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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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