Examples

Example 1. A company markets an eight week long weight loss program
and claims that at the end of the program on average a participant will have
lost 5 pounds. On the other hand, you have studied the program and you believe
that their program is scientifically unsound and shouldn’t work at all. With some
limited funding at hand, you want test the hypothesis that the weight loss
program does not help people lose weight. Your plan is to get a random sample of
people and put them on the program. You will measure their weight at the
beginning of the program and then measure their weight again at the end of the
program.  Based on some previous research, you believe that the standard
deviation of the weight difference over eight weeks will be 5 pounds.
You now want to know how many people you should enroll in the program to test
your hypothesis.

Example 2. A human factors researcher wants to study the difference between
dominant hand and the non-dominant hand in terms of manual dexterity. She designs an
experiment where each subject would place 10 small beads on the table in a bowl,
once with the dominant hand and once with the non-dominant hand. She measured
the number seconds needed in each round to complete the task. She has also decided
that the order in which the two hands are measured should be counter balanced. She
expects that the average difference in time would be 5 seconds with the dominant
hand being more efficient with standard deviation of 10. She collects her
data on a sample of 35 subjects. The question is, what is the statistical power of her
design with an N of 35 to detect the difference in the magnitude of 5 seconds.

Prelude to the Power Analysis

In both of the examples, there are two measures on each subject, and we are
interested in the mean of the difference of the two measures. This can be done
with a t-test for paired samples (dependent samples). In a power analysis, there are always a pair of
hypotheses:  a specific null hypothesis and a specific alternative hypothesis.
For instance, in Example 1, the null hypothesis is that the mean weight loss
is 5 pounds and the alternative is zero pounds. In Example 2, the null hypothesis
is that mean difference is zero seconds and the alternative hypothesis is that the mean
difference is 5 seconds.

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

Both of these calculations depend on the Type I error rate, the significance
level. The significance level (called alpha), or the Type I error rate, is the
probability of rejecting H₀ when it is actually true. The smaller the
Type I error rate, the larger the sample size required for the same power.
Likewise, the smaller the Type I error rate, the smaller the power for the same
sample size. This is the trade-off between the reliability and sensitivity of
the test.

Power Analysis

In Stata, it is fairly straightforward to perform a power analysis for
the paired sample t-test using Stata’s power command.

For the calculation of Example 1, we can set the power at different
levels and calculate the sample size for each level. For example, we can
set the power to be at the .80 level at first, and then reset it to be at the .85 level, and so on.
First, we specify that we have paired means. Next, we specify the two means, the mean for the
null hypothesis and the mean for the alternative hypothesis. Then we specify the
standard deviation for the difference in the means. The default significance level (alpha
level) is set at .05, so we will not specify it for the initial runs.

power pairedmeans 0 5, sddiff(5)

Performing iteration ...

Estimated sample size for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0500          ma1 =    0.0000
        power =    0.8000          ma2 =    5.0000
        delta =    1.0000
           d0 =    0.0000
           da =    5.0000
         sd_d =    5.0000

Estimated sample size:

            N =        10

power pairedmeans 0 5, sddiff(5) power(.85)

Performing iteration ...

Estimated sample size for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0500          ma1 =    0.0000
        power =    0.8500          ma2 =    5.0000
        delta =    1.0000
           d0 =    0.0000
           da =    5.0000
         sd_d =    5.0000

Estimated sample size:

            N =        12

power pairedmeans 0 5, sddiff(5) power(.9)

Performing iteration ...

Estimated sample size for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0500          ma1 =    0.0000
        power =    0.9000          ma2 =    5.0000
        delta =    1.0000
           d0 =    0.0000
           da =    5.0000
         sd_d =    5.0000

Estimated sample size:

            N =        13

Next, let’s change the level of significance to .01 with a power of .85. What does this
mean for our sample size calculation?

power pairedmeans 0 5, sddiff(5) power(.85) alpha(.05)
Performing iteration ...

Estimated sample size for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0500          ma1 =    0.0000
        power =    0.8500          ma2 =    5.0000
        delta =    1.0000
           d0 =    0.0000
           da =    5.0000
         sd_d =    5.0000

Estimated sample size:

            N =        12

power pairedmeans 0 5, sddiff(5) power(.85) alpha(.01)

Performing iteration ...

Estimated sample size for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0100          ma1 =    0.0000
        power =    0.8500          ma2 =    5.0000
        delta =    1.0000
           d0 =    0.0000
           da =    5.0000
         sd_d =    5.0000

Estimated sample size:

            N =        17

As you can see, the sample size goes up from 12 to 17 for specified power of .85
when alpha drops from .05 to .01.

This
means if we want our test to be more reliable, i.e., not rejecting the null hypothesis in
case it is true, we will need a larger sample size. If we think that we want a
lower alpha at 0.01 level and a high power at .90 then we would need 15 subjects
as shown below. Remember this is under the normality assumption. If the
distribution is not normal, then 15 subjects are, in general, not enough for
this t-test.

power pairedmeans 0 5, sddiff(5) power(.9) alpha(.01)

Performing iteration ...

Estimated sample size for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0100          ma1 =    0.0000
        power =    0.9000          ma2 =    5.0000
        delta =    1.0000
           d0 =    0.0000
           da =    5.0000
         sd_d =    5.0000

Estimated sample size:

            N =        19

Now, let’s now turn our calculation around the other way. Let’s look at Example 2.
In this example, our researcher has already collected data on 35
subjects. How much statistical power does her design have to detect the
difference of 5 seconds with standard deviation of 10 seconds?

Again we use the
power command to calculate the power. We enter the first mean
as 0 and the second mean as 5 since the only thing we know is the difference of
the two means is 5 seconds. In terms of hypotheses, this is the same way of
saying that the
null hypothesis is that the difference is zero, and the alternative hypothesis is that
the mean difference is 5. Then we enter the standard deviation for the
difference and the number of subjects. Again we specify pairedmeans since the
design is a paired-sample t-test.

power pairedmeans 0 5, sddiff(10) n(35)

Estimated power for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d != d0

Study parameters:

        alpha =    0.0500          ma1 =    0.0000
            N =        35          ma2 =    5.0000
        delta =    0.5000
           d0 =    0.0000
           da =    5.0000
         sd_d =   10.0000

Estimated power:

        power =    0.8195

This means that the researcher would detect the
difference of 5 seconds about 82 percent of the time.  Notice we did this
as two-sided test. Since it is believed that our dominant hand is always better
than the non-dominant hand, the researcher actually could conduct a one-tailed
test. Now, let’s recalculate the power for one-tailed paired-sample t-test.

power pairedmeans 0 5, sddiff(10) n(35) onesided

Estimated power for a two-sample paired-means test
Paired t test
H0: d = d0  versus  Ha: d > d0

Study parameters:

        alpha =    0.0500          ma1 =    0.0000
            N =        35          ma2 =    5.0000
        delta =    0.5000
           d0 =    0.0000
           da =    5.0000
         sd_d =   10.0000

Estimated power:

        power =    0.8950

Discussion

The way to conduct the power analysis for
paired-sample t-test is the same as for the one-sample t-test. This is due to
the fact that in the paired-sample t-test we compute the difference in the two
scores for each subject and then compute the mean and standard deviation of the
differences. This turns the paired-sample t-test into a
one-sample t-test.

The other 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. It
might not even be a good idea to do a t-test on a small sample to begin with.

What we really need to know is the difference between the two means, not the
individual values. In fact, what really matters, is the difference of the means
over the standard deviation. We call this the effect size. It is usually not an
easy task to determine the effect size. It usually comes
from studying the existing literature or from pilot studies. A good estimate of the effect size
is the key to a successful power analysis.

See Also