Biomedical Engineering Reference
In-Depth Information
ability
that the result would occur randomly when we assume that the sample has
no activity. To do this, we compare the net count rate with its estimated standard
deviation
σ
nr
, given by Eq. (11.51):
r
g
53
t
g
+
r
b
10
+
50
σ
nr
=
t
b
=
30
= 2.64 cpm.
(11.64)
The observed net rate differs from 0 by 3/2.64
1.14 standard deviations. As found
in Table 11.1, the area under the standard normal curve to the right of this value is
0.127. Assuming that the activity
A
is zero, as shown in Fig. 11.4, we conclude that an
observation giving a net count rate greater than the observed
r
n
=
=
3cpm
would occur randomly with a probability of 0.127. This single set of measurements,
gross and background, is thus consistent with the conclusion that the sample likely
contains little or no activity. However, one does not know where the bell-shaped curve
in Fig. 11.4 should be centered. Based on this single measurement, the most likely
place is
r
n
= 3 cpm, with the sample activity corresponding to that value of the net
count rate.
(b) Assigning a maximum probability for type-I errors enables one to give a more
definitive answer, with that proviso, for reporting the presence or absence of activity
in a sample. When
A
= 0, as in Fig. 11.4, the net rate that leaves 5% of the area under
the normal curve to its right is
r
1
=
1.14
σ
nr
=
1.65
σ
nr
(Table 11.2). Using Eq. (11.51), we write
1.65
r
g
1.65
r
1
+50
10
t
g
+
r
b
+
50
r
1
=
t
b
=
30
,
(11.65)
where the substitution
r
g
=
r
1
+
r
b
has been made. This equation is quadratic in
r
1
.
After some manipulation, one finds that
r
1
- 0.272
r
1
- 18.2
=
0.
(11.66)
The solution is
r
1
=
4.40 cpm. The corresponding gross count rate is
r
g
=
r
1
+
r
b
=
4.40 + 50
=
54.4 cpm, and so the critical number of gross counts is
n
g
=
r
g
t
g
=
(54.4 min
-1
)
544. Thus, a sample giving
n
g
> 544 (i.e., a minimum of
545 gross counts) can be reported as having significant activity, with a probability no
greater than 0.05 of making a type-I error.
×
(10 min)
=
We can generalize the last example to compute decision levels for arbitrary
choices of
t
g
,
t
b
, and the maximum acceptable probability
α
for type-I errors. As
in Table 11.2, we let
k
α
represent the number of standard deviations of the net
count rate that gives a one-tail area equal to
α
. Then, like Eq. (11.65), we can write
for
r
1
, the minimum significant measured net count rate,
r
1
+
r
b
t
g
+
r
b
2
br
gr
+
(11.67)
r
1
=
k
α
σ
σ
=
k
α
t
b
.
Solving for
r
1
, we obtain (Problem 38)
k
2
α
t
g
+4
r
b
t
g
+
t
b
.
k
2
α
2
t
g
+
k
2
r
1
=
(11.68)
t
g
t
b
