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or

If the absolute value of t is greater than the critical value of t, we reject the null

hypothesis and accept the research hypothesis
.

Since we are using a type of t-test, we need to take the absolute value of t. Since

the absolute value of

─

3.80 is greater than the critical t-value of 2.042, we reject

the null hypothesis (that the population means of the two groups are equal) and

accept the research hypothesis (that the population means of the two groups are

signiﬁcantly different from one another).

This means that our conclusion, in plain English, is as follows:

The average number of sound bursts per cycle for SUBSPECIES C was signiﬁ-

cantly greater than the average number of sound bursts per cycle for SUBSPECIES

B (16.75 vs. 14.66).

8.4.2 Performing an ANOVA t-test Using Excel commands

Now, let’s do these calculations for the ANOVA t-test using Excel with the ﬁle you

created earlier in this chapter: BEE12

A36:

SUBSPECIES B vs. SUBSPECIES C

A38:

1/n C

A40: s.e. B vs. C

A42: ANOVA t-test

B38:

1/n B

þ

(1/10

1/12)

¼

þ

B40:

SQRT(D32*B38)

¼

B42:

(D25

─

D26)/B40

¼

You should now have the following results in these cells when you round off all

these ﬁgures in the ANOVA t-test to two decimal points.:

B38:

0.18

B40:

0.55

B42:

─

3.78

S
ave this ﬁnal result under the ﬁle name: BEES6

Print out the resulting spreadsheet so that it ﬁts onto one page like
Figure 8.5

(Hint: Reduce the Page Layout / Scale to Fit to
85%
).

For a more detailed explanation of the ANOVA t-test, see Black (2010).

Important note:
You are only allowed to perform an ANOVA t-test comparing the

means of two groups when the F-test produces a signiﬁcant difference between the

means of all of the groups in your study.It is improper to do any ANOVA t-test when

the value of F is less than the critical value of F. Whenever F is less than the critical

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