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发信人: happyzhb (不再年青), 信区: Biology
标 题: Re: 问大家一个数据统计的问题
发信站: Unknown Space - 未名空间 (Thu Nov 11 11:21:00 2004) WWW-POST
【 在 stlstl (射天狼) 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.
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 her question,
just an example of demonstrating that how the test work and how to interpret
it. the suggestion to her question is in the middle portion of my post.
: 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.
Yes, I agree, as I said, if cell culture among the 4 experiments are
comparable, paired t-test is suitable. But if they are not, then you cannot
use paired t-test without a reasonable normalization. For example, you have 2
affy chips with same gene expression profile(same expression level
distribution, t-test should say they are same), but one signal happen to be
stronger overall. Without normalization, t-test will say they are different
distribution, because its the fact that one chip's signal are stronger,
meaning, signal wise, they are actually different, but if you normalize, then
the t-test will tell you they are the same.
In gardenia's case, one cannot tell from the data if they are normalizable,
but as she said in her own post, the statistics professor said its hard to do
normalization. That make me think its might not be comparable.
:
:
: 【 在 happyzhb (不再年青) 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 column-wise paired t-test. In you
case,
: : if you are sure that your cell count from 4 different experiments are
: : comparable from each other, then you can use pair-wise t-test. otherwise,
: you
: : can use anova, but I am not sure what is the model.
: :
: : 【 在 gardenia (十六字令) 的大作中提到: 】
: : :
: : : 打个比方我做一个实验，3种treatment A B C
: : : 做了4次实验
: : :
: : : 结果分别是：
: : :
: : : A B C
: : : 1 1 5 9
:
--
//昨天晚上我作了个梦，梦见我变成一只大蜜蜂
//嗡嗡嗡，嗡嗡嗡
//飞到西来飞到东
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