# 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 #