#This is an 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 18.6
# 3) we have an alternative hypothesis that the mean
# of the population is less than 18.6
# 4) we want to test the null hypothesis against the
# alternative hypothesis at the 0.025 level of
# significance.
# 5) to do this we take a random sample of size 37
# 6) compute the sample mean and standard deviation;
# sample mean = 17.21 sample standard deviation=4.8
# -----------------------
# 7) at this point we could use the critical value
# approach and find the critical low value for the
# sample mean, the value that has probability that a
# sample mean will be less than that value = 0.025
# 8) if the sample mean is less than the critical low
# 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 low or lower than we jsut got.
# 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 0.025 to its left
# for 36 degrees of freedom
low_t <- qt( 0.025,36 )
low_t
#
# our critical low will be
18.6 + low_t*4.8/sqrt( 37 ) # remember that low_t is negative
# Because our sample mean, 17.21 is not less than the
# critical low value we do not reject H0 in favor of H1
# or use the attained significance approach
# First normalize the sample mean sor we can use
# the pt() function.
#
standard_t <- (17.21-18.6)/(4.8/sqrt(37))
standard_t # this becomes our test statistic
# find the probability of getting that value or lower
pt( standard_t, 36 )
# that answer, 0.0433, is not less than our level of
# significance, 0.025 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 -1 to indicate that H1: mean < 18.6
hypoth_test_unknown( 18.6, -1, 0.025, 37, 17.21, 4.8)