# Descriptive test statistics v.s. Probability

n00b question:

Lets say I am looking at a sample of records, say, sales data from several McDonald's restaurants.

Lets say I detect a difference between two groups of McDonalds. For example, McDonald's stores located in Oregon sell fewer McRibs per month than do stores in Tennessee.

Some descriptive stats (let's assume large outliers have been removed and that the sample size of each group is 2000):

Quantity of McRibs sold in the month of July

Oregon stores: 450 mean, 420 median

Tennessee stores: 700 mean, 520 median

I can conduct significance tests to see if the difference between the two groups is real.

However, is there another way to look at this: based on probability? For example, the odds that a given Store in Tennessee sells more McRibs than a given store in Oregon?

For example, if 40% of the Tennessee stores actually sell below the Tennessee median, say, 450 McRibs each, meanwhile, 40% of the Oregon stores sell at or above the Oregon mean, also 450 McRibs.

In this scenario, I would write a report concluding that McRibs are much more popular in Tennessee. This conclusion would be supported by descriptive statistics and tests of significance.

However, in reality, There is a 40% chance that a store in Tennessee sells no more McRibs than the average store in Oregon, weakening the validity of my report.

Thoughts?

edit: grammar

We can reconstruct your data into a 2 by 2 contingency table. Something like this:

| City      | Above mean (%) | Below mean (%) |
|-----------|:--------------:|:--------------:|
| Tennessee |       60       |       40       |
| Oregon    |       40       |       60       |


Given this, we can calculate an odds ratio (OR), or preferably a relative risk/risk ratio (RR).

The OR would be the odds of Tennessee divided by the odds of Oregon: $\frac{60/40}{40/60}\approx2.24$

The RR would be the risk of Tennessee divided by the risk of Oregon: $\frac{60/(60+40)}{40/(40+60)}=1.5$

Conclusion from RR: sales at or above the mean are 1.5 times more likely to occur in Tennessee than they are to occur in Oregon.

Of course, this is a sample estimate, for which a confidence interval can be obtained. This can be done in the epitools package in R or online using this web calculator.

• quick follow-up: When 60% of Tennessee is above mean, that is above the Tennessee mean, correct? – Jesse Downing Jun 21 '17 at 17:48
• I think I misunderstood your question. One simple solution to make the above meaningful is a single mean for both cities. Another idea is to find a cut-off unique to each city that is meaningful. Saying 60% of stores sold above the mean for a city simply means you have some extremely poor sales in that city. Saying 40% of stores sold below the mean in a city simply means you have some extremely high sales in that city. – Heteroskedastic Jim Jun 21 '17 at 17:57
• Additionally, even the single overall mean should have some meaningfulness to it. It is better thought of as a cut-off point that has a substantive importance, – Heteroskedastic Jim Jun 21 '17 at 18:08