# Statistics - Part 1 with Lyes LAKHAL
###### tags: emPLANT+, 第一年 法國 UniLaSalle, 上學期

**Use R Studio**

R程式 轉換語言
https://metalanguageinuse.com/how-to-change-language-setting-to-english-in-r-rstudio-a-temporary-measure/

課程時間表

#read csv files from my dell computer
x = read.csv(file.choose())
x


Practice 1
* 題目: [https://www.r-bloggers.com/2016/09/examining-data-exercises](https://www.r-bloggers.com/2016/09/examining-data-exercises/)

Class 7
* 不同檢定方式的差別整理: [https://mropengate.blogspot.com/2015/03/hypothesis-testing-p-value.html](https://mropengate.blogspot.com/2015/03/hypothesis-testing-p-value.html)
* 統計假說檢定參考資料:[https://slidesplayer.com/slide/16909216/](https://slidesplayer.com/slide/16909216/)

Class 11 統計分析
* dnorm, pnorm, and qnorm 的差異: [https://www.jianshu.com/p/a24eb1b94177](https://www.jianshu.com/p/a24eb1b94177)


**1. 虛無假設 (null hypothesis, H0):**
Null = nothing! 虛無假設表示 “什麼也沒發生” 的假設。通常虛無假設是表示觀測值完全反應隨機變異的結果

**2. 對立假設 (alternaive hypothesis, Ha):**
某種效應的確存在! 因此觀測值是此效應加上隨機變異的結果

**3. 連續變數資料分析**

D: 單獨
P: 連續
如果要計算連續的不能用很多個 d 相加會有誤差
如果要計算連續的區間要相減的話要記得減一個數字不能用要的數值它會涵蓋在裡面

quantitative (定量的)
qualitative (定性的)

---

## <font color="red"><font size=5>Code in Class </font></font><br>

### **1. Class 1 & 2**

# 2021-2022 Univariate statistics – emPLANT & 4A Plant Breeding

# vector (向量), numeric (數字)
x1 <- c(1, 2, 3)
x1
# numeric (數字)
x2 <- 10.5
x2             
class(x2)
# logical(邏輯，是/否)
x3 <- 1
y3 = 2
z3 = x3 > y3
z3
class(z3)
# "&" means and, "|" means or, and "!" means negation(否定)
# Character
fname <- "Judy"
fname
class(fname)
lname <- "Lin"
lname
paste(fname, lname)
sprintf("%s has %d dollars", "Sam", 100)
# %s mean give character
# %d mean give numeric
# 輸入數值到指定空格
substr("Mary has a little lamb.", start=3, stop=12)
# replace
sub("little", "big", "Mary has a little lamb.")
# 計算長度
length(c("aa", "bb", "cc", "dd", "ee"))
#合併
n1 <- c("aa", "bb", "cc")
s1 <-  c(2,3,5) 
s2 = c("aa", "bb", "cc", "dd", "ee")
c1 <- c(n1, s1)
c1
# 乘法計算
a1 = c(1, 3, 5, 7) 
b1 = c(1, 2, 4, 8)
d1 <- 5 * a1
d1
# 加法計算
u1 = c(10, 20, 30) 
v1 = c(1, 2, 3, 4, 5, 6, 7, 8, 9) 
e1 <- u1 + v1 
e1
# 不太可能有辦法將兩個擁有不同數量的數值相加，u會一直重複直到達到v的量值
# 指定
s3 = c("aa", "bb", "cc", "dd", "ee") 
s3[3]
s3[c(2, 3)]
s3[c(2, 3, 3)]
s[c(2, 1, 3)] 
s[2:4]
# 減掉特定值
s3[-3]
s3[10]
# Matrix (矩陣)
A1 = matrix( c(2, 4, 3, 1, 5, 7), nrow=2, ncol=3,   byrow = TRUE)
A1
# nrow直行3，ncol橫列2 
# Fill the byrow，並非必要(可以選擇性的輸入)，如果不輸入的話系統會自動帶入Fill the bycol
A1[2, 3]
# element at 2nd row, 3rd column 
A1[2, ]
# the 2nd row
# []裡面的數值，前面代表直行row，後面代表橫列column
A1[ ,3]
# the 3rd column 
A1[ ,c(1,3)]
# 命名直行row與橫列column
dimnames(A1) = list( c("row1", "row2"),  c("col1", "col2", "col3"))  
A1
# print A1 
A1["row2", "col3"]
# element at 2nd row, 3rd column
B1 = matrix( c(2, 4, 3, 1, 5, 7), nrow=3, ncol=2) 
B1
# B1 has 3 rows and 2 columns 
t(B1)
# transpose of B (矩陣轉向)
C1 = matrix( c(7, 4, 2), nrow=3, ncol=1) 
C1
# C has 3 rows 
# 矩陣結合(相同直行row)
f1 <- cbind(B1, C1)
f1
# cbind將矩陣連結，兩矩陣需要有相同的直行row，若是不符合規定則無法將兩不同矩陣連結
D1 = matrix( c(6, 2),  nrow=1, ncol=2) 
D1
# 矩陣結合(相同橫列column)
g1 <- rbind(B1, D1) 
g1
# rbind將矩陣連結，兩矩陣需要有相同的橫列column，若是不符合規定則無法將兩不同矩陣連結
# 將矩陣轉成向量
h1 <- c(B1) 
h1
# List (表)
n2 = c(2, 3, 5) 
s4 = c("aa", "bb", "cc", "dd", "ee") 
b2 = c(TRUE, FALSE, TRUE, FALSE, FALSE) 
x1 = list(n2, s4, b2, 3)
# x contains copies of n2, s4, b2
x1
# 指定
x1[2]
x1[c(2, 4)]
# 取代
x1[[2]][1] = "ta" 
x1[[2]] 
# 如果需要將數值置換掉記得要加上””
v1 = list(bob=c(2, 3, 5), john=c("aa", "bb")) 
v1 
v1["bob"]
v1$bob
v1[c("john", "bob")]
# Data frame (數據框)
# 必須要有相同的數字值
n3 = c(2, 3, 5) 
s5 = c("aa", "bb", "cc") 
b3 = c(TRUE, FALSE, TRUE) 
df1 = data.frame(n3, s5, b3)
df1 
mtcars
# 呼叫資料可以用直行橫列或是資料項目的名稱
nrow(mtcars)
# number of data rows 
ncol(mtcars)
# number of columns 
head(mtcars)
# 橫列用雙引號"[[]]"
mtcars[[9]]
mtcars[["am"]]
mtcars$am
mtcars[,"am"]
# 直行用單引號"[]" 
mtcars[1]
mtcars["mpg"]
mtcars[c("mpg", "hp")]
mtcars[24,]
mtcars[c(3, 24),]
mtcars["Camaro Z28",] 
mtcars[c("Datsun 710", "Camaro Z28"),]
# 1比對值
L1 = mtcars$am == 0
L1
mtcars[L1,]
mtcars[L1,]$mpg

### **2. Class 3**

head(faithful)
duration = faithful$eruptions
range(duration) 
step=0.1
r = range(duration)
minval = r[1]
maxval = r[2]
minval
maxval
breaks = seq(minval-step, maxval+step, by=step)
breaks
duration.cut = cut(duration, breaks, right=FALSE)
breaks
duration  
duration.cut
duration.freq = table(duration.cut)
duration.freq
cbind(duration.freq)
duration = faithful$eruptions 
hist(duration, right=FALSE)
colors = c("red", "yellow", "green", "violet", "orange", "blue", "pink", "cyan") 
hist(duration,   right=FALSE,    col=colors,     main="Old Faithful Eruptions", xlab ="Duration minutes")       
duration = faithful$eruptions 
breaks = seq(1.5, 5.5, by=0.5) 
duration.cut = cut(duration, breaks, right=FALSE) 
duration.freq = table(duration.cut)
SS = nrow(faithful)
duration.relfreq = duration.freq / SS
duration.relfreq 
duration.relfreq
cbind(duration.freq, duration.relfreq)
duration = faithful$eruptions 
breaks = seq(1.5, 5.5, by=0.5) 
duration.cut = cut(duration, breaks, right=FALSE) 
duration.freq = table(duration.cut)
duration.cumfreq = cumsum(duration.freq)
duration.cumfreq
cbind(duration.cumfreq) 
duration = faithful$eruptions 
breaks = seq(1.5, 5.5, by=0.5) 
duration.cut = cut(duration, breaks, right=FALSE) 
duration.cut
duration.freq = table(duration.cut)
duration.cumfreq = cumsum(duration.freq) 
duration.cumfreq
cumfreq0 = c(0, duration.cumfreq) 
cumfreq0
plot(breaks, cumfreq0,  main="Old Faithful Eruptions",  xlab="Duration minutes",  ylab ="Cumulative eruptions") 
lines(breaks, cumfreq0)
duration = faithful$eruptions 
breaks = seq(1.5, 5.5, by=0.5) 
duration.cut = cut(duration, breaks, right=FALSE) 
duration.freq = table(duration.cut)
duration.freq
duration.cumfreq = cumsum(duration.freq)
ss = nrow(faithful) 
duration.cumrelfreq = duration.cumfreq /ss
duration.cumrelfreq
cbind(duration.cumrelfreq)
duration=faithful$eruptions
waiting=faithful$waiting
plot(duration, waiting, xlab="Eruption duration", ylab="Time waiting")


### **3. Class 4**

duration = faithful$eruptions 
a1 <- mean(duration) 
duration = faithful$eruptions
a2 <- median(duration) 
a1
a2
duration = faithful$eruptions 
a3 <- quantile(duration)
a3
duration = faithful$eruptions 
quantile(duration, c(.32, .57, .98)) 
duration = faithful$eruptions
max(duration)
min(duration)
a4 <- max(duration) - min(duration)
a4
duration = faithful$eruptions
boxplot(duration, horizontal=TRUE)
duration = faithful$eruptions
a5 <- var(duration)
a5
duration = faithful$eruptions
a6 <- sd(duration)
a6
z <- (1/3)*((4-2.5)*(7-4.5)+(2-2.5)*(4-4.5)+(3-2.5)*(2-4.5)+(1-2.5)*(5-4.5))
z
duration = faithful$eruptions 
waiting = faithful$waiting
a7 <- cov(duration, waiting)
a7
library(e1071)
duration = faithful$eruptions
moment(duration, order=3, center=TRUE)

### **4. Class 5**

d1 <- dbinom(4, size=12, prob=0.2) 
d1
d2 <- dbinom(0, size=12, prob=0.2) + dbinom(1, size=12, prob=0.2) +  dbinom(2, size=12, prob=0.2) + dbinom(3, size=12, prob=0.2) + dbinom(4, size=12, prob=0.2) 
pbinom(4, size=12, prob=0.2)
d2
p1 <- ppois(16, lambda=12)
p2 <- ppois(16, lambda=12, lower=FALSE)
p1
p2
e1 <- pexp(2, rate=1/3)
e1
n1 <- pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) 
n1
n2 <- pnorm(84, mean=72, sd=15.2) 
n2
c1 <- qchisq(.95, df=7)
c1
t1 <- qt(c(.025, .975), df=5)
t1
f1 <- qf(.95, df1=5, df2=2)
f1

### **5. Class 6 & 7**

library(MASS)
survey
help(survey)
height.survey <- survey$Height
height.survey
xbar <- mean(height.survey, na.rm=TRUE)
xbar

library(MASS)
height.response = na.omit(survey$Height) 
n = length(height.response)
sigma <- 9.48
# population standard deviation 
sem = sigma/sqrt(n)
sem
# standard error of the mean
alpha <- 0.05
E = qnorm(1- alpha/2)*sem
E
# margin of error 
mean1 = mean(height.response)
mean1 + c(-E, E)
# sample mean

library(MASS)
height.response = na.omit(survey$Height)
n1 = length(height.response)
s = sd(height.response)
n1
s
SE = s/sqrt(n)
SE
E = qt(.975, df=n-1)*SE
E
# qt function 是學生式 t 檢定
xbar2 = mean(height.response)
xbar2 + c(-E, E)

library(MASS)
height.response = na.omit(survey$Height)
n2 = length(height.response)
s2 = sd(height.response)
n2
s2
SE1 = s/sqrt(n)
SE1
alpha2 = 0.01
E2 = qt(.99, df=n-1)*SE
E2
xbar3 = mean(height.response)
xbar3 + c(-E, E)

t.test(height.response)

alpha3 <- 0.05
zstar = qnorm(.975)
sigma = 9.48
E = 1.2
zstar^2 * sigma^2/ E^2

library(MASS)
gender.response = na.omit(survey$Sex)
n4 = length(gender.response)
k4 = sum(gender.response == "Female")
n4
k4
pbar = k/n
pbar
SE = sqrt(pbar*(1-pbar)/n) 
SE
prop.test(k, n)

zstar = qnorm(.975) 
zstar
p = 0.5 
E = 0.05 
p1 <- zstar^2*p*(1-p)/E^2
p1

sigma5 <- 120
mu05 <- 10000
xbar5 <- 9900
alpha5 <- 0.05
n5 <- 30
a5 <- (xbar5-mu05)/(sigma5/sqrt(n5))
a5
alpha5 = .05 
z.alpha5 = qnorm(1-alpha5) 
z.alpha5 
pval5 = pnorm(a5) 
pval5

sigma6 <- 0.25
mu06 <- 2
xbar6 <- 2.1
alpha6 <- 0.05
n6 <- 35
a6 <- (xbar6-mu06)/(sigma6/sqrt(n6))
a6
alpha6 = .05 
z.alpha6 = qnorm(1-alpha6) 
z.alpha6
pval6 = pnorm(a6, lower.tail=FALSE)  
pval6

sigma7 <- 2.5
mu07 <- 15.4
xbar7 <- 14.6
alpha7 <- 0.05
n7 <- 35
a7 <- (xbar7-mu07)/(sigma7/sqrt(n7))
a7
alpha7 = .05 
z.alpha7 = qnorm(1-alpha7/2)
c(-z.alpha7, z.alpha7)

sigma8 <- 125
mu08 <- 10000
xbar8 <- 9900
alpha8 <- 0.05
n8 <- 30
a8 <- (xbar8-mu08)/(sigma8/sqrt(n8))
a8
t.alpha8 = qt(1-alpha8, df = n8-1)
t.alpha8
c(-t.alpha8, t.alpha8)

### **6. Class 8**

xbar1 <- 9900 
mu01 <- 10000
sd1 <- 125
n1 <- 30
t1 <- (xbar1-mu01)/(sd1/sqrt(n1)) 
t1              
alpha1 <- 0.05
t.alpha1 <- qt(1-alpha1, df=n1-1)
t.alpha2 <- c(t.alpha1, -t.alpha1)
t.alpha2

xbar2 <- 2.1 
mu02 <- 2.0
sd2 <- 0.3
n2 <- 35
t2 <- (xbar2-mu02)/(sd2/sqrt(n2)) 
t2              
alpha2 <- 0.05
t.alpha2 <- qt(1-alpha2, df=n2-1)
t.alpha3 <- c(t.alpha2, -t.alpha2)
t.alpha3

xbar3 <- 14.6
mu03 <- 15.4
sd3 <- 2.5
n3 <- 35
t3 <- (xbar3-mu03)/(sd3/sqrt(n3))
t3
t.alpha3 <- 0.05
t.alpha4 <- qt(1-(t.alpha3/2), df = n3-1)
t.alpha5 <- c(t.alpha4, -t.alpha4)
t.alpha5

pbar <- 85/148
p04 <- 0.6
n4 <- 148
z4 <- (pbar-p04)/sqrt(p04*(1-p04)/n4)
z4
z.alpha6 <- 0.05
z.alpha7 <- qnorm(1-t.alpha6)
z.alpha8 <- c(z.alpha7, -z.alpha7)
z.alpha8

### **7. Class 9**

library(MASS)
head(immer)
t.test(immer$Y1, immer$Y2, paired=TRUE)

mtcars
mtcars$mpg 
mtcars$am 
L = mtcars$am == 0
L
mpg.auto = mtcars[L,]$mpg 
mpg.auto                
mpg.manual = mtcars[!L,]$mpg 
mpg.manual
t.test(mpg.auto, mpg.manual)
t.test(mpg ~ am, data=mtcars)

---

## <font color="red"><font size=5>Code of Exercises </font></font><br>

### **1. Exercises 1**

#question 1
islands
N <- length(islands)
N
M1 <- mean(islands)
M1
M2 <- median(islands)
M2
BP <- boxplot(islands,horizontal=TRUE)
R <- range(islands)
R
max(islands)
min(islands)
SD <- sd(islands)
SD
A <- quantile(islands, c(.0005, .95))
A

### **2. Exercises 2**

#question 1
#A car dealer knows that from past experience he can make a sale to 20% of the customers that he interacts with. What is the probability that, in five randomly selected interactions, he will make a sale to: 
#a.) Exactly three customers?  
#b.) At most one customer?  
#c.) At least one customer?  
#d.) Calculate the probability for each number of sales ? (k = 0, 1, 2, 3, 4, 5)
n1 <- 5
p01 <- 0.2
x1 <- 3
x2 <- c(0,1)
x3 <- c(1,2,3,4,5)
d1 <- dbinom(x1, size=n1, prob=p01) 
d1
d2 <- dbinom(x2, size=n1, prob=p01) 
d2
#sum together
d2f <- 0.32768+0.40960
d2f
d3 <- dbinom(x3, size=n1, prob=p01) 
d3
#sum together
d3f <- 0.40960+0.20480+0.05120+0.00640+0.00032
d3f
d4 <- dbinom(0, size=n1, prob=p01)
d4
d5 <- dbinom(1, size=n1, prob=p01)
d5
d6 <- dbinom(2, size=n1, prob=p01)
d6
d7 <- dbinom(3, size=n1, prob=p01)
d7
d8 <- dbinom(4, size=n1, prob=p01)
d8
d9 <- dbinom(5, size=n1, prob=p01)
d9

#question 2
#A machine is set up such that the average content of juice per bottle equals µ. A sample of 100 bottles yields an average content of 48cl. Calculate a 90% and a 95% confidence interval for the average content:
#(i) Assuming that the population standard deviation σ = 5 cl. 
#(ii) Assuming that the population standard deviation σ is unknown. We suppose that the standard divination of the sample is equal to **s = 7**.
n2 <- 100
xbar2 <- 48
sigma2 <- 5
alpha21 <- 0.1
alpha22 <- 0.05
sem2 <- sigma2/sqrt(n2)
sem2
E21 <- qnorm(1-alpha21/2)*sem2
E21
E22 <- qnorm(1-alpha22/2)*sem2
E22
IE21 <- xbar2+c(-E21, E21)
IE21
IE22 <- xbar2+c(-E22, E22)
IE22
#we suppose that the standard divination of the sample is equal to s = 7 ml
n2 <- 100
xbar2 <- 48
s2 <- 7
alpha21 <- 0.1
alpha22 <- 0.05
sem2 <- sigma2/sqrt(n2)
sem2
E23 <- qt(1-alpha21/2, df = n2-1)*sem2
E23
E24 <- qt(1-alpha22/2, df = n2-1)*sem2
E24
IE23 <- xbar2+c(-E23, E23)
IE23
IE24 <- xbar2+c(-E24, E24)
IE24

#question 3
#A machine is set up such that the average content of juice per bottle equals μ. A sample of 100 bottles yields an average content of 48.8 cl. Test the hypothesis that the average content per bottle is 50 cl. at the 5% significance level: H0: n=50
#(i) Assuming that the population standard deviation σ = 5 cl.
#(ii) Assuming that the population standard deviation σ is unknown. We suppose that the standard divination of the sample is equal to **s = 7 cl**.
xbar3 <- 48.8
mu03 <- 50
n3 <- 100
sigma31 <- 5
z3 <- (xbar3-mu03)/(sigma31/sqrt(n3)) 
z3
alpha3 <- .05
z.alpha3 <- qnorm(1-alpha3/2)
z.alpha3
z.alpha4 <- c(z.alpha3, -z.alpha3)
z.alpha4
#teacher's answer
xbar3 <- 48.8
# sample mean
mu03 <- 50
# hypothesized value 
n3 <- 100
# sample size
sigma32 <- 7
z3 <- (xbar3-mu03)/(sigma32/sqrt(n3)) 
z3
alpha3 <- .05
z.alpha5 <- qt(1-alpha3/2, df=n3-1)
z.alpha5
z.alpha6 <- c(z.alpha5, -z.alpha5)
z.alpha6

#question 4
#A machine is set up such that the average content of juice per bottle equals μ. A sample of 36 bottles yields an average content of 48.5 cl. Can you reject the hypothesis that the average content per bottle is less than or equal to 45 cl (upper tail) in favor of the alternative that it exceeds 45 cl (5% significance level) ? Assume that the population standard deviation σ = 5 cl. H0 <= 45
n4 <- 36
# sample size
xbar4 <- 48.5
# sample mean
mu04 <- 45
# hypothesized value 
sigma4 <- 5
# population standard deviation 
alpha4 <- 0.05
z4 <- (xbar4-mu04)/(sigma4/sqrt(n4)) 
z4
# u0
alpha4 <- 0.05
t.alpha5 <- qnorm(1-alpha4)
t.alpha5
t.alpha6 <- c(t.alpha5, -t.alpha5)
t.alpha6
# u

#question 5
#60 customers reply they are satisfied with the service they received. The sample size is 80 customers. Calculate a 95% confidence interval for the proportion of satisfied customers. 
n5 <- 80
k5 <- 60
pbar <- k5/n5
alpha5 <- 0.05
z.half.alpha7 <- qnorm(1-alpha5/2)
A5 <- sqrt(pbar*(1-pbar)/n5)
B5 <- z.half.alpha7*A5
IE5 <- pbar + c(-B5, +B5)
IE5


### **3. Exercises 3**

#question 1
#In the past the average length of an outgoing telephone call from a business office has been 143 seconds. A manager wishes to check whether that average has **decreased** after the introduction of policy changes. A sample of 100 telephone calls produced a mean of 133 seconds, with a standard deviation of 35 seconds. (from the sample not from the population) Perform the relevant test at the 1% level of significance. H0: mu <= mu0 (lower tail) ? 
n1 <- 100
# sample size
xbar1 <- 133
# sample mean
mu01 <- 143
# hypothesized value 
s1 <- 35
z1 <- (xbar1-mu01)/(s1/sqrt(n1)) 
z1
alpha1 <- 0.01
t.alpha1 <- qt(1-alpha1, df = n1-1)
t.alpha1

#question 2
#The recommended daily calorie intake for teenage girls is 2 200 calories/day. A nutritionist at a state university believes the average daily caloric intake of girls in that **state to be lower**. Test that Hypothesis, at the 5% level of significance, against the null hypothesis that the population average is 2 200 calories/day using the following sample data:n = 36 (sample size) ;  xbar = 2150 (sample mean) ;  s = 203 (sample standard deviation) H1: u < u0 = 2200, H0: u > u0 (lower tail) ?
n2 <- 36
xbar2 <- 2150
mu02 <- 2200
s2 <- 203
z2 <- (xbar2-mu02)/(s2/sqrt(n2)) 
z2
alpha1 <- 0.05
t.alpha1 <- qt(1-alpha1, df = n2-1)
t.alpha2 <- c(-t.alpha1, t.alpha1)
#since t > -t.alpha, then H0 is to be accepted

#question 3
#A random sample is drawn from a population of known standard deviation 11,3. Construct a 90% confidence interval for the population mean based on the information given (not all the information given need be used). 
#a.) n = 36 ;  xbar = 105.2 ;  s = 11.2 ? 
#b.) n = 100 ;  xbar =105.2 ; s = 11.2 ? 
n31 <- 36
xbar31 <- 105.2
sd31 <- 11.3
se31 <- 11.2
a31 <- qnorm(0.9,sd31)
a31
answer1 <- mean(n31*xbar31/sd31*se31)
answer1

n32 <- 100
xbar32 <- 105.2
sd32 <- 11.3
se32 <- 11.2
a32 <- qnorm(0.9, sd32)
a32
answer2 <- mean(n32*xbar32/sd32*se32)
answer2

#question 4
#A government agency was charged by the legislature with estimating the length of time it takes citizens to fill out various forms. Two hundred randomly selected adults were timed as they filled out a particular form. The times required had mean 12.8 minutes with standard deviation 1.7 minutes. Construct a 90% confidence interval for the mean time taken for all adults to fill out this form ? 
n4 <- 200
xbar4 <- 12.8
sigma4 <- 1.7
sem4 <- sigma4/sqrt(n4)
alpha4 <- 0.1
a4 <- qnorm(1-alpha4/2)*sem4
a4
answer4 <- xbar4 + c(a4, -a4)
answer4

#question 5
#The amount X of beverage in a can labeled 12 ounces is normally distributed with mean 12.1 ounces and standard deviation 0.05 ounce. A can is selected at random. 
#a.) Find the probability that the can contains at least 12 ounces ? 
#b.) Find the probability that the can contains between 11.9 and 12.1 ounces ? 
d5111 <- 1- pnorm(12, mean=12.1, sd=0.05)
d5112 <- pnorm(12, mean=12.1, sd=0.05, lower.tail = FALSE)
d511 <- c(d5111, d5112)
d511
d5121 <- pnorm(12.1, mean=12.1, sd=0.05)
d5122 <- pnorm(11.9, mean=12.1, sd=0.05)
d512 <- d5121-d5122
d512

*pnorm calculate the left side (左尾檢定)

### **4. Exercises 4**

# question 1
# total number of the population is 80, 60 people reply yes to the project, find the 95% ? 
n1 <- 80
k1 <- 60
pbar1 <- k1/n1
SE1 <- sqrt(pbar1*(1-pbar1)/n1)
alpha1 <- 0.05
z.alpha1.by2 <- qnorm(1-alpha1/2)
a1 <- pbar1*(1-pbar1)
b1 <- sqrt(a1/n1)
c1 <- z.alpha1.by2*b1
a11 <- c(pbar-c1, pbar+c1)
a11

#question 2
# Cereals offer gifts 1/6, father buy 20, 
#a.) the probability find 4 in 20 ? 
#b.) the probability find 0 in 20 ? 
#c.) 3 toys in the package. the probability that find 2 toys in 5 package ? 
n2 <- 20
pbar2 <- 1/6
a21 <- dbinom(4,size=n2, prob=pbar2)
a21
a22 <- dbinom(0,size=n2, prob=pbar2)
a22
n23 <- 5
pbar23 <- 3/20
a23 <- dbinom(2,size=n23, prob=pbar23)
a23

### **5. Exercises 5**

# question 1
# 15 multiple choice need to get 10 or more to find. exactly equal to 10? more than 10 ?
n1 <- 15
pbar1 <- 1/4
a <- dbinom(10,size=n1, prob=pbar1)
a
b1 <- pbinom(15, size=n1, prob=pbar1)
b1
b2 <- pbinom(10, size=n1, prob=pbar1)
b2
b <- b1-b2
b
c <- pbinom(10,size=n1, prob=pbar1, lower.tail = FALSE)
c
d <- 1 - pbinom(10,size=n1, prob=pbar1)
d

# question 2
# average 10 person in 2 hours. exactly have 9 customers? have 10~15 ? 
e <- dpois(9, lambda=10)
e
f1 <- ppois(9, lambda=10)
f1
f2 <- ppois(15, lambda=10)
f2
f <- f2-f1
f

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