#  Line 1:  a small demonstration of hypothesis testing,
#   in this case for a population with unknown standard deviation
#
#  First, we will get a population
#  In this case we will get a large population
source("../gnrnd5.R")
gnrnd5(177938412404,184000676)
#let us look at the head and tail values
head(L1)
tail(L1)
min(L1)
max(L1)
# just a quick look at L1
hist(L1)
boxplot(L1, horizontal=TRUE)
source("../assess_normality.R")
assess_normality( L1 )
#
#  L1 sure looks like a Normal distribution.
#
# ##########################
# ##  Problem: test the null hypothesis
# ##      that the population mean is equal to 60.0
# ##      against the alternative hypothesis that
# ##      the population mean is greater than 60.0.
# ##      Run the test at the 0.05 level of significance.
# ##########################

# take a simple random sample of size 23
#
#  Be careful:  Every time we do this we get 
#               a different random sample
#
L2 <- as.integer( runif(23, 1, 4126) )
# L2  holds the index values of our simple random sample
L2
L3 <- L1[ L2 ]   # L3 holds the simple random sample
L3
# we will get the mean of L3
xbar <- mean(L3)
xbar
# and we need to get the standard deviation of 
#  the sample
sx <- sd( L3 )
sx
#  The long way to do the test is to find the probability 
#  getting this mean or higher if the true mean is 60.0 
#  given that the standard deviation of the population is
#  unknown and that we have a sample of size 23 
#  from a population that we know to be normally distributed.
#
#  Since the means of samples of size 23 are 
#  distributed  as a Student's-t with 22 degrees
#  of freedom and with mean = population mean and standard
#  deviation = sx/sqrt( n )  for the 
#  attained significance approach we just need to find
pt( (xbar-60)/( sx/sqrt(23)), 22,
       lower.tail=FALSE)
#  If that value is less than 0.05 then we reject the 
#  null hypothesis in favor of the alternative

#  For the critical value approach we first need to find the
#  value of t that has P(X>t)=0.05
t <- qt( 0.05,  22,  lower.tail=FALSE)
t
#  Then transform that to our sample
t*sx/sqrt(23)+60.0
# Then if xbar is greater than this value we reject the
#    null hypothesis in favor of the alternative
#
#   Of course, we could use the function
#   hypoth_test_unknown() to do both of these approaches in
#  one easy step.
#
source("../hypo_unknown.R")
hypoth_test_unknown( 60.0, 1, 0.05,
                   23, xbar, sx)
#
#
#################################
# go back and execute lines 34-77 many more times.
#    Each time you get a different random sample.
#    Keep track of the number of times that you reject or
#    do not reject the null hypothesis.  By the way, the 
#    true mean of the population is 67.63552.  
#################################

#
#  now we will do the same thing for a different population
#
gnrnd5(146723412404,184000600)
#let us look at the head and tail values
head(L1)
tail(L1)
min(L1)
max(L1)

sigma <- pop_sd( L1 )
sigma
# just a quick look at L1
hist(L1)
boxplot(L1, horizontal=TRUE)

assess_normality( L1 )
#
#  L1 sure looks like a Normal distribution.
#
# ##########################
# ##  Problem: test the null hypothesis
# ##      that the population mean is equal to 60.0
# ##      against the alternative hypothesis that
# ##      the population mean is greater than 60.0.
# ##      Run the test at the 0.05 level of significance.
# ##########################

# take a simple random sample of size 23
#
#  Be careful:  Every time we do this we get 
#               a different random sample
#
L2 <- as.integer( runif(23, 1, 4126) )
# L2  holds the index values of our simple random sample
L2
L3 <- L1[ L2 ]   # L3 holds the simple random sample
L3
# we will get the mean of L3
xbar <- mean(L3)
xbar
# and we need to get the standard deviation of 
#  the sample
sx <- sd( L3 )
sx
#  The long way to do the test is to find the probability 
#  getting this mean or higher if the true mean is 60.0 
#  given that the standard deviation of the population is
#  unknown and that we have a sample of size 23 
#  from a population that we know to be normally distributed.
#
#  Since the means of samples of size 23 are 
#  distributed  as a Student's-t with 22 degrees
#  of freedom and with mean = population mean and standard
#  deviation = sx/sqrt( n )  for the 
#  attained significance approach we just need to find
pt( (xbar-60)/( sx/sqrt(23)), 22,
    lower.tail=FALSE)
#  If that value is less than 0.05 then we reject the 
#  null hypothesis in favor of the alternative

#  For the critical value approach we first need to find the
#  value of t that has P(X>t)=0.05
t <- qt( 0.05,  22,  lower.tail=FALSE)
t
#  Then transform that to our sample
t*sx/sqrt(23)+60.0
# Then if xbar is greater than this value we reject the
#    null hypothesis in favor of the alternative
#
#   Of course, we could use the function
#   hypoth_test_unknown() to do both of these approaches in
#  one easy step.
#
source("../hypo_unknown.R")
hypoth_test_unknown( 60.0, 1, 0.05,
                     23, xbar, sx)
#
#################################
# go back and execute lines 121-164 many more times.
#    Each time you get a different random sample.
#    Keep track of the number of times that you reject or
#    do not reject the null hypothesis.  By the way, the 
#    true mean of the population is 60.03343.  
#################################