Geoscience Reference
In-Depth Information
Table 2.2 Potassium/argon
age determinations, Grenville
Province, Canadian Shield
Biotite
Muscovite
Age class
No. ages
Age class
No. ages
(m.y)
(m.y)
770
2
830
1
850
6
870
5
870
3
890
6
910
7
910
1
930
12
930
3
950
11
970
5
970
2
990
2
990
4
1,010
3
1,010
2
1,030
3
1,050
4
1,050
1
1,070
1
1,070
1
1,090
2
1,110
1
1,130
1
1,170
1
1,190
2
1,210
1
1,210
1
n
1
¼76
n
2
¼18
Source: Agterberg (
1974
, Table XIII)
(Table
2.2
). They are mainly for micas in granitic rocks. A distinction was made
between biotite and muscovite ages. The measurements were grouped into classes
which are 20 million years wide. Midpoints of these classes are given in the table.
For example, there are six ages of 890 Ma meaning that six measured dates were
between 880 and 900 Ma. This example originally was given in Agterberg (
1974
).
The average biotite age for 76 single dates is 954 Ma, and that for muscovite is
976 Ma. The null hypothesis that the biotite and muscovite population means are
equal, can be tested by Student's
t
-test. Strictly speaking this test is based on the
assumption that the variance of the population for biotite ages is equal to that for
muscovite ages. A sample variance is estimated as
s
2
2
(
cf
. Box
2.1
). The sum of squared deviations from the sample mean is di
v
ided by
n
¼
½
1
=
ð
n
1
Þ
Σ
ð
x
x
Þ
1 instead of by
n
to account for the variance of the deviation between
x
and its
corresponding population mean (
). The hypothesis of equality of variance will be
tested first by using an
F
-test. Both
t
- and
F
-test are based on the assumption that
the underlying populations are normal. The normality assumption can be tested
separately by using a chi-square test.
Using the subscript 1 for biotite and 2 for muscovite, the sample variances are
μ
s
1
2
8,337 and
s
2
2
6,861. The
F
-test for comparing two variances consists of
calculating the ratio
F
(75, 17)
¼
¼
s
1
2
/
s
2
2
1.22. It is customary to divide the larger
variance by the smaller one, so that the
F
-ratio is greater than one. If the two
population variances are equal to one another, the estimated
F
-ratio would be close
¼
¼
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