Use multiple T-Test vs Anova

mrclavet

New Member
Hi,
I am doing a research projet and I need to interpret some data. More specifically, I need to compare 10 groups where our treatment worked, and find if one of the group statistically comes out from the rest. Here's the data. I am wondering what test I should use. I could do a lot of student t tests but then from what I understand the accuracy drops. Would Anova work in this case? Prism doesn't seem to let me do a one way anova with this data right now so I'm not sure I understand how I'd be able to do it.

Thank you so much!

obh

Active Member
One way anova compare the means of all the groups. H0 will says all means are equal. But you have binary variables.
May be you can run Chi squared goodness of fit expected equal probability for each dog

mrclavet

New Member
One way anova compare the means of all the groups. H0 will says all means are equal. But you have binary variables.
May be you can run Chi squared goodness of fit expected equal probability for each dog
I will try that, thank you

obh

Active Member
Hi Mrclavet,

I tried to run your example over the expected cure numbers (if I understand it correctly.)

Total cured / total = 230/242=0.950413223

Group Observed Expected
1 60 58.92561983
2 22 27.56198347
3 20 19.95867769
4 15 14.25619835
5 11 10.45454545
6 10 10.45454545
7 9 9.504132231
8 9 8.553719008
9 30 28.51239669
10 44 41.81818182

p-value equals 0.997359, ( p(x≤χ²) = 0.00264080 ). This means that if we would reject H0, the chance of type I error (rejecting a correct H0) would be too high: 0.9974 (99.74%).

http://www.statskingdom.com/310GoodnessChi.html

PS if you choose to test each dog separately using the proportion test, and compare to 0.950413223 then you must take a lower significant level since you run several tests and one of them can be randomly correct so your combine significant level is bigger then each test'significant level.

mrclavet

New Member
Hi Mrclavet,

I tried to run your example over the expected cure numbers (if I understand it correctly.)

Total cured / total = 230/242=0.950413223

Group Observed Expected
1 60 58.92561983
2 22 27.56198347
3 20 19.95867769
4 15 14.25619835
5 11 10.45454545
6 10 10.45454545
7 9 9.504132231
8 9 8.553719008
9 30 28.51239669
10 44 41.81818182

p-value equals 0.997359, ( p(x≤χ²) = 0.00264080 ). This means that if we would reject H0, the chance of type I error (rejecting a correct H0) would be too high: 0.9974 (99.74%).

http://www.statskingdom.com/310GoodnessChi.html

PS if you choose to test each dog separately using the proportion test, and compare to 0.950413223 then you must take a lower significant level since you run several tests and one of them can be randomly correct so your combine significant level is bigger then each test'significant level.
Thank you! Those were not the real numbers, but with your example I get the idea!

obh

Active Member
Great noetsi

Fortran must die
To answer the initial question always do ANOVA as compared to multiple independent t tests. There are ANOVA tests called I think post hoc tests that tells you which level of the IV differs when you know it does from the model F test. Any good psychology text should cover them.

obh

Active Member
Thanks Noetsi The post hoc is Tukey HSD. (As Noesti wrote, this for your initial question, I assume not for your example data)