---
title: "Mood's Median Test"
editor:
markdown:
wrap: 72
---
## Packages
```{r}
library(tidyverse)
library(smmr)
```
## Two-sample test: What to do if normality fails
- If normality fails (for one or both of the groups), what do we do
then?
- Again, can compare medians: use the thought process of the sign
test, which does not depend on normality and is not damaged by
outliers.
- A suitable test called Mood's median test.
- Before we get to that, a diversion.
## The chi-squared test for independence
Suppose we want to know whether people are in favour of having daylight
savings time all year round. We ask 20 males and 20 females whether they
each agree with having DST all year round ("yes") or not ("no"). Some
randomly chosen data:
```{r inference-5-R-1, message=F}
my_url <- "http://ritsokiguess.site/datafiles/dst.txt"
dst <- read_delim(my_url," ")
dst %>% slice_sample(n = 10)
```
## ... continued
Count up individuals in each category combination, and arrange in
contingency table:
```{r inference-5-R-2}
tab <- with(dst, table(gender, agree))
tab
```
- Most of the males say "yes", but the females are about evenly split.
- Looks like males more likely to say "yes", ie. an association
between gender and agreement.
- Test an $H_0$ of "no association" ("independence") vs. alternative
that there is really some association.
- Done with `chisq.test`.
## ...And finally
```{r inference-5-R-3}
chisq.test(tab, correct=FALSE)
```
- Reject null hypothesis of no association (P-value 0.008)
- therefore there is a difference in rates of agreement between (all)
males and females (or that gender and agreement are associated).
- This calculation gives same answers as you would get by hand.
(Omitting `correct = FALSE` uses "Yates correction".
## Mood's median test
- Earlier: compare medians of two groups.
- Sign test: count number of values above and below something (there,
hypothesized median).
- Mood's median test:
- Find "grand median" of all the data, regardless of group
- Count data values in each group above/below grand median.
- Make contingency table of group vs. above/below.
- Test for association.
- If group medians equal, each group should have about half its
observations above/below grand median. If not, one group will be
mostly above grand median and other below.
## Mood's median test for reading data
```{r inference-5-R-4, echo=FALSE, message=FALSE}
my_url <- "http://ritsokiguess.site/datafiles/drp.txt"
kids <- read_delim(my_url," ")
```
- Find overall median score:
```{r inference-5-R-5}
kids %>% summarize(med=median(score)) %>% pull(med) -> m
m
```
- Make table of above/below vs. group:
```{r inference-5-R-6}
tab <- with(kids, table(group, score > m))
tab
```
- Treatment group scores mostly above median, control group scores
mostly below, as expected.
## The test
- Do chi-squared test:
```{r inference-5-R-7}
chisq.test(tab, correct=FALSE)
```
- This test actually two-sided (tests for any association).
- Here want to test that new reading method *better* (one-sided).
- Most of treatment children above overall median, so do 1-sided test
by halving P-value to get 0.017.
- This way too, children do better at learning to read using the new
method.
## Or by smmr
- `median_test` does the whole thing:
```{r inference-5-R-8}
median_test(kids,score,group)
```
- P-value again two-sided.
## Comments
- P-value 0.013 for (1-sided) t-test, 0.017 for (1-sided) Mood median
test.
- Like the sign test, Mood's median test doesn't use the data very
efficiently (only, is each value above or below grand median).
- Thus, if we can justify doing *t*-test, we should do it. This is the
case here.
- The *t*-test will usually give smaller P-value because it uses the
data more efficiently.
- The time to use Mood's median test is if we are definitely unhappy
with the normality assumption (and thus the t-test P-value is not to
be trusted).