Hi there,
I have design a experiment, which will last for 8 weeks. At the
beginning of each week, e.g., Monday, I add some substrate into the >experimental unit, and then measured a response. Now, I get a data set,
which is composed of variable of response, and time.
Now, I hope to test the effects of time on response. What's method I
should used? ANOVA is the first method I try. However, I don't think the >response is independent, after all, I add the substrate into the
experimental unit every week.
Any helps will be really appreciated. Thanks in advance.
On Thu, 12 Dec 2019 13:20:36 +0800, Jinsong Zhao <jszhao@yeah.net>
wrote:
Hi there,
I have design a experiment, which will last for 8 weeks. At the
beginning of each week, e.g., Monday, I add some substrate into the
experimental unit, and then measured a response. Now, I get a data set,
which is composed of variable of response, and time.
Now, I hope to test the effects of time on response. What's method I
should used? ANOVA is the first method I try. However, I don't think the
response is independent, after all, I add the substrate into the
experimental unit every week.
Any helps will be really appreciated. Thanks in advance.
Please note that "ANOVA" in general specifies a variety of
methods dealing with "variance". Another name is "method of
least squares." The other usual method is "maximum likelihood"
(ML). For the simplest designs, the analyses are sometimes
exactly the same in computation (thus, result).
For some designs, the difference in "error measurement" leads to
slight differences in computations. ML methods are sometimes
easier to generalize to complex circumstances, which is mainly
why some computer programs for statisitics use ML.
As to your question about what method ... What outcome
do you expect?
I don't know what you refer to as "substrate" so I only guess
that you might expect a uniform growth from week to week.
Or you might expect an exponential growth.
Or you might expect an early growth, followed by plateau.
In any case: This is a "repeated measure" on some unit.
Is there just one unit? If so, that places a limit - you don't
have the degrees of freedom for error in order to do a
simple test between "weeks", as you might be tempted
to do if there were multiple units. For multiple units, you
want something called "repeated measures."
My suggestion of the hypotheses suggests that the test
would be for linear trend, as opposed to all "non-linear"
components - 1 d.f. for linear, the rest for non-linear.
"Linear trend" would make use of the actual "size" or whatever
at each week, rather than a "response" computed as a
difference or a ratio. If an exponential growh is expected,
the analysis might start with the scores for each week
after the baseline as its ratio over the previous week.
summary(lme(fungi ~ week, random = ~1|unit, data = xxw))Linear mixed-effects model fit by REML
anova(lme(fungi ~ week, random = ~1|unit, data = xxw))numDF denDF F-value p-value
On 2019/12/13 6:42, Rich Ulrich wrote:
On Thu, 12 Dec 2019 13:20:36 +0800, Jinsong Zhao <jszhao@yeah.net>
wrote:
Thank you very much for the explanation and the suggestions. We have 3
units in the experiment. In the experiments, we added a tracers into the >soil, and measured the tracers in fungi. At the beginning of each week,
we added the same amount of tracers into the soil. Finally, we got the >following data:
week fungi unit
1 0.70 1
1 0.96 2
1 0.71 3
2 0.92 1
2 0.91 2
2 0.92 3
3 1.50 1
3 2.13 2
3 2.07 3
4 1.88 1
4 1.58 2
4 1.86 3
5 3.83 1
5 4.06 2
5 3.97 3
6 6.82 1
6 5.91 2
6 6.56 3
7 10.05 1
7 10.38 2
7 8.12 3
8 11.17 1
8 11.29 2
8 11.11 3
It's some kind of "repeated measure". We have try to test the effects of
time on the residual of tracers in fungi by linear mixed model in R. The >results are:
summary(lme(fungi ~ week, random = ~1|unit, data = xxw))Linear mixed-effects model fit by REML
Data: xxw
AIC BIC logLik
90.95309 95.31726 -41.47655
Random effects:
Formula: ~1 | unit
(Intercept) Residual
StdDev: 3.439275e-05 1.328719
Fixed effects: fungi ~ week
Value Std.Error DF t-value p-value
(Intercept) -2.489643 0.5977476 20 -4.165041 5e-04
week 1.566310 0.1183717 20 13.232133 0e+00
Correlation:
(Intr)
week -0.891
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-1.65241557 -0.80463759 -0.06303065 0.87239442 1.43407057
Number of Observations: 24
Number of Groups: 3
and
anova(lme(fungi ~ week, random = ~1|unit, data = xxw))numDF denDF F-value p-value
(Intercept) 1 20 282.5119 <.0001
week 1 20 175.0893 <.0001
However, we don't have the full confidence about the results...
Sysop: | Keyop |
---|---|
Location: | Huddersfield, West Yorkshire, UK |
Users: | 296 |
Nodes: | 16 (2 / 14) |
Uptime: | 46:55:24 |
Calls: | 6,648 |
Files: | 12,198 |
Messages: | 5,329,893 |