ability” (Aronson, Lustina, Good, Keough, Steele, & Brown, 1998). This feeling of stereo-
type threat is then hypothesized to affect performance, generally by lowering it from what
it would have been had the individual not felt threatened. Considerable work has been done
with ethnic groups who are stereotypically reputed to do poorly in some area, but Aronson
et al. went a step further to ask if stereotype threat could actually lower the performance of
white males—a group that is not normally associated with stereotype threat.
Aronson et al. (1998) used two independent groups of college students who were
known to excel in mathematics, and for whom doing well in math was considered impor-
tant. They assigned 11 students to a control group that was simply asked to complete a
difficult mathematics exam. They assigned 12 students to a threat condition, in which
they were told that Asian students typically did better than other students in math tests,
and that the purpose of the exam was to help the experimenter to understand why this
difference exists. Aronson reasoned that simply telling white students that Asians did
better on math tests would arousal feelings of stereotype threat and diminish the stu-
dents’ performance.
The data in Table 7.7 have been constructed to have nearly the same means and stan-
dard deviations as Aronson’s data. The dependent variable is the number of items correctly
solved.
First we need to specify the null hypothesis, the significance level, and whether we will
use a one- or a two-tailed test. We want to test the null hypothesis that the two conditions
perform equally well on the test, so we have. We will set alpha at a 5 .05, in
line with what we have been using. Finally, we will choose to use a two-tailed test because
it is reasonably possible for either group to show superior math performance.
Next we need to calculate the pooled variance estimate.Finally, we can calculate t using the pooled variance estimate:For this example we have n 11 n 22 25 21 degrees of freedom. From Appendix t we
find. Because 2.37 .2.080, we will reject and conclude that the two pop-
ulation means are not equal.t.025=2.080 H 0t=(X 12 X 2 )
D
s^2 p
n 11
s^2 p
n 2=
(9.64 2 6.58)
B
9.5942
11
1
9.5942
12
=
3.06
1 1.6717
=
3.06
1.2929
=2.37
=
10(10.0489) 1 11(9.1809)
21
=
201.4789
21
=9.5942
s^2 p=(n 12 1)s^211 (n 22 1)s^22
n 11 n 222=
10(3.17^2 ) 1 11(3.03^2 )
1111222
H 0 : m 1 =m 2212 Chapter 7 Hypothesis Tests Applied to Means
Table 7.7 Data from Aronson et al. (1998)
Control Subjects Threat Subjects
49128 7872
9131213 69 710
1376 50108
Mean 5 9.64 Mean 5 6.58
St. Dev 5 3.17 St. Dev 5 3.03
n 1511 n 2512