#  Topic 11e
#  Look at the Student's-t distribution
# 
#  First we will look at the standard 
#      Student's-t, mean=0, standard deviation=1
#
#  We will just repeat the kind of questions that
#  we saw in earlier distributions.
#
#  Remember, that for the Student's-t we need to
#  know the number of degrees of freedom to use
#

############################################
#   Using the table, find
#For the standard Student's-t distribution
# for df=6   P(X < -0.47 ) = 
# for df=18  P(X < -0.47 ) = 
# for df=60  P(X < -0.47 ) = 
# for df=32  P(X <= 2.38 ) = 
# for df=19  P(X < 1.3765) =
# for df=5   P( X > 2.24 ) =
# for df=15 P( X > -1.39 ) =
# for df=28 P( -0.62 < X < 1.26 ) =
# for df=14 P( X < 0.34  or X>0.98 ) =
######################################

################
#  Now go back and try to do those problems
#  with the "Critical Values for the Student's-t"
#  table.
################

######################################
# now do the same using the pt() function

# for df=6   P(X < -0.47 ) =  
pt( -0.47, 6 )

# for df=18  P(X < -0.47 ) = 
pt( -0.47, 18 )

# for df=60  P(X < -0.47 ) = 
pt( -0.47, 60 )

# for df=32  P(X <= 2.38 ) = 
pt( 2.38, 32 )

# for df=19  P(X < 1.3765) =
pt( 1.3765, 19 )

# for df=5   P( X > 2.24 ) =
1 - pt( 2.24, 5 ) # Is one approach
pt( 2.24, 5, lower.tail=FALSE )  # another way

# for df=15 P( X > -1.39 ) =
1 - pt( -1.39, 15 ) # Is one approach
pt( -1.39, 15, lower.tail=FALSE )  # another way

# for df=28 P( -0.62 < X < 1.26 ) =
pt(1.26, 28) - pt(-0.62, 28)

# for df=14 P( X < 0.34  or X>0.98 ) =
pt(0.34,14) + (1-pt(0.98, 14))  # is one way
pt(0.34, 14) + 
    pt( 0.98, 14, lower.tail=FALSE)# is another way

#######################################
##  go back to the table and use it to solve:
#
#  find the y value such that
#  for df=8    P(X < y ) = 0.2374    y=
#  for df=5    P(X < y ) = 0.7620    y=
#  for df=13   P(X > y ) = 0.3550    y=
#  for df=22   P(X > y ) = 0.9181    y=
#  for df=9    P(X < y ) = 0.6409    y=
#        because Student's-t is symmetric
#  for df=12   P(X< -y or X>y ) = 0.0348  y=?
######################################

######################################
# now do the same using the qt() function

#  for df=8    P(X < y ) = 0.2374    y=
qt( 0.2374, 8 )

#  for df=5    P(X < y ) = 0.7620    y=
qt( 0.7620, 5 )

#  for df=13   P(X > y ) = 0.3550    y=
qt( 1 - 0.3550, 13 )  # is one way
qt( 0.3550, 13,lower.tail=FALSE ) # is another way

#  for df=22   P(X > y ) = 0.9181    y=
qt( 1 - 0.9181, 22 )  # is one way
qt( 0.9181, 22, lower.tail=FALSE ) # is another way

#  for df=9    P(X < y ) = 0.6409    y=
qt( 0.6409, 9 )

#        because Student's-t is symmetric
#  for df=12   P(X< -y or X>y ) = 0.0348  y=?
qt( 0.0348/2, 12, lower.tail=FALSE)


#############################################
##  Now we will look at non-standard Student's-t
##  distribution.  For each such non-standard
##  Student's-t distribution we need to give the 
##  mean and the standard deviation of that
##  distribution.  We will do this in the 
##  St(mean, standard deviation) syntax.  Thus
##  St(25,7) is a Student's-t distribution with 
##  mean=25 and standard deviation=7.
###############################################
##

# for a St(25,7) with 18 df find P(X < 32)
# the long way is to compute the equivalent t value
# for a standard Student's-t with 18 df
t <- (32 - 25)/7
t
#  and then find P(X < t ) for df=18 and St(0,1)
pt( t, 18 )


# for a St(9,13) with 7 df,  find P(X<32)
#  the long way
t <- (32-9)/13
t
pt( t, 7 )


# for St(13.7, 4.6) and df=26 find P( X > 16.2 )
# the long way
t <- (16.2 - 13.7)/4.6
t
pt( t, 26, lower.tail=FALSE)
#  or 
1 - pt( t, 26)


# for St(-6.2, 1.34) with df=4 find P( -7 < x < -5)
# the long way
t1 <- (-7 - -6.2)/1.34
t1
t2 <- (-5 - -6.2)/1.34
t2
pt( t2, 4 ) - pt(t1,4)



##############################################
###   Now we will go the other way.  Find the 
###   y value such that p is the desired 
###   probability.

# for a St(45.7,8.6 ) with df=16, find y such
#     that P(X < y ) = 0.333
# the long way...
#   what is the t value in a standard Student's-t 
#   with df=16 such that P(X < t) = 0.333
t <- qt( 0.333, 16 )
t
# then translate that back to our St(45.7, 8.6)
t*8.6+45.7


# for a St(-9.2, 1.6 ) with df=34, find y such
#     that P(X > y ) = 0.219
#   what is the t value in a standard Student's-t 
#   such that P(X > t) = 0.219
t <- qt( 1 - 0.219, 34 )
t
# then translate that back to our St(-9.2, 1.6 )
t*1.6 + -9.2


# for a St(134.2, 12.43 ) with df=7 distribution, 
#     find y1 and y2 such
#     that P(X < y1 or X>y2 ) = 0.075
#     and where abs(mean - y1 ) = abs(mean-y2)
#   What is the t value in a standard Student's-t 
#   such that P(-z < X < z) = 0.075
#
# this works because the Student's-t distribution
#     is symmetric
t <- qt( 0.075/2, 7, lower.tail=FALSE )
t
# then translate t back to our N(134.2, 12.43 )
y2 <- t*12.43 + 134.2  #  this is our y2
y2
# and translate -z back to our N(134.2, 12.43 )
y1 <- -t*12.43 + 134.2  # this is our y1
y1
#  note
abs(134.2-y1)
abs(134.2-y2)