Question

We are planning an experiment comparing three fertilizers. We will have six experimental units per fertilizer and will do our test at the 5% level. One of the fertilizers is the standard and the other two are new; the standard fer- tilizer has an average yield of 10, and we would like to be able to detect the situation when the new fertilizers have average yield 11 each. We expect the error variance to be about 4. What sample size would we need if we want power .9?

Answer 1

0 From the information,s-0,05,1-p2090 B-o.lo. the e.yoð Vað iance is ơ 4 The average af standaxaael fes tilslo, the aveage 약 new fertilizer is.u,-11 -the sample size & g U 21-28116 2 - 6-48 a9 Thos, theYegoİyed Sample sizes 42.

Answer 2

> group_means <- c(10,11,11)

> 

> p <- power.anova.test(groups = length(group_means), between.var = var(group_means), within.var = 4, power = 0.9, sig.level = 0.05, n=NULL)

> p

     Balanced one-way analysis of variance power calculation 

         groups = 3

              n = 76.93183

    between.var = 0.3333333

     within.var = 4

      sig.level = 0.05

          power = 0.9

NOTE: n is the number in each group