MEGN 537 – Probabilistic Biomechanics

Ch.5 – Determining Distributions and Parameters from Observed Data

Anthony J Petrella, PhD

Determination of Distribution

• The underlying distribution can be established in one of the following ways:• Drawing a frequency diagram• Plotting the data on probability paper• Conducting statistical tests known as

goodness-of-fit tests for distribution

Probability Paper• Gumbel (1954)

• N observations (X1, X2, X3…XN)

• Arrange Data in increasing order• ith value is plotted at the CDF of i/(N+1)

m E(ksi) m/(N+1) m E(ksi) m/(N+1) m E(ksi) m/(N+1)1 25900 0.02381 14 29000 0.33333 28 30200 0.666672 27400 0.04762 15 29200 0.35714 29 30200 0.690483 27400 0.07143 16 29300 0.38095 30 30300 0.714294 27500 0.09524 17 29300 0.40476 31 30500 0.738105 27600 0.11905 18 29300 0.42857 32 30500 0.761906 28100 0.14286 19 29300 0.45238 33 30600 0.785717 28300 0.16667 20 29300 0.47619 34 31100 0.809528 28300 0.19048 21 29400 0.50000 35 31200 0.833339 28400 0.21429 22 29400 0.52381 36 31300 0.85714

10 28400 0.23810 23 29500 0.54762 37 31300 0.8809511 28700 0.26190 24 29600 0.57143 38 31300 0.9047612 28800 0.28571 25 29600 0.59524 39 32000 0.9285713 28900 0.30952 26 29900 0.61905 40 32700 0.95238

27 30200 0.64286 41 33400 0.97619

Probability Paper

Plotted versus Normal Dist

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20000 22000 24000 26000 28000 30000 32000 34000 36000

Young's Modulus

F(x

)

Experimental

Normal

Goodness of Fit

• Question: • Whether two independent samples come from

identical continuous distributions?• Dataset compared to the theoretical distribution• Restated: Is the theoretical distribution an

acceptable representation of the dataset?

• Chi Square based on PDF• Kolmogorov-Smirnov based on the CDF

• Based on error between the observed and assumed PDF of the distribution

• Methodology:• Arrange N data points in increasing order• Break data into m intervals• Determine:

• ni – observed frequency of data points in interval “i”• ei – theoretical Frequency of data points in interval “i”

Chi-Square Test (c2)

m

i i

ii

e

enfactor

1

2

(n)i) interval( ie

• Methodology:• Determine c1-a,f

• a = Significance Level (usually between 1% and 10%)

• f = degrees of freedom = m – 1 – k• m = # of intervals• k = # of distribution parameters (= 2 for normal or lognormal)

• Obtain c1-a,f from Appendix 3

• The assumed distribution is acceptable at the significance level a if:

Chi-Square Test (c2)

NOTE: m shouldbe > = 5 to obtain satisfactory results

f

m

i i

ii ce

enfactor ,1

1

2

Example (Haldar 5.2)

Example (Haldar 5.2)

a) Uniform distributed random variablesOrdinary graph paper can be prob. paper

b)

Example (Haldar 5.2)

c)

f = m – 1 - k f

m

i i

ii ce

enfactor ,1

1

2

(n)i) interval( ie

Example (Haldar 5.5)

• Perform Chi-square test on the data from Problem 3.1.

• Can the underlying distribution be accepted as normal at a 5% significance level?

• f = degrees of freedom = m – 1 – k• m = # of intervals• k = # of distribution parameters

Solution (Haldar 5.5a)

Kolmogorov-Smirnov (K-S) Test

• Based on the error between the observed and assumed CDF of the distribution

• Methodology:• Arrange data in increasing order and assign

index, m to each data point where m = 1,2,…,n• Determine Sn(xi) = manual CDF:

• Sn(xi) = 0; x < x1

• Sn(xi) = m/n; xm ≤ x ≤ xm+1

• Sn(xi) = 1; x ≥ xn

• Determine FX(xi) = Assumed distribution

K-S Test

• Methodology:• Determine Dn = max| Fx(xi) – Sn(xi) |

• Determine Dna

• a = Significance Level• Dn

a value found in Appendix 4

• The assumed distribution is acceptable at the significance level a if the maximum difference Dn is less than or equal to the tabulated value of Dn

a

Example (Haldar 5.8)

• Perform K-S test on the data fromProblem 3.1.

• Can the underlying distribution be accepted as normal at a 5% significance level?

Solution (Haldar, 5.8)

Parameter Estimation

Method of Moments

• Moments are statistical parameters of a dataset• 1st moment (mean = E(X))• 2nd central moment (Var(X))• 3rd central moment (skewness)

• Distribution parameters are derived from the moments

• PDF forms and parameters for distributions in Table 5.6 on page 118• All are based on first two moments, E(x) and Var(X)

Method of Maximum Likelihood

• For a random variable X, if x1, x2, … xn are the n observations or sample values, the estimated value of the parameter is the value most likely to produce these observed values.

• L(x1, x2, … xn;p) = fx(x1;p) fx(x2;p)…fx(xn;p)• p = distribution parameter that needs to be established

• Maximize likelihood L by setting dL/dp = 0• Solve for p• Solve for multiple parameters with simultaneous equations

• Results are very similar to method of moments

Interval Estimation

• Differences exist between expected values of populations and samples

• Distribution parameters ( , m s) are typically • Estimated from samples• Applied to populations

• Intervals estimate the range of possible values for the parameter to a specified level of confidence

Confidence Intervals• Distributions can be linked to probability – making possible

predictions and evaluations of the likelihood of a particular occurrence

• In a normal distribution, the number of standard deviations from the mean tells us the percent distribution of the data and thus the probability of occurrence

• x = Mean• s = Standard Deviation• n = Sample Size• (1 – a) = Confidence Interval• ka/2 = value of the standard normal variate (z)

= F-1(p) (found using Appendix 1)

Interval Estimation for the Mean with Known Variance

n

kxn

kx 2/2/1

;

Two tailed interval!

Lower Confidence Limit for m

•

Upper Confidence Limit for m

•

Lower and Upper Confidence Limit for the Mean with Known Variance

n

kx 1

Each is a one tailed interval!

n

kx 1

Interval Estimation for the Mean with Unknown Variance

ta/2,n-1 = value of Student’s t distribution – found using Appendix 5

n

stx

n

stx nn 1,2/1,2/1

;

• Standard normal distribution valid for…• Known population variance• Large n ( > 30)

• If n is small (< 10), s ≠ ,s use Student’s t

Student’s t distribution

f = n – 1 = DOF

Interval Estimation for Variance

1,2/

2

1,2/1

2

1

2 )1(;

)1(

nn c

sn

c

sn

• C*,n-1 = value of Chi Square distribution – found using Appendix 3