Peter,
Thank you very much! That did the trick…
Regards,
Tom
Peter Ehlers wrote:
Tom,
t.test(MAE ~ type, data=data, subset=type %in% c('hpc','rfc'))
-Peter Ehlers
Thomas Adams wrote:
Dennis,
Thank you for the suggestion, but I get this error:
> t.test(MAE ~ type,data=data)
Error in t.test.formula(MAE ~ type, data = data) :
grouping factor must have exactly 2 levels
Tom
Dennis Murphy wrote:
Hi:
On Thu, Feb 4, 2010 at 11:07 AM, Thomas Adams <thomas.ad...@noaa.gov
<mailto:thomas.ad...@noaa.gov>> wrote:
I am trying to use t.test on the following data:
date type INTERVAL nCASES MTF SDF MTO SDO nFST MF nOBS MO MB BIASCV
BIASEV ME MAE RMSE CRCF
2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385
1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056
2001-06-15 avn 0.00LT0.01 852225 0.018 0.066 0.000 0.001 708406
0.001 852225 0.000 0.000 1.663 71.664 0.018 0.018 0.068 0.176
2001-06-15 avn 0.01LT0.10 77643 0.097 0.151 0.039 0.025 176129 0.040
77643 0.039 0.040 2.331 2.486 0.058 0.086 0.162 0.096
2001-06-15 avn 0.10LT0.25 29388 0.145 0.186 0.162 0.043 74164 0.160
29388 0.162 0.160 2.493 0.897 -0.017 0.129 0.189 0.056
2001-06-15 avn 0.25LT0.50 17592 0.177 0.208 0.353 0.070 25189 0.336
17592 0.353 0.343 1.365 0.503 -0.175 0.238 0.279 0.033
2001-06-15 avn 0.50LT1.00 10503 0.208 0.245 0.693 0.138 6481 0.666
10503 0.693 0.683 0.593 0.300 -0.485 0.517 0.560 0.017
2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385
1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056
2001-06-15 eta GE1.00 4385 0.242 0.308 1.502 0.556 577 1.338 4385
1.502 1.483 0.117 0.161
-1.261 1.272 1.398 0.111
2001-06-15 eta 0.00LT0.01 852225 0.013 0.055 0.000 0.001 799424
0.000 852225 0.000 0.000 1.368 50.193 0.013 0.013 0.057 0.175
2001-06-15 eta 0.01LT0.10 77643 0.079 0.139 0.039 0.025 113987 0.043
77643 0.039 0.041 1.617 2.013 0.040 0.079 0.144 0.083
2001-06-15 eta 0.10LT0.25 29388 0.116 0.169 0.162 0.043 47461 0.160
29388 0.162 0.161 1.596 0.719 -0.045 0.139 0.178 0.055
2001-06-15 eta 0.25LT0.50 17592 0.147 0.197 0.353 0.070 23284 0.345
17592 0.353 0.348 1.296 0.417 -0.205 0.258 0.291 0.040
2001-06-15 eta 0.50LT1.00 10503 0.180 0.230 0.693 0.138 7003 0.643
10503 0.693 0.673 0.619 0.260 -0.513 0.532 0.576 0.041
2001-06-15 eta GE1.00 4385 0.242 0.308 1.502 0.556 577 1.338 4385
1.502 1.483 0.117 0.161
-1.261 1.272 1.398 0.111
2001-06-15 hpc GE1.00 4385 0.339 0.345 1.502 0.556 1326 1.265 4385
1.502 1.447 0.255 0.225 -1.163 1.172 1.314 0.144
2001-06-15 hpc 0.00LT0.01 852225 0.014 0.057 0.000 0.001 777147
0.000 852225 0.000 0.000 0.823 54.824 0.014 0.014 0.059 0.195
2001-06-15 hpc 0.01LT0.10 77643 0.092 0.148 0.039 0.025 123342 0.048
77643 0.039 0.045 1.967 2.346 0.053 0.085 0.156 0.109
2001-06-15 hpc 0.10LT0.25 29388 0.147 0.190 0.162 0.043 56107 0.161
29388 0.162 0.161 1.896 0.908 -0.015 0.137 0.192 0.077
2001-06-15 hpc 0.25LT0.50 17592 0.195 0.219 0.353 0.070 25677 0.344
17592 0.353 0.348 1.424 0.552 -0.158 0.237 0.276 0.057
2001-06-15 hpc 0.50LT1.00 10503 0.251 0.265 0.693 0.138 8137 0.659
10503 0.693 0.678 0.737 0.362 -0.442 0.480 0.529 0.066
2001-06-15 hpc GE1.00 4385 0.339 0.345 1.502 0.556 1326 1.265 4385
1.502 1.447 0.255 0.225 -1.163 1.172 1.314 0.144
2001-06-15 ngm GE1.00 4385 0.157 0.199 1.502 0.556 297 1.119 4385
1.502 1.478 0.050 0.105
-1.345 1.345 1.474 -0.062
2001-06-15 ngm 0.00LT0.01 852225 0.017 0.063 0.000 0.001 771901
0.000 852225 0.000 0.000 0.703 65.457 0.017 0.017 0.065 0.132
2001-06-15 ngm 0.01LT0.10 77643 0.070 0.127 0.039 0.025 133779 0.041
77643 0.039 0.040 1.803 1.784 0.031 0.073 0.131 0.073
2001-06-15 ngm 0.10LT0.25 29388 0.100 0.152 0.162 0.043 54850 0.161
29388 0.162 0.161 1.859 0.620 -0.061 0.137 0.168 0.050
2001-06-15 ngm 0.25LT0.50 17592 0.130 0.177 0.353 0.070 24526 0.344
17592 0.353 0.348 1.360 0.369 -0.222 0.263 0.291 0.047
2001-06-15 ngm 0.50LT1.00 10503 0.152 0.196 0.693 0.138 6383 0.643
10503 0.693 0.674 0.564 0.219 -0.541 0.551 0.591 0.025
2001-06-15 ngm GE1.00 4385 0.157 0.199 1.502 0.556 297 1.119 4385
1.502 1.478 0.050 0.105
-1.345 1.345 1.474 -0.062
2001-06-15 rfc GE1.00 4385 0.343 0.349 1.502 0.556 1192 1.239 4385
1.502 1.446 0.224 0.228 -1.159 1.168 1.310 0.157
2001-06-15 rfc 0.00LT0.01 852225 0.014 0.055 0.000 0.001 773777
0.000 852225 0.000 0.000 0.719 53.984 0.014 0.014 0.056 0.200
2001-06-15 rfc 0.01LT0.10 77643 0.091 0.141 0.039 0.025 123689 0.047
77643 0.039 0.044 1.899 2.333 0.052 0.084 0.150 0.114
2001-06-15 rfc 0.10LT0.25 29388 0.148 0.184 0.162 0.043 58569 0.159
29388 0.162 0.160 1.957 0.913 -0.014 0.134 0.186 0.081
2001-06-15 rfc 0.25LT0.50 17592 0.197 0.214 0.353 0.070 26386 0.340
17592 0.353 0.345 1.448 0.558 -0.156 0.232 0.271 0.055
2001-06-15 rfc 0.50LT1.00 10503 0.253 0.262 0.693 0.138 8123 0.643
10503 0.693 0.671 0.718 0.365 -0.440 0.476 0.525 0.074
2001-06-15 rfc GE1.00 4385 0.343 0.349 1.502 0.556 1192 1.239 4385
1.502 1.446 0.224 0.228 -1.159 1.168 1.310 0.157
2001-07-15 avn GE1.00 3258 0.194 0.233 1.399 0.400 1323 1.440 3258
1.399 1.410 0.418 0.139 -1.204 1.209 1.287 0.039
2001-07-15 avn 0.00LT0.01 879285 0.021 0.073 0.000 0.001 736915
0.001 879285 0.000 0.000 1.541 73.048 0.020 0.020 0.075 0.137
2001-07-15 avn 0.01LT0.10 84628 0.081 0.139 0.039 0.025 179228 0.040
84628 0.039 0.040 2.200 2.104 0.043 0.078 0.146 0.079
This wouldn't read for me:
Error: unexpected string constant in:
"79285 0.000 0.000 1.541 73.048 0.020 0.020 0.075 0.137
2001-07-15 avn 0.01LT0.10 84628 0.081 0.139 0.039 0.025
of which this is just a small portion of the data. What I want to
do is to test the difference between the MAE values for those that
are, for example, 'hpc' vs those that are 'rfc', that is, by
'type' in the header.
t.test(MAE ~ type, data = yourdf, ...)
By default, t.test uses var.equal = FALSE and paired = FALSE. If you
want to
assume equal population variances, set var.equal = TRUE. Since the
sample sizes
are going to be large, this is essentially a Z-test.
I have looked for many examples and have tried to construct the
correct syntax, but no luck so far. If possible, I would further
like to break down the test, not only by type, but type and INTERVAL.
If you want this type of breakdown, you're going to be doing a
two-way ANOVA.
Individual t-tests in this case would be an extremely inefficient
use of the data.
HTH,
Dennis
-- Thomas E Adams
National Weather Service
Ohio River Forecast Center
1901 South State Route 134
Wilmington, OH 45177
EMAIL: thomas.ad...@noaa.gov <mailto:thomas.ad...@noaa.gov>
VOICE: 937-383-0528
FAX: 937-383-0033
______________________________________________
R-help@r-project.org <mailto:R-help@r-project.org> mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Thomas E Adams
National Weather Service
Ohio River Forecast Center
1901 South State Route 134
Wilmington, OH 45177
EMAIL: thomas.ad...@noaa.gov
VOICE: 937-383-0528
FAX: 937-383-0033
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
