#########################################################################
###########                     Hypotesis test               ############ 
#########################################################################


################### examples on the slides #############################

# 1 variance in the pop is known 

xbar = 2.84            # sample mean 
mu0 = 3            # hypothesized value 
sigma = 0.8            # population standard deviation 
n = 100                # sample size 

z = (xbar-mu0)/(sigma/sqrt(n)) #Z-stat 
z   #Z-stat 


alpha = 0.05 

# critical values z (two tails test) 
# qnorm:  quantile function  for the normal distribution 
qnorm(alpha/2) # negative value (only 0.025 of the distrib)
qnorm(1-alpha/2) # positive value (0.925 of the distrib)

# conclusion: Z-stat is bigger than 1.96=> reject null hypothesis

# p-value computation 
pval = 2 * pnorm(z)
pval

lower_bound<-xbar-(1.959964*(sigma/sqrt(n)))
upper_bound<-xbar+(1.959964*(sigma/sqrt(n)))

ci<-c(lower_bound, upper_bound)
ci

################################################################################

# 2 - # 1 variance in the pop is unknown - Use S (st.dev in the sample)
xbar = 172.5             # sample mean 
mu0 = 168              # hypothesized value 
s = 15.4               # sample standard deviation 
n = 25                # sample size 
t = (xbar-mu0)/(s/sqrt(n)) 
t                      # test statistic 
#t-stat - 1.46


df=n-1
alpha = 0.05 
t.alpha = qt(1-alpha/2, df=n-1) 
t.alpha                # critical value 2.06

# Conclusion: do not reject Ho (Accept Ho)

pt(t,df)

############# proportion case #################################################


pbar = 0.05           # sample proportion 25/500
p0 = 0.08             # hypothesized value in the Pop
n = 500                # sample size 
z = (pbar-p0)/sqrt(p0*(1-p0)/n) 
z                      # test statistic 
#[1] 0.89443

#We then compute the critical values at .05 significance level.

alpha = .05 
z.half.alpha = qnorm(1-alpha/2) 
c(-z.half.alpha, z.half.alpha) 
#[1] ???1.9600  1.9600

# conclusion: reject Ho 

pval = 2 * pnorm(z)
pval


#############################################################################################
########################### Hypotesis test with dataset ######################################

data<-mtcars

summary(data$mpg)

mu=22 # valore di Ho 

#t.test(x, y = NULL,
#       alternative = c("two.sided", "less", "greater"), # choose an  altern. hyp on the  basis of the test 
#       conf.level = 0.95 )


mpg<-data$mpg

t.test(data$mpg, alternative = "two.sided", mu=22, conf.level = 0.95)
t.alpha = qt(0.975, df=31)  # crit.val. 
t.alpha
# accept ho

##### VARIANCE KNOWN 
# es. mean  mpg
# sigma=5 (conf.int 0.95) --> variance in the pop is known (mu=22 is the mean in the popolation)

# Ho: mu=22
# H1: mu != 22 


library("BSDA")
z.test(mtcars$mpg, alternative = "two.sided", mu=22, sigma.x = 5, conf.level = 0.95)


########## proportion. test  
#  variable am, con p_0=0.45

# Ho: p0=0.45
# H1: p0 != 0.45 

# test bilaterale --> two.sided
table(data$am) #13 units with value equal to 1

prop.test(x = 13, n = 32,  p=0.45, alternative = "two.sided", correct = FALSE)




########### exercise to do at home ####################

# load dataset marketing campagin 

# A) compute an hypotesis test in which the food exp is equal to  25 
# against the hypotesis that the mean is greater by considering a conf. level equal to 90%


# B) compute an hypotesis test in which the percentage of complaint is equal to  3% 
#against the hypotesis that the prop is lower than those hypotized in the null hyp 
# consider a conf. level equal to  90%