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##Exemplo 1 Resolução do exemplo 1 da aula de 09/11 ###Inserindo os dados no R
od <- c(1.2, 1.4, 1.4, 1.3, 1.2, 1.35, 1.4, 2.0, 1.95, 1.1, 1.75, 1.05, 1.05, 1.4)
od
## [1] 1.20 1.40 1.40 1.30 1.20 1.35 1.40 2.00 1.95 1.10 1.75 1.05 1.05 1.40
str(od)
## num [1:14] 1.2 1.4 1.4 1.3 1.2 1.35 1.4 2 1.95 1.1 ...
Estatistica descritivas
summary(od)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.050 1.200 1.375 1.396 1.400 2.000
sort(od)
## [1] 1.05 1.05 1.10 1.20 1.20 1.30 1.35 1.40 1.40 1.40 1.40 1.75 1.95 2.00
Visualisando a distribuição de dados
boxplot(od - 1.2, col = "lightgrey", ylab = "Oxigênio Dissolvido (mg/L)")
###3.Teste de shapiro-wilks de normalidade
shapiro.test(od - 1.2)
##
## Shapiro-Wilk normality test
##
## data: od - 1.2
## W = 0.86929, p-value = 0.041
#Diferença observada
mean(od)-1.2
## [1] 0.1964286
As evidencias do boxplot e do teste de Shapiro- wilks indicam que os dados não tem distribuição aproximadamente normal. Entretanto, como exercicios, iremos utilizar o teste e o intervalo de confiança baseado na distribuição t para responder, inicialmente, a questão.
###4. Teste e intervalo de Confiança-t Student
# Teste com alpha = 5%
t.test(od - 1.2, alternative = c("two.sided"), mu = 0, conf.level = 0.95)
##
## One Sample t-test
##
## data: od - 1.2
## t = 2.4068, df = 13, p-value = 0.03168
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 0.02010878 0.37274836
## sample estimates:
## mean of x
## 0.1964286
###5.Teste e intervalo de confiança-t-Student
#teste com alpha=5%
wilcox.test(od - 1.2, mu = 0, conf.int = TRUE)
##
## Wilcoxon signed rank test with continuity correction
##
## data: od - 1.2
## V = 70, p-value = 0.01627
## alternative hypothesis: true location is not equal to 0
## 95 percent confidence interval:
## 0.0249788 0.4749679
## sample estimates:
## (pseudo)median
## 0.1999742
t.test(x, y = NULL,
alternative = c("two.sided", "less", "greater"),
mu = 0,
paired = FALSE,
var.equal = FALSE,
conf.level = 0.95, ...)
library(pwr)
pwr.p.test(h = ES.h(p1 = 0.75, p2 = 0.50),
sig.level = 0.05, power = 0.80,
alternative = "greater")
##
## proportion power calculation for binomial distribution (arcsine transformation)
##
## h = 0.5235988
## n = 22.55126
## sig.level = 0.05
## power = 0.8
## alternative = greater
Definindo \(H_a\) como bilateral
pwr.p.test(h = ES.h(p1 = 0.75, p2 = 0.50),
sig.level = 0.05,
n = 23)
##
## proportion power calculation for binomial distribution (arcsine transformation)
##
## h = 0.5235988
## n = 23
## sig.level = 0.05
## power = 0.7092308
## alternative = two.sided
Reduzindo o tamanho do efeito
pwr.p.test(h = ES.h(p1 = 0.65, p2 = 0.50),
sig.level = 0.05,
power = 0.80)
##
## proportion power calculation for binomial distribution (arcsine transformation)
##
## h = 0.3046927
## n = 84.54397
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
pwr.t.test(d = c(0.2, 0.5, 0.8),
n = 14,
sig.level = 0.05,
type="one.sample",
alternative="two.sided")
##
## One-sample t test power calculation
##
## n = 14
## d = 0.2, 0.5, 0.8
## sig.level = 0.05
## power = 0.1068087, 0.4102363, 0.7900878
## alternative = two.sided
Exemplo do od
##Bootstrap
#dados=
gas= c(64, 65, 75, 67, 65, 74, 75)
xbar = c()
# inicializando o vetor
for (i in 1:1999) {
amostras = sample(gas, size = length(gas), replace = TRUE)
xbar[i] = mean(amostras) }
hist(xbar)
# Estimativa de IC via Boostrapp
quantile(xbar, c(.025, .975))
## 2.5% 97.5%
## 66 73