# Help selecting test

#### JP712111

##### New Member
Hi all, so im doing a research project and stuck on which statistical test to use. Essentially I want to prove that once people receive training, the difference between them and the control(professional) opinion is reduced. So i have data from people before they received training, and after they received training and want to compare this to the control in which case i will take as the correct value.
Does anyone have any idea how id show this?

Thanks!

#### Archidamus

##### Member
How are you measuring the end result? I think a little more information about how you are measuring will help us lead you to the correct test.

Also you mentioned a professional opinion as the 'correct value'. I think you might be interested in two things, the training's impact on getting closer to the professionals opinion, and reduction in the variation between people after training.

#### JP712111

##### New Member
So it’s called body condition score in cats.
It’s on a scale of 1-9 so say a cat is a 6, the owner before advice guesses 4 but after advice they guess 7. I’d want to prove statistically that when advice is given, the owners are “better able” to body condition score (in the sense that the difference between (post advice - vets guess) is less than (pre advice - vets).
I thought I could do two non paired t tests and then state the variance and standard deviation?

#### Archidamus

##### Member
Okay, so the 1-9 scale is ordinal data, meaning '1'<'2'<'3'<...<'9' but it does not mean a cat with a score of '6' is twice as good as a cat with score '3'. A T-test is used on continuous data like time, length, etc. where there is an infinitely amount of possibilities. Also, let think about your null and alternative hypothesis, that sometimes helps me figure out the right test.

Null: Advice has no impact on a persons ability to score a cat closer to the Vet's score.

Depending on how many data points you have, you might be able to construct a contingency table and perform a Persons Chi-squared. The way I would set this up in on one axis I would groups the difference between the unadvised score and the vet into 0,1,2,3+ and on the other axis the difference between the advised score and the vets score into the same 0,1,2,3+ to create yourself a 4x4 contingency table.

Do you have a statistical software package? R, MiniTab, SPSS, SAS. If not you can do this pretty easily in excel. The only caution with this test is when you don't have a lot of data, the page has some details on it. Good luck and let me know if you have any other questions.

#### Karabiner

##### TS Contributor
The way I would set this up in on one axis I would groups the difference between the unadvised score and the vet into 0,1,2,3+ and on the other axis the difference between the advised score and the vets score into the same 0,1,2,3+
But doing so would only be possible if you consider the data as interval scaled,
i.e. if you assume that the difference between levels 4 and 3 is the same as
between 8 or 7, or that the difference between levels 8 and 6 is the same as
between 3 and 1. In that case, one could just calculate the individual absolute
differences before training and after training, respectively, and compare them
using t-test for dependent observations, or the Wilcoxon signed rank test.

With kind regards

Karabiner

#### Karabiner

##### TS Contributor
The way I would set this up in on one axis I would groups the difference between the unadvised score and the vet into 0,1,2,3+ and on the other axis the difference between the advised score and the vets score into the same 0,1,2,3+
But doing so would only be possible if you consider the data as interval scaled,
i.e. if you assume that the difference between levels 4 and 3 is the same as
between 8 or 7, or that the difference between levels 8 and 6 is the same as
between 3 and 1. In that case, one could just calculate the individual absolute
differences before training and after training, respectively, and compare them
using t-test for dependent observations, or the Wilcoxon signed rank test.

With kind regards

Karabiner

#### JP712111

##### New Member
So I have 100 sets of data, and the access to graphpad prism 7.0. My initial impression was to perform unpaired t test between the vets score and the pre/post advice but just the differences so making the vets score 0 and only the difference for pre and post (1,2,3 (no +/- differences)) and then performing a t test and the value I should get assuming that advice makes a difference is a reduction in difference so the unpaired t test value should be closer to p<0.05 if that makes sense? I’d take any advice I can get!

#### JP712111

##### New Member
I thought of an unpaired t test as the owners do not know what the vet thinks their pets BCS is until after the survey so the owners wont have any biased data or any influence

#### Karabiner

##### TS Contributor
Your dependent variable is the difference between owner's rating and the veterinary's rating? You have measured this for the very same owners at two time points? Therefore, an analysis for dependent measures is appropriate.

With kind regards

Karabiner

#### JP712111

##### New Member
Your dependent variable is the difference between owner's rating and the veterinary's rating? You have measured this for the very same owners at two time points? Therefore, an analysis for dependent measures is appropriate.

With kind regards

Karabiner
So T test is appropriate?
Ive done paired t tests for the differences before and after training and obtained a value but doesnt this just tell me the relationship between each other and not in relation to the vets score? So it would tell me that the two data sets are different from each other but without a comparison to the vets score it would be pointless? (the attached pictures are of t tests i had performed, one pic of a paired t tests between the two values and another pic comparing the differences to the vets (which i had made 0))

Sorry to seem like a pain but to me making two independent t tests i can then go on to say the t stat value for pre advice is > than t stat value after advice showing weak agreement that giving advice helps reduce difference, but im not sure if im explaining this right to even my academic tutor who is also a bit confused

Thank you to all who have helped

#### Karabiner

##### TS Contributor
Your ideas are wrong.

With kind regards

Karabiner

#### noetsi

##### No cake for spunky
If you are comparing two people (or whatever) at different points in time than that is not an independent t test. It is a pooled one. I agree with Karabiner.

#### Archidamus

##### Member
You just conducted a 1-Sample T test and you didn't even know it. You essentially gave evidence that there is a difference in scoring between a non-trained person and a vet in the first one. And in the second one you again gave evidence to show a difference in the scoring on a trained person and a vet. Both which do not answer the question you are asking, does training make a difference on how close they grade to a vet?

You need to create two sets of a data(100 observations each) which are the difference between the person's score and the vet. Lets U = untrained score, T = trained Score, and V = vet Score. Your Null hypothesis is |U-V| = |T-V| ; you are wanting to know the absolute difference and are not caring about the direction.

It looks like you are using excel, if you want to attach it, and erase any personal information like medical numbers, names, dates of birth, etc. I can set up a new tab with notes.

#### JP712111

##### New Member
You just conducted a 1-Sample T test and you didn't even know it. You essentially gave evidence that there is a difference in scoring between a non-trained person and a vet in the first one. And in the second one you again gave evidence to show a difference in the scoring on a trained person and a vet. Both which do not answer the question you are asking, does training make a difference on how close they grade to a vet?

You need to create two sets of a data(100 observations each) which are the difference between the person's score and the vet. Lets U = untrained score, T = trained Score, and V = vet Score. Your Null hypothesis is |U-V| = |T-V| ; you are wanting to know the absolute difference and are not caring about the direction.

It looks like you are using excel, if you want to attach it, and erase any personal information like medical numbers, names, dates of birth, etc. I can set up a new tab with notes.
Thank you Arch, i did in the second pic what i thought was a paired t test and was enough?
Please find attached the excel file- if you could reply also with an explanation so i actually know what on earth im looking at! Have a meeting with my tutor about this on wednesday so need to look like i know what im doing!

Ive attached the file and made two more columns with the difference between (Pre advice and vet) and (Post advice and vet). These values are with ABS formula just so you know, again like you said the absolute difference and not caring about negative or positive.
(Talkstats doesnt let me attach excel file so just attached the dropbox link to download) - If you could reply to my email address with the new tab and or any stats formulas then it would be amazing. My email address is jay_p1997@hotmail.co.uk
https://www.dropbox.com/s/rwu74kxv9552uw0/Trial12.xlsx?dl=0
Any help will be appreciated. Many thanks