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Lecture 16
Statistical Analysis in Biomaterials Research (Part II)
C. F Distribution
¾
Allows comparison of variability of behavior between populations
using test of hypothesis:
σ
x
=
σ
x’
Named for British statistician
Sir Ronald A. Fisher.
Define a statistic:
2
2
χ
2
=
ν
1
S
/
σ
where
ν
1
is degrees of freedom.
1
2
/
v
1
then
F
=
2
/
ν
1
v
2
2
2
S
x
F
For
σ
x
=
σ
x’
⇒
F
=
S
2
x
'
Procedure to test variability hypothesis:
1. Calculate
S
x
2
and
S
x’
2
(with
ν
1
=
Ν
−1
and
ν
2
=
N’
1, respectively)
2. Compute
F
3. Look in
F
distribution tables for critical
F
for
ν
1
, ν
2
, and desired confidence
level
P
F
1
−
P
<<
4. For
FF
1
+
P
⇒ σ
x
=
σ
x’
2
2
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Case Example
:
Measurements of C5a production for blood exposure to an
extracorpeal filtration device and tubing gave same means, but different
variabilities. Are the standard deviations different within 95% confidence?
Control (tubing only):
S
x’
2
= 26 (
µ
g/ml)
2
,
ν
2
=9
Filtration device: S
x
2
= 32 (
µ
g/ml)
2
,
ν
1
= 7
2
1. Calculate
S
x
2
and
S
x’
(provided)
2. Compute
F
2
S
x
F
=
S
2
= 32/26 = 1.231
x
'
3. Determine critical
F
values from
F
distribution chart
ν
1
=7
and
ν
2
=9
(m, n for use with tables)
1
−
P
=
0.025
⇒
F
0.025
=
0.207
2
1
+
P
=
0.975
⇒
F
0.975
=
4.20
2
For 0.207
≤
F
≤
4.20
⇒ σ
x
=
σ
x’
F
=1.231 falls within this interval.
Conclude
σ
values for two systems are the same!
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D. Other distributions of interest
¾
Radioactive decay
⇒
Poisson
distribution
¾
Only 2 possible outcomes
⇒
Binomial
distribution
3. Standard deviations of computed values
¾
If quantity
z
of interest is
a function
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This note was uploaded on 11/11/2011 for the course BIO 2.797j taught by Professor Matthewlang during the Fall '06 term at MIT.
 Fall '06
 MatthewLang

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