## 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 H_{0} 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 SAS, it is fairly straightforward to perform a power analysis for
the paired sample t-test using **proc power**.

For the calculation of Example 1, we can set the power at different levels and calculate the sample size for each level. We will specify the difference in means, which is 5-0 = 5, and the standard deviation for the difference. One thing that SAS requires is the correlation between the two measures, pre and post. In this example, we don’t know the magnitude of the correlation of the pre and post measures, we will set it to be .5, a medium strength of correlation. This way, the standard deviation can be considered to be the pooled standard deviation from the standard deviation of the two measures. We set the power level from .6 to .9 and look for sample size for each level of power.

proc power; pairedmeans test=diff meandiff = 5 std = 5 corr = .5 npairs = . power = 0.6 to .9 by .1; run;Paired t Test for Mean Difference Fixed Scenario Elements Distribution Normal Method Exact Mean Difference 5 Standard Deviation 5 Correlation 0.5 Number of Sides 2 Null Difference 0 Alpha 0.05 Computed N Pairs Nominal Actual N Index Power Power Pairs 1 0.6 0.600 7 2 0.7 0.748 9 3 0.8 0.803 10 4 0.9 0.911 13

Next, let’s change the level of significance to .01 with a power of .911. What does this mean for our sample size calculation?

proc power; pairedmeans test=diff meandiff = 5 std = 5 corr = .5 npairs = . alpha =.01 power = 0.911; run;Paired t Test for Mean Difference Fixed Scenario Elements Distribution Normal Method Exact Alpha 0.01 Mean Difference 5 Standard Deviation 5 Correlation 0.5 Nominal Power 0.911 Number of Sides 2 Null Difference 0 Computed N Pairs Actual N Power Pairs 0.915 19

As you can see, the sample size goes up from 13 to 19 for specified power of .911 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. Remember all the calculation is under the normality assumption. If the distribution is not normal, then 19 subjects are, in general, not enough for this t-test.

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
**proc power** to calculate the power.

proc power; pairedmeans test=diff meandiff = 5 std = 10 corr = .5 npairs = 35 power = .; run;Paired t Test for Mean Difference Fixed Scenario Elements Distribution Normal Method Exact Mean Difference 5 Standard Deviation 10 Correlation 0.5 Number of Pairs 35 Number of Sides 2 Null Difference 0 Alpha 0.05 Computed Power Power 0.820

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. But 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.

proc power; pairedmeans test=diff meandiff = 5 std = 10 corr = .5 npairs = 35 sides = 1 power = .; run;Paired t Test for Mean Difference Fixed Scenario Elements Distribution Normal Method Exact Number of Sides 1 Mean Difference 5 Standard Deviation 10 Correlation 0.5 Number of Pairs 35 Null Difference 0 Alpha 0.05 Computed Power Power 0.895

Recall that we set the correlation between the two measures at .5 for all the calculations we have done. Let’s take a look at how the strength of correlation affects the sample size.

proc power; pairedmeans test=diff meandiff = 5 std = 10 corr = -.9 to .9 by .1 npairs = . power = .8; run;Paired t Test for Mean Difference Fixed Scenario Elements Distribution Normal Method Exact Mean Difference 5 Standard Deviation 10 Nominal Power 0.8 Number of Sides 2 Null Difference 0 Alpha 0.05 Computed N Pairs Actual N Index Corr Power Pairs 1 -0.9 0.802 122 2 -0.8 0.800 115 3 -0.7 0.801 109 4 -0.6 0.802 103 5 -0.5 0.804 97 6 -0.4 0.801 90 7 -0.3 0.802 84 8 -0.2 0.804 78 9 -0.1 0.806 72 10 -0.0 0.802 65 11 0.1 0.804 59 12 0.2 0.806 53 13 0.3 0.801 46 14 0.4 0.804 40 15 0.5 0.808 34 16 0.6 0.814 28 17 0.7 0.803 21 18 0.8 0.812 15 19 0.9 0.835 9

We can see clearly that the more positively correlated the two measures are, the smaller the sample size needs to be.

Also, we don’t have to assume that the standard deviation is the same for both groups. If we
know the standard deviation for each measure and the correlation between the two
measures, we can supply that information to **proc power**.
For instance, the standard deviation for the measure of weight before the
program might be smaller than the standard deviation for the measure of weight
after the weight-loss program. Let’s say, the standard deviation for the first
measure (before the program) is 7 and the standard deviation for the second
measure (after the program) is 12 and the correlation between the two is .5. We
can calculate the sample size in this setting as well.

proc power; pairedmeans test=diff meandiff = 5 pairedstddevs = (7 12) corr = .5 npairs = . power = .6 to .9 by .05; run;Paired t Test for Mean Difference Fixed Scenario Elements Distribution Normal Method Exact Mean Difference 5 Standard Deviation 1 7 Standard Deviation 2 12 Correlation 0.5 Number of Sides 2 Null Difference 0 Alpha 0.05 Computed N Pairs Nominal Actual N Index Power Power Pairs 1 0.60 0.613 24 2 0.65 0.651 26 3 0.70 0.702 29 4 0.75 0.760 33 5 0.80 0.809 37 6 0.85 0.858 42 7 0.90 0.901 48

## Discussion

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 between the means over the pooled standard deviation. We call this the effect size (Cohen’s d). 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

- References

- D. Moore and G. McCabe, Introduction to the Practice
of Statistics, Third Edition, Section 6.4.