SMARTPsych SPSS Tutorial
t-tests in SPSS
To demonstrate SPSS’s one sample t-test
command, we have the data set to the left.
These data were collected in an experiment
in which students (10 groups of 5
individuals) solved a number of
complicated anagrams in groups. One
person in each group was randomly
assigned anagrams that were far more
difficult than those that the others received.
Participants were asked to allocate a $10
prize among the members of the group in
any manner that they wished. The
dependent variable is the amount of money
allocated to the “least helpful” partner in
each of the groups (almost always, the
person who had the most difficult
anagrams to solve).
The null hypothesis is that all members of
the group would receive equal allocation of
the $10 prize. Thus, the null and
alternative hypotheses are:
Ho: μ = 2 (they allocate the $10 equally, so
each person gets $2)
H1: μ ≠ 2
We want to subject these data to a one
sample t-test of the null hypothesis above.
Use a two-tailed test, α = .05.
Click here for the data set (you must save it
to disk in order to access the file).
Once you have the data, it is fairly simple to compute a one-sample t-test in SPSS.
Choose, from the menu,
ANALYZE / COMPARE MEANS / ONE SAMPLE T TEST
You will get the dialog box that appears below:
Your test variable is amount; click this over into the test variable column.
What is your test value? It is the value of the mean under the null, in this case, 2.
Enter 2 into the test value box.
You may want to check out the “options” box to see what is available to you. The
box is below.
Note that will automatically get the SPSS default 95% confidence interval for the
data. The missing values command tells SPSS how to deal with missing data; the
SPSS default is fine for most purposes (this will become more critical when you
are dealing with more variables and have missing data in your set).
Click OK once you have set up your dialog boxes correctly, and SPSS will run the
test.
The output that you will receive is below:
One-Sample Statistics
N
AMOUNT
10
Mean
1.1770
Std. Deviation
.9499
Std. Error
Mean
.3004
One-Sample Test
Test Value = 2
AMOUNT
t
-2. 740
df
9
Sig. (2-tailed)
.023
Mean
Difference
-.8230
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
-1. 5026
-.1434
Take a moment to see what you have here. Of course, your critical information is
the t value (-2.74), your degrees of freedom (9), and the significance of your test
(.023; notice that you can get exact p-values in SPSS). You also get your sample
mean, the standard deviation (this is the estimated standard deviation), and the
estimated standard error of the mean. You could compute the t-test yourself with
this information:
t = 1.177 – 2 = -2.74
.9499 / √10
Also note that your 95% confidence interval of the difference ranges from –1.50 to
-.14. It’s strange that SPSS computes the confidence intervals this way, but you
can easily get the confidence intervals around the mean as follows:
If you were to compute the confidence intervals by hand, you would calculate
Mean +/– tcrit (α=.05, 2 tailed) * σ hat / √N
Or 1.177 +/– 2.262 * .9499 / √10, the confidence interval is .4975 to 1.8565
Because you use +/- .6795 to give the bounds of your interval. Using these values
around the mean difference (-.8230) you get the SPSS values; using these values
around the mean (1.177) you get .4975 to 1.8565. These are the values we would
use.
What would a confidence interval look like if you were using a lower alpha level
(i.e., having a greater % confidence)? Let’s try using a 99% confidence interval.
Run the t-test as before, but select in the options that you want a 99% c.i..
One-Sample Test
Test Value = 2
AMOUNT
t
-2. 740
df
9
Sig. (2-tailed)
.023
Mean
Difference
-.8230
99% Confidenc e
Int erval of t he
Difference
Lower
Upper
-1. 7993
.1533
You will get the table above. Notice that your confidence interval is now larger
(this makes intuitive sense; to have a greater degree of confidence, you would need
a larger range of possible values). This time, the confidence interval of the
difference contains 0 (this is essentially the same as stating that the confidence
interval around the mean contains 2 (c.i. around the mean would be .201 to 2.153).
In other words, if you use an α level of .01, you will fail to reject the null
hypothesis, and your confidence interval would contain the null value. You can
see that your exact p-value is .023, too high to reject the null if α is .01.
Now let’s do a paired-groups t-test. The example for this problem comes from the
Implicit Assumptions Test (IAT). If you are interested in this test (developed by
Tony Greenwald at UW), you can go to the following link:
http://buster.cs.yale.edu/implicit/index.html .
The IAT method is used to assess automatic evaluations (i.e., evaluations that
occur outside of conscious awareness and individuals’ control). The purpose of
this study was to see if the IAT method could be used to assess automatic
associations toward specific people who play significant roles in our lives (e.g.,
mother, romantic partner). The hypothesis was that if the IAT is able to assess
automatic attitudes, it should be able to discriminate between a “positive”
significant person and a “negative” significant person.
Subjects in the study were asked to nominate a positive person from their lives and
a negative person, and then two IATs were performed to assess their automatic
associations toward each person. The two variables represent the strength of
positive automatic associations toward each person (larger numbers indicating
more liking). Use a two-tailed test, α = .05.
The data set is below (click here for the data):
Why is this a paired-samples t test (also called a repeated measures t test)?
Because each participant is involved in both independent variable manipulations
(in this case, both positive and negative significant person associations).
The key to this type of test is that a line of data represents each participant’s
performance. Thus, the two dependent variables are paired in that they apply to
each participant.
To run this test, choose the commands
ANALYZE / COMPARE MEANS / PAIRED-SAMPLES T TEST
from the menu system. You will get the following dialog box:
Here, you need to click on both variables (the positive and negative dependent
variables). Click one, hold down the SHIFT key, and click the other. Then select
them to move to the test variable box. Your options are the same as they were for
the one-sample t test.
The output follows:
Paired Samples Statistics
Mean
Pair
1
IAT effect for pos itive
significant person
IAT effect for negative
significant person
N
Std. Deviation
Std. Error
Mean
371.1356
20
195.1234
43.6309
114.8433
20
177.8288
39.7637
Pa ired Sa mples Correlations
N
Pair
1
IAT effect for positive
signific ant pers on &
IAT effect for negat ive
signific ant pers on
Correlation
20
.418
Sig.
.067
Pa ired Sa mpl es Test
Paired Differenc es
Mean
Pair
1
IAT effect for positive
signific ant pers on IAT effect for negative
signific ant pers on
256.2923
St d. Deviat ion
St d. Error
Mean
201.7501
45.1127
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
161.8703
350.7142
t
5.681
df
Sig. (2-tailed)
19
For this study, the null hypothesis is that there is NO DIFFERENCE between
conditions (positive or negative), so Ho: μpositive = μnegative. The alternative
hypothesis is H1: μpositive ≠ μnegative.
The first table of your output gives you the mean, standard deviation, and standard
error of the mean of the two conditions. Don’t worry about the second table for
now, other than knowing that it shows how the two dependent variables correlate
with one another.
The third table is the t test itself. Notice that you have a t of 5.681, with 19 DFs,
and that the significance is listed as .000. What kind of p-value is that? Well,
since SPSS only displays three decimal places, you know that p is at least less than
.005.
Your 95% confidence interval is constructed around the difference between the
means. If the null hypothesis is that there is no difference, this value is 0. Since 0
is not in your 95% confidence interval, you can reject the null at the alpha level of
.05 (actually even less than that, as you know from the exact p-value).
SO, a paired samples t test is essentially the same as a one-sample t test of the
difference scores. Let’s prove that you would get the same result if you were to
run the test that way:
First, compute a variable of the difference scores. Select:
TRANSFORM / COMPUTE
You will get the following dialog box:
.000
Select that your target variable (diff) is equal to pmsiat – nmsiat. Then click OK.
Now, run the
ANALYZE / COMPARE MEANS / ONE SAMPLE T TEST
command again. This time, your test variable is “diff.” You will get the
following:
One-Sample Test
Test Value = 0
DIFF
t
5.681
df
19
Sig. (2-tailed)
.000
Mean
Difference
256.2923
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
161.8703
350.7142
Notice that this is the same t, with the same dfs, significance level, and confidence
intervals as when you ran the paired-samples t test.
Independent-samples t test: Now, let’s get a new data set that applies to
independent groups.
In this study, the IAT is again used, this time to assess automatic associations with
“mother.” The variable of interest is gender; that is, do males and females differ in
their automatic associations with their mothers?
Ho: μmales = μfemales. The alternative hypothesis is H1: μmales ≠ μfemales.
The data set is as follows (click here for the data set):
Notice that you still have two variables,
but here the first variable is the
independent variable (sex) and the second
is the dependent variable (momiatms, or a
measure of automatic associations to
“mother.”
This should look quite different from the
paired-samples t test; put simply, nothing
is paired in this study.
To run this t test, select:
ANALYZE / COMPARE MEANS /
INDEPENDENT-SAMPLES T TEST
The dialog box that you will get follows
on the next page.
Your test variable (the DV) is momiatms; the grouping variable is sex. You need
to define the groups, which are 0 and 1 (what the experimenter has used, but you
can use anything; here 1 is male and 0 is female).
Once you click OK, you will get the following:
Group Statistics
mom IAT effect (in ms)
subjects' sex
females
males
N
Mean
246.3593
373.1423
13
13
Std. Deviation
158.4815
151.6543
Std. Error
Mean
43.9549
42.0613
Independent Samples Test
Levene's Test for
Equality of Variances
F
mom IAT effect (in ms)
Equal variances
as sumed
Equal variances
not ass umed
.001
Sig.
.975
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-2.084
24
.048
-126.7830
60.8374
-252.3452
-1.2209
-2.084
23.954
.048
-126.7830
60.8374
-252.3581
-1.2080
Not much is new here; Levene’s test is a measure that tests an important
assumption of the independent-samples t-test; that is, the two groups have equal
variances. Here we’ll assume that the variances are equal (use the first line in the
second table for your t).
What can you say about this study? It looks like your t, with 24 degrees of
freedom (N1 + N2 – 2), is –2.084. The p-value is .048. You can reject the null
hypothesis at the alpha level of .05, but what do you conclude? It appears that
males have more positive automatic associations with “mother” than females.