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Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Estimation of Means
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
1
Estimation
• Estimation of the Mean - Normal Distribution
• Estimation of the Difference Between two Means Normal Distribution
• Estimation of the Mean - Infinite Population - Type
Unknown
• Estimation of the Mean - Finite Population
• Estimation of Lognormal Distribution Parameters
• Estimation of Weibull Distribution Parameters
2
Estimation of Means
3
Estimation of the Mean - Normal Distribution
• X1, X2, …, Xn is a random sample of size n from
N(, ), where both  &  are unknown.
• Point Estimate of 
n
1
μ^   X i  X
n i 1
• (1 - )  100% Confidence Interval for the mean μ L , μ U 
μ L  X  Δμ
and
μ U  X  Δμ
where
Δμ  t α
2
, n 1
s
n
4
Example
X = diameter of a rod where X ~ N(, )
A random sample of 26 randomly selected rods
resulted in x = 1.030 inches and s = 0.020 inches.
Estimate  in terms of a point estimate and 98%
and 95% confidence intervals.
5
Example - Solution
μ^  x  1.030
For  = 0.02,
t
2
, n 1
 t0.01, 25  2.485
so that
 L  x  t
2
, n 1

s
0.020
 1.030  2.485
n
26
 1.030  0.0097
 1.0202
6
Example - Solution
and
U  x  t
2
, n 1

s
0.020
 1.030  2.485
n
26
 1.030  0.0097
 1.0397
Therefore a 98% confidence interval for  is
(1.020, 1.040)
For  = 0.05,
t
2
, n 1
 t0.025, 25  2.060
7
Example - Solution
μL  x  t α
2
, n 1

s
0.020
 1.030  2.060
n
26
 1.030  0.0081
 1.0219

s
0.020
μU  x  t α
 1.030  2.060
, n 1
n
26
2
 1.030  0.0081
 1.0381
so that a 95% confidence interval for  is (1.022,
1.038).
8
Estimation of Difference Between Two Means
Normal Distribution
• Let X11, X12, …, X1n, and X21, X22, …,
X 2 n2be
1
random samples from N(1, 1) and N(2, 2),
respectively, where 1, 1, 2 and 2 are all unknown
• Point estimation of  = 1 - 2
^
^
^
Δ μ  μ1  μ 2
 X1  X 2
9
Estimation of Difference Between Two Means
Normal Distribution
• An approximate (1 - )  100% Confidence
Interval for  = 1 - 2 is
ΔμL , ΔμU  ,
Where

 s12 s 22
ΔμL  Δμ̂   t α 

 2 , ν  n1 n 2
and

 s12 s 22
ΔμU  Δμ̂   t α 

,ν
 2  n1 n 2
10
Estimation of Difference Between Two Means
Normal Distribution
where ν = degrees of freedom
2
s
s 
  
n1 n2 


2
2
2
2
 s1   s2 
   
 n1    n2 
n1  1 n2  1
2
1
2
2
11
Example - Estimation of 
1
-

2
Records for the past 15 years have shown the
average rainfall in a certain region of the country
for the month of May to be 4.93 cm, with a standard
deviation of 1.14 cm. A second region of the country
has had an average rainfall in may of 2.64 cm, with
a standard deviation of 0.66 cm during the past 10
years. Find a 95% confidence interval for the
difference of the true average rainfalls in these two
regions, assuming that the observations came from
normal populations.
12
Example - Solution
For the first region we have X1= 4.93, s1 = 1.14,
and n1 = 15. For the second region X 2= 2.64,
s2 = 0.66, and n2 = 10. We wish to find a 95%
confidence interval for 1 - 2. Our point estimate
of
Δμ  μ1  μ 2
is
Δμ̂  X1  X 2
= 4.93 - 2.64
= 2.29.
Since we have no information concerning the
population variances and our sample sizes are not
the same, we can only find an approximate 95%
confidence interval based on the t distribution with
ν degrees of freedom, where
13
Example - Solution


s
2
1

s
2
1
n1  s n2
2
2


2

n1 / n1  1  s n2 / n2  1
2
1.14
2
1.14 15  0.66 10
15 / 15  1  0.66 10 / 10  1
2
2
2
2
2
2
2
2
2
 22.7  23
Using  = 0.05, we find that t0.025 = 2.069 for  = 23
degrees of freedom.
14
Example - Solution
Therefore, substituting in the formula
ΔμL  x1  x 2   t α
2
s12 s 22

n1 n 2
1.14 2 0.66 2
 2.29  2.069

15
10
 1.5434
15
Example - Solution
ΔμU  x1  x 2   t α
2
s12 s 22

n1 n 2
1.14 2 0.66 2
 2.29  2.069

15
10
 3.037
Hence, we are approximately 95% confident that
the interval from 1.54 to 3.04 contains the true
difference of the average rainfall for the two regions.
16
Estimation of Means
Infinite Population - Type Unknown
17
Estimation of the Mean
Infinite Population - Type Unknown
• X1, X2, …, Xn is a random sample of size n
• Point Estimate of 
1 n
μ̂   X i  X
n i 1
18
Estimation of the Mean
Infinite Population - Type Unknown
• An approximate
(1 - )  100% Confidence Interval for the mean μL , μU 
based on the Central Limit Theorem is:
where
μL  X  Δμ
μU  X  Δμ
and
μ  t α
2
, n 1
s
n
19
Estimation of Means Finite Population
20
Estimation of Means - Finite Populations
• X1, X2, ... , Xn is a random sample of size n from
a population of size N with parameters  and 
• Point Estimate of : μ̂  X
• An approximate (1 - )  100% Confidence
Interval for  is: μL , μU 
where
μL  x  Δμ
μU  x  Δμ
and Δμ  t α
2
, n 1
s
n
Nn
N 1
21
Estimation of Means - Finite Populations
where
n

1
S 
Xi  X

n  1 i 1
2

2

 
t


• ,n 1 is the value of T ~ tdf for which P T  t   

2
, n 1 
2

 2
•
N  n is the finite population correction factor
N 1
22
Estimation - Lognormal Distribution
23
Estimation of Lognormal Distribution
• Random sample of size n, X1, X2, ... , Xn from
LN (, )
• Let Yi = ln Xi for i = 1, 2, ..., n
• Treat Y1, Y2, ... , Yn as a random sample from
N(, )
• Estimate  and  using the Normal Distribution
Methods
24
Graphical Estimation of Lognormal Distribution Parameters
• Take logarithms of sample values
• Prepare Normal Probability Plot
• μ̂= 50th percentile
μ̂  x 0.1
• σ̂ 
1.28
25
Estimation of the Mean of a Lognormal Distribution
• Mean or Expected value
of
2
mean  E(x)  e
μ
σ
2
• Point Estimate of mean
^
^
mean  E(x)  e
σ̂ 2
μ̂ 
2
where μ̂ and σ̂ are point estimates of μ and σ respectively.
26
Estimation - Weibull Distribution
27
Estimation of Weibull Distribution
• Random sample of size n, T1, T2, …, Tn, from
W(, ), where both  &  are unknown.
• Point estimates
^
• β is the solution of g() = 0
n
where
gβ  
β
T
 i lnT i
i 1
n
β
T
 i
1 1 n
   lnT i
β n i 1
i 1
1
 1 n ^β ^β
• θ    Ti 
 n i 1 
^
28
Graphical Estimation of Weibull Distribution Parameters
• Prepare Weibull probability Plot
• θ̂  63.2th percentile
ln - ln 1  F(x i ) 
• β̂ 
 x
ln  
 θ̂ 
for a selected value of x.
29
Estimation of the Mean of a Weibull Distribution
• Mean or Expected value of
1 
μ  E(x)  θΓ  1
β 
• Point Estimate of mean
1 
 θ̂Γ  1
 β̂ 
where β̂ and θ̂ are point estimates of β and θ respectively.
^
μ
30
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