#This is some work for Topic 26
#
# First input the matrix that has perfect independence
matrix_perfect <- matrix(
c( 24, 20, 12, 36, 30, 18, 30, 25, 15, 24, 20, 12),
ncol=3, nrow=4, byrow=TRUE)
matrix_perfect
# Now, generate all the related tables
source("../crosstab.R")
crosstab( matrix_perfect )
#
# Now, repeat the same thing, but change 1 value
matrix_perfect <- matrix(
c( 25, 20, 12, 36, 30, 18,
30, 25, 15, 24, 20, 12),
ncol=3, nrow=4, byrow=TRUE)
matrix_perfect
# and get the new tables
crosstab( matrix_perfect )
# rather than look at the small print in the View tabs
row_percent
#
# Now let us look at the other data on the video
matrix_other <- matrix(
c( 26, 23, 13, 27, 40, 16, 35, 19, 10, 21, 13, 14),
ncol=3, nrow=4, byrow=TRUE)
matrix_other
# and get the new tables
crosstab( matrix_other )
#
# And a quick look at the Chi-squared distribution
# First verify the last attaained result
pchisq( 11.77495, 6, lower.tail=FALSE)
# Then find the value that has 5% of the area to
# the right
qchisq( 0.05, 6, lower.tail=FALSE)
#
# Finally, let us genetate a new matrix
source("../gnrnd4.R")
gnrnd4( key1=895863908, key2=75864894754 )
matrix_A
# See if the rows and columns of this are independent,
# that is the null hypothesis, versus that they are
# not independent, that is the alternative hypothesis,
# and do this test at the 0.05 level of significance.
crosstab( matrix_A )
#
#
# This was such fun, let us generate a new matrix
gnrnd4( key1=114863908, key2=75864894754 )
matrix_A
crosstab( matrix_A )