一个朋友跟我咨询这个问题，自己也想不出什么好的方法，请各位大侠多指点！



Dear All,



Sorry I cannot type chinese here but I have a question to ask. I would like to test the significance between two group measurements. What I have is the RNA measurement of two strain rats at different time points. For each time point, I have 6 measurement.Two strains and each strain has three measurements from three rats. And the rats are kill after the measurement.



I have attached the data in the back.



If I plot RNA measurement vs time then I would have two curves represent the progression of that RNA in two strains. What I want is to test if the two strains

have different progression on this particular RNA measurement.



I was wondering if there is anyway besides ANOVA to test that. The thing I don't like ANOVA is that it treats the time as discontinuous variable and what I want to test is the progression of the RNA vs time curve.



Any suggestion is appreciated.Thanks



Jing



[img]file:///C:/DOCUME~1/yuanye/LOCALS~1/Temp/moz-screenshot-1.jpg[/img]

以下是数据和图片



strain    time (week)     mRNA (fmol/g tissue)

gk    4    3.457556317

gk    4    3.770057966

gk    4    4.41222839

gk    4    2.25564172

gk    4    5.697042369

gk    4    5.499910792

wky    4    1.495201124

wky    4    1.5808647

wky    4    3.179360818

wky    4    3.113355782

wky    4    1.672961823

wky    4    2.51877681

gk    8    1.986966807

gk    8    1.817359067

gk    8    2.607335716

gk    8    2.521751939

gk    8    3.54783496

gk    8    1.603732451

wky    8    0.57066157

wky    8    2.039664195

wky    8    1.582738336

wky    8    1.521576999

wky    8    0.982305373

wky    8    1.453794745

gk    12    2.580166887

gk    12    2.680766965

gk    12    2.327498637

gk    12    2.782070612

gk    12    2.478199531

gk    12    1.772934765

wky    12    1.259066597

wky    12    2.143513256

wky    12    1.945891739

wky    12    2.020674981

wky    12    0.718499774

wky    12    1.188810565

gk    16    1.33091868

gk    16    2.279603818

gk    16    2.404676131

gk    16    2.225135797

gk    16    3.0657563

gk    16    2.136957405

wky    16    1.604713246

wky    16    1.469989954

wky    16    0.12755652

wky    16    1.168993362

wky    16    1.808343899

wky    20    3.387474245

gk    20    2.349570002

gk    20    2.405931947

gk    20    1.686711135

gk    20    2.346464483

gk    20    2.112037923

gk    20    3.150802465

wky    20    2.205191972

wky    20    2.69184351

wky    20    1.459197824

wky    20    1.340014551

wky    20    1.226826031
"For each time point, I have 6 measurement.Two strains and each strain has three measurements from three rats. .."



2 strains*6measures*5 time points=60 measures;

so how to understand the latter sentence 'Two strains and each strain has three measurements from three rats'?



since each rat had multiple measures, it is necessary to give the IDs of those rats measured. we need to differentiate

measures from different mouse at the same time and from the same mouse at different time.
5 天 后
不好意思，那个地方我朋友写错了。如图，应该是对于每个时间点来说，有两种老鼠，每一种老鼠有六只，所以一共有12个measurements.每一次测量需要杀了老鼠，所以每个时间点的老鼠都是不同的，因此不需要ID.请问大家对于这种问题有什么好的方法吗？
ANOVA需要很多条件。可以用 kruskal- wallis。
谢谢楼上的答复，但无论是ANOVA还是kruskal- wallis，都是把时间当成一个因素，把对老鼠的处理当成另一个因素，有没有什么方法能检验mRNA的含量随时间的变化呢？
一个简单的办法是对老鼠的处理分组讨论。
有时间因素吗？ 也许要用纵向数据分析  分层模型

还不太懂背景  只是随便说说 不一定正确
试一下minitab中GLM，结果见下面：

一般线性模型: mRNA 与 strain



因子    类型  水平数  值

strain  固定       2  gk, wky





mRNA 的方差分析，在检验中使用调整的 SS



来源    自由度  Seq SS  Adj SS  Adj MS      F      P

time         1   5.244   5.461   5.461   7.41  0.009

strain       1  15.071  15.071  15.071  20.45  0.000

误差        56  41.278  41.278   0.737

合计        58  61.592





S = 0.858544   R-Sq = 32.98%   R-Sq（调整） = 30.59%





项        系数  系数标准误      T      P

常量    2.8470      0.2600  10.95  0.000

time  -0.05356     0.01968  -2.72  0.009





mRNA 的异常观测值



                           拟合值           标准化

观测值     mRNA   拟合值   标准误     残差    残差

     5  5.69704  3.13829  0.22216  2.55875    3.09 R

     6  5.49991  3.13829  0.22216  2.36162    2.85 R

    48  3.38747  1.27024  0.22597  2.11724    2.56 R



R 表示此观测值含有大的标准化残差
请问纵向数据分析，分层模型的英文名是什么？我可以自己去查查
不太知道minitab里的GLM是怎样运行的，选择了什么link呢？
从大家的回复中得到点启发，可以用ANCOVA来做吗？把老鼠类型当成qualitative factor,把时间当成quantitative factor,这样有道理吗？
想到一种比较复杂的方法

方程一：y=a1+a2x1+b1x2+b2x2^2+b3x1x2+b4x1x2^2

x1 代表老鼠的种类，for gk, x1=0, for wky, x1=1

x2代表时间，

when x1=0, y=a1+b1x2+b2x2^2

when x1=1,y=a1+a2+b1x2+b2x2^2+b3x2+b4x2^2

这样要判断这两条曲线是否相同，可以用所有数据fit方程一，检验whether a2=0, b3=0, b4=0

我比较担心的是，变量太多，数据较少，用此种方法可行吗？可信度如何？
10 个月 后