Note: You must be registered in order to post a reply. To register, click here. Registration is FREE!
T O P I C R E V I E W
rldc2
Posted - 01/07/2011 : 07:08:57 AM Origin Ver. and Service Release (Select Help-->About Origin): 8.1.88.89 (Academic) Operating System: Windows XP
Hi, I'm attempting to fit some data with both x and y errors linearly using the 'Ripley' technique. This appears to give convincing intercept and slope values, but gives unfeasible values for the standard errors which are important for us to know. The actual data is: X XErr Y YErr --------------------------------------------- 185.01172 17.53857 542.5 201 218.15645 20.37692 830 411.5 242.4442 24.2924 761.5 63.5 306.3891 33.87535 1025 75.5 350.38988 31.47779 1202 202.5 371.85471 35.43873 1223.5 147.5 397.61866 34.50598 1197 230.5 442.85956 39.02208 1292.5 167.5 464.1512 42.77052 1291.5 94.5 520.11983 47.9279 1818 503.5 568.47473 55.99731 1982.5 213 518.55814 54.24961 1434.5 369.5 623.59091 69.87129 2246 242 721.37299 63.82286 2586 527 951.53452 95.24863 2349.5 353.5 851.85754 73.99066 2779.5 1132 628.54034 53.34334 1663.5 500 838.98543 70.65024 2714.5 601 1385.44779 166.61908 4425 1174
Which appears to give:
Intercept = -5.67818 St Err = 0.02723 Slope = 3.16769 St Err = 13.27179
Which can't be right. I'm fitting via the Dialog box and using the Ripley method as the most appropriate. I can't see what I might be doing wrong, or if there is a bug with the output how to get around this (possible use of an X-function in LabTalk? fitLRXErr? But I'm not clear in syntax for this function...)
Any help would be greatly appreciated.
Many thanks,
Robert
1 L A T E S T R E P L I E S (Newest First)
Sam Fang
Posted - 01/11/2011 : 02:29:48 AM We will check it further.