s****l
发帖数: 10462 | 1 我来胡诌一把
看你的结果四个实验之间没有重复性
但是A1和B1和C1是有可比性的
用A1-4 B1-4可以做一个two-tail paired T-test, p= 0.005
A1-4 and C1-4, p=0.006
B1-4 and C1-4, p=0.012
T-Test are not good enough to analyze all three groups together. However, it
is okay if you want to know whether treatment C is better than treatment A. Of
course, if you have many treatments (say A, B, C... XYZ), it is not good to do
every possible pairs test.
Good luck.
it | h******b
发帖数: 312 | 2 You are right that you cannot compare between 4 different experiment, because
in you case, cell counts in each experiment could vary so much that they are
not come from the same distribution, that is why you cannot use pair-wised
t-test between column, because 1, 9, 21, 15, are not from within the same
distribution. For example, you are testing weight loss of animals, weight from
mouse and from lion are different, if you test 4 different animals, and they
cannot be normalized, you cannot use col | s****l
发帖数: 10462 | 3 I have to respectfully disagree.
Gardenia is comparing 3 treatments, not 4 experiments. Each treatment gets 4
results which are not consistent. However, in each experiment the treatments
are comparable. In another word, A1~B1~C1 is comparable, so is A4~B4~C4. If I
only want to know whether treatment A and B have significant difference, I
would say paired t-test is suitable.
because
from
you
【在 h******b 的大作中提到】
: You are right that you cannot compare between 4 different experiment, because
: in you case, cell counts in each experiment could vary so much that they are
: not come from the same distribution, that is why you cannot use pair-wised
: t-test between column, because 1, 9, 21, 15, are not from within the same
: distribution. For example, you are testing weight loss of animals, weight from
: mouse and from lion are different, if you test 4 different animals, and they
: cannot be normalized, you cannot use col
| h******b
发帖数: 312 | 4
Thanks for pointing that out! I might be wrong, but my understanding is from
gardenia's original post:
>打个比方我做一个实验，3种treatment A B C
>做了4次实验
my understanding is that gardenia did the first experiment, treat cell with A,
B, and C, then record counts, then she repeats, culture cells again, treat
with A, B, and C then record counts, and she(?) did it 4 times.
I guess I might confused you by use the whole data to do the chisq and fisher
test in the last portion of my post. That's not the answer to
【在 s****l 的大作中提到】
: I have to respectfully disagree.
: Gardenia is comparing 3 treatments, not 4 experiments. Each treatment gets 4
: results which are not consistent. However, in each experiment the treatments
: are comparable. In another word, A1~B1~C1 is comparable, so is A4~B4~C4. If I
: only want to know whether treatment A and B have significant difference, I
: would say paired t-test is suitable.
:
: because
: from
: you
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