Are there several plants in each replicate? Is the data actually count of live plants and count of dead plants for each replicate?

Suggest you look at Cox regression and Kaplan- Meier plots. many references are available. Best, wishes, David Booth

Salvatore S. Mangiafico Each replicate has 10 plants. The value is the percentage of plants surviving the treatment (e.g. 80% means 8 plants lived and 2 died).

David Eugene Booth Thank you Mr. Booth. But is there any common and straighforward statistical test I can apply here instead of regression models?

One idea to simplify: If you don't find any meaningful difference among the replicates within each treatment, you can simply pool the data across replicates within each treatment.

In this case, you have two variables: Concentration, which is really ordinal, but could also be treated as nominal. And Alive/Dead which is dichotomous nominal.

You will probably want to arrange this data as a table of counts.

Because Concentration is ordinal, the best approach is to use a test that treats it as ordinal. You might look at Cochran-Armitage or the linear-by-linear test. The latter treats Alive/Dead as ordinal, but that's okay because there are only two levels. Neither of these two tests identify an independent or a dependent variable.

Another test would be a chi-square test of independence (or association). This is less advantageous because it treats Concentration as nominal and not ordinal. It does have the advantage, tho, that it is likely to be more recognizable by your audience.

For an effect size measure, reporting the proportions (% Dead) or each Concentration will be sufficient. If you want to get fancy, you could determine a confidence interval for each of these proportions. This would make a very compelling plot. Be sure to use a confidence interval calculation appropriate for proportions.

Thank you professor Salvatore. However I am confused. If I simply pool the data across replicates, then that means these statistic tests do not consider replication like common tests such as ANOVA? Then how can we calculate the confidence interval?

Duy Minh Pham , for the simple chi-square or Cochan-Armitage tests, each plant counts as an observation, so you have lots of replication with that approach. In this approach, your "replicates" are just for convenience. That is, you are just clumping them into groups of 10 to make it easy. There's no meaning to the replicates. That is, it's not the case that e.g. one replicate is from one clone and another is from another.

In an alternative approach, you could use each replicate as an observation. Here you might use correlation of Survival(%) vs. Concentration. Because the trend in your data is so clear, this will also work for you, even though that gives you only 9 observations.

I attached some code in R just to make these ideas a little more concrete. A couple of notes: 1) I used Spearman correlation for the correlation. 2) I used Monte Carlo simulation in the Cochran-Armitage test due to low cell counts.

I have applied ANOVA for your data. Hope the results will help.