. 34
( 137 .)



Use the 5 percent lowest and 5 percent highest value customers for the

challenger, and everyone else for the champion.
Use the 10 percent most recent customers for the challenger, and every­

one else for the champion.
Use the customers with telephone numbers for the telemarketing cam­

paign; everyone else for the direct mail campaign.
All of these are biased ways of splitting the population into groups. The pre­
vious results all assume that there is no such systematic bias. When there is
systematic bias, the formulas for the confidence intervals are not correct.
Using the formula for the confidence interval means that there is no system­
atic bias in deciding whether a particular customer receives the champion or
the challenger message. For instance, perhaps there was a champion model
that predicts the likelihood of customers responding to the champion offer. If
this model were used, then the challenger sample would no longer be a ran­
dom sample. It would consist of the leftover customers from the champion
model. This introduces another form of bias.
Or, perhaps the challenger model is only available to customers in certain
markets or with certain products. This introduces other forms of bias. In such
a case, these customers should be compared to the set of customers receiving
the champion offer with the same constraints.
Another form of bias might come from the method of response. The chal­
lenger may only accept responses via telephone, but the champion may accept
them by telephone or on the Web. In such a case, the challenger response may
be dampened because of the lack of a Web channel. Or, there might need to be
special training for the inbound telephone service reps to handle the chal­
lenger offer. At certain times, this might mean that wait times are longer,
another form of bias.
The confidence interval is simply a statement about statistics and disper­
sion. It does not address all the other forms of bias that might affect results,
and these forms of bias are often more important to results than sample varia­
tion. The next section talks about setting up a test and control experiment in
marketing, diving into these issues in more detail.
The Lure of Statistics: Data Mining Using Familiar Tools 147

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 cham­
pion). This section talks about ensuring that the test and control are large
enough for the purposes at hand. The previous section talked about determin­
ing the confidence interval for the sample response rate. Here, we turn this
logic inside out. Instead of starting with the size of the groups, let™s 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 differ­
ence 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 con­
fidence of 95 percent, then what is the corresponding standard error? A confi­
dence 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, let™s
assume that the test and control are the same size:
p ) (1 - p) (1 - p - d)
0.2% )
1.96 N + (p + d)

Plugging in the values just described (p is 5% and d is 0.2%) results in:

0.102% = 5% ) 95% + 5.2% ) 94.8% = 0.
N = 0.0963 2 = 66, 875
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
148 Chapter 5

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 inter­
preted. However, this effect can be determined in advance, before the test. It is
worthwhile determining the acuity of the test and control groups before run­
ning the test, to be sure that the test can produce useful results.

T I P 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 con­
trol 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.
Occam™s 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
The Lure of Statistics: Data Mining Using Familiar Tools 149

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

Bonferroni™s 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 per­
cent, 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 confi­
dence of 1 “ p for all comparisons.

Chi-Square Test
The difference of proportions method is a very powerful method for estimat­
ing 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).
150 Chapter 5

The appeal of the chi-square test is that it readily adapts to multiple test
groups and multiple outcomes, so long as the different groups are distinct
from each other. This, in fact, is about the only important rule when using this
test. As described in the next chapter on decision trees, the chi-square test is
the basis for one of the earliest forms of decision trees.

Expected Values
The place to start with chi-square is to lay data out in a table, as in Table 5.5.
This is a simple 2 — 2 table, which represents a test group and a control group
in a test that has two outcomes, say response and nonresponse. This table also
shows the total values for each column and row; that is, the total number of
responders and nonresponders (each column) and the total number in the test
and control groups (each row). The response column is added for reference; it
is not part of the calculation.
What if the data were broken up between these groups in a completely unbi­
ased way? That is, what if there really were no differences between the
columns and rows in the table? This is a completely reasonable question. We
can calculate the expected values, assuming that the number of responders
and non-responders is the same, and assuming that the sizes of the champion
and challenger groups are the same. That is, we can calculate the expected
value in each cell, given that the size of the rows and columns are the same as
in the original data.
One way of calculating the expected values is to calculate the proportion of
each row that is in each column, by computing each of the following four
quantities, as shown in Table 5.6:
Proportion of everyone who responds

Proportion of everyone who does not respond

These proportions are then multiplied by the count for each row to obtain
the expected value. This method for calculating the expected value works
when the tabular data has more columns or more rows.

Table 5.5 The Champion-Challenger Data Laid out for the Chi-Square Test


Champion 43,200 856,800 900,000 4.80%
Challenger 5,000 95,000 100,000 5.00%

TOTAL 48,200 951,800 1,000,000 4.82%
The Lure of Statistics: Data Mining Using Familiar Tools 151

Table 5.6 Calculating the Expected Values and Deviations from Expected for the Data in
Table 5.5


Champion 43,200 856,800 900,000 43,380 856,620 “180 180

Challenger 5,000 95,000 100,000 4,820 95,180 180 “180

TOTAL 48,200 951,800 1,000,000 48,200 951,800

PROPORTION 4.82% 95.18%

The expected value is quite interesting, because it shows how the data
would break up if there were no other effects. Notice that the expected value is
measured in the same units as each cell, typically a customer count, so it actu­
ally has a meaning. Also, the sum of the expected values is the same as the sum
of all the cells in the original table. The table also includes the deviation, which
is the difference between the observed value and the expected value. In this
case, the deviations all have the same value, but with different signs. This is
because the original data has two rows and two columns. Later in the chapter
there are examples using larger tables where the deviations are different.
However, the deviations in each row and each column always cancel out, so
the sum of the deviations in each row is always 0.


. 34
( 137 .)