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summary(cars)
##      speed           dist       
##  Min.   : 4.0   Min.   :  2.00  
##  1st Qu.:12.0   1st Qu.: 26.00  
##  Median :15.0   Median : 36.00  
##  Mean   :15.4   Mean   : 42.98  
##  3rd Qu.:19.0   3rd Qu.: 56.00  
##  Max.   :25.0   Max.   :120.00

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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 ...

2. Análise exploratoria

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

6. Teste de hipotese e IC para duas amostras

6.1 Teste e IC t para duas amostras independentes no R

t.test(x, y = NULL,
       alternative = c("two.sided", "less", "greater"), 
       mu = 0, 
       paired = FALSE, 
       var.equal = FALSE, 
       conf.level = 0.95, ...)

7. Determinando o tamanho da amostra

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

diversos tamanhos do efeito

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