SPSS Tutorial 3—Binomial Tests (Sign Test) in SPSS
November 16, 2000
While studying for his statistics
midterm, a UW student notices that his
understanding of the concepts increases
while he listens to Limp Biskit. He
wonders if other students would also
benefit by listening to Limp Biskit
while they study stats.
In order to decide if this intervention
might work, he takes 25 students from
his class who agree to be part of his
study, and assigns each student to a “no
music” condition and to a “music”
condition (a within-subjects design); in
the latter condition, Limp Biskit is
playing through a set of headphones
that each student wears. In both
conditions, students work through 25
difficult statistics problems.
The data that are collected are the
number of problems that are correct
(out of 25) in each condition. These
data appear to the left.
Now that the study has been conducted,
we would of course want to know if
Limp Biskit truly does make a
difference in one’s statistics
performance. This can be
accomplished (among other ways that
you will discover later) with the
BINOMIAL TEST (also known as the
sign test) in SPSS.
Below I’ve constructed a boxplot of the data in SPSS so that you can see how
scores for the two conditions are distributed.
(If you wanted to do this, you would start by selecting GRAPHS / BOXPLOT
from the menu in SPSS. You would then do a simple boxplot of the separate
variables “music” and “nomusic.”)
30
It appears that the median values
are about the same for both
conditions, though the Q1 and Q3
scores are higher in the ‘music’
condition. It’s difficult to
determine if there’s a difference
without using a statistical test…
20
10
0
N=
25
25
MUSIC
NOMUSIC
Well, you can’t run a binomial test on these data until you create a variable that
meets the binomial criteria. That is, you need to have a variable that identifies
“successes” and “failures.” Here, let’s use 0 to represent a failure (the student’s
performance in the ‘no music’ condition is better than performance in the ‘music’
condition) and a 1 to represent a success (superior performance in the ‘music’
condition).
You may ask at this point, what do you do if the performance under the two
conditions is equal? Well, the convention is to throw out cases in which this
happens. Ideally, your measurement would be so sensitive that you would always
have a difference between the scores, but if you don’t, you usually reduce your
sample size by the number of “ties” that you find.
At this point, you will need to compute a new variable that is the difference
between the two conditions. We’ll do this through the menu system, though you
can do it through syntax.
Start by selecting TRANSFORM / COMPUTE. You now want to compute a new
variable (here I used ‘diff’) that is the score in the ‘no music’ condition subtracted
from the score in the ‘music’ condition.
In this dialog box, you can click the variables over from the list to the left, and
perform functions on them using the calculator. Notice that there are many
functions from which to choose; here, we’re just doing simple subtraction.
When you have your dialog box set up as above, click “OK” to run the command
and to create a new variable. You can also click “Paste” to place the command
into syntax (the best option if you are going to run the command more than once;
e.g., if you had a data entry error earlier).
Now, we need to make another variable that denotes successes (1) and failures (0).
The best way to do this in SPSS is using the RECODE command. This feature
allows you to take the values in one variable and recode them (either to the same or
to a different variable). We want all values in the “diff” variable that are positive
to be recoded as 1, and all values that are negative to be recoded as 0. We will
exclude cases with a difference of 0.
Select TRANSFORM / RECODE / INTO DIFFERENT VARIABLES from the
menu in SPSS.
You will get the dialog box above. Click over the “diff” variable, because you
want to recode it. The “output variable,” the new variable, we’ll call “outcome.”
You must click the “CHANGE” button for the new variable to be recoded.
Now, select the values that you want to recode by clicking “OLD AND NEW
VARIABLES.” You will get the dialog box that follows:
This box has the information you need already filled in appropriately. You want
the old values of range –25 to –1 recoded as 0 (these are failures, representing the
entire range of possible difference scores that would be failures). You also want 1
through 25 recoded as 1 for successes. Recode “Else” (which will only be the 0’s)
as system-missing, so that they will be excluded from the analysis.
When your dialog box has the OldNew column filled in exactly as above, click
“Continue.”
Next, click “OK” to recode the variable. If “OK” is not highlighted, you probably
need to click the “Change” button to indicate that you want the variable changed.
Your new variable, outcome, should have 0’s, 1’s, and .’s (representing systemmissing) as values only. Now, you’re ready to run your binomial test!!!
Select
Analyze /
Nonparametric Tests /
Binomial.
The dialog box to the
left should then appear.
Notice that you can specify the test proportion (this is the same as p). Here, we are
using .5, because if our null hypothesis is that there is no difference between
conditions, we would expect half of the population to be successes, and half to be
failures, by chance alone.
Your test variable (the one on which you want to run the binomial test) is
“outcome.” Click it over to the left, then click OK.
Binomial Test
succes s or failure
Group 1
Group 2
Total
Category
1.00
.00
N
18
5
23
Observed
Prop.
.78
.22
1.00
Test Prop.
.50
Exact Sig.
(2-tailed)
.011
Your output is relatively simple. First, notice that your N is 23; this is because you
had two cases in which there was no difference between conditions, which were
removed from the analysis. There are 18 successes (78%) and 5 failures (22%).
The significance (exact sig.) is the most interesting part here: it is .011.
What does this mean? If there really were no difference between the conditions,
you would obtain results as extreme as those you obtained (18 out of 23 successes,
or failures) or more extreme 1.1% of the time. So, it’s unlikely that there is no
difference between these two conditions.
Let’s check out the binomial probability calculator on the web to see if we get the
same result: http://faculty.vassar.edu/~lowry/binom_stats.html
Your N here is 23; your k is 18, and p is .5. When you calculate the results, you
find that the probability of 18 or more successes is 0.00531. If you multiply this
by 2, you will get .011, the same result as in SPSS. Why is the value that the web
calculator finds half of the SPSS value?
The answer lies in the fact that SPSS gives a 2-tailed probability value. That is,
SPSS gives you the probability of finding 18 or more successes or 18 or more
failures (5 or fewer successes). Because the binomial distribution is symmetrical
when p is .5, these two scenarios are equally likely.
Normal Approximation to the Binomial
As you know, you can use the normal approximation to the binomial distribution to
find the probability that you would find a certain number of successes (or a more
extreme number).
To do this, you need to obtain a z score for the number of successes obtained (18
out of 23 trials). In order to do this, you need to know both the mean (expected
value) and the standard deviation of the population.
Mean = np = 23 * .5 = 11.5
Standard Deviation = sqrt (npq) = sqrt (23 * .5 * .5) = 2.40
Your z score is: 18 – 11.5 = 2.71
2.4
Looking at the probabilities under the normal curve in the back of your textbook,
you will find that the probability of having this number of successes, or more, is
.0034. This is relatively close to the .0053 value we found through the calculator.
What if you were to use the correction for continuity?
If you used the correction for continuity (which is typically used for samples of
size N=20 or less), you would use the score of 17.5 (the lower real limit of the
interval containing 18) in the calculation of your z-value. Here is what you would
get:
z score is: 17.5 – 11.5 = 2.50
2.4
The area under the normal curve beyond a z value beyond 2.5 is .0062. This is
even closer to the value that we would expect. You can see that the normal
approximation to the binomial distribution is a fairly accurate way of determining
the probability of obtaining a certain number of successes.
Now, it’s your turn to do a binomial test. The assignment begins on the following
page.
Assignment
You have conducted a cognitive psychology experiment in which you are
interested in how subjects react to being “primed” with a word that is relevant to
the target word to which they respond. You measure their reaction time to the
target word and obtain the following data:
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Primed
430
440
520
500
400
480
420
520
600
500
540
480
520
420
Not Primed
450
440
580
420
500
500
440
680
650
480
590
580
560
480
Now, you should determine if there is an effect of priming. Analyze and interpret
the data. For extra credit, use the normal approximation to the binomial
distribution in addition to your test.
You should indicate:
1) Whether or not there is an effect of priming (if so, the reaction time should
be less for priming, indicating that they responded quicker in the primed
condition).
2) How you made your decision (what test you used, what was the likelihood
of obtaining these results or those more extreme, etc.)
3) Include your SPSS printout or hand calculations for the data.