# A quick review of using pt and qt
#
#   For the Student's t distribution
#
# find P(X<1.34) for 6 degrees of freedom
pt( 1.34, 6 )
# find P( X > 1.34) for 9 degrees of freedom
1 - pt( 1.34, 9 )  # first way to do this
pt( 1.34, 9, lower.tail=FALSE)
#
# find P( -0.43 < X < 1.73 ) for 7 degrees of freedom
pt( 1.73, 7 ) - pt( -0.43, 7 )
#
# find P( X < -0.75 or X > 1.14 ) for 19 degrees of freedom
#            first way
pt( -0.75, 19 ) + ( 1 - pt(1.14, 19))
#            second way
pt( -0.75, 19 ) + pt(1.14, 19, lower.tail=FALSE)
#            third way
1 - ( pt(1.14, 19)-pt( -0.75, 19) )
#
# find the value of x such that 
#    P( X < x ) = 0.23 for 6 degrees of freedom
qt( 0.23, 6 )
# find the value of x such that 
#    P( X > x ) = 0.23 for 10 degrees of freedom
qt( 1 - 0.23, 10 )  # first way
qt( 0.23, 10, lower.tail=FALSE)  # second way

# there is another type of problem that we can
# do because the t distribution is symmetric.
# find x such that 
#  P( -x < X < x ) = 0.58 for 13 degrees of freedom
#  we can find -x by
qt( (1 - 0.58)/2, 13)
# we can find x by 
qt( 0.21, 13, lower.tail= FALSE)
# or 
qt( 0.79, 13 )

#
#  or find x such that 
#  P( X <-x or X>x ) = 0.075 for 4 degrees of freedom
# Because we have a symmetric distribution we 
# also know that this means 
# that P( X < -x ) = 0.075/2 = 0.0375
qt( 0.0375, 4 )  # which gives the value for -x
# or we could do
qt(0.0375, 4, lower.tail=FALSE) # to get x

######################################################
#  Now, let us try some for non-standard             #
#  t distributions.                                  #
######################################################
#
#  For a t population with mean=54.3 and
#  standard deviation 12.7, 
#  find P( X < 63 ) for 3 degrees of freedom?
#  Convert to a t-score
t <- (63 - 54.3)/ 12.7
t
#  then use pt
pt( t, 3 )
#


#######################
# for a mean=74.2 and standard deviation=6.54)find
# P( X > 68) for 8 degrees of freedom
#
# convert to t-score
t <- (68 - 74.2)/12.7
t
#  use pt
1 - pt( t, 8 )  # looking to the left,
# or
pt( t, 8, lower.tail = FALSE)  # looking to the right

#
########################
# for a mean=3.2 and standard deviation= 0.73 distribution
#  find P( 2.8 < X < 3.6 ) for 7 degrees of freedom
#
# convert both values to t-scores
t_left <- (2.8 - 3.2)/ 0.73
t_left
t_right <- ( 3.6 - 3.2 )/ 0.73
t_right
# use the pt functions
pt( t_right, 7 ) - pt( t_left, 7 )
#



#########################
# For a mean=153, standard deviation=14  distribution find
# P( X < 140 or X > 170) for 8 degrees of freedom
#
# convert both values to t-scores
t_left <- ( 140 - 153 ) / 14
t_left
t_right <- ( 170 - 153 ) / 14
t_right
# use pt
pt( t_left, 8) + (1-pt( t_right,8) )
#    or we coud do that as
pt( t_left, 8 ) + pt( t_right, 8, lower.tail=FALSE)
#
#
################### qt
#
# for a mean = 13.2 and standard deviation=8.4 distribution 
# find the value of x such that
# P( X < x ) is 0.315 for 23 degrees of freedom
# for a standard t value would be
t <- qt( 0.315, 23 )
t
# so we convert that to our mean=13.2 and standard deviation=8.4
# via
x <- t*8.4 + 13.2
x


################################
#  for a mean=4.3 and standard deviation= 8.2 distribution
# find the value of x such  that 
#  P( X > x ) = 0.29 for 12 degrees of freedom
# for a standard t value would be
t1 <- qt( 1 - 0.29, 12 )
t1
#  which we could have found by 
t2 <- qt( 0.29, 12, lower.tail=FALSE)
t2
# and then we need to move that value back 
# to our mean=4.3 and standard deviation=8.2
x <- t1*8.2 + 4.3
x
#