### SOLUTIONS TO LING280 W13 Problem Set 1
### M. Wagers, January 18, 2013

### Problem 1

# median of each condition (i.e. component of the data frame)
median(islands$that.matrix)
median(islands$that.embedded)
median(islands$whether.matrix)
median(islands$whether.embedded)
# OR
apply(islands, 2, median)

# standard deviation
sd(islands$that.matrix)
# etc … OR
apply(islands, 2, sd)

# mean
mean(islands$that.matrix)
# etc … OR
apply(islands, 2, mean)
# OR
colMeans(islands)
  
# inter-quartile range
IQR(islands$that.matrix)
# etc … OR
apply(islands, 2, IQR)

### Problem 2

# The function apply applies the function table (its third argument)
# to the columns (its second argument, 2 for columns) of the data 
# frame judge (its first argument). The function table gives the count
# for each level.
apply(judge, 2, table)

### Problem 3
# There are numerous ways to solve this problem; here are 3

# APPROACH 1
# Create a table of Counts per rating category and condition
apply(islands, 2, table) -> islands.tab

# Sum up just the 6/7 category 
apply(islands.tab[6:7, ], 2, sum) -> islands.hi.sum

# Sum up each condition
apply(islands.tab, 2, sum) -> islands.bycondition.sum

# Take the ratio
islands.hi.prop <- islands.hi.sum / islands.bycondition.sum

# APPROACH 2
# Use the prop.table function; the second arg, '2' means to proportionalize by column
prop.table(islands.tab, 2) -> islands.tab.prop
islands.hi.prop <- apply(islands.tab.prop[6:7,], 2, sum)

# APPROACH 3
# Calculate the empirical cumulative distribution function; it gives us the 
# fraction of judgments less than or equal to its argument (5 in this case);
# subtract from 1 to obtain the proportion of judgements assigned 6 or 7.
Fn.tm <- ecdf(islands$that.matrix)
1-Fn.tm(5)

# Repeat for each condition


### Problem 4

# boxplot
quartz()
boxplot(islands)


### Problem 5

# freq=TRUE produces a histogram with the counts
quartz()
hist(islands[,'whether.matrix'], breaks=7, freq=TRUE)

# freq=FALSE produces a histogram with the probability density
quartz()
hist(islands[,'whether.matrix'], breaks=7, freq=FALSE)


### Problem 6

# This code fragment performs 5 different experiments which differ in the
# number of observation that each experiment involves (i.e. how many observations
# are sampled). Each experiment is replicated 1000 times and the mean for each
# experiment is plotted (i.e. boxplot of the means). The resulting boxplots show
# that the higher the nunmber of observations the means are based on the higher
# the precision of the estimate of the random variable.

rep.experiment <- matrix(nrow=1000, ncol=5)

islands[,'whether.embedded'] -> island.judgments

for(i in 1:1000){
	
	sample(island.judgments,5,replace=TRUE) -> n5;
	mean(n5) -> rep.experiment[i,1]; 
	
	sample(island.judgments,10,replace=TRUE) -> n10;
	mean(n10) -> rep.experiment[i,2];
	
	sample(island.judgments,20,replace=TRUE) -> n20;
	mean(n20) -> rep.experiment[i,3];
	
	sample(island.judgments,40,replace=TRUE) -> n40;
	mean(n40) -> rep.experiment[i,4];
	
	sample(island.judgments,80,replace=TRUE) -> n80;
	mean(n80) -> rep.experiment[i,5];
}

quartz()
boxplot(rep.experiment~col(rep.experiment), 
	names=c("n = 5", "n = 10", "n = 20", "n = 40", "n = 80"), 
	ylim=c(1,7), ylab="mean rating", xlab="sample size", main="Simulated outcome of 1000 experiments")


### Problem 7

# This code fragment computes and plots the average squared deviations of the data from the numbers along the interval [0-7] - the smallest 'average squared deviation' is from the mean of the data set -- this is the variance -- i.e., the mean is the value for which the average deviation is smallest. The mean minimizes variance!

# create a vector numbers containing the numbers from 0 to 7  in 0.05 increments
numbers <- seq(0, 7, by=0.05)

# create a vector with the length of the newly created numbers vector
squared.deviation <- vector(length=length(numbers))

# initialize k
k <- 1

# loop for every number in numbers
for (n in numbers) {
	# for each number in the numbers vector, subtract the vector island.judgements from n;
	# square the resulting vector and sum its elements storing the value in sum.squares
	sum((n - island.judgments)^2) -> sum.squares;
	# divide the sum of the squares by the number of elements in island.judgements minus 1
	# store the result in the current position in the squared.deviation vector
	sum.squares / (length(island.judgments) - 1) -> squared.deviation[k];
	# increment k
	k <- k+1;
	}
	
quartz()
# plot the values in squared.deviation with the numbers on the x axis
plot(numbers, squared.deviation, pch="+")

# draw a horizontal line through the min value of squared.deviation
abline(h=min(squared.deviation))