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Algorithms :: Statistics :: comparing 2 experimentally determined means
By: pdanese	Date: 19/09/2003 00:00:00	Points: 125	Status: Answered Quality : Excellent
let's say i perform 2 measurements under condition X, and my average measurement for condition X turns out to be 100 with an estimated standard deviation of 7. let's say i perform 2 more measurements under condition Y, and under this condition my average measurement turns out to be 70, with an estimated standard deviation of 11. finally, let's say i perform 2 more measurements under condition Z, and under this condition my average measurement turns out to be 26 with an estimated standard deviation of 5. Is there a way of expressing the likelihood that the true mean under condition Y is different than the true mean under condition X? Same Question when comparing the means observed under condition Z to those under condition X. How does one generalize this to more than 2 measurements? Thank you for any information you can provide!
By: aburr	Date: 20/09/2003 17:34:00		Type : Answer
The quick and somewhat dirty way of detecting differences is to use 3 times the standard deviation. If everything is normally distributed, two values which differ by three sds have a 99% chance of being different. In your case measurement Y has such a large sd that little can be concluded about what it represents. However it is abundantly clear that measurement Z differs from measurement X. The above can be extended to many more measurements that two. Indeed it is hard to get a useful sd from just two meaurements.

By: VGR	Date: 22/09/2003 06:59:00		Type : Assist
welcome in the wonderful land of "experience plans" :D Also, the laws followed by "means of means" are known and there are tests designed for them

By: rfr1tz	Date: 24/09/2003 04:27:00		Type : Assist
This is completely from memory, but this is a standard problem that is in every elementary statistics book. X: mean = 100 s.d. = 7 variance = 49 Y: mean = 70 s.d. = 11 variance = 121 So add up the variances to get the variance of X-Y = 170 implies s.d. = 13. So then you have to look up in a book to see what the probability that X and Y are equal. In this case, it's 30/13 = 2.3 standard deviations. Look that up in a table. It's not very likely. Probably 1 or 2 %.

By: VGR	Date: 24/09/2003 04:56:00		Type : Comment
more than that (from memory ttoo :D ) 1 sigma = 66% 2 sigma = 90% 3 sigma = 99%

By: alexcvt	Date: 30/09/2003 08:34:00		Type : Comment
Google for "confidence intervals"

By: VGR	Date: 30/09/2003 14:50:00		Type : Assist
absolutely, but it depeds on th test used, ie on the law underlying

By: rfr1tz	Date: 06/10/2003 00:29:00		Type : Comment
VGR - It's 68%, 95%, 99.7% but I'm thinking it's a one-tailed test, so the perecentages might have to be halfed!?

By: VGR	Date: 06/10/2003 02:13:00		Type : Comment
mine were for two-tail but I can make memory errors (GPFs ;-)

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