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MECH 373
Instrumentation and Measurements
Lecture 13/A
Statistical Analysis of Experimental Data
(Chapter 6)
• Introduction
• General Concepts and Definitions
• Probability
• Probability Distribution Function
• Parameter Estimation
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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PDF with Engineering Applications
A number of distribution functions are used in
engineering applications. We will discuss briefly about
some common distribution functions.
►Binomial (values either true/false)
►Poisson
►Normal (Gaussian)
►Student’s t
► χ2 (Chi-squared)
►Weibull
►Exponential
►Uniform
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Probability Distribution Function
 Binomial Distribution
• The binomial distribution is a distribution which describes discrete random
variables that can have only two possible outcomes: “success” or “failure”.
• This distribution has applications in production quality control, when the
quality of a product is either acceptable or unacceptable.
• The binomial distribution provides the probability (P) of finding exactly r
successes in a total of n trials and is expressed as
n r
P(r )    p (1  p) n r
r
where, p is the probability of success which remains constant throughout the
experiment, and  n 
n!
  
 r  r!(n  r )!
is called as n combination r, which is the number of ways that we can choose r
identical items from n items.
• The expected number of successes in n trials for binomial distribution is   np
• The standard deviation of the binomial distribution is   np(1  p)
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Probability Distribution Function
 Poisson Distribution
• The Poisson distribution is used to estimate the number of random occurrences
of an event in a specified interval of time or space if the average number of
occurrences is already known.
• The probability (P) of occurrence of x events is given by
e   x
P( x) 
x!
where, λ is the expected or mean number of occurrences during the interval of interest.
• The expected value of x for the Poisson distribution, the same as the mean (μ),
is given by
E (x)    
• The standard deviation is given by
 
• Probability that the number of occurrences is less than or equal to k is given by
e  i
P( x  k )  
i!
i 0
k
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Probability Distribution Function
 Normal (Gaussian) Distribution
• The Normal or Gaussian distribution is a simple distribution function that is useful for
a large number of common problems involving continuous random variables.
1
 ( x  ) 2 / 2  2
f ( x) 
e
 2
• Symmetric about μ
• Bell-shaped
• Mean μ : the peak of
the density occurs
• Standard deviation σ:
indicates the spread of
the bell curve.
-1
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0
1
2
2
Lecture Notes on MECH 373 – Instrumentation and Measurements
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4
5
5
Standard Normal Distribution
(mean=0, standard deviation=1)
For a given population, the probability of having a single value of x between a
lower limit of x1 and upper limit of x2 is
x2
z2
x1
z1
P( x1  x  x2 )   f ( x)dx  
1
 ( x  ) 2 / 2  2
e
dx
 2
Since f(x) is in the form of an error function, the integration must be
performed numerically. To simplify the integration process, the integrand is
modified with a change of variable. A non-dimensional variable z is defined
as:
x
z

A function f(z) which is called the standard normal density function is defined as
1 z2 / 2
f ( z) 
e
2
This function represents the normal probability density function for a random variable
z, with mean μ = 0, and standard deviation σ = 1. The function f(z) is plotted below
versus z.
Lecture Notes on MECH 373 – Instrumentation and Measurements
Lecture 14
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Standard Normal Distribution
(mean=0, standard deviation=1)
x2
z2
x1
z1
P( x1  x  x2 )   f ( x)dx   f ( z )dz
The probability that x is between x1 and x2 is the same as the
probability that the transformed variable z is between z1 and z2.
That is
P( x1  x  x2 )  P( z1  z  z 2 )  P(
Lecture 14
x1  


Lecture Notes on MECH 373 – Instrumentation and Measurements
x


x2  

)
7
Confidence Intervals in Sample Population
It is also of interest to determine the probability that a measurement will fall within one
or more standard deviations of the mean. The values are listed in the following table:
This implies that if a measurement is performed in an environment where the deviation from
the mean value is totally influenced by random variables, we are
• 68.26% confident that the value of the random variable will fall within one standard
deviation (1σ) from the mean.
• 95.44% confident that the value of the random variable will fall within two standard
deviation (2σ) from the mean.
• 99.74% confident that the value of the random variable will fall within three standard
deviations (3σ) from the mean.
• 99.96% confident that the value of the random variable will fall within 3.5 standard
deviations (3.5σ) from the mean.
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Standard Normal Distribution
(mean=0, standard deviation=1)
z
x
1 z2 / 2
f ( z) 
e
2

x2
z2
x1
z1
P( x1  x  x2 )   f ( x)dx   f ( z )dz
P( x1  x  x2 )  P( z1  z  z 2 )  P(
1
Z[1,1]
Lecture 14


x


x2  

)
3
Z [-3,3]
2
Z [-2,2]
68%
x1  
95%
Lecture Notes on MECH 373 – Instrumentation and Measurements
99.7%
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Normal Distribution Example 1
• The distribution of heights of American women aged 18 to 24 is
approximately normally distributed with mean 65.5 inches and
standard deviation 2.5 inches.
• 68% of these American women have heights between 65.5 –
1(2.5) and 65.5 + 1(2.5) inches, or between 63 and 68 inches.
• 95% of these American women have heights between 65.5 2(2.5) and 65.5 + 2(2.5) inches, or between 60.5 and 70.5 inches.
• 99.7% of these American women have heights between 65.5 3(2.5) and 65.5 + 3(2.5) inches, or between 58 and 73 inches.
68%
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95%
Lecture Notes on MECH 373 – Instrumentation and Measurements
99.7%
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Normal Distribution Example 2
•
•
•
•
•
•
•
•
•
•
•
Normal distribution mean value u = 10.0, Standard deviation σ = 1.0
A, Probability a single value between 9 and 12?
Z1 = (x – u)/σ = (9 - 10) / 1 = -1
P(-1 ≤ Z ≤ 0) = 0.3413 (Area from 0 to -1)
Z2 = (12 – 10) / 1 = 2
P(0 ≤ Z ≤ 2) = 0.4772 ( area from 0 to 2)
P(-1 ≤ Z ≤ 2) = 0.3413 + 0.4772 = 0.8185 (82%).
B, Probability less than 9?
Z1 = -1
P(-1 ≤ Z ≤ 0) = 0.3413 (Area from 0 to -1)
P (-∞ ≤ Z ≤ -1) = 0.5 – 0.3413 = 0.1587 (from -1 to negative infinity)
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Normal Distribution Example 3
•
•
•
•
•
•
•
•
•
•
Diameter tolerance of cylinders – normal distribution.
Average diameter u = 4in, Standard deviation = 0.002.
A, Probability of D ≤ 4.002.
Z = (D – u) /σ = (4.002 – 4) / 0.002 = 1
P(0 ≤ Z ≤ 1) = 0.3413 (area from 0 to 1)
P (-∞ ≤ Z ≤ 0) = 0.5 (area from 0 to negative infinity)
P(-∞ ≤ Z ≤ 1) = 0.5 + 0.3413 = 0.8413 (84%)
B, D > 2 σ rejected?
P(0 ≤ Z ≤ 2) = 0.4772, also P(-2 ≤ Z ≤ 0) = 0.4772
P(-2 ≥ Z ≥ 2) = 1 – 2*0.4772 = 0.0456 (about 5%).
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Parameter Estimation
In most experiments the sample size is small relative to the
population, yet the principle intention from the statistical
aspect is to estimate the parameters describing the entire
population. The most common estimated parameter is the
population mean and in some cases it is also necessary to
estimate the population standard deviation.
 Estimate population mean using sample mean
x
 Estimate population standard deviation using sample
standard deviation
S 
 However, different samples from the sample population
will generate different values. Therefore, it is necessary
to determine the uncertainty intervals of estimated
parameters. Lecture Notes on MECH 373 – Instrumentation and Measurements
Lecture 14
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Interval Estimation of Population Mean
• Estimate of population mean
  x 
• Confidence interval
x     x 
• Confidence level (degree of confidence)
P( x      x   )
• Level of significance
Confidence level  1 a
where α is the probability that the mean will fall
outside the confidence interval.
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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From Sample to Population
 Choose many samples from population
 Find the mean of each sample
 Determine the uncertainly range of the means
(Sample mean is a variable !!!)
How to estimate the statistics of a population from
only one single sample?
The central limit theorem makes it possible to make an
estimate of the confidence interval with a suitable
confidence level.
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Central Limit Theorem (a)
 Consider a population of the random variable x with a mean of μ
and standard deviation of σ
 Several different samples, each of size n, mean of x
i
 If sample size n is sufficiently large (>30), then

x 
n
(standard error of the mean)
 For the central limit theorem to apply, the sample size n must be
large (typically greater than 30)
 If original population is normal, the distribution for the xi is normal.
We can use the following equation
z
x
x

x
/ n
 If original population is not normal and n is large (n > 30) the
distribution for the xi is normal
 If original population is not normal and n < 30 the distribution for the
is only approximately normal
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
xi
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Interval Estimation of Population
P[ za / 2  z  za / 2 ]  1  a
P[ x  za / 2

n
  x  za / 2
   x  za / 2

z

n
x
x
]  1a
Area  a / 2
Area  a / 2
n
with confidence level 1-a
Confidence Interval
 za / 2
Lecture 14
x

/ n
Lecture Notes on MECH 373 – Instrumentation and Measurements
z 0
za / 2
17
Interval Estimation of Population
• If z is zero, this means that x  
• However, we only expect the true value of μ to lie
somewhere within the confidence interval, which is  za / 2
• If the sample size is less than 30, the assumption that
the population standard deviation can be represented by
the sample standard deviation may not be accurate.
• In case of small samples, a statistic called Student’s t
distribution is used. That is
x 
t
S/ n
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Normal Distribution Table
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example 6.11
We would like to determine the confidence interval of the mean of a batch of
resistor made in a certain process. Based on 36 measurements, the average
resistance is 25  and the sample stanadard deviation is 0.5 . Determine the
90% confidence interval of the mean resistance of the batch.





n = 36
Average = 25 
Sample standard deviation S = 0.5 
Determine 90% confidence interval of the mean
Solution:
 1-α=90%, -> α = 0.1
 0.5- α/2=0.5-0.05=0.45
 Table 6.3 -> Z α/2=1.645
S
S
0.5
0.5
   x  za / 2
 25  1.645 
   25  1.645 
n
n
36
36
 (24.86    25.14)
x  za / 2
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Student’s t-distribution (n<30)
In case of small samples, a statistic called Student’s t distribution
is used. That is
 1
x
t
S/ n
(
f (t , ) 

2
 ( )(1 
2
)
t2

) ( 1) / 2
• Student’s t, degree of freedom ν = n-1
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Interval Estimation of Population
Mean (n<30)
P[ta / 2  t  ta / 2 ]  1  a
P[ x  ta / 2
x
t
S/ n
S
S
   x  ta / 2
]  1a
n
n
  x  ta / 2
S
n
with confidence level 1-a
Area  a / 2
Confidence Interval
 ta / 2
Lecture 14
Area  a / 2
Lecture Notes on MECH 373 – Instrumentation and Measurements
t 0
ta / 2
22
Student’s t-Distribution Table
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example 6.12
•
•
•
•
n=6
1250, 1320, 1542, 1464, 1275, and 1383 h
Estimate population mean and 95% confidence interval on the mean
Solution:
– Mean=(1250+1320+1542+1464+1275+1383)/6=1372 h
– S=[(1250-1372)2+(1320-1372)2+ (1542-1372)2+ (1464-1372)2+
(1275-1372)2+(1383-1372)2]/(6-1)]1/2 = 114 h
– ν=n-1=6-1=5, α=1-95%=0.05, α/2=0.05/2=0.025
– Table 6.6 -> t α/2=2.571
S
S
   x  ta / 2
n
n
114
114
1372  2.571
   1372  2.571
6
6
 (1252    1492)h  1372  120h
x  ta / 2
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example 6.13
•
•
•
•
•
•
•
•
•
•
Lecture 14
Reduce the 95% confidence interval to ±50 h from ±120 h
Determine how many more systems should be tested
Solution: Assume n>30,



   x  za / 2
n
n
x  50    x  50
x  za / 2
za / 2
n
n  ( za / 2
α=1-95%=0.05, 0.5-α/2=0.5-0.05/2=0.475
Table 6.3 -> z α/2=1.96
n = (1.96x114/50)2=20
<30 => t-distribution
ν=n-1=20-1=19, α/2=0.05/2=0.025
S
x  ta / 2
Table 6.6 -> t α/2=2.093
n
n = (2.093x114/50)2=23
 50

50
)2
   x  ta / 2
S
n
x  50    x  50
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Chi-squared Distribution Function
n
 
2
2
(
x


)
 i
i 1
ν=4
2
f ( 2 )
n
S2 
2
2
(
x


)
 i
Area=α/2
ν = 10
i 1
Area=α/2
n 1
S2
 ( n  1) 2

2,1a / 2
•Relates the sample variance to
the population variance
2
( )
e
f ( ) 
2 / 2 ( / 2)
2
(  2 ) / 2
2
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
 2 / 2
2,a / 2
for 2  0
26
Interval Estimation of Population
Variance
P[ 
2
 ,1a / 2
 
2
2
 ,a / 2
P[ 2,1a / 2  (n  1)
S2

2
]  1a
  (n  1)
2
S2
2
 2,a / 2 ]  1  a
ν=4
(n  1) S

2
 ,a / 2
2
 
2
(n  1) S
2
Area=α/2
ν = 10

2
 ,1a / 2
With confidence level 1-a
Lecture 14
f ( 2 )
Area=α/2
2,1a / 2
Lecture Notes on MECH 373 – Instrumentation and Measurements

2
2,a / 2
27
Chi-squared Distribution Table
Lecture 14
Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example 6.14
• n = 20, mean = 0.32500 in, S = 0.00010 in
• obtain 95% confidence interval for the standard deviation
• Solution:
– ν=n-1=20-1=19, α=1-95%=0.05, α/2=0.05/2=0.025, 1α/2=1-0.05/2=0.975
– Table 6.7 -> χ2ν, α/2=32.825, χ2ν, 1-α/2=8.9065
(n  1) S

2
 ,a / 2
Lecture 14
2
 
2
(n  1) S
2,1a / 2
2
(20  1)0.00012
(20  1)0.00012
2
 
32.825
8.9065
0.000076   2  0.00015
Lecture Notes on MECH 373 – Instrumentation and Measurements
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