Statistical Methods for Psychology

so our conclusions will be the same as theirs.^11 The dependent variable is the degree of

arousal at the end of the 4-minute video, with larger values indicating greater arousal.

Before we consider any statistical test, and ideally even before the data are collected,

we must specify several features of the test. First we must specify the null and alternative

hypotheses:

H 0 : m 1 5 m 2

H 1 : m 1 m 2

The alternative hypothesis is bi-directional (we will reject H 0 if m 1 ,m 2 or if m 1 .m 2 ),

and thus we will use a two-tailed test. For the sake of consistency with other examples in

this book, we will let a 5 .05. It is important to keep in mind, however, that there is noth-

ing particularly sacred about any of these decisions. (Think about how Jones and Tukey

(2000) would have written this paragraph. Where would they have differed from what is

here, and why might their approach be clearer?)

Given the null hypothesis as stated, we can now calculate t:

Because we are testing H 0 , m 1 2 m 25 0, the m 1 2 m 2 term has been dropped from the

equation. We should pool our sample variances because they are so similar that we do not

have to worry about a lack of homogeneity of variance. Doing so we obtain

Notice that the pooled variance is slightly closer in value to than to because of the

greater weight given in the formula. Then

For this example, we have n 12 15 34 dffor the homophobic group and n 22 15 28 df

for the nonhomophobic group, making a total of n 12 11 n 22 15 62 df. From the sam-

pling distribution oft in Appendixt, t.025(62) 6 2.003 (with linear interpolation). Since

the value of tobtfar exceeds ta/2, we will reject H 0 (at a 5 .05) and conclude that there is a

difference between the means of the populations from which our observations were drawn.

In other words, we will conclude (statistically) that m 1 m 2 and (practically) that m 1 .m 2.

In terms of the experimental variables, homophobic subjects show greater arousal to a

homosexual video than do nonhomophobic subjects. (How would the conclusions of Jones

and Tukey (2000) compare with the one given here?)

Z



t=

X 12 X 2

D

s^2 p

n 1

1

s^2 p

n 2

=

(24.00 2 16.50)

B

144.48

35

1

144.48

29

=

7.50

1 9.11

=2.48

s^21

s^21 s^22

=

34(148.87) 1 28(139.16)

3512922

=144.48

s^2 p=

(n 12 1)s^211 (n 22 1)s^22

n 11 n 222

t=

X 12 X 2

sX 12 X 2

=

X 12 X 2

C

s^2 p

n 1

1

s^2 p

n 2

=

X 12 X 2

C

s^2 pa

1

n 1

1

1

n 2

b

Z

208 Chapter 7 Hypothesis Tests Applied to Means

(^11) I actually added 12 points to each mean, largely to avoid many negative scores, but it doesn’t change the results
or the calculations in the slightest.