Displaying 5 results from an estimated 5 matches for "stde".
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2012 May 03
1
Proposal: model.data
...tudying termplot code, I saw the "carrier" function approach
to deducing the "raw" predictors. However, it does not always work.
Here is one problem. termplot mistakes 10 in log(10 + x1) for a variable
Example:
dat <- data.frame(x1 = rnorm(100), x2 = rpois(100, lambda=7))
STDE <- 10
dat$y <- 1.2 * log(10 + dat$x1) + 2.3 * dat$x2 + rnorm(100, sd = STDE)
m1 <- lm( y ~ log(10 + x1) + x2, data=dat)
termplot(m1)
## See the trouble? termplot thinks 10 is the term to plot.
Another problem is that predict( type="terms") does not behave
sensibly sometimes....
2010 Jul 30
4
Programming Statistical Functions
Hello,
I'm new in R. I'm meteorological modeller and i will calculate some
statistics for my model results.
These statistis are the follow:
ANB: Average Normalized Absolute BIAS
MNB: Mean Normalized BIAS
MNE: Mean Normalised Error
STDE: Standard Deviation of Error
FB: Fractional BIAS
MG: Geometric Mean BIAS
VG: Geometric Variance
SKVAR: Skill Variance
RMSE: Root Mean Square Error
NMSE: Normalized Mean Square Error
r: Correlation Coefficient
CV: Coeficient of Variation
FAC2: Fractional Predictions within a factor of two of observa...
2003 Nov 15
0
computing a p-value for a one-way ANOVA given only means and std dev of 5 factor levels
...n a
p-value indicating whether there is evidence of a difference between these
means.
Mean Std Dev N
Level A mA stdA nA
Level B mB stdB nB
Level C mC stdC nC
Level D mD stdD nD
Level E mE stdE nE
My current plan is to assume normally distributed responses and use the
known mean and std dev to generate data points which can be combined to
create a data set similar to the "true" (but unavailable) data set. Then I
can use a standard 1-way ANOVA to determine the p-value.
D...
2003 Nov 15
0
FW: computing a p-value ...
...that you need in order to do the usual ANOVA. Most elementary
statistics texts will tell you how.
The ``standard 1-way ANOVA'' assumes that the population standard
deviations are equal for the various levels. You thus form
SSE by ``pooling'':
SSE = (nA-1)*stdA^2 + ... + (nE-1)*stdE^2
You form the sum of squares for the factor as
SSF = nA*(mA-mBar)^2 + ... + nE*(mE-mBar)^2
where ``mBar'' is the `grand mean'':
mBar = (nA*mA + ... + nE*mE)/n where in turn n = nA + ... + nE
Then form
MSF = SSF/4 and MSE = SSE/(n - 5)
(4 because 4 = 5-1 and the factor has...
2004 Aug 06
2
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