Hello Cliff. You have found the probability (not odds) that X = 5. But p-values are tail probabilities. The two-tailed p-value CROSSTABS is showing you is p(x <= 5) + p(x >= 15) = 0.041389. Here is a little demo you can play around with, if you like.
NEW FILE.
DATA LIST FREE / X (F5.0).
BEGIN DATA
0 1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20.
END DATA.
DATASET NAME ds2.
COMPUTE pX = PDF.BINOM(X,20,.5).
COMPUTE LeftTail = CDF.BINOM(X,20,.5).
COMPUTE RightTail = CDF.BINOM(20-X,20,.5).
IF ANY(X,5,15) p2tail = 2*MIN(LeftTail,RightTail).
FORMATS pX LeftTail RightTail p2tail (F8.6).
LIST.
Here is the output.
X pX LeftTail RightTail p2tail
0 .000001 .000001 1.000000 .
1 .000019 .000020 .999999 .
2 .000181 .000201 .999980 .
3 .001087 .001288 .999799 .
4 .004621 .005909 .998712 .
5 .014786 .020695 .994091 .041389
6 .036964 .057659 .979305 .
7 .073929 .131588 .942341 .
8 .120134 .251722 .868412 .
9 .160179 .411901 .748278 .
10 .176197 .588099 .588099 .
11 .160179 .748278 .411901 .
12 .120134 .868412 .251722 .
13 .073929 .942341 .131588 .
14 .036964 .979305 .057659 .
15 .014786 .994091 .020695 .041389
16 .004621 .998712 .005909 .
17 .001087 .999799 .001288 .
18 .000181 .999980 .000201 .
19 .000019 .999999 .000020 .
20 .000001 1.000000 .000001 .
Number of cases read: 21 Number of cases listed: 21
For confirmation, here are results from Stata's -bitesti- command.
. bitesti 20 5 .5
Binomial probability test
N Observed k Expected k Assumed p Observed p
----------------------------------------------------------------
20 5 10 0.50000 0.25000
Pr(k >= 5) = 0.994091 (one-sided test)
Pr(k <= 5) = 0.020695 (one-sided test)
Pr(k <= 5 or k >= 15) = 0.041389 (two-sided test)
I hope this helps.
------------------------------
Bruce Weaver
------------------------------
Original Message:
Sent: Thu June 12, 2025 09:29 AM
From: Clifton Hamilton
Subject: McNemar Test Results
When conducting the McNemar test in SPSS, with discordant cells of b = 5 and c = 15, the software produces a p value of .041. Using the actual binomial table, the odds of 5/20 with a 50/50 proposition are reported as p = .0148, for a one-tailed test. Can anyone explain the discrepancy between .041 and .0296 (two-tailed test)?
Thanks,
Cliff