# line 1
#  Look at finding a confidence interval for
#  the mean difference of  PAIRED data
#  when the population standard deviation is unknown.
#
#  For this example get a 95% confidence interval for
#  the mean  the difference of the pairs
#
#  We will start by generating a population, but the
#  data appears as paired values
source("../gnrnd5.R")
gnrnd5( 282091499910, 41001525010176)

head(L1,10)
head(L2,10)
tail(L1,10)
tail(L2,10)
length(L1)

#
#  Now that we have our population of pairs 
#  we need to take random samples of the pairs 
#  of values
#
#        ############################################
#        ##  Each time we do the following steps   ##
#        ##  we will get a different sample and as ##
#        ##  such we will get different confidence ##
#        ##  interval.                            ##
#        ############################################
n_1 <- 38  # get a sample of size 38 from L1 and L2
index_1 <- as.integer( runif( n_1, 1, 5001))
index_1
samp_1 <- L1[ index_1 ]
samp_1
samp_2 <- L2[ index_1 ]
samp_2
#
#  We really want to look at a confidence interval for L2-L1
#  so we will form a new data set based upon samp_2 - samp_1
samp_3 <- samp_2 - samp_1
samp_3
#
# then our best point estimate is the mean(samp_3)
pnt_est <- mean( samp_3 )
pnt_est
#
#  Our problem now resolves to finding the 95% confidence
#  interval for the mean on samp_3, but samp_3 is 
#  disributed as a Student's-t with n_1 - 1 degrees of
#  freedom.
#
# find the t value with half of 5% to its right
t <- qt( 0.025, n_1 - 1, lower.tail=FALSE )
std_dev <- sd( samp_3 )
std_dev
# then the confidence interval will have a low value of 
pnt_est - t*std_dev/sqrt( n_1 )
# and the high value of 
pnt_est + t*std_dev/sqrt( n_1 )
#
# rather than take the long way for this we could just 
# jump to doing ci_unknown.
# 
source("../ci_unknown.R")
ci_unknown( sd( samp_3), length(samp_3),
            mean( samp_3), cl=0.95)
#
#    ################################################
#    ##   Now, highlight and rerun, over and over, ##
#    ##   lines 31-67, to get repeated samples and ##
#    ##   thus, repeated 95% confidence intervals. ##
#    ##   While you do this be aware that the true ##
#    ##   mean of the paired values in the two     ##
#    ##   populations is 4.6501. How often do you r##
#    ##   intervals contain the true mean?         ##
#    ################################################

#  Finally, we could get a picture of the change between
# samp_1 and samp_2 via the following commands.
#  This will be for the last repetition of lines 31-67
plot(1:38,samp_1, col="darkgreen",
     xlim=c(0,45), xaxp=c(0,45,9),
     xlab="index values",
     ylim=c(40,160), yaxp=c(40,160, 12),
     ylab="item values",pch=22, las=1, cex.axis=0.7,
     main="Pairs of Values from last sample"
)

points(1:38,samp_2, col="darkred", pch=20)
for(i in 1:38)
{ lines(c(i,i),c(samp_1[i],samp_2[i]))}
abline(h=seq(40,160,10),lty=3,col="darkgray")
abline(v=seq(0,40,5),lty=3,col="darkgray")
legend("topright", 
       legend = c("samp_1","samp_2"),
       pch=c(22,20),
       col=c("darkgreen","darkred"),
       inset=0.03)