#This is a second example from Topic 16
#  The problem is:
#   1) we have a population with an unknown mean
#      and unknown standard deviation.
#   2) we have a null hypothesis that the mean of the 
#      population is 135.3
#   3) we have an alternative hypothesis that the mean
#      of the population is not equal to 135.3
#   4) we want to test the null hypothesis against the 
#      alternative hypothesis at the 0.03 level of
#      significance.
#   5) to do this we take a random sample of size 32
#   6) compute the sample mean and standard deviation;
#      sample mean = 138.8   sample standard deviation=9.2
#   -----------------------
#   7) at this point we could use the critical value
#      approach and find the critical low value and critical
#      high values for the sample mean, the values that have
#      probability that a sample mean will be either less than 
#      the critical low value or higher than the critical high
#      value = 0.03, The probabilities need to be split evenly.
#   8) if the sample mean is less than the critical low
#      or higher than the critical high
#      value then we reject H0 in favor of H1, if not then 
#      we do not have enough evidence to reject H0 in favor 
#      of H1.
#   ------- alternatively  -----
#   9) we could use the attained significance approach and
#      determine the probability of getting a sample mean 
#      as extreme or more extreme than we just got.  Note 
#      that we need to specify extreme or more extreme, not
#      just lower or higher.
#  10) if that probability is less than the level of 
#      significance then we reject H0 in favor of H1, 
#      if not the we do not have enough evidence to reject
#      H0 in favor of H1
################################################

#  for the critical value approach first
#  find the t value with 0.03/2 to its right
#      for 31 degrees of freedom
high_t <- qt( 0.03/2, 31, lower.tail=FALSE )
high_t
#  by symmetry, the low t values will be the opposite
low_t <- -high_t
low_t
#
#   our critical low will be 
135.3 + low_t*9.2/sqrt( 32 )  # remember that low_t is negative
#   our critical high value will be
135.3 + high_t*9.2/sqrt( 32 )  # remember that high_t is positive

#  Because our sample mean, 138.8 is not less than the
#  critical low value  and not higher than the 
#  critical high we do not reject H0 in favor of H1

#  or use the attained significance approach
#  First normalize the sample mean so that we can use 
#  the pt() function.
#
standard_t <- (138.8-135.3)/(9.2/sqrt(32))
standard_t  # this becomes our test statistic
# find the probability of getting that value or more extreme.
# because this is a high value we will use lower.tail=FALSE
p_val <- pt( standard_t, 31, lower.tail=FALSE )
#  That answer now needs to be doubled to account for
#  being this extreme on the other side (lower side in this case).
p_val*2
#   that answer, 0.03929236, is not less than our level of 
#   significance, 0.03 so we do not reject H0 
#   in favor of H1

#
#   Or we could have just used our 
#       hypoth_test_unknown function to compute both
#       approaches
source("../hypo_unknown.R")
#              note the     0   to indicate that H1: mean != 135.3
hypoth_test_unknown( 135.3, 0, 0.03, 32, 138.8, 9.2)