R: is my t.test and reasoning correct?

CABAL

New Member
#1
Hi all. My first post here :) I am doing my master's, and I have samples from two different places. I have been measuring aerobic bacterial
counts, 7 from place C and 4 from place f. I want to ascertain if there is a difference :)



Code:
C <- c(6.9, 8.2, 9.4, 9.2, 7.3, 6, 8.6)

f <- c(5.4, 7.1, 5.6, 7.6)
To find out if C and f are similar or not, I do a t.test:


Code:
t.test(C, f, var.equal=TRUE, paired=FALSE)
This returns:
Two Sample t-test

data: C and f
t = 2.0115, df = 9, p-value = 0.07515
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.1891484 3.2248627
sample estimates:
mean of x mean of y
7.942857 6.425000

Am I correct in assuming that this means that I fail to reject H0, and this
means that I cannot say that they are different (like they seem to be in
the boxplot)


Upon doing a boxplot of the values, it is seen that the median seem very far apart.

Thanks!
 

hlsmith

Not a robit
#2
You just assumed equal variance, you should test for it. Also, describe why these are counts. Count data can be examined with ttest, but also using a Poisson model.
 

CABAL

New Member
#3
I did a f.test before the t.test:

Code:
> var.test(C, f)
Welch Two Sample t-test


data: C and f

t = 2.099, df = 7.2065, p-value = 0.07285

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-0.1822245 3.2179388

sample estimates:

mean of x mean of y

7.942857 6.425000