Question

Suppose you observe 1Qs of 77, 88, and 95 from a group that did not take NZT. Suppose you also observe lQs of 90 and 110 from a group that did take NZT. The sample size is only five, so we shouldn't expect a low p-value. Compute the p-value based on a Wilcoxon rank-sum test with a two-sided alternative..

Answer 1

Here the sample is

No NZT	NZT
77	90
88	110
95

H₀ : $\mu _1=\mu _2$

H_a : $\mu _1\neq \mu _2$

Here writing the samples in a column in ascending order

Sample	Belong to	Rank
77	No NZT	1
88	No NZT	2
90	NZT	3
95	No NZT	4
110	NZT	5

for NZT, total rank = 3 + 5 = 8

for No NZT, total rank = 1 + 2 + 4 = 7

There is one method for small samples

Method one:

For comparing two small sets of observations, a direct method is quick, and gives insight into the meaning of the U statistic, which corresponds to the number of wins out of all pairwise contests (see the tortoise and hare example under Examples below). For each observation in one set, count the number of times this first value wins over any observations in the other set (the other value loses if this first is larger). Count 0.5 for any ties. The sum of wins and ties is U for the first set. U for the other set is the converse.

for first set U = 1 (as only one number of set 1 is greater than any number of set 2)

for second set U = 2 + 3 = 5

We can't approximate it to normal so here n = 6 and p = 0.5

so

p - value = Pr(x <= 1 ; 6 ; 0.5) = BINOMDIST(x < = 1 ; 6 ; 0.5) = 0.1094

now, two sided test is here so

p - value = 2 * 1.094 = 0.2188