ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Binomial Distribution:
P( x ) =
n!
p x (1 − p)n − x
x ! (n − x )!
x∈ℵ
P(x) is the probability of obtaining “x successes” in “n
independent trials” of a “discrete event” for which “p”
and “1-p” are respective probabilities of “success” and
“no success” in a single trial.
Mean = n.x
S.D. = np(1 − p)
When “p” in the binomial distribution is too small and
“n” is too large, the calculation of the binomial
distribution becomes quite complex and approaches
“Poisson” distribution.
Poisson Distribution:
e− µ µ x
x∈ℵ
P( x ) =
x!
P(x) is the probability of obtaining “x successive
successes” of a “discrete event” during an interval of
time T where “µ” is the “average number of successes”
in the same interval T.
Mean = µ
S.D. = µ
When n becomes large in binomial distribution, or µ
becomes large in Poison distribution, the envelope f(x)
of the resulting continuous distribution of a
continuous variable x is called the “Gaussian/Normal
distribution” or “Gaussian Probability Density
Function”.
Page 44
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Gaussian/Normal Distribution or
Gaussian Probability Density Function:
f (x ) =
1
e
σ 2π
1⎛ x−µ⎞
− ⎜
⎟
2⎝ σ ⎠
2
x∈ℜ
where
Mean = µ
S.D. = σ
or in standardized form
f (z ) =
1
e
2π
−
z2
2
with
z=
x−µ
σ
where
Mean = 0
S.D. = 1
Its “bell-shaped” curve for the standardized normal
distribution looks like as follows:
0.4
f m ax = 1 2 π
=0.3989
0.3
f(z)
0.2
0.1
0.0
-∞ Å -3
-2
-1
0
1
2
3 Æ +∞
z
Page 45
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Properties of Gaussian Distribution:
1. f(x)>0 for all finite values of x and f(x) Æ 0 as x Æ ∞
2. f(x) is symmetrical about its mean.
3. Mean ( x ) determines its location and S.D. (σ)
determines its amount of dispersion.
f(x)
f1(x)
x1 < x 2
σ1 < σ 2
f2(x)
x
0
4.
+∞
+∞
−∞
−∞
∫ f (x ) dx = ∫ f (z) dz = 1
z2
5.
P(z1 ≤ z ≤ z2) =
∫ f (z) dz
or geometrically
z1
f(z)
Area
is
P(z1≤z≤z2)
0 z1
z2
z
where P(z1 ≤ z ≤ z2) is the “probability” of having z
between z1 and z2.
Page 46
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Integrals of the Normalized
Gaussian Error Function
1
2π
z
∫e
−
z2
2
f (z )
dz
0
z
0.00
0.0 0.0000
0.1 0.0398
0.2 0.0793
0.3 0.1179
0.4 0.1554
0.5 0.1915
0.6 0.2257
0.7 0.2580
0.8 0.2881
0.9 0.3159
1.0 0.3413
1.1 0.3643
1.2 0.3849
1.3 0.4032
1.4 0.4192
1.5 0.4332
1.6 0.4452
1.7 0.4554
1.8 0.4641
1.9 0.4713
2.0 0.4772
2.1 0.4821
2.2 0.4861
2.3 0.4893
2.4 0.4918
2.5 0.4938
2.6 0.4953
2.7 0.4965
2.8 0.4974
2.9 0.4981
3.0 0.4987
3.1 0.4990
3.2 0.4993
3.3 0.4995
3.4 0.4997
3.5 0.4998
4.0 0.4999683
4.5 0.4999966
5.0 0.4999997
0
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
0.4991
0.4993
0.4995
0.4997
0.4998
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
0.4991
0.4994
0.4995
0.4997
0.4998
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
0.4991
0.4994
0.4996
0.4997
0.4998
0.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
0.4992
0.4994
0.4996
0.4997
0.4998
0.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
0.4992
0.4994
0.4996
0.4997
0.4998
0.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
0.4992
0.4994
0.4996
0.4997
0.4998
z
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
0.4992
0.4995
0.4996
0.4997
0.4998
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.4993
0.4995
0.4996
0.4997
0.4998
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
0.4993
0.4995
0.4997
0.4998
0.4998
Page 47
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Normal Distribution Based Probability
Estimates of Error (Confidence Intervals):
If a data population abides by Gaussian theory, the
probability that a single data item will fall into an
interval in terms of the standard deviation of this
population can be obtained by using the tabulated
integral given in the previous page.
Example 1:
P(µ+σ ≤ x ≤ µ+1.2σ) = P(0 ≤ x ≤ µ+1.2σ) - P(0 ≤ x ≤ µ+σ)
= P(0≤(x- µ)/σ ≤ 1.2) - P(0 ≤ (x- µ)/σ ≤ 1)
= 0.3849 - 03413 = 0.0436 = 4.36 %
Implying that the probability of a measurement to be
in an interval [ µ+σ , µ+1.2σ ] is 4.36 %
Example 2:
P(µ-2σ ≤ x ≤ µ+2σ) = 2*P(0 ≤ x ≤ µ+2σ)
= 2*P(0 ≤ (x- µ)/ σ ≤ 2)
= 2*0.4772 = 0.9544 = 95.44 %
Implying that x value will be as close to µ as ±2σ with a
probability (certainty) greater than 95 %; i.e. with an
uncertainty less 5 %.
This information may be shown as:
x = µ ± 2σ (95.44 %)
or if unbiased estimates are used:
x = x ± 2s (95.44 %)
Example 3:
P(x < µ-3σ) = P(x > µ+3σ)
= P(0 ≤ x ≤ + ∞) - P(0 ≤ x ≤ µ+3σ)
= 0.5 - P(0 ≤ (x- µ)/σ ≤ 3)
= 0.5 - 0.4987 = 0.0013 = 1.3 ‰
Implying that x values 3σ less than µ are only likely
with a probability of 1.3 ‰.
Page 48
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Example 4:
Let the result of measurements to determine the
spring constants of a sample drawn from a very large
number of valve springs manufactured be obtained as:
n = 40, x = 152.5 N/cm, s = 0.889 N/cm
a) Determine the range of spring constants with a
“confidence level” of ±95 %.
±0.95 confidence level Æ P( x -zs ≤ x ≤ x +zs) = 0.95
Æ P(0 ≤ (x- x )/s ≤ z) = 0.475
Æ z = 1.96 (from table)
x = 152.5 ± 1.96*0.889 = 152.5 ± 1.74 N/cm (95 %)
b) Determine the confidence interval of the mean
value with a “confidence level” of ±95 %.
Standard deviation of the mean:
s x = s / n = 0.889 / 40 = 0.141 N/cm
µ = 152.5 ± 1.96*0.141 = 152.5 ± 0.28 N/cm (95 %)
c) Determine the % probability of having springs with
spring constants greater than 154 N/cm.
z = (x- x )/s = (154-152.5)/0.889 = 1.69
P(x > 154) = P(z > 1.69)
= P(0 ≤ z ≤ + ∞) - P(0 ≤ z ≤ 1.69)
= 0.5 - 0.4545 = 0.0455 = 4.6 %
Some commonly used confidence levels are:
Common Name
Confidence
Level
Probable Error
±0.6754 σ
Standard Deviation
±σ
90 % Error
±1.65 σ
95 % Error
±1.96 σ
Two Sigma Error
±2 σ
99 % Error
±2.58 σ
Three Sigma Error
±3 σ
Maximum Error
±3.29 σ
Certainty P(|x- µ|>zσ)
P(| x - µ|>zσ x )
%
50
68.26
90
95
95.44
99
99.74
99.9+
1 in 2
1 in 3
1 in 10
1 in 20
1 in 22
1 in 100
1 in 369
1 in 1000
Page 49
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Student’s t-Distribution:
When the sample size is small (n < 20-30), then s
becomes a biased estimate of σ, and the standardized
variable
x−µ
t =
s/ n
is used to determine confidence levels of the
estimation of the mean, based on so-called “Student’s
t-Distribution” rather than the “Normal Distribution”.
Therefore, the estimation results of the mean must be
presented as
σ
µ = x ± t
n
The “t value” corresponds to the limits of the integral
t
∫ f (t, ν)dt = Desired Certainty (Confidence)
−t
where f(t,ν) is the Student’s t-distribution defined as
f ( t, ν ) =
ν +1
−
t2 ⎞ 2
Γ[( ν + 1) / 2] ⎛
⎜1 +
⎟
ν⎠
Γ ( ν / 2) νπ ⎝
which is independent of µ and σ, but its shape is
determined only by the degrees of freedom ν=n-1,
therefore by the number of data points, n.
It has almost the same characteristics as the normal
distribution, except that it has more probability
concentrated in tails and less in the central part.
ν
Mean = 0
& S.D. =
ν>2
ν−2
It converges to the normal distribution very fast as n
gets large.
Page 50
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Table of t-values
Corresponding to Various Confidence Levels
as a Function of Number of Data Points
n
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
40
50
100
250
500
∞
0.5
1.000
0.816
0.765
0.741
0.727
0.718
0.711
0.706
0.703
0.700
0.697
0.695
0.694
0.692
0.691
0.690
0.689
0.688
0.688
0.687
0.686
0.686
0.685
0.685
0.684
0.684
0.684
0.683
0.683
0.681
0.680
0.677
0.675
0.675
0.674
0.6
1.376
1.061
0.978
0.941
0.920
0.906
0.896
0.889
0.883
0.879
0.876
0.873
0.870
0.868
0.866
0.865
0.863
0.862
0.861
0.860
0.859
0.858
0.858
0.857
0.856
0.856
0.855
0.855
0.854
0.851
0.849
0.845
0.843
0.842
0.842
0.7
1.963
1.386
1.250
1.190
1.156
1.134
1.119
1.108
1.100
1.093
1.088
1.083
1.079
1.076
1.074
1.071
1.069
1.067
1.066
1.064
1.063
1.061
1.060
1.059
1.058
1.058
1.057
1.056
1.055
1.050
1.048
1.042
1.039
1.038
1.036
CONFIDENCE LEVEL
0.8
0.9
0.95
0.98
3.078 6.314 12.706 31.821
1.886 2.920 4.303 6.965
1.638 2.353 3.182 4.541
1.533 2.132 2.776 3.747
1.476 2.015 2.571 3.365
1.440 1.943 2.447 3.143
1.415 1.895 2.365 2.998
1.397 1.860 2.306 2.896
1.383 1.833 2.262 2.821
1.372 1.812 2.228 2.764
1.363 1.796 2.201 2.718
1.356 1.782 2.179 2.681
1.350 1.771 2.160 2.650
1.345 1.761 2.145 2.624
1.341 1.753 2.131 2.602
1.337 1.746 2.120 2.583
1.333 1.740 2.110 2.567
1.330 1.734 2.101 2.552
1.328 1.729 2.093 2.539
1.325 1.725 2.086 2.528
1.323 1.721 2.080 2.518
1.321 1.717 2.074 2.508
1.319 1.714 2.069 2.500
1.318 1.711 2.064 2.492
1.316 1.708 2.060 2.485
1.315 1.706 2.056 2.479
1.314 1.703 2.052 2.473
1.313 1.701 2.048 2.467
1.311 1.699 2.045 2.462
1.304 1.685 2.023 2.426
1.299 1.677 2.010 2.405
1.290 1.660 1.984 2.365
1.285 1.651 1.970 2.341
1.283 1.648 1.965 2.334
1.282 1.645 1.960 2.326
0.99
0.999
63.656 636.578
9.925 31.600
5.841 12.924
4.604
8.610
4.032
6.869
3.707
5.959
3.499
5.408
3.355
5.041
3.250
4.781
3.169
4.587
3.106
4.437
3.055
4.318
3.012
4.221
2.977
4.140
2.947
4.073
2.921
4.015
2.898
3.965
2.878
3.922
2.861
3.883
2.845
3.850
2.831
3.819
2.819
3.792
2.807
3.768
2.797
3.745
2.787
3.725
2.779
3.707
2.771
3.689
2.763
3.674
2.756
3.660
2.708
3.558
2.680
3.500
2.626
3.391
2.596
3.330
2.586
3.310
2.576
3.290
Page 51
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
0.4
Gaussian
Student t
0.2
0.0
-∞ Å -3
-2
-1
0
z, t
1
2
3 Æ +∞
Example 4b (revised):
Let the result of measurements to determine the
spring constants of a sample drawn from a very large
number of valve springs manufactured be obtained as:
n = 10, x = 152.5 N/cm, s = 0.889 N/cm
Determine the confidence interval of the mean value
with a “confidence level” of ±95 %.
Standard deviation of the mean:
s x = s / n = 0.889 / 10 = 0.281 N/cm
i) If normal distribution is used, z = 1.96
µ = 152.5 ± 1.96*0.281 = 152.5 ± 0.55 N/cm
i) If t-distribution is used, t = 2.262 (from table)
µ = 152.5 ± 2.262*0.281 = 152.5 ± 0.64 N/cm
If the data were:
n = 5, x = 152.5 N/cm, s = 0.889 N/cm
Standard deviation of the mean:
s x = s / n = 0.889 / 5 = 0.398 N/cm
i) If normal distribution is used, z = 1.96
µ = 152.5 ± 1.96*0.398 = 152.5 ± 0.78 N/cm
i) If t-distribution is used, t = 2.776 (from table)
µ = 152.5 ± 2.776*0.398 = 152.5 ± 1.10 N/cm
Page 52
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Rejection of bad data (Chauvenet's criterion)
The question is here that “Whether a loner or outlier
which is thought to be faulty is to be rejected or
eliminated?”
If one considers n measurements, then the
Chauvenet’s criterion states that:
“A reading may be rejected if the probability
of obtaining it deviation from the mean is
less than 1/2n”
Example 4 (revisited):
Let the result of measurements to determine the
spring constants of a sample drawn from a very large
number of valve springs manufactured be obtained as:
n = 40, x = 152.5 N/cm, s = 0.889 N/cm
It is desired to determine the range of the spring
constant value to be used to eliminate a measurement
(out of 40 measurements) if it happens to be located
out of this range.
P( x +zs ≤ |x|) < 1/2n = 1/(2*40) = 0.0125
P( x +zs ≤ x) < (1/2n)/2 = 0.0125/2 = 0.00625
0.5 - P(0 ≤ x ≤ x +zs) < 0.00625
P(0 ≤ x ≤ x +zs) > 0.5 - 0.00625 = 0.49375
Æ z ≈ 2.50 (from table)
Therefore, a measurement may be rejected if it lies
outside the range:
x = x ± zs = 152.5 ± 2.5*0.889 = 152.5 ± 2.22 N/cm
Page 53
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
The following table lists values of maximum acceptable
normalized deviations (z) for various values of number
of data points (n) according to this criterion:
n
3
4
5
6
7
10
15
25
50
100
300
500
1000
zmax
1.38
1.54
1.65
1.73
1.80
1.96
2.13
2.33
2.57
2.81
3.14
3.29
3.48
Note that when applying the Chauvenet’s criterion:
1. n should be large.
2. End points of a curve must not be eliminated.
3. If a data point is rejected, then x and s must be
recomputed.
4. Successive applications more than once are not
acceptable.
Page 54
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Example:
The following readings (n=10) are taken of a certain
physical length in cm:
5.30, 5.73, 6.77, 5.26, 4.33, 5.45, 6.09, 5.64, 5.81,
5.75
Then the best estimates of the mean and standard
deviation of the true length can be computed as:
x = 5.613 cm and s = 0.627 cm
Now, let us test our data points for a possible
inconsistency in readings using Chauvenet’s criterion:
i
1
2
3
4
5
6
7
8
9
10
xi
5.30
5.73
6.77
5.26
4.33
5.45
6.09
5.64
5.81
5.75
zi
0.499
0.187
1.845
0.563
2.046
0.260
0.761
0.043
0.314
0.219
In accordance with the Table given in previous page,
for n=10, a data point with zi > 1.96 may be eliminated.
Hence, let us decide to reject the data point number 5
with z5 = 2.046 < 1.96. When this point is eliminated,
the new estimates of the mean and the standard
deviation of the true length becomes:
x = 5.756 cm and s = 0.462 cm
Note that the elimination of this data point has
resulted in a 26.5 % reduction in s (from 0.627 to
0.462).
Page 55
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
Randomness test (χ2 Chi-square test).
If we want to determine how well a given set of data fit
to an assumed distribution, chi-square test is used. In
this test the quantity chi-square is defined as:
N
[(no )i − (ne )i ]2
i =1
(ne )i
χ = ∑
2
where
N : the number of cells or groups of observations
(no)i : number of observed occurrences in group i
(ne)i : number of expected occurrences in group i (in
other words, the value which would be
obtained if the measurements matched the
expected distribution perfectly)
A plot of the chi-square function is given below.
Chi-Square Function
50
P=0.001
0.01
0.05
45
40
35
0.50
χ2
30
25
0.95
20
0.99
15
10
5
0
1 3 5 7 9 11131517192123252729313335
Number of degrees of freedom, F
Page 56
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
The following is tabulated data of this plot:
Probability
F
0.99
0.95
0.50
0.05
0.01
0.001
1
0.0002
0.004
0.45
3.84
6.63
10.83
2
0.02
0.10
1.39
5.99
9.21
13.82
3
0.11
0.35
2.37
7.81
11.34
16.27
4
0.30
0.71
3.36
9.49
13.28
18.47
5
0.55
1.15
4.35
11.07
15.09
20.51
6
0.87
1.64
5.35
12.59
16.81
22.46
7
1.24
2.17
6.35
14.07
18.48
24.32
8
1.65
2.73
7.34
15.51
20.09
26.12
9
2.09
3.33
8.34
16.92
21.67
27.88
10
2.56
3.94
9.34
18.31
23.21
29.59
11
3.05
4.57
10.34
19.68
24.73
31.26
12
3.57
5.23
11.34
21.03
26.22
32.91
13
4.11
5.89
12.34
22.36
27.69
34.53
14
4.66
6.57
13.34
23.68
29.14
36.12
15
5.23
7.26
14.34
25.00
30.58
37.70
16
5.81
7.96
15.34
26.30
32.00
39.25
17
6.41
8.67
16.34
27.59
33.41
40.79
18
7.01
9.39
17.34
28.87
34.81
42.31
19
7.63
10.12
18.34
30.14
36.19
43.82
20
8.26
10.85
19.34
31.41
37.57
45.31
21
8.90
11.59
20.34
32.67
38.93
46.80
22
9.54
12.34
21.34
33.92
40.29
48.27
23
10.20
13.09
22.34
35.17
41.64
49.73
24
10.86
13.85
23.34
36.42
42.98
51.18
25
11.52
14.61
24.34
37.65
44.31
52.62
26
12.20
15.38
25.34
38.89
45.64
54.05
27
12.88
16.15
26.34
40.11
46.96
55.48
28
13.56
16.93
27.34
41.34
48.28
56.89
29
14.26
17.71
28.34
42.56
49.59
58.30
30
14.95
18.49
29.34
43.77
50.89
59.70
31
15.66
19.28
30.34
44.99
52.19
61.10
32
16.36
20.07
31.34
46.19
53.49
62.49
33
17.07
20.87
32.34
47.40
54.78
63.87
34
17.79
21.66
33.34
48.60
56.06
65.25
35
18.51
22.47
34.34
49.80
57.34
66.62
Page 57
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
In this plot and table, F represents the number of
degrees of freedom in the measurements and is given
by
F=N-k
where N is the number of cells and k is the number of
imposed conditions on the expected distribution.
Obviously, the smaller the χ2-value is the better is the
agreement between the assumed distribution and the
observed values, because it corresponds to a larger
probability value for this match.
Example:
An equal number of motors are purchased from a
company A and company B. At the end of a year,
records show 5 failures for A-type motors and 9
failures for B-type motors. Does this result imply that
A-type motors are more reliable?
In this problem, we have two cases (N=2): A-type
failures and B-type failures; that is, (no)A = 5 & (no)B = 9.
If we make the hypothesis that failures are random,
then for a total of 14 failures one would expect 7
failures for A and 7 failures for B. So, (ne)A = (ne)B = 7,
giving
(5 - 7)2 (9 - 7)2
2
χ =
+
= 114
.
7
7
The only imposed condition in this problem is
(no)A + (no)B = 14 hence k=1 Æ F=2-1=1
If these values are used, it is found that the probability
of the difference in failure rates being coincidental has
a probability of 0.286 (≈ 30 %). This is a reasonably
high probability and does not allow us to say directly
that A-types are more reliable than B-types.
Page 58
ME 410 MECHANICAL ENGINEERING
SYSTEMS LABORATORY
If the same failure rates continues and at the end of a
longer period we end up 50 failures for A and 90
failures for B, χ2 becomes 11.4 giving a probability
value of 0.000734 (< 0.1 %) for the difference to be
coincidental. In this case, we can say without a doubt
that A-type motors are more reliable than B-type
motors.
Usually, a probability of less than 5 % allows rejection
of a random difference hypothesis.
If we have a series of measurements and the goodness
of the fit of the measurement data to a normal
distribution is in question, then the measurement
range is divided somewhat arbitrarily into sampling
intervals such that each interval contains at least 5
data points. Then using the number of occurrences in
each interval, the hypothesis of the data distribution
is normal is tested by using χ2 method. Note that k=3
for this case since
• A fixed number of data points are used (1 constraint)
• To estimate the expected occurrences mean and
standard deviation of the sample are used (2 more
constraints)
Page 59