as calculator

ci90

80% ci using R as calculator

ci80

50% ci using R as calculator

ci50

using t.test()

object <- t.test(d, conf.level = .80)
object$conf.int

chisq

#95
xsq1_95=qchisq(0.975,29) #1-a/2
xsq2_95=qchisq(0.025,29) #a/2
#90
xsq1_90=qchisq(0.95,29) #1-a/2
xsq2_90=qchisq(0.05,29) #a/2
#80
xsq1_80=qchisq(0.9,29) #1-a/2
xsq2_80=qchisq(0.1,29) #a/2
#50
xsq1_50=qchisq(0.75,29) #1-a/2
xsq2_50=qchisq(0.25,29) #a/2

95% ci for $\sigma^2$

ci=c()
ci[1]=29*var(d)/xsq1_95
ci[2]=29*var(d)/xsq2_95
ci

90% ci for $\sigma^2$

ci=c()
ci[1]=29*var(d)/xsq1_90
ci[2]=29*var(d)/xsq2_90
ci

80% ci for $\sigma^2$

ci=c()
ci[1]=29*var(d)/xsq1_80
ci[2]=29*var(d)/xsq2_80
ci

50% ci for $\sigma^2$

ci=c()
ci[1]=29*var(d)/xsq1_50
ci[2]=29*var(d)/xsq2_50
ci

Task 3

# n = 20
blue=c(21.65, 17.48, 20.1, 21.57, 14.82, 19.17, 21.08, 18.23, 22.93, 
15.66, 20.89, 21.66, 18.5, 20.59, 18.63, 18.91, 19.53, 17.7, 
16.5, 19.03)
# n = 15
snapper=c(31.65, 27.48, 30.1, 31.57, 24.82, 29.17, 31.08, 28.23, 32.93, 
25.66, 30.89, 31.66, 28.5, 30.59, 28.63)

95% ci using R calc

n1=length(snapper)
n2=length(blue)
spsq=((n1-1)*var(snapper)+(n2-1)*var(blue))/(n1+n2-2)
t=qt(0.975,n1+n2-2)
ci=c()
ci[1]=mean(snapper)-mean(blue)-t*sqrt(spsq*(1/n1+1/n2)) 
ci[2]=mean(snapper)-mean(blue)+t*sqrt(spsq*(1/n1+1/n2))
ci

The mean difference will fall into the above interval with 95% confidence

95% $\mu_{snapper} - \mu_{blue}$

t.test(snapper,blue,conf.level=0.95,var.equal=TRUE)$conf.int

90% $\mu_{snapper} - \mu_{blue}$

t.test(snapper,blue,conf.level=0.90,var.equal=TRUE)$conf.int

80% $\mu_{snapper} - \mu_{blue}$

t.test(snapper,blue,conf.level=0.80,var.equal=TRUE)$conf.int

50% $\mu_{snapper} - \mu_{blue}$

t.test(snapper,blue,conf.level=0.50,var.equal=TRUE)$conf.int

As the confidence level decreases, so does the interval

Task 4

#Paired exams
Exam1=c(40.98, 59.36, 46.69, 41.8, 61.63, 65.31, 62.96, 60.21, 56.89, 
78.41, 53.44, 75.2, 60.54, 52.43, 41.41, 70.79, 73.55, 55.65, 
61.43, 63.84, 58.07, 53.79, 54.45, 67.18, 44.46)

Exam2=c(50.22, 66.19, 58.75, 51.88, 66.61, 70.86, 74.25, 70.23, 69.55, 
87.18, 63.62, 81.7, 70.5, 66.02, 51.35, 80.92, 85.65, 65.44, 
74.37, 75.28, 67.86, 59.92, 64.42, 73.57, 57.15)

95% ci for $\mu_d = \mu_1-\mu_2$

alpha = .05
n = length(Exam1)
dbar = mean(Exam1 - Exam2)
t = qt(1- alpha / 2, n -1)
sd = sd(Exam1-Exam2)
mp = c(-1,1)
diff = dbar + mp * t *sd / sqrt(n)
diff

The difference in means from Exam1 - Exam2 falls into the interval with 95% confidence

use t.test() for 90% ci for $u_d$

t.test(Exam1, Exam2, paired = T, conf.level = .90)$conf.int

use t.test() for 80% ci for $u_d$

t.test(Exam1, Exam2, paired = T, conf.level = .80)$conf.int

use t.test() for 70% ci for $u_d$

t.test(Exam1, Exam2, paired = T, conf.level = .70)$conf.int

use t.test() for 60% ci for $u_d$

t.test(Exam1, Exam2, paired = T, conf.level = .60)$conf.int

use t.test() for 10% ci for $u_d$

t.test(Exam1, Exam2, paired = T, conf.level = .10)$conf.int

Task 5

birds <- read.csv("NZBIRDS.csv")

addmargins(with(birds, table(Extinct, Flight)))

The data agrees with the given table

Use R as a calculator for 95% ci

n1 = 38 # total not extinct
n2 = 78 #total extinct

p1hat = 21 / n1 #nonflying and NOT extinct / total not extinct
q1hat = 1 - p1hat
p2hat = 7 / n2
q2hat = 1 - p2hat

mp = c(-1,1)
ci <- (p1hat - p2hat) + mp * qnorm(1-.05/2,0,1)*sqrt((p1hat*q1hat/n1) + (p2hat*q2hat/n2))
ci

Task 6

set.seed(35); sam1=rnorm(25,mean=10, sd=5)
set.seed(45); sam2=rnorm(34, mean=40, sd=8)

95% CI for ratio

a = .05
s1 = sd(sam1)
nu1 = length(sam1) - 1
s2 = sd(sam2)
nu2 = length(sam2) - 1
fl = qf(1 - a/2,nu1,nu2)
fr = qf(1-a/2,nu2, nu1)

c(s1^2/s2^2/fl, s1^2/s2^2*fr)

Use var.test for ci 80%

var.test(sam1,sam2, conf.level = .80)$conf.int

Use var.test for ci 70%

var.test(sam1,sam2, conf.level = .70)$conf.int

Use var.test for ci 60%

var.test(sam1,sam2, conf.level = .60)$conf

Use var.test for ci 50%

var.test(sam1,sam2, conf.level = .50)$conf.int

Task 7

set.seed(23)
x = rnorm(30, mean=10, sd=12)
grac0009MATH4753::myci(x)

agracy2246/MATH4753grac0009 documentation built on April 26, 2020, 9:39 a.m.