Statistical Methods for Psychology

Control 0.1 mg 0.5 mg1 mg2 mg

34.00 50.80 60.33 48.50 38.10

with a grand mean of 46.346 and an average sample variance (MSerror) of 240.35.

Power Calculations

I can illustrate the calculations of and by assuming that the population values corre-

spond exactly to those that Conti and Musty found in their experiment. I will simplify the

problem slightly by assuming that we plan to run 10 subjects in each group, rather than the

unequal numbers of subjects they had.

I defined

Then

Each of our samples will contain 10 subjects, so n 5 10. Then

To use the table of the noncentral Fdistribution (Appendix ncF) we must enter it with

,dftanddfe, where dftis the dffor treatments and dfeis the dffor error. For our example,

dft 5 4, dfe 5 45, and 5 1.91. Because tables of the noncentral Fdistribution are very

coarse, they do not contain all possible values of , dft, and dfe. We either have to inter-

polate or else round off to the nearest value. For purposes of illustration, we will round off

every value in the conservative direction. Thus we will take ,dft 5 4, anddfe 5 30.

The entry in the table for F(dft,dfe; ) 5 F(4, 30; 1.8) is .14 at 5 .05. This value is , the

probability of a Type II error. Power 512512 .14 5 .86, which is a conservative esti-

mate given the way we have rounded off.

Perhaps we are willing to sacrifice some power to save on the number of subjects we

use. To calculate the required sample sizes for a different degree of power, we simply need

to work the problem backwards. Suppose that we would be satisfied with power 5 .80.

Then 5 .20, and we simply need to find that value of for which 5 .20. A minor com-

plication arises because we cannot enter Appendix ncFwithout and we cannot calculate

without knowing n. This is not a serious problem, however, because whether dfe is 30,

50, 180, or whatever will not make any really important difference in the tables. We will

therefore make the arbitrary decision that dfe 5 30, because we already know that it will

have to be less than 45, and 30 is the closest value. With dft 5 4, dfe 5 30, and 5 .20, we

find from the table that will have to be 1.68 (by interpolation).

Given

then

n=f^2 >f¿^2

f=f¿ 2 n

f

b

fe

fe

b f b

b

f a b

f=1.8

f

f

f

f=f¿ 2 n=0.6054 210 =1.91

=

B

88.0901

240.35

= 2 0.3665=0.6054

=

B

(34.00 2 46.346)^2 1 Á 1 (38.10 2 46.346)^2 > 5

240.35

f¿=

B

a(mj2m)

(^2) >k
s^2 e
f¿=
st
se

=

B

a(mj2m)

(^2) >k
s^2 e
f f¿
350 Chapter 11 Simple Analysis of Variance