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 10:26:12 2004) WWW-POST

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.

My method will be chi square test or fisher exact test, I think what you don't
have here is your control counts or those counts that not desired. So for each
experiments you should have a contigency table like this:

A B C
desired phenotype 1 5 9
not desired 56 45 36

then you use either chi square test or fisher exact test, that will tell you
whether their is an association between different treatments and the desired
phenotype. The difference btw chi and fisher is that if most counts in the
contingency table is less than 5, chi square is not accurate, then you got use
fisher exact test. On the contrary, if the numbers are too big, fisher exact
test will take a huge computation power to comput.

Now, if you treat the table you gave below as an contigency table, do a chisq
test or fisher test, you are basically test if treatment have any association
with your different experiment repeats. Another word, the 4 time treatment
should not affects the treatments of ABC. You null hypothesis is that there is
no association(you did nothing wrong between repeats), if the p value of chisq
or fisher test is significant, you reject the null hypothesis and say, hey, I
must 've done sth wrong, otherwise, repeats should not affect the effects of
treatment
ABC have on cells.

the p value turn out to be 0.8225 and 0.8417, so you keep you null
hypothesis.



【 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
: 2 9 15 18
: 3 21 29 36
: 4 15 22 28
:
: 显然ABC三种treatment对结果有明显的不同。
: 但是如果直接用mean+SEM表示，显得就没有太大的不同了。
: 我问了一个统计学教授，他说我这个实验的问题是4次实验彼此间无法
: normalize，因此是一个randomized block design，然后教我用ANOVA算
: 了一下，的确ABC彼此之间是有显著差别的。
: 但是我现在要present给我老板，如果就用mean+SEM的形式表示，error
: bar显得很大。不知道他能不能构理解了。。。
:
:
:


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※ 修改:·happyzhb 於 Nov 11 10:26:12 修改本文·[FROM: 130.219.]
※ 来源:．Unknown Space - 未名空间 mitbbs.com．[FROM: 130.219.]