##### SCA-31b
##### 
##### Analysis of Variance, II
##### 

### This gives a few more examples of ANOVA procedure



### Preamble

source("http://rfs.kvasaheim.com/stat200.R")
library(agricolae)

dt = read.csv("http://rfs.kvasaheim.com/data/stat4013rxtimes.csv")

dt$attempt = as.factor(dt$attempt)
dt$group   = as.factor(dt$group)

summary(dt)
attach(dt)


### Example 1: Distance by group

shapiroTest(distance~group)

kruskal.test(distance,group)
print( kruskal(distance,group) )

#   Conclusion:
#   Because the p-value of 0.0000 is less than our usual alpha
#   of 0.05, we reject the null hypothesis. We did detect a
#   difference in the average distance fallen across the 13
#   groups.
#
#   In fact, Group A fell more (on average) of any other group.




### Example 2: Distance by Person Number

shapiroTest(distance~person)

kruskal.test(distance~person)
#   Conclusion:
#   Because the p-value of 0.6050 is greater than our usual
#   alpha of 0.05, we cannot reject the null hypothesis. We
#   did not detect a difference in the avearge distance
#   fallen by the person number within the group.






### Example 3: Distance by Attempt Number

shapiroTest(distance~attempt)

kruskal.test(distance~attempt)

print(kruskal(distance,attempt))

#   Conclusion:
#   Because the p-value of 0.0319 is less than our usual alpha
#   value of 0.05, we did detect a difference in the average
#   drop distance based on attempt number. 
#
#   In fact, the first attempt was significantly slower than 
#   attempts 2, 3, and 4. 



### Example 4: Distance by treatment

shapiroTest(distance~treatment)

kruskal.test(distance~treatment)

print(kruskal(distance,treatment))

#   Conclusion: 
#   Because the p-value of 0.0000 is less than the usual alpha 
#   of 0.05, we did detect a difference in average drop 
#   distance between the control and treatment groups. 
#

boxplot(distance~treatment)
aggregate(distance, by=list(treatment), mean)

#
#   In fact, we are able to conclude that the control group (singing)
#   tends to allow the ruler to drop an average of 1.5 inches more
#   than the control group.
#

#
#   By the way, we can use math and physics to estimate just how much
#   this means in terms of car lengths...
#

#   Since d = 1/2 gt^2, we know t = sqrt(2d/g)

time = sqrt(2*distance*2.54/9.80665)
aggregate(time, by=list(treatment), mean)

#   This tells us that the singing group is slower by an average
#   of 0.23s.
#
#   If a car it traveling 65 mph, this is

65*0.23/60/60         # miles
65*0.23/60/60*5280    # feet

#   The difference in stopping distances at highway speeds is
#   22 feet. Is this a lot? It depends on what is found in 
#   those 22 feet.