t test or ANOVA, ? normal distribution

#1
I have an experiment on cancer cell lines.
There are 7 cell lines.
Each cell line is treated with a different concentration of the same drug.
The average concentration that killed half the cells in each cell line is calculated, along with a standard deviation as follows:


Cell Line ////// Avg Conc ////// SD

1 ////// 0.03 ////// 0.00
2 ////// 7.84 ////// 0.50
3 ////// 6.72 ////// 2.86
4 ////// 10.78 ////// 1.56
5 ////// 10.68 ////// 9.07
6 ////// 15.14 ////// 2.87
7 ////// 32.27 ////// 0.35

Question 1
To say that cell line 1's average concentration is statistically smaller than all the others is it best to use an ANOVA comparing across all 7 groups?

Or do I use a t test and compare Cell line 1 to cell line 2.... cell line 1 to cell line 3...etc? My understanding from reading is this is the wrong way to do this because I am bound to find a statistical difference somewhere doing this.

Question 2
I know to use either the t-test or ANOVA I need normally distributed data. Because cell line 5 has a SD that is quite high, do I assume that it is not a normally distributed result and therefore comparing across cell lines, I should use the Kruskal Wallis test instead?
 
Last edited:

noetsi

Fortran must die
#2
If you do a t test you will have to conduct multiple ones (comparing each of the six cell lines to different cell lines for all comparisons you want). That will generate serious errors in a standard t-test because its alpha level assumes a single t test is being done. Your real chances of making a type I error will be much higher than the formal (nominal it is called) alpha level you are using.

ANOVA does t test for multiple test at one time (it uses an adjustment to address this problem). When doing multiple t test it is always preferable.

ANOVA formally requires normality. But it is highly robust to violations of it as long as you have at least 30 cases because of the central limit theorem. You could do both ANOVA and Kruskal Wallis and see if there is much difference.