Contoh Kode

Sebaran Binomial

n <- 5
p <- 1/6
y <- 3

prob <- dbinom(y, size=n, prob=p)
print(prob)
## [1] 0.03215021

Sebaran Multinomial

n <- 20
prob <- c(0.7, 0.25, 0.05)
k <- c(15, 3, 2)

prob_result <- dmultinom(k, size=n, prob=prob)
print(prob_result)
## [1] 0.02875242

Uji Proposi

# Data
n <- 2000
x <- 1120
p0 <- 0.5  # hypothesized proportion

# Sample proportion
phat <- x / n

# One-sample proportion test
z <- (phat - p0) / sqrt(p0 * (1 - p0) / n)

# P-value for two-sided test
p_value <- 2 * pnorm(-abs(z))

z
## [1] 5.366563
p_value
## [1] 8.025111e-08

Uji Wald

# Data
n <- 60000
x <- 2347
p0 <- 0.04 # churn rate yang dihipotesiskan

# Sample proportion
phat <- x / n

# Wald test
var_hat <- (phat * (1 - phat)) / n
wald <- (phat - p0)^2 / var_hat

# P-value for Wald test
p_value <- 1 - pchisq(wald, df = 1)

wald
## [1] 1.24557
p_value
## [1] 0.2644004

Latihan Soal

1

# Data
n1 <- 10000
x1 <- 8700
pro1 <- 0.87

# Sample proportion
phat1 <- x1 / n1

# Wald test
var_hat1 <- (phat1 * (1 - phat1)) / n1
wald1 <- (phat1 - pro1)^2 / var_hat1

# P-value for Wald test
p_value1 <- 1 - pchisq(wald1, df = 1)

wald1
## [1] 0
p_value1 
## [1] 1

karena p-value > 0,05, maka tidak tolak \(H_0\), artinya retensi pengguna tetap sama dengan 87%

2

# Data
n2 <- 150
x2 <- 120
pro2 <- 0.8

# Sample proportion
phat2 <- x2 / n2

# Wald test
var_hat2 <- (phat2 * (1 - phat2)) / n2
wald2 <- (phat2 - pro2)^2 / var_hat2

# P-value for Wald test
p_value2 <- 1 - pchisq(wald2, df = 1)

wald2
## [1] 0
p_value2 
## [1] 1

karena p-value > 0,05, maka tidak tolak \(H_0\), artinya tingkat keberhasilan obat sama dengan 80%

3

# Data
n3 <- 50000
x3 <- 2500
pro3 <- 0.05

# Sample proportion
phat3 <- x3 / n3

# Wald test
var_hat3 <- (phat3 * (1 - phat3)) / n3
wald3 <- (phat3 - pro3)^2 / var_hat3

# P-value for Wald test
p_value3 <- 1 - pchisq(wald3, df = 1)

wald3
## [1] 0
p_value3  
## [1] 1

karena p-value > 0,05, maka tidak tolak \(H_0\), artinya tingkat konversi pengguna sama dengan 5%

4

# Data
n4 <- 30000
x4 <- 1800
pro4 <- 0.06

# Sample proportion
phat4 <- x4 / n4

# Wald test
var_hat4 <- (phat4 * (1 - phat4)) / n4
wald4 <- (phat4 - pro4)^2 / var_hat4

# P-value for Wald test
p_value4 <- 1 - pchisq(wald4, df = 1)

wald4
## [1] 0
p_value4 
## [1] 1

karena p-value > 0,05, maka tidak tolak \(H_0\), artinya tingkat pengangguran sama dengan 6%

5

# Data
n5 <- 1200000
x5 <- 840000
pro5 <- 0.7

# Sample proportion
phat5 <- x5 / n5

# Wald test
var_hat5 <- (phat5 * (1 - phat5)) / n5
wald5 <- (phat5 - pro5)^2 / var_hat5

# P-value for Wald test
p_value5 <- 1 - pchisq(wald5, df = 1)

wald5
## [1] 0
p_value5 
## [1] 1

karena p-value > 0,05, maka tidak tolak \(H_0\), artinya tingkat partisipasi sama dengan 70%