between-group multinomial test

Question

I'm having some trouble to decide what is the best method to analyze a between-groups / multinomial problem.

For simplicity, I'll use an example: Let's say I want to compare basketball shooting abilities between boys and girls. Each child has one minute to throw as many balls as possible to the basket. Scoring is given for each throw as -1, 0 or 1: did not reach basket, reached but missed, or made the shot - respectively. So in the end I have a table such as this one:

subject	group	-1	0	1
1	boy	3	8	2
2	girl	2	6	9
...

My initial though was to sum all outcomes for each group, and then use a chi^2 test. I wonder if there's another method which is preferable?

Thanks!

Answer 1

If you want to test whether there is an overall difference in shooting abilities between boys and girls you could use the improved Wilcoxon-Mann-Whitney test by Brunner & Munzel that allows tied values. In R you can do this with the function bmp from the WRS package:

# sample dataset:
> basket <- list("boy" = c(rep(-1, 3), rep(0, 8), rep(1,2)), 
                 "girl" = c(rep(-1, 2), rep(0, 6), rep(1, 9)))

# load the package
> source("https://dornsife.usc.edu/assets/sites/239/docs/Rallfun-v38.txt")

# test:
> bmp(basket$boy, basket$girl)
$n1
[1] 13

$n2
[1] 17

$test.stat
[1] 2.113882

$phat
[1] 0.6923077

$dhat
[1] -0.3846154

$s.e.
[1] 0.09097372

$p.value
[1] 0.04356367

$ci.p
[1] 0.5059565 0.8786589

$df
[1] 27.99999

$summary.dval
        P(X<Y)    P(X=Y)    P(X>Y)
[1,] 0.5294118 0.3257919 0.1447964

You see this test tells us that the distributions of shooting capabilities between girls and boys do indeed differ. Specifically, we are more likely to observe lower values among boys because P = 0.69 and the confidence interval does not contain 0.50 so we can reject H0: P = 0.5.

Another approach of testing general difference is to use Cliff's improvement of the Wilcoxon-Mann-Whitney test again allowing tied values. This test has a better control over Type 1 error probability than the former test and keeps it below 5%.

> cidv2(basket$boy, basket$girl)
$n1
[1] 13

$n2
[1] 17

$d.hat
[1] -0.3846154

$d.ci
[1] -0.671372675  0.002307866

$p.value
[1] 0.06

$p.hat
[1] 0.6923077

$p.ci
[1] 0.4988461 0.8356863

$summary.dvals
        P(X<Y)    P(X=Y)    P(X>Y)
[1,] 0.5294118 0.3257919 0.1447964

With this test we would fail to reject H0 that the distributions are identical (P = 0.5) because although P = 0.69 the confidence interval for P contains in this case 0.5 and the confidence interval for the difference between the probability of observing higher values among boys and the probability of observing higher values among girls contains 0.