Transcript ch8_L3_i

Estimates Made Using Sx
• Two different statistical estimates can be made using Sx.
[1] the value of the next sample value, xi, where
xi  x  t , P S x
[2] the value of the true mean, x‘, where
x  x  t , P
Sx
N
Example 8.5
[1] p range that contains the next measurement
with P = 95 %
t , P  t18,95  2.101
pi  p  t , P S p  4.97  0.10
p  4.97
Sp  0.046
[2] same but for N = 5
t ,P  t 4,95  2.770  p i  4.97  0.13
[3] for N = 19 and P = 50 %
t ,P  t18,50  0.668  p i  4.97  0.03
[4] p range that contains the true mean
p  p  t ,P
Sp
N
 4.97 
(2.101)( 0.046 )
19
 4.97  0.02
c2 and the c2 distribution
• In the same way we used the student t distribution to
estimate the range containing the population mean…
• We can use the statistical variable c2 is and the c2
distribution to estimate the range containing the
population standard deviation
• The statistical variable c2 is defined as
2 we could calculate c2 directly from a sample
N
N
( xi  x)
2
2
if we assumed the population mean and
c   zi  
2

standard deviation (e.g., assumed they equaled
i 1
i 1
the sample mean and standard deviation)
then
since
Sx 


2
1 N
xi  x

N  1 i 1
2

S
c 2  ( N  1) 2  2x

 2
2
2
2
• Thus, as N→∞, S x    c  
S x2
c
 
S x2
2
The
c2 distribution: how c2 behaves for normally distributed data
c2 = f()  infinite number of c2 distributions (like for Student’s t).
pdf
PDF
Figure 8.10
Figure 8.12
Determine Pr[c2≤10] for N = 11:
56 %
Determine Pr[c2≤10] for N = 5:
96 %
Determine c2 for P = 50 % and N = 5:
consider a sample:
N = 5,
c2 = 11 (calculated)
3.36
meaning?
subscript α often used:
2
c
a
a
“level of
significance”
P+a=1
P
α is directly related
to the % confidence
P used, e.g., with
the student t
distribution
a
Figure 8.11
c2 Table
For N = 13, find a
when c2 = 21.0
a=5%
For P = 5 %, find
c2 if N = 20
c2 = 10.1
Table 8.8
see also Excel or LabVIEW help files
Uses of the c2 distribution
•
To infer  from Sx
•
To establish a rejection criterion (e.g., when to
stop making something and fix the equipment; see
ex 8.9)
•
To compare a sample to an assumed population
(see ex 8.10)
• Back to the top bullet, the true variance, 2, estimated
with P % confidence, is in the range
2
S2x

S
2
x



c (2a / 2)
c12(a / 2)
noting a = 1 – P and  = N -1.
In-Class Example (x’ and  Inference)
• Given the mean and standard deviation are 10 and
1.5, respectively, for a sample of 16, estimate with
95 % confidence the ranges within which are the
true mean and true standard deviation (assuming a
normally distributed population).
range for true mean = sample mean ± t,PSx/√N
= 10 ± (2.131)(1.5)/√16 >> between 9.2 and 10.8
range for true variance: Sx2/ca/22 ≤ 2 ≤ Sx2/c1a/22
>> (15)(1.5)2/27.5 ≤ 2 ≤ (15 )(1.5)2/6.26 for a = 0.05
>> 1.23 ≤ 2 ≤ 5.39
>> 1.11 ≤  ≤ 2.32
Using the c2 Table
 = 15 and P = 0.95
>> a (= 1 – P) = 0.05
a/2 = 0.025
gives c2 = 27.5
1-(a/2) = 0.975
gives c2 = 6.26
In-Class Example (cont’d)
What happens to the range which contains the true standard
deviation when P is reduced from 95 % to 90 %?
To be less confident, we would expect the extent of the
range to decrease.
Let’s see what happens.
P = 90 % → a = 10 % → a/2 = 0.05 and 1-(a/2) = 0.95
For  = 15: c20.05 (lower bound) = 25.0 (vs 27.5)
and c20.95 (upper bound) = 7.26 (vs 6.26).
So, the extent of the range does decrease.
For the P = 90 % case:
1.16 (vs 1.11) ≤  ≤ 2.16 (vs 2.32)