Power calculation for dose-response or trend test

[Bill H]

Hi

I find your website incredibly useful. I am designing an experiment where I want to know if there is a 'dose-response'. Essentially, is there a trend for a variable (weight) across different doses? The study design is a parallel group trial with more subjects getting placebo than each individual drug dose. For example (though the number of subjects will change based on the power calculation - placebo will always be somewhat more):

Placebo: n = 10

Dose 0.01 mg: n = 6

Dose 0.03 mg: n = 6

Dose 0.1 mg: n = 6

Dose 0.3 mg: n = 6

Dose 1 mg: n = 6

The SD is 0.715 for the variable and we would like to detect an effect of 1.0. How many subjects do I need to see an overall trend (80% power at 0.05)? I do NOT need to know if an individual dose is better than placebo or another dose, just is there a trend?

Do I use some kind of ANOVA? When i use a one way ANOVA with multiple treatments the power goes down as the number of treatments goes up, which seems intuitively wrong for my question.

Thanks in advance

[Lenth, Russell V]

This seems like it calls for some sort of regression analysis. The trick is deciding what the x variable is. It probably shouldn’t be dose itself, as the values you give are on a kind of logarithmic scale. I am going to make a couple of assumptions for illustration:

1.       I’ll model these dose values as a variable x with x=0 (placebo), x=1(dose 0.01), …, x=5(dose=1)

2.       I’ll assume that the expected response is approximately linear in x

With this structure, the other needed task is to identify a useful effect size, quantified as the slope of the line of response versus x. Suppose I decide that a slope of .2 is of clinical interest – i.e., if I increase the dose one step on the x scale, the expected response increases by .2.

 

To do the power for the linear regression, we need the SD of the x values. Here is the scenario described with 10 obs at x=0 and 6 at each other dose:

 

X = 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5

sd(x) = 1.808101

n = 40

Then with an error SD of 0.72 as described, we get a power of about .87 as shown in this screen shot:

 

 

Now, suppose instead that we skip the experiments at x = 2 and x = 4, so that

x2 = 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 3 3 3 3 3 3 5 5 5 5 5 5

sd(x2) = 1.980059

n = 28

 

Then the power drops down to about .80, as shown in this second screen shot:

 

 

This meets your correct intuition. The key is treating it as a regression, keeping thigs and a consistent scale, and defining an appropriate slope. Hope this helps.

 

PS – as I recall, there is something special you have to do in Google Groups to actually see the images I pasted-in. But they are there if you do the right thing.

 

Russ

 

Russell V. Lenth  -  Professor Emeritus

Department of Statistics and Actuarial Science  

The University of Iowa  -  Iowa City, IA 52242  USA  

Voice (319)335-0712 (Dept. office)  -  FAX (319)335-3017

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