# Most appropiate statistical test - (Jan/30/2010 )

Hi, I could do with a bit of help from any statistic experts out there. My experiment was that I transfected cells with either normal wildtype protein, mutant 1, mutant 2 or mutant 3, and then looked at the number of colonies formed by them in 3 seperate experiments. The data is:

Exp1 Exp2 Exp3

Wt 0.6 0 0.2

Mut1 5.8 8.2 14

Mut2 7.6 8.1 7

Mut3 1.1 1.1 2

I've done a unpaired t-test comparing each of the mutants to the wildtype, although for mutant 1 I did a welchs corrected test as the stdev of Mut1 is significantly different from the stdev of wt. Is this the most appropiate test, or should I have done an one-way ANOVA test (although I dont know if the differences in stdev may cause a problem with that test too)? Thanks for your help!

suziekins on Jan 30 2010, 10:30 AM said:

Exp1 Exp2 Exp3

Wt 0.6 0 0.2

Mut1 5.8 8.2 14

Mut2 7.6 8.1 7

Mut3 1.1 1.1 2

I've done a unpaired t-test comparing each of the mutants to the wildtype, although for mutant 1 I did a welchs corrected test as the stdev of Mut1 is significantly different from the stdev of wt. Is this the most appropiate test, or should I have done an one-way ANOVA test (although I dont know if the differences in stdev may cause a problem with that test too)? Thanks for your help!

I would do ANOVA. You can run a Brown-Forsythe and/or Levene's test to look at homogeneity of variance (homoscedasiticity test). In that test, you want a P value > 0.05 to show the variances are not significantly different. If the data is homoscedastic, you can use ANOVA, if it's heteroscedastic (unequal variances) then you need to transform the data somehow to get it homoscedastic. Once you run the ANOVA, you can do pairwise comparisons with Tukey's HSD, Bonferroni's, or some other pairwise comparison to compare the means of each treatment to each other treatment. Alternatively, you can use Dunnett's test to compare each mutant to the WT, but it will not compare mut 1 to mut 2. Also consider that the more comparisons you run, the more difficult it is to achieve significance. Therefore, Dunnett's will be less conservative than Tukey's or Bonferroni's. Also, if you do pairwise among all treatments, there will be differences in calculating the P value. I think Bonferroni's or Scheffe's tests are more conservative than HSD.