1.1 Questao 01

xq1 <- c(6,12,10,3,5,1,6,
         8,1,5,2,2,5,8,1)

indices = 1:length(xq1)
plot_ly(x = ~indices, y = ~xq1)

1.2 Questao 02

xq2 = c(2,4,29,6,3,1,1,
        2,3,9,10,11,5,5,1)
labs = c(0,0,1,1,rep(0,5),1,
         0,0,1,0,0)
df = data.frame(xq2,
               labs)
df

##    xq2 labs
## 1    2    0
## 2    4    0
## 3   29    1
## 4    6    1
## 5    3    0
## 6    1    0
## 7    1    0
## 8    2    0
## 9    3    0
## 10   9    1
## 11  10    0
## 12  11    0
## 13   5    1
## 14   5    0
## 15   1    0

indices = 1:(dim(df)[1])
plot_ly(df, x = ~indices, y = ~xq2,
        color=df$labs)

1.3 Questao 03

df03 = data.frame(
  Y = c(4,6,12,3,6,9),
  cens = c(1, 0,0,0,1,0),
  row.names = 1:6
)
df03

##    Y cens
## 1  4    1
## 2  6    0
## 3 12    0
## 4  3    0
## 5  6    1
## 6  9    0

indices = 1:(dim(df03)[1])
plot_ly(df03, x = ~indices, y = ~Y,
        color=df03$cens)

1.4 Questão 04

Curva de sobrevivência

1.5 Questão 05 (duvida)

sendo \(h(t) = c\), sabendo que:

\[ h(t) = \frac{f(t)}{S(t)} \]

então,

\[ S(t) = \frac{f(t)}{h(t)} = \frac{1}{c}\cdot f(t) \]

1.6 Questão 06

St = 0.3
Ht = -log(St)
Ht

## [1] 1.203973

1.7 Questão 07

temos as seguintes formulas a serem aplicadas:

\[ \hat{f}(t) = \frac{N(t)}{n\cdot \Delta t} \]

\[ \hat{S}(t) = \frac{R(t)}{n} \]

\[ \hat{h}(t) = \frac{N(t)}{R(t)\cdot \Delta t} \] com \(t\geq 0\)

xt = seq(0,800, by=100)
yt = c(2,5,10,16,9,7,4,1)
intervals = cut(c(0,800), xt,right=F)
inter.freq = cbind(table(intervals), 'freq' = yt)
inter.freq = inter.freq[,-c(1)]
(freqs = cbind('freq' = inter.freq))

##           freq
## [0,100)      2
## [100,200)    5
## [200,300)   10
## [300,400)   16
## [400,500)    9
## [500,600)    7
## [600,700)    4
## [700,800)    1

rt = c(54, 52, 47, 37, 21, 12, 5, 1)
dt = rep(100, length.out = length(rt))
nt = yt
n = rt[1]
freqs.f = cbind(freqs,'rt' = rt, 'nt' = nt, 'dt' = dt)
fhatt = nt/(n*dt)
shatt = rt/(n)
hhatt = nt/(rt*dt)
cbind(freqs.f,
      'fhatt' = round(fhatt, 5),
      'shatt' = round(shatt, 5),
      'hhatt' = round(hhatt, 5)
      )

##           freq rt nt  dt   fhatt   shatt   hhatt
## [0,100)      2 54  2 100 0.00037 1.00000 0.00037
## [100,200)    5 52  5 100 0.00093 0.96296 0.00096
## [200,300)   10 47 10 100 0.00185 0.87037 0.00213
## [300,400)   16 37 16 100 0.00296 0.68519 0.00432
## [400,500)    9 21  9 100 0.00167 0.38889 0.00429
## [500,600)    7 12  7 100 0.00130 0.22222 0.00583
## [600,700)    4  5  4 100 0.00074 0.09259 0.00800
## [700,800)    1  1  1 100 0.00019 0.01852 0.01000

1.8 Questão 08

Demonstração da questão 08

1.9 Questão 09

Demonstração da questão 08

1.10 Questão 10

min = 0; max = 9;by = 1
interval = seq(min, max, by=by)
rt = c(1100,860,680,496,358,240,180,128,84,52)
nt = c(240,180,184,138,118,60,52,44,32,28)
dt = rep(by, length.out=length(rt))
n = rt[1]
intervals = cut(c(min, max), interval,right=F)
inter.freq = cbind(table(intervals))
inter.freq = rbind(inter.freq, '[9,+)' = 0)
inter.freq = inter.freq[,-c(1)]
(freqs.f = cbind(inter.freq,
              'rt' = rt,
              'nt' = nt,
              'dt' = by))

##         rt  nt dt
## [0,1) 1100 240  1
## [1,2)  860 180  1
## [2,3)  680 184  1
## [3,4)  496 138  1
## [4,5)  358 118  1
## [5,6)  240  60  1
## [6,7)  180  52  1
## [7,8)  128  44  1
## [8,9)   84  32  1
## [9,+)   52  28  1

fhatt = nt/(n*dt)
shatt = rt/(n)
hhatt = nt/(rt*dt)
cbind(freqs.f,
      'fhatt' = round(fhatt, 5),
      'shatt' = round(shatt, 5),
      'hhatt' = round(hhatt, 5))

##         rt  nt dt   fhatt   shatt   hhatt
## [0,1) 1100 240  1 0.21818 1.00000 0.21818
## [1,2)  860 180  1 0.16364 0.78182 0.20930
## [2,3)  680 184  1 0.16727 0.61818 0.27059
## [3,4)  496 138  1 0.12545 0.45091 0.27823
## [4,5)  358 118  1 0.10727 0.32545 0.32961
## [5,6)  240  60  1 0.05455 0.21818 0.25000
## [6,7)  180  52  1 0.04727 0.16364 0.28889
## [7,8)  128  44  1 0.04000 0.11636 0.34375
## [8,9)   84  32  1 0.02909 0.07636 0.38095
## [9,+)   52  28  1 0.02545 0.04727 0.53846

1.11 Questão 11

Demonstração da questao 11

1.12 Questão 12

Demonstração da questao 12

1.13 Questão 13

Demonstração da questao 13

1.14 Questão 14 (Duvida)

1.14.1 A) e B)

### C)

### D)

Demonstração da questao 14, letra D

1.15 Questão 15 (Duvida)

1.15.2 B)

Demonstração da questao 15, letra B

1.16 Questão 17 (Duvida)

df17 = data.frame('id'=c(1,1,2,2,3,3),
                  'inicio'=c(25,324,39,131,1,131),
                  'fim'=c(323,768,130,345,56,145),
                  'status'=c(0,1,0,0,0,1),
                  'CD4'=c(0,1,0,1,0,1))
df17

##   id inicio fim status CD4
## 1  1     25 323      0   0
## 2  1    324 768      1   1
## 3  2     39 130      0   0
## 4  2    131 345      0   1
## 5  3      1  56      0   0
## 6  3    131 145      1   1

1.17 Questão 18 (Duvida)

2.1 Questão 01

Demonstração da questão 01

2.2 Questão 02

n = 48
xt = c(36.3, 41.7, 43.9, 49.9, 50.1,
       50.8, 51.9, 52.1, 52.3, 52.3,
       52.4, 52.6, 52.7, 53.1, 53.6,
       53.6, 53.9, 53.9, 54.1, 54.6,
       54.8, 54.8, 55.1, 55.4, 55.9,
       56.0, 56.1, 56.5, 56.9, 57.1,
       57.1, 57.3, 57.7, 57.8, 58.1,
       58.9, 59.0, 59.1, 59.6, 60.4, 
       60.7)
tps = c(26.8, 29.6, 33.4,
        35,40,41.9, 42.5)
cens = NULL
m = length(xt) - length(tps)
for (i in 1:length(xt)){
  if (i >= m){
    cens[i] = 1
  }else{
    cens[i] = 0
  }
}
df02 = data.frame(
  xt, cens
)
df02

##      xt cens
## 1  36.3    0
## 2  41.7    0
## 3  43.9    0
## 4  49.9    0
## 5  50.1    0
## 6  50.8    0
## 7  51.9    0
## 8  52.1    0
## 9  52.3    0
## 10 52.3    0
## 11 52.4    0
## 12 52.6    0
## 13 52.7    0
## 14 53.1    0
## 15 53.6    0
## 16 53.6    0
## 17 53.9    0
## 18 53.9    0
## 19 54.1    0
## 20 54.6    0
## 21 54.8    0
## 22 54.8    0
## 23 55.1    0
## 24 55.4    0
## 25 55.9    0
## 26 56.0    0
## 27 56.1    0
## 28 56.5    0
## 29 56.9    0
## 30 57.1    0
## 31 57.1    0
## 32 57.3    0
## 33 57.7    0
## 34 57.8    1
## 35 58.1    1
## 36 58.9    1
## 37 59.0    1
## 38 59.1    1
## 39 59.6    1
## 40 60.4    1
## 41 60.7    1

indices = 1:(dim(df02)[1])
plot_ly(df, x = ~indices, y = ~xt,
        color=df02$cens)

2.3 Questão 03 (DUVIDA TOTAL)

xt = c(2,29,6,3,1,1,
       2,3,9,10,5,5,1)
cens = c(0,1,1,rep(0,5),
         1,0,1,0,0)
df03 =  data.frame(xt,cens)

indices = 1:(dim(df03)[1])
plot_ly(df03, x = ~indices, y = ~xt,
        color=df03$cens)

ekm<- survfit(Surv(df03$xt,df03$cens)~1)
ss = summary(ekm)
plot(ekm,
     xlab="Tempo (semanas)",ylab="S(t) estimada")

ss$surv

## [1] 0.8333333 0.6250000 0.4166667 0.0000000

2.4 Questão 05

g1 = c(7,8,8,8,8,12,12,17,18,22,rep(30,6))
censg1 = c(rep(0,10), rep(1,6))
g2 = c(8,8,9,10,10,14,15,15,18,19,21,22,23,25)
censg2 = c(rep(0, 14))
g3 = c(rep(8,6), 9,10,10,10,11,17,19)
censg3 = c(rep(0, 13))

ekm1<- survfit(Surv(g1,censg1)~1)
ekm2<- survfit(Surv(g2,censg2)~1)
ekm3<- survfit(Surv(g3,censg3)~1)

par(mfrow=c(2,2))
plot(ekm1, xlab="Tempo (semanas)",ylab="S(t) estimada", conf.int=F,
     main='Grupo 01')#sem intervalo de confianca
plot(ekm2, xlab="Tempo (semanas)",ylab="S(t) estimada", conf.int=F,
     main='Grupo 02')#sem intervalo de confianca
plot(ekm3, xlab="Tempo (semanas)",ylab="S(t) estimada", conf.int=F,
     main='Grupo 03')#sem intervalo de confianca

grupos = c(rep(1,length(g1)), rep(2, length(g2)), rep(3,length(g3)))
cens = c(censg1,censg2,censg3)
cens = cens + 1
cens[cens == 2] = 0
data = c(g1,g2,g3)
df03 = data.frame(grupos,cens,'temp'=data)

survdiff(Surv(df03$temp,
              df03$cens)~df03$grupos,rho=0)

## Call:
## survdiff(formula = Surv(df03$temp, df03$cens) ~ df03$grupos, 
##     rho = 0)
## 
##                N Observed Expected (O-E)^2/E (O-E)^2/V
## df03$grupos=1 16       10     17.1    2.9328    6.7157
## df03$grupos=2 14       14     13.2    0.0456    0.0829
## df03$grupos=3 13       13      6.7    5.9240    9.6123
## 
##  Chisq= 11.9  on 2 degrees of freedom, p= 0.003

subdf03 = df03[df03$grupos==1|df03$grupos==2,]
survdiff(Surv(subdf03$temp,
              subdf03$cens)~subdf03$grupos,rho=0)

## Call:
## survdiff(formula = Surv(subdf03$temp, subdf03$cens) ~ subdf03$grupos, 
##     rho = 0)
## 
##                   N Observed Expected (O-E)^2/E (O-E)^2/V
## subdf03$grupos=1 16       10     13.7      1.02      2.67
## subdf03$grupos=2 14       14     10.3      1.37      2.67
## 
##  Chisq= 2.7  on 1 degrees of freedom, p= 0.1

subdf03 = df03[df03$grupos==2|df03$grupos==3,]
survdiff(Surv(subdf03$temp,
              subdf03$cens)~subdf03$grupos,rho=0)

## Call:
## survdiff(formula = Surv(subdf03$temp, subdf03$cens) ~ subdf03$grupos, 
##     rho = 0)
## 
##                   N Observed Expected (O-E)^2/E (O-E)^2/V
## subdf03$grupos=2 14       14    19.09      1.36      6.63
## subdf03$grupos=3 13       13     7.91      3.27      6.63
## 
##  Chisq= 6.6  on 1 degrees of freedom, p= 0.01

subdf03 = df03[df03$grupos==1|df03$grupos==3,]
survdiff(Surv(subdf03$temp,
              subdf03$cens)~subdf03$grupos,rho=0)

## Call:
## survdiff(formula = Surv(subdf03$temp, subdf03$cens) ~ subdf03$grupos, 
##     rho = 0)
## 
##                   N Observed Expected (O-E)^2/E (O-E)^2/V
## subdf03$grupos=1 16       10    15.34      1.86      7.86
## subdf03$grupos=3 13       13     7.66      3.72      7.86
## 
##  Chisq= 7.9  on 1 degrees of freedom, p= 0.005

2.5 Questão 06

2.6 Questão 10 (Duvida)

2.7 Questão 12 (Duvida)

2.8 Questão 13 (Duvida)

3.1 Relações

\[ S(t) = 1 - F(t) \] \[ h(t) = \frac{f(t)}{S(t)} \]

\[ H(t) = -ln(S(t)) \]

\[ S(t) = exp(-H(t)) \]

3.2 Tabelas

Tabela 01

3.3 Kaplan meier

\[ \hat{S}(t) = \Pi_{j:t_j \leq t}\bigg(1 - \frac{d_j}{n_j}\bigg) \]

############## Kaplan meier s(t) ###########
ekm<- survfit(Surv(g1,censg1)~1)
summary(ekm)

## Call: survfit(formula = Surv(g1, censg1) ~ 1)
## 
##  time n.risk n.event survival std.err lower 95% CI upper 95% CI
##    30      6       6        0     NaN           NA           NA

plot(ekm, xlab="Tempo (semanas)",ylab="S(t) estimada")

plot(ekm, xlab="Tempo (semanas)",ylab="S(t) estimada", conf.int=F)#sem intervalo de confian?a

3.4 Teste de hipotese

############## Teste de hipotese ###########
##### Gehan
gehan.wilcoxon.test(Surv(df03$temp,df03$cens)~df03$grupos,
                    data=df03)

## 
##  Gehan-Wilcoxon
## 
## data:  
## = 7.0584, p-value = 0.02933
## alternative hypothesis: two-sided

##### Log-rank
survdiff(Surv(df03$temp,df03$cens)~df03$grupos,rho=0)

## Call:
## survdiff(formula = Surv(df03$temp, df03$cens) ~ df03$grupos, 
##     rho = 0)
## 
##                N Observed Expected (O-E)^2/E (O-E)^2/V
## df03$grupos=1 16       10     17.1    2.9328    6.7157
## df03$grupos=2 14       14     13.2    0.0456    0.0829
## df03$grupos=3 13       13      6.7    5.9240    9.6123
## 
##  Chisq= 11.9  on 2 degrees of freedom, p= 0.003

##### Peto
survdiff(Surv(df03$temp,df03$cens)~df03$grupos,rho=1)

## Call:
## survdiff(formula = Surv(df03$temp, df03$cens) ~ df03$grupos, 
##     rho = 1)
## 
##                N Observed Expected (O-E)^2/E (O-E)^2/V
## df03$grupos=1 16     6.91     9.63     0.769     2.201
## df03$grupos=2 14     6.79     8.37     0.299     0.764
## df03$grupos=3 13     9.74     5.44     3.401     6.965
## 
##  Chisq= 7.1  on 2 degrees of freedom, p= 0.03

3.5 Função de risco

h(t) table

############## Função de Risco #############
library(muhaz)
risco<- pehaz(df03$temp,df03$cens)

## 
## max.time= 30
## width= 1.82134
## nbins= 17

riscoS <- muhaz(df03$temp,df03$cens, bw.smooth=200, max.time=3647)
objects(risco)

## [1] "At.Risk"  "call"     "Cuts"     "Events"   "F.U.Time" "Hazard"   "Width"

risco$Hazard

##  [1] 0.00000000 0.00000000 0.00000000 0.01281522 0.22221343 0.10785783
##  [7] 0.07763437 0.02788928 0.06285091 0.14230222 0.09255855 0.05214129
## [13] 0.21154839 0.08162850 0.00000000 0.00000000 0.00000000

plot(risco)
lines(riscoS, col="red")

3.6 IC para S(t)

• \(n_j\): Nº sob risco

• \(d_j\): Nº Falhas

• \(c_j\): Nº Censuras

\[ \hat{q}_j = \frac{d_j}{n_j - \frac{c_j}{2}} \] \[ \hat{S}(t) = \Pi(1 - \hat{q}_{i-1}) \]

ou,

\[ \hat{S}(t) = \Pi_{j:t\leq t}\bigg(1 - \frac{d_j}{n_j}\bigg) \]

\[ IC(\hat{S}(t) | 1-\alpha \%) = \bigg[\hat{S}(t) +- \sqrt{\hat{Var}(\hat{S}(t))}\bigg] \]