Marketing – as much as cashflow – is the lifeblood of any business. No matter how good your product or service may be, it’s worthless if you can’t get it in front of your customers and get them to buy it. So all businesses, large and small, must engage in marketing. And we see countless types of marketing promotions or tactics being tried: radio and TV commercials, magazine and newspaper advertisements, public relations, coupons, email blasts, and so forth. But are our promotions working? The merchant John Wannamaker, often dubbed the father of modern advertising is said to have remarked, “Half the money I spend on advertising is wasted; the trouble is I don’t know which half.”
Some basic statistics can help you evaluate the effectiveness of your marketing and take away much of the mystique Wannamaker complained about. When deciding whether to do a promotion, managers and business owners have no way of knowing whether it will succeed; in fact, in today’s economy, budgets are still tight. The cost to roll out a full promotion can wipe out an entire marketing budget if it proves to be a fiasco. This is why many businesses do a test before doing a complete rollout. The testing helps to reduce the amount of uncertainty involved in an all-out campaign.
Quite often, large companies need to choose between two or more competing campaigns for rollout. But how do they know which will be effective? Consider the example of Jenny Kaplan, owner of K-Jen, a New Orleans-style restaurant. K-Jen serves up a tasty jambalaya entrée, which is priced at $10.00. Jenny believes that the jambalaya is a draw to the restaurant and believes that by offering a discount, she can increase the average amount of the table check. Jenny decides to issue coupons via email to patrons who have opted-in to receive such promotions. She wants to knock a dollar off the price of the jambalaya as the offer, but doesn’t know whether customers would respond better to an offer worded as “$1.00 off” or as “10% off.” So, Jenny decides to test the two concepts.
Jenny goes to her database of nearly 1,000 patrons and randomly selects 200 patrons. She decides to send half of those a coupon for $1.00 off for jambalaya, and the other half a coupon for 10% off. When the coupon offer expires 10 days later, Jenny finds that 10 coupons were redeemed for each offer – a redemption rate of 10% each. Jenny observes that either wording will get the same number of people to respond. But she wonders which offer generated the largest table check. So she looks at the guest checks to which the coupons were stapled. She notices the following:
Guest Check Amounts |
|||
Offer |
|||
$1.00 off |
10% Off |
||
$38.85 |
$50.16 |
||
$36.97 |
$54.44 |
||
$35.94 |
$32.20 |
||
$54.17 |
$32.69 |
||
$68.18 |
$51.09 |
||
$49.47 |
$46.18 |
||
$51.39 |
$57.72 |
||
$32.72 |
$44.30 |
||
$22.59 |
$59.29 |
||
$24.13 |
$22.94 |
Jenny quickly computes the average for each offer. The “$1.00 off” coupon generated an average table check of $41.44; the “10% off” coupon generated an average of $45.10. At first glance, it appears that the 10% off promotion generated a higher guest check. But is that difference meaningful, or is it due to chance? Jenny needs to do further analysis.
Hypothesis Testing
How does Jenny determine if the 10% off coupon really did better than the $1.00 off coupon? She can use statistical hypothesis testing, which is a structured analytical method for comparing the difference between two groups – in this case, two promotions. Jenny starts her analysis by formulating two hypotheses: a null hypothesis, which states that there is no difference in the average check amount for either offer; and an alternative hypothesis, which states that there is, in fact, a difference in the average check amount between the two offers. The null hypothesis is often denoted as H_{0}, and the alternative hypothesis is denoted as H_{A}. Jenny also refers to the $1.00 off offer as Offer #1, and the 10% off offer as Offer #2. She wants to compare the means of the two offers, the means of which are denoted as μ_{1} and μ_{2}, respectively. Jenny writes down her two hypotheses:
H_{0}: The average guest check amount for the two offers is equal.
H_{A}: The average guest check amount for the two offers is not equal.
Or, more succinctly:
H_{0}: μ_{1}=μ_{2 }
H_{A}: μ_{1}≠μ_{2}
Now, Jenny is ready to go to work. Note that the symbol μ denotes the population she wants to measure. Because Jenny did her test on a portion – a sample – of her database, the averages she computed were the sample average, which is denoted as . As we stated earlier, the average table checks for the “$1.00 off” and “10% off” offers were _{1}=$41.44 and _{2}=$45.10, respectively. Jenny needs to approximate μ using . She must also compute the sample standard deviation, or s for each offer.
Computing the Sample Standard Deviation
To compute the sample standard deviation, Jenny must subtract the mean of a particular offer from each of its check amounts in the sample; square the difference; sum them up; divide by the total observations minus 1(9) and then take the square root:
$1.00 Off |
|||
Actual Table Check |
Average Table Check |
Difference |
Difference Squared |
$38.85 |
$41.44 |
-$2.59 |
$6.71 |
$36.97 |
$41.44 |
-$4.47 |
$19.99 |
$35.94 |
$41.44 |
-$5.50 |
$30.26 |
$54.17 |
$41.44 |
$12.73 |
$162.03 |
$68.18 |
$41.44 |
$26.74 |
$714.97 |
$49.47 |
$41.44 |
$8.03 |
$64.46 |
$51.39 |
$41.44 |
$9.95 |
$98.98 |
$32.72 |
$41.44 |
-$8.72 |
$76.06 |
$22.59 |
$41.44 |
-$18.85 |
$355.36 |
$24.13 |
$41.44 |
-$17.31 |
$299.67 |
Total |
$1,828.50 |
||
S^{2}_{1}= |
$203.17 |
||
S_{1}= |
$14.25 |
10% Off |
|||
Actual Table Check |
Average Table Check |
Difference |
Difference Squared |
$50.16 |
$45.10 |
$5.06 |
$25.59 |
$54.44 |
$45.10 |
$9.34 |
$87.22 |
$32.20 |
$45.10 |
-$12.90 |
$166.44 |
$32.69 |
$45.10 |
-$12.41 |
$154.03 |
$51.09 |
$45.10 |
$5.99 |
$35.87 |
$46.18 |
$45.10 |
$1.08 |
$1.16 |
$57.72 |
$45.10 |
$12.62 |
$159.24 |
$44.30 |
$45.10 |
-$0.80 |
$0.64 |
$59.29 |
$45.10 |
$14.19 |
$201.33 |
$22.94 |
$45.10 |
-$22.16 |
$491.11 |
Total |
$1,322.63 |
||
S^{2}_{2}= |
$146.96 |
||
S_{2}= |
$12.12 |
Notice the denotation of S^{2}. That is known as the variance. The variance and the standard deviation are used to measure the average distance between each data point and the mean. When data are normally distributed, about 95% of all observations fall within two standard deviations from the mean (actually 1.96 standard deviations). Hence, approximately 95% of the guest checks for the $1.00 off offer should fall between $41.44 ± 1.96*($14.25) or between $13.51 and $69.37. All ten fall within this range. For the 10% off offer, about 95% will fall between $45.10 ± 1.96*($12.12), or between $21.34 and $68.86. All 10 observations also fall within this range.
Degrees of Freedom and Pooled Standard Deviation
Jenny noticed two things immediately: first, that the 10% off coupon has the higher sample average, and second each individual table check is closer to it mean than it is for the $1.00 off coupon. Also notice that when we were computing the sample standard deviation for each offer, Jenny divided by 9, and not 10. Why? Because she was making estimates of the population standard deviation. Since samples are subject to error, we must account for that. Each observation gives us information into the population’s actual values. However, Jenny had to make an estimate based on that sample, so she gives up one observation to account for the sampling error – that is, she lost a degree of freedom. In this example, Jenny has 20 total observations; since she estimated the population standard deviation for both offers, she lost two degrees of freedom, leaving her with 18 (10 + 10 – 2).
Knowing the remaining degrees of freedom, Jenny must pool the standard deviations, weighting them by their degrees of freedom. This would be especially evident if the sample sizes of the two offers were not equal. The pooled standard deviation is given by:
FYI – n is simply the sample size. Jenny then computes the pooled standard deviation:
S^{2}_{p} = ((9 * $203.17) + (9 * $146.96))) / (10 + 10 – 2)
= ($1,828.53 + $1,322.64)/18
= $3,151.17/18
= $175.07
Now take the square root: $13.23
Hence, the pooled standard deviation is $13.23
Computing the t-Test Statistic
Now the fun begins. Jenny knows the sample mean of the two offers; she knows the hypothesized difference between the two population means (which we would expect to be zero, if the null hypothesis said they were equal); she knows the pooled standard deviation; she knows the sample size; and she knows the degrees of freedom. Jenny must now calculate the t-Test statistic. The t-Test Statistic, or the t-value, represents the number of estimated standard errors the sample average is from that of the population. The t-value is computed as follows:
So Jenny sets to work computing her t-Test Statistic:
t = (($41.44 – $45.10) – (0)) / ($13.23) * SQRT(1/10 + 1/10)
= -$3.66 / ($13.23 * SQRT(1/5))
=-$3.66 / ($13.23 * .45)
=-$3.66/$5.92
= -0.62
This t-statistic gives Jenny a basis for testing her hypothesis. Jenny’s t-statistic indicates that the difference in sample table checks between the two offers is 0.62 standard errors below the hypothesized difference of zero. We now need to determine the critical t – the value that we get from a t-distribution table that is available in most statistics textbooks and online. Since we are estimating with a 95% confidence interval, and since we must account for a small sample, our critical t-value is adjusted slightly from the 1.96 standard deviations from the mean. For 18 degrees of freedom, our critical t is 2.10. The larger the sample size, the closer to 1.96 the critical t would be.
So, does Jenny Accept or Reject her Null Hypothesis (Translation: Is the “10% Off” Offer Better than the “$1.00 Off” Offer)?
Jenny now has all the information she needs to determine whether one offer worked better than the other. What does the critical t of 2.10 mean? If Jenny’s t-statistic is greater than 2.10, or (since one offer can be lower than the other), less than -2.10, then she would reject her null hypothesis, as there is sufficient evidence to suggest that the two means are not equal. Is that the case?
Jenny’s t-statistic is -0.62, which is between -2.10 and 2.10. Hence, it is within the parameters. Jenny should not reject H_{0}, since there is not enough evidence to suggest that one offer was better than the other at generating higher table checks. In fact, there’s nothing to say that the difference between the two offers is due to anything other than chance.
What Does Jenny Do Now?
Basically, Jenny can conclude that there’s not enough evidence that the “$1.00 off” coupon was worse/better than the “10% off” coupon in generating higher table check amounts, and vice-versa. This does not mean that our hypotheses were true or false, just that there was not enough statistical evidence to say so. In this case, we did not accept the null hypothesis, but rather, failed to reject it. Jenny can do a few things:
- She can run another test, and see if the same phenomenon holds.
- Jenny can accept the fact that both offers work equally well, and compare their overall average table checks to those of who ordered jambalaya without the coupons during the time the offer ran; if the coupons generated average table checks that were higher (using the hypothesis testing procedures outlined above) than those who paid full price, then she may choose to rollout a complete promotion using either or both of the offers described above.
- Jenny may decide that neither coupon offer raised average check amounts and choose not to do a full rollout after all.
So Why am I Telling You This?
The purpose of this blog post was to take you step-by-step into how you can use a simple concept like t-tests to judge the performance of two promotion concepts. Although a spreadsheet like Excel can run this test in seconds, I wanted to walk you through the theory in laymen’s terms, so that you can grasp the theory, and then apply it to your business. Analysights is in the business of helping companies – large and small – succeed at marketing, and this blog post is one ingredient in the recipe for your marketing success. If you would like some assistance in setting up a promotion test or in evaluating the effectiveness of a campaign, feel free to contact us at www.analysights.com.
Tags: ad testing, alternative hypothesis, Analysights, Analysights Small Business Solutions, campaign effectiveness, campaign measurement, coupons, degree of freedom, degrees of freedom, Direct marketing, E-mail marketing, email marketing, estimation for two population means, hypothesis, hypothesis testing, Marketing Analytics, null hypothesis, pooled standard deviation, promotional assessment, promotional evaluation, sampling error, standard deviation, statistical analysis, statistical sampling, statistical significance, t-Test, t-value, variance
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