## 1

die <- c(87,96,108,89,122,98)

probs.null <- c(1/6,1/6,1/6,1/6,1/6,1/6)

expec.counts <- sum(die)*probs.null; print (expec.counts)

chisq.test(die, p=probs.null)

###############################################
        Chi-squared test for given probabilities

data:  die 
X-squared = 8.58, df = 5, p-value = 0.127
###############################################

# 2

Oj <- c(0,6,8,12)

expect.counts <- c(26/4,26/4,26/4,26/4)

# Test statistic:

test.stat <- sum(Oj^2 / expect.counts) - sum(Oj); print(test.stat)

p.val <- pchisq(test.stat, df=3, lower=F); print(p.val)

##############################
> test.stat <- sum(Oj^2 / expect.counts) - sum(Oj); print(test.stat)
[1] 11.53846
> 
> p.val <- pchisq(test.stat, df=3, lower=F); print(p.val)
[1] 0.009143631
##############################

# 3


prop.test(matrix(c(23,7,27,3),nrow=2,ncol=2,byrow=TRUE), alternative="less", correct=FALSE)

#################################################
data:  matrix(c(23, 7, 27, 3), nrow = 2, ncol = 2, byrow = TRUE) 
X-squared = 1.92, df = 1, p-value = 0.08293
alternative hypothesis: less 
#################################################



# 4

fisher.test(matrix(c(10,11,14,49),nrow=2,ncol=2,byrow=TRUE), alternative="greater")

#######################
data:  matrix(c(10, 11, 14, 49), nrow = 2, ncol = 2, byrow = TRUE) 
p-value = 0.02774
########################


# 5

# chi-square test for independence

horse.data <- matrix(c(8,6,8,16,3,6,5,28), nrow=2, ncol=4, byrow=TRUE, 
dimnames = list(c("1-4", "5-9"), c("1", "2", "3", "Other")))

as.table(horse.data)

# A quick way to get expected cell counts and verify large-sample assumption is met:

expected.counts <- (apply(horse.data,1,sum) %o% apply(horse.data,2,sum))/sum(horse.data)
print(expected.counts)


chisq.test(horse.data, correct=FALSE)

####################################################
> print(expected.counts)
        1   2     3 Other
1-4 5.225 5.7 6.175  20.9
5-9 5.775 6.3 6.825  23.1

        Pearson's Chi-squared test

data:  horse.data 
X-squared = 6.0529, df = 3, p-value = 0.1091
####################################################


## 6


drill.2c <-  matrix(c(11,8,8,15,9,14,12,7), nrow=2, ncol=4, byrow=TRUE, 
dimnames = list(c("> GM", "<= GM"), c("A", "B", "C", "D")))

# Printing the 2 by c table:

as.table(drill.2c)

# A quick way to get expected cell counts:

expected.counts <- (apply(drill.2c,1,sum) %o% apply(drill.2c,2,sum))/sum(drill.2c)
print(expected.counts)

# Row sums:

> apply(drill.2c,1,sum)
 > GM <= GM 
   42    42 

#Column sums:

> apply(drill.2c,2,sum)
 A  B  C  D 
20 22 20 22 

# Test statistic:

test.stat <- ((84^2)/(42*42))*(11^2 /20 + 8^2/22 + 8^2/20 + 15^2/22) - (84*42/42); print(test.stat)

# P-value:

pchisq(5.55,df=3,lower=F)

##################################

> test.stat <- ((84^2)/(42*42))*(11^2 /20 + 8^2/22 + 8^2/20 + 15^2/22) - (84*42/42); print(test.stat)
[1] 5.545455
> 
> # P-value:
> 
> pchisq(5.55,df=3,lower=F)
[1] 0.1356785
##################################