Examples

Example 1. A clinical dietician wants to compare two different diets, A
and B, for diabetic patients. She hypothesizes that diet A (Group 1) will be better than
diet B (Group 2), in terms of lower blood glucose. She plans to get a random sample of
diabetic patients and randomly assign them to one of the two diets. At the end
of the experiment, which lasts 6 weeks, a fasting blood glucose test will be
conducted on each patient. She also expects that the average difference in
blood glucose measure between the two group will be about 10 mg/dl. Furthermore,
she also assumes the standard deviation of blood glucose distribution for diet
A to be 15 and the standard deviation for diet B to be 17. The dietician wants to know
the number of subjects needed in each group assuming equal sized groups.

Example 2. An audiologist wanted to study the effect of gender on the
response time to a certain sound frequency. He suspected that men were better at
detecting this type of sound then were women.
He took a random sample of 20 male and 20 female subjects
for this experiment. Each subject was be given a button to press
when he/she heard the sound. The audiologist then measured the response time – the time
between the sound was emitted and the time the button was pressed.
Now, he
wants to know what the statistical power is based on his total of 40
subjects to detect the gender difference.

Prelude to The Power Analysis

There are two different aspects of power analysis. One is to calculate the necessary
sample size for a specified power as in Example 1. The other aspect is to calculate the power when
given a specific sample size as in Example 2. Technically, power is the probability of rejecting
the null hypothesis when the specific alternative hypothesis is true.

For the power analyses below, we are going to focus on Example 1, calculating
the sample size for a given statistical power of testing the difference in the
effect of diet A and diet B. Notice the assumptions that the dietician has made in order
to perform the power analysis. Here is the information we have to know or have
to assume in order to perform the power analysis:

Notice that in the first example, the dietician didn’t specify the mean for each
group, instead she only specified the difference of the two means. This is
because that she is only interested in the difference, and it does not matter
what the means are as long as the difference is the same.

Power Analysis

In Stata, it is fairly straightforward to perform power analysis for
comparing means. For example, we can use Stata’s power command for our
calculation as shown below. We first specify that we have two means. Next, we specify the two means, the mean for Group 1
(diet A) and the mean for Group 2 (diet B). Since what really matters is the
difference, instead of means for each group, we can enter a mean of zero for Group 1
and 10 for the mean of Group 2, so that the difference in means will be 10. Next, we specify the
standard deviation for the first population and standard deviation for the
second population. The default significance level (alpha level) is .05.
For this example we will set the power to be at .8, which is the default value.

power twomeans 0 10, sd1(15) sd2(17)
	
Performing iteration ...

Estimated sample sizes for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1  versus  Ha: m2 != m1

Study parameters:

        alpha =    0.0500
        power =    0.8000
        delta =   10.0000
           m1 =    0.0000
           m2 =   10.0000
          sd1 =   15.0000
          sd2 =   17.0000

Estimated sample sizes:

            N =        84
  N per group =        42

The calculation results indicate that we need 42 subjects for diet A and
another 42 subject for diet B in our sample in order the effect. Now, let’s use
another pair of means with the same difference. As we have discussed earlier,
the results should be the same, and they are.

power twomeans 5 15, sd1(15) sd2(17)
	
Performing iteration ...

Estimated sample sizes for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1  versus  Ha: m2 != m1

Study parameters:

        alpha =    0.0500
        power =    0.8000
        delta =   10.0000
           m1 =    5.0000
           m2 =   15.0000
          sd1 =   15.0000
          sd2 =   17.0000

Estimated sample sizes:

            N =        84
  N per group =        42

Now the dietician may feel that a total sample size of 84 subjects is beyond her
budget. One way of reducing the sample size is to increase the Type I error
rate, or the alpha level. Let’s say instead of using alpha level of .05 we will
use .07. Then our sample size will reduce by 4 for each group as shown below.

power twomeans 5 15, sd1(15) sd2(17) alpha(.07)
	
Performing iteration ...

Estimated sample sizes for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1  versus  Ha: m2 != m1

Study parameters:

        alpha =    0.0700
        power =    0.8000
        delta =   10.0000
           m1 =    5.0000
           m2 =   15.0000
          sd1 =   15.0000
          sd2 =   17.0000

Estimated sample sizes:

            N =        76
  N per group =        38

Now suppose the dietician can only collect data on 60 subjects with 30 in each
group. What will the statistical power for her t-test be with respect to alpha
level of .05?

power twomeans 0 10, sd1(15) sd2(17) n(60)
	
Estimated power for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1  versus  Ha: m2 != m1

Study parameters:

        alpha =    0.0500
            N =        60
  N per group =        30
        delta =   10.0000
           m1 =    0.0000
           m2 =   10.0000
          sd1 =   15.0000
          sd2 =   17.0000

Estimated power:

        power =    0.6610

What if she actually collected her data on 60 subjects but with 40 on diet A
and 20 on diet B instead of equal sample sizes in the groups?

power twomeans 0 10, sd1(15) sd2(17) n(60) nratio(2)
	
Estimated power for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1  versus  Ha: m2 != m1

Study parameters:

        alpha =    0.0500
            N =        60
           N1 =        20
           N2 =        40
        N2/N1 =    2.0000
        delta =   10.0000
           m1 =    0.0000
           m2 =   10.0000
          sd1 =   15.0000
          sd2 =   17.0000

Estimated power:

        power =    0.6232

As you can see the power goes down from .66 to .62 even though the total number of subjects
is the same. This is why we always say that a balanced design is more efficient.

Discussion

An important technical assumption is the normality assumption. If the
distribution is skewed, then a small sample size may not have the power shown in
the results, because the value in the results is calculated using the method
based on the normality assumption. We
have seen that in order to compute the power or the sample size, we have to make
a number of assumptions. These assumptions are used not only for the purpose of
calculation, but are also used in the actual t-test itself. So one important side
benefit of performing power analysis is to help us to better understand our designs
and our hypotheses.

We have seen in the power calculation process that what matters in the
two-independent sample t-test is the difference in the means and
the standard deviations for the two groups. This leads to the concept of effect
size. In this case, the effect size will be the difference in means over the
pooled standard deviation. The larger the effect size, the larger the power
for a given sample size. Or, the larger the effect size, the smaller
sample size needed to achieve the same power. So, a good estimate of effect
size is the key to a good power analysis. But it is not always an easy task to
determine the effect size. Good estimates of effect size come from the existing literature
or from pilot studies. One may also want to consider using the minimum effect size of interest.

See Also