Assuming independence of errors, zero mean and normally distributed error, analysis of variance (ANOVA) would be appropriate for to determine if the mean levels of ethanol are different across the yeast types...(though I am not sure if those levels you posted are mean or just observed levels or we might resort to Chi-Square Tests). Anyway, if I understood you correctly, ANOVA would be okay and would also lead you to carrying out post-hoc analyses among them include: Tukey Honestly Significant Differences, Fisher's Least Significant Differences, Boniferron tests, etc to tell which among yeast types are significantly different from another. Such procedures are easily done in SAS using proc mixed, proc glm, etc. I hope this answers your question.

To give you an answer, is necessary to know how many data you have. To perform ANOVA, you will need at least 10 data by yeast. I think that the best way to compare one by one is to use the "t of student", you only will need to define the freedom grades, to have the means and standard deviation to calculate an experimental t value. Then, you compare with tabulated t value and you define if differences are statistically significant. 

Hi John Ng'ombe and Christian Segura, thanks for the reply. I only have three replicates for each yeast (from which the values above were calculated). Would an ANOVA still be appropriate? Regarding the chi-square test, I am unsure if it would be appropriate for my data set, especially regarding the calculation of the expected values.

Hello Justin, yes I would still go for ANOVA, that is fine and just as I explained above. For Christian Segura's argument for t tests, for groups more than 2, it is not advisable to use the t tests because doing so would increase the probability of committing type I error which is not acceptable. (thus why a t test is for comparison of means of two samples or categories). With ANOVA, these errors are accounted for and probability of committing type I error is constant at 5% which gives you more confidence when you have statistically significant results. Hopefully this helps.

Thanks John. Much appreciated.