I am conducting a study where I am collecting prices from two groups of funeral homes in a particular state: in the first group, the homes I collected prices from are owned by families, and in the second group, the homes are owned by a massive consolidator. I have not collected the same number of prices (data points) for each group, so I have unequal sets of data which vary greatly in size (one group has 17 points, while the other has over 40). Using a Shapiro-Wilk test, I have determined that my groups do not reflect normal distributions. I am trying to figure out whether the homes in the consolidator group have higher prices generally than those in the family-owned group. Is there a test I can use to determine whether the difference between the means/medians of the prices are significant?
-
$\begingroup$ Welcome to Cross Validated! You’ve tagged this with wilcoxon-mann-whitney-test. Are you thinking of that test? $\endgroup$– DaveCommented Mar 21, 2022 at 23:52
-
$\begingroup$ @Dave I am thinking of that test! It seems to be the only one that makes sense considering my parameters, at least with the research I've done on the subject. Does that seem fitting? $\endgroup$– NolaCommented Mar 22, 2022 at 0:26
-
1$\begingroup$ Why simplify your study into a test of difference in means/medians? Do you have more information about the funeral homes? For example: rural/urban location or even better exact location (which can be linked to data on population density or median income). I'd guess that these conditions might be associated with price differences as well and then it's better to account for them in the analysis. $\endgroup$– dipetkovCommented Mar 22, 2022 at 23:13
-
$\begingroup$ Have you done a quick look at the price data? I'd be interested to see what the histograms (or some other representation of density) look like. That might help point folks in the right direction. $\endgroup$– noNameTedCommented Jul 18 at 18:13
1 Answer
In a class on basic statistics, you would be taught that a non-parametric test would be required in a situation such as this. Then, you would be taught to perform the Wilcoxon Rank Sum (Mann-Whitney U) test. There's nothing wrong with applying that approach here.
I suggest that you consider applying Mood's median test, which is different from the Wilcoxon Rank Sum test. The median test determines whether the medians of two independent samples are equal, which is what I interpret your reason for applying a test to be. Wikipedia has a nice little write-up: https://en.wikipedia.org/wiki/Median_test. (The Wikipedia entry also describes why what most are taught in basic statistics about the Wilcoxon Rank Sum test and its comparison of medians is erroneous.)
The test is quite easy to implement.
Step 1. Lump all your observations and calculate the overall median.
Step 2. Produce a 2x2 table that counts the number of observations that are greater than the Step 1 median and the number of observations as large as the median, but group. For example:
Group 1 | Group 2 | |
---|---|---|
> Median | 3 | 26 |
<= Median | 14 | 14 |
Step 3. Perform a chi-square test or Fisher's exact test on these counts.
For example, in Stata
. tabi 3 26 \ 14 14, exact
| col
row | 1 2 | Total
-----------+----------------------+----------
1 | 3 26 | 29
2 | 14 14 | 28
-----------+----------------------+----------
Total | 17 40 | 57
Fisher's exact = 0.001
1-sided Fisher's exact = 0.001
In this example, the two-sided p-value of the Fisher's text is 0.001 indicating that the null hypothesis is rejected at the 5% level of significance.
If you don't have access to a statistical package, you can also run this in Excel. Here is a neat little worked example: https://www.real-statistics.com/non-parametric-tests/moods-median-test-two-samples/
EDIT: As pointed out by @dipetkov in the comments, the median test has low power. This means that it might not be able to detect a difference between groups if one does exist. I suggest trying it out and, if the result is not statistically significant, using another test such as the Wilcoxon Rank Sum test, to confirm.
Good luck to you.
-
$\begingroup$ The median test is easy to implement but has low power. Both the wikipedia page and related SE answers mention this drawback. $\endgroup$– dipetkovCommented Mar 22, 2022 at 21:45
-
$\begingroup$ Good point. I'll add that to the response. $\endgroup$ Commented Mar 22, 2022 at 22:41
-
2