Two Sample t-Test
Example with Wide Format Data
# read in our data (wide format)
Cd.BeetBarley<- data.frame(
redbeet= c(18, 5, 10, 8, 16, 12, 8, 8, 11, 5, 6, 8, 9, 21, 9),
barley= c(8, 5, 10, 19, 15, 18, 11, 8, 9, 4, 5, 13, 7, 5, 7))
# three ways to look at the data structure
str(Cd.BeetBarley)
'data.frame': 15 obs. of 2 variables:
$ redbeet: num 18 5 10 8 16 12 8 8 11 5 ...
$ barley : num 8 5 10 19 15 18 11 8 9 4 ...
summary(Cd.BeetBarley)
redbeet barley
Min. : 5.00 Min. : 4.0
1st Qu.: 8.00 1st Qu.: 6.0
Median : 9.00 Median : 8.0
Mean :10.27 Mean : 9.6
3rd Qu.:11.50 3rd Qu.:12.0
Max. :21.00 Max. :19.0
head(Cd.BeetBarley)
redbeet barley
1 18 8
2 5 5
3 10 10
4 8 19
5 16 15
6 12 18
with(Cd.BeetBarley,boxplot(redbeet,barley,
col= "lightgrey",
main= "Phytoremediation Efficiency of Crop Plants",
xlab= "Crop type", ylab= "Cadmium reduction (%)",
names= c("Redbeet","Barley"),
ylim= c(0,25), las= 1,
boxwex=0.6))
with(Cd.BeetBarley, var.test(redbeet, barley))
F test to compare two variances
data: redbeet and barley
F = 0.97888, num df = 14, denom df = 14, p-value = 0.9687
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.3286377 2.9156685
sample estimates:
ratio of variances
0.9788762
with(Cd.BeetBarley, t.test(redbeet, barley, var.equal = TRUE))
Two Sample t-test
data: redbeet and barley
t = 0.38658, df = 28, p-value = 0.702
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.865852 4.199185
sample estimates:
mean of x mean of y
10.26667 9.60000
Example with Long Format Data
# read in our data (long format)
Cd.CabbageMaize <- data.frame(remed.pcnt = c(46, 50, 44, 44, 43, 52, 48, 24, 51,
29, 53, 32, 61, 59, 35, 34, 26, 44, 17, 34, 19, 34, 34, 43, 18, 34, 27, 27, 53,
30), plt.typ = c(rep("cabbage", times = 15), rep("maize", times = 15)))# read in our data (long format)
Cd.CabbageMaize <- data.frame(remed.pcnt = c(46, 50, 44, 44, 43, 52, 48, 24, 51,
29, 53, 32, 61, 59, 35, 34, 26, 44, 17, 34, 19, 34, 34, 43, 18, 34, 27, 27, 53,
30), plt.typ = c(rep("cabbage", times = 15), rep("maize", times = 15)))
# get summary & check data structure
str(Cd.CabbageMaize)
'data.frame': 30 obs. of 2 variables:
$ remed.pcnt: num 46 50 44 44 43 52 48 24 51 29 ...
$ plt.typ : chr "cabbage" "cabbage" "cabbage" "cabbage" ...
summary(Cd.CabbageMaize)
remed.pcnt plt.typ
Min. :17.00 Length:30
1st Qu.:29.25 Class :character
Median :34.50 Mode :character
Mean :38.17
3rd Qu.:47.50
Max. :61.00
head(Cd.CabbageMaize)
remed.pcnt plt.typ
1 46 cabbage
2 50 cabbage
3 44 cabbage
4 44 cabbage
5 43 cabbage
6 52 cabbage
# Note we don't NEED to give names for boxes if the data is in long format
# This is an advantage of the long format approach
with(Cd.CabbageMaize,boxplot(remed.pcnt~plt.typ,
col= "lightgrey",
main= "Phytoremediation Efficiency of Crop Plants",
xlab= "Crop type", ylab= "Cadmium reduction (%)",
ylim= c(10,70), las= 1, boxwex=.6))
with(Cd.CabbageMaize, var.test(remed.pcnt ~ plt.typ)) # long format, use ~ not ,
F test to compare two variances
data: remed.pcnt by plt.typ
F = 1.1449, num df = 14, denom df = 14, p-value = 0.8037
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.3843653 3.4100823
sample estimates:
ratio of variances
1.144866
with(Cd.CabbageMaize, t.test(remed.pcnt ~ plt.typ, var.equal = TRUE))
Two Sample t-test
data: remed.pcnt by plt.typ
t = 3.4687, df = 28, p-value = 0.00171
alternative hypothesis: true difference in means between group cabbage and group maize is not equal to 0
95 percent confidence interval:
5.377502 20.889165
sample estimates:
mean in group cabbage mean in group maize
44.73333 31.60000