Size of Test and Control for an Experiment

The champion-challenger model is an example of a two-way test, where a new method (the challenger) is compared to business-as-usual activity (the champion). This section talks about ensuring that the test and control are large enough for the purposes at hand. The previous section talked about determining the confidence interval for the sample response rate. Here, we turn this logic inside out. Instead of starting with the size of the groups, lets instead consider sizes from the perspective of test design. This requires several items of information:

Estimated response rate for one of the groups, which we call p

Difference in response rates that we want to consider significant (acuity of the test), which we call d

Confidence interval (say 95 percent)

This provides enough information to determine the size of the samples needed for the test and control. For instance, suppose that the business as usual has a response rate of 5 percent and we want to measure with 95 percent confidence a difference of 0.2 percent. This means that if the response of the test group greater than 5.2 percent, then the experiment can detect the difference with a 95 percent confidence level.

For a problem of this type, the first step this is to determine the value of SEDP. That is, if we are willing to accept a difference of 0.2 percent with a confidence of 95 percent, then what is the corresponding standard error? A confidence of 95 percent means that we are 1.96 standard deviations from the mean, so the answer is to divide the difference by 1.96, which yields 0.102 percent. More generally, the process is to convert the p-value (95 percent) to a z-value (which can be done using the Excel function NORMSINV) and then divide the desired confidence by this value.

The next step is to plug these values into the formula for SEDP. For this, lets assume that the test and control are the same size:

0% I p * (1 - p) (1 - p - d) 1.96 у N + (p + d) N

Plugging in the values just described (p is 5% and d is 0.2%) results in: 0.ю2% = /5% ** 95% + 5.2% * 94.8% /°.0%3

N NN

XI 0.0963 0-7c

N = (0 00102)2

So, having equal-sized groups of of 92,561 makes it possible to measure a 0.2 percent difference in response rates with a 95 percent accuracy. Of course, this does not guarantee that the results will differ by at least 0.2 percent. It merely

says that with control and test groups of at least this size, a difference in response rates of 0.2 percent should be measurable and statistically significant.

The size of the test and control groups affects how the results can be interpreted. However, this effect can be determined in advance, before the test. It is worthwhile determining the acuity of the test and control groups before running the test, to be sure that the test can produce useful results.

Before running a marketing test, determine the acuity of the test by calculating the difference in response rates that can be measured with a high confidence (such as 95 percent).

Multiple Comparisons

The discussion has so far used examples with only one comparison, such as the difference between two presidential candidates or between a test and control group. Often, we are running multiple tests at the same time. For instance, we might try out three different challenger messages to determine if one of these produces better results than the business-as-usual message. Because handling multiple tests does affect the underlying statistics, it is important to understand what happens.

The Confidence Level with Multiple Comparisons

Consider that there are two groups that have been tested, and you are told that difference between the responses in the two groups is 95 percent certain to be due to factors other than sampling variation. A reasonable conclusion is that there is a difference between the two groups. In a well-designed test, the most likely reason would the difference in message, offer, or treatment.

Occams Razor says that we should take the simplest explanation, and not add anything extra. The simplest hypothesis for the difference in response rates is that the difference is not significant, that the response rates are really approximations of the same number. If the difference is significant, then we need to search for the reason why.

Now consider the same situation, except that you are now told that there were actually 20 groups being tested, and you were shown only one pair. Now you might reach a very different conclusion. If 20 groups are being tested, then you should expect one of them to exceed the 95 percent confidence bound due only to chance, since 95 percent means 19 times out of 20. You can no longer conclude that the difference is due to the testing parameters. Instead, because it is likely that the difference is due to sampling variation, this is the simplest hypothesis.

The confidence level is based on only one comparison. When there are multiple comparisons, that condition is not true, so the confidence as calculated previously is not quite sufficient.

Bonferronis Correction

Fortunately, there is a simple correction to fix this problem, developed by the Italian mathematician Carlo Bonferroni. We have been looking at confidence as saying that there is a 95 percent chance that some value is between A and B. Consider the following situation:

X is between A and B with a probability of 95 percent.

Y is between C and D with a probability of 95 percent.

Bonferroni wanted to know the probability that both of these are true. Another way to look at it is to determine the probability that one or the other is false. This is easier to calculate. The probability that the first is false is 5 percent, as is the probability of the second being false. The probability that either is false is the sum, 10 percent, minus the probability that both are false at the same time (0.25 percent). So, the probability that both statements are true is about 90 percent.

Looking at this from the p-value perspective says that the p-value of both statements together (10 percent) is approximated by the sum of the p-values of the two statements separately. This is not a coincidence. In fact, it is reasonable to calculate the p-value of any number of statements as the sum of the p-values of each one. If we had eight variables with a 95 percent confidence, then we would expect all eight to be in their ranges 60 percent at any given time (because 8 * 5% is a p-value of 40%).

Bonferroni applied this observation in reverse. If there are eight tests and we want an overall 95 percent confidence, then the bound for the p-value needs to be 5% / 8 = 0.625%. That is, each observation needs to be at least 99.375 percent confident. The Bonferroni correction is to divide the desired bound for the p-value by the number of comparisons being made, in order to get a confidence of 1 - p for all comparisons.

Chi-Square Test

The difference of proportions method is a very powerful method for estimating the effectiveness of campaigns and for other similar situations. However, there is another statistical test that can be used. This test, the chi-square test, is designed specifically for the situation when there are multiple tests and at least two discrete outcomes (such as response and non-response).