Selected Statistics Examples from Lectures

Matlab: Histograms

>> y = [1 3 5 8 2 4 6 7 8 3 2 9 4 3 6 7 4 1 5 3 5 8 9 6 2 4 6 1 5 6 9 8 7 5 3 4 5 2 9 6 5 9 4 1 6 7 8 5 4 2 9 6 7 9 2 5 3 1 9 6 8 4 3 6 7 9 1 3 4 7 5 2 9 8 5 7 4 5 4 3 6 7 9 3 1 6 9 5 6 7 3 2 1 5 7 8 5 3 1 9 7 5 3 4 7 9 1]’;

>> mean(y) ans = 5.1589

>> var(y) ans = 6.1726

>> std(y) ans = 2.4845

>> hist(y,9) histogram plot with 9 bins

>> n = hist(y,9) store result in vector n

>> x = [2 4 6 8]’

>> n = hist(y,x) create histogram with bin centers specified by vector x 2 4 6 8

0

5

10

15

20

25

30

35

1 2 3 4 5 6 7 8 90

2

4

6

8

10

12

14

16

Probability Examples

Probability that at least one coin will turn heads up from five tossed coins» Number of outcomes: 25 = 32» Probability of each outcome: 1/32» Probability of no heads: P(AC) = 1/32» Probability at least one head: P(A) = 1-P(AC) = 31/32

Probability of getting an odd number or a number less than 4 from a single dice toss» Probability of odd number: P(A) = 3/6» Probability of number less than 4: P(B) = 3/6» Probability of both:» Probability of either:

3/26/26/36/3)()()()( BAPBPAPBAP

6/2)( BAP

complement set

Permutation Examples

Process for manufacturing polymer thin films» Compute the probability that the first 6 films will be too thin

and the next 4 films will be too thick if the thickness is a random variable with the mean equal to the desired thickness

» n1 = 6, n2 = 4

» Probability

Encryption cipher» Letters arranged in five-letter words: n = 26, k = 5

» Total number of different words: nk = 265 = 11,881,376

» Total number of different words containing each letter no more than once:

%5.0005.0!10

!4!6

!!!

1

21

nnn

P

600,893,7)!526(

!26

)!(

!

kn

n

Combination Examples

Effect of repetitions» Three letters a, b, c taken two at a time (n = 3, k = 2)

» Combinations without repetition

» Combinations with repetitions

500 thin films taken 5 at a time» Combinations without repetitions

bcacabk

n3

)!23(!2

!3

2

3

ccbbaa

bcacab

k

kn6

)!24(!2

!4

2

41

600,686,244,255)!5500(!5

!500

5

500

k

n

Matlab: Permutations and Combinations

>> perms([2 4 6]) all possible permutations of 2, 4, 6

6 4 26 2 44 6 24 2 62 4 62 6 4

>> randperm(6) returns one possible permutation of 1-65 1 2 3 4 6

>> nchoosek(5,4) number of combinations of 5 things taken 4 at a time without repetitions

ans = 5

>> nchoosek(2:2:10,4) all possible combinations of 2, 4, 6, 8, 10 taken 4 at a time without repetitions

2 4 6 82 4 6 102 4 8 102 6 8 104 6 8 10

Continuous Distribution Example

Probability density function

Cumulative distribution function

Probability of events

1)1(75.0)(otherwise0

11)1(75.0)(

1

1

22

dvvdvvfxx

xf

11

1125.075.05.0

10

)(

25.075.05.0)1(75.0)()(

3

3

1

2

x

xxx

x

xF

xxdvvdvvfxFxx

73.025.075.05.095.0)()(

%75.68)1(75.0)5.0()5.0()5.05.0(3

5.0

5.0

2

xxxxFxXP

dvvFFxP

Matlab: Normal Distribution

Normal distribution: normpdf(x,mu,sigma)» normpdf(8,10,2) ans = 0.1210» normpdf(9,10,2) ans = 0.1760» normpdf(8,10,4) ans = 0.0880

Normal cumulative distribution: normcdf(x,mu,sigma)» normcdf(8,10,2) ans = 0.1587» normcdf(12,10,2) ans = 0.8413

Inverse normal cumulative distribution: norminv(p,mu,sigma)» norminv([0.025 0.975],10,2) ans = 6.0801 13.9199

Random number from normal distribution: normrnd(mu,sigma,v)» normrnd(10,2,[1 5]) ans = 9.1349 6.6688

10.2507 10.5754 7.7071

Matlab: Normal Distribution Example

The temperature of a bioreactor follows a normal distribution with an average temperature of 30oC and a standard deviation of 1oC. What percentage of the reactor operating time will the temperature be within +/-0.5oC of the average?

Calculate probability at 29.5oC and 30.5oC, then calculate the difference:» p=normcdf([29.5 30.5],30,1)p = [0.3085 0.6915]» p(2) – p(1)0.3829

The reactor temperature will be within +/- 0.5oC of the average ~38% of the operating time

25 26 27 28 29 30 31 32 33 34 350

0.1

0.2

0.3

0.4

Temperature

Discrete Distribution Example

Matlab function: unicdf(x,n),

>> x = (0:6);

>> y = unidcdf(x,6);

>> stairs(x,y)

Poisson Distribution Example

Probability of a defective thin polymer film p = 0.01 What is the probability of more than 2 defects in a

lot of 100 samples? Binomial distribution: = np = (100)(0.01) = 1 Since p <<1, can use Poisson distribution to

approximate the solution.

%03.8)2(1)2(

9197.0!

)()2()2(2

0!2

1!1

1!0

112

0

210

XPXP

eex

xfFXPj j

x

jj

j

Matlab: Maximum Likelihood

In a chemical vapor deposition process to make a solar cell, there is an 87% probability that a sufficient amount of silicon will be deposited in each cell. Estimate the maximum likelihood of success in 5000 cells.

s = binornd(n,p) – randomly generate the number of positive outcomes given n trials each with probability p of success» s = binornd(5000,0.87)

s=4338

phat = binofit(s,n) – returns the maximum likelihood estimate given s successful outcomes in n trials» phat = binofit(s,5000)

phat = 0.8676

Mean Confidence Interval Example

Measurements of polymer molecular weight (scaled by 10-5)

Confidence interval

26.131.127.127.112.133.119.122.136.125.1

31.121.1050.010

0049.0)26.2(

0049.026.1

26.2975.0)(911095.0

95.0

2

CONFk

sx

ccFm

Variance Confidence Interval Example

Measurements of polymer molecular weight (scaled by 10-5)

Confidence interval

26.131.127.127.112.133.119.122.136.125.1

016.0002.0

0023.002.19

)0049.0)(9(0163.0

70.2

)0049.0)(9(0049.0

02.19975.0)(70.2025.0)(995.0

295.0

212

2211

CONF

kks

ccFccFm

Matlab: Confidence Intervals

>> [muhat,sigmahat,muci,sigmaci] = normfit(data,alpha) data: vector or matrix of data alpha: confidence level = 1-alpha muhat: estimated mean sigmahat: estimated standard deviation muci: confidence interval on the mean sigmaci: confidence interval on the standard deviation

>> [muhat,sigmahat,muci,sigmaci] = normfit([1.25 1.36 1.22 1.19 1.33 1.12 1.27 1.27 1.31 1.26],0.05)

muhat = 1.2580sigmahat = 0.0697muci = 1.2081

1.3079sigmaci = 0.0480

0.1273

Mean Hypothesis Test Example

Measurements of polymer molecular weight

Hypothesis: 0 = 1.3 instead of the alternative 1 < 0

Significance level: = 0.10 Degrees of freedom: m = 9 Critical value

Sample t

Reject hypothesis

0049.0258.1 2 sx

38.1~10.0)(

38.1~90.01)~(

0

0

cccTP

ccTP

38.1897.1100049.0

3.1258.1

ct

Polymerization Reactor Control

Lab measurements» Viscosity and monomer content of polymer every four hours» Three monomer content measurements: 23%, 18%, 22%

Mean content control» Expected mean and known variance: 0 = 20%, = 1» Control limits:

» Sample mean: 21% process is in control

MonomersCatalysts

SolventHydrogen

On-linemeasurements

Labmeasurements

Polymer

Monomer ContentViscosity

%5.2158.2%5.183

158.22058.2

2

0

2

0 n

UCLn

LCL

Goodness of Fit Example

Maximum likelihood estimates

n

jj

n

jj

xxn

xxn

1

22

1

9.712)(1

ˆ

7.3641

ˆ

Goodness of Fit Example cont.

Linear Regression Example

Reaction rate data

Sample variances and covariance

Linear regression

Confidence interval for = 0.95 and m = 6

Experiment 1 2 3 4 5 6 7 8

Reactant Concentration

0.1 0.3 0.5 0.7 0.9 1.2 1.5 2.0

Rate 2.3 5.7 10.7 13.1 18.5 25.4 32.1 45.2

206.94.2084086.0 22 xyyx sss

15.15.225.2215.1 10 xykk

53.1114045.2 0 Kqc }0.240.21{ 195.0 CONF

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

5

10

15

20

25

30

35

40

45

50

Rea

ctio

n R

ate

Reactant Concentration

Data

Prediction

Matlab: Linear Regression Example

>> c = [0.1 0.3 0.5 0.7 0.9 1.2 1.5 2];

>> r = [2.3 5.6 10.7 13.1 18.5 25.4 32.1 45.2];

>> [k, kint] = regress(r’, [ones(length(c), 1), c’], 0.05)k = -1.1847 22.5524kint = -2.8338 0.4644 21.0262 24.0787

Correlation Analysis Example

Polymerization rate data

Correlation coefficient

= 5%, m = 6 c = 1.94 Compute t-statistic

Reject hypothesis that the hydrogen concentration and the polymerization rate are uncorrelated

Experiment 1 2 3 4 5 6 7 8

Hydrogen Concentration

0 0.1 0.3 0.5 1.0 1.5 2.0 3.0

Polymerization rate

9.7 9.2 10.7 10.1 10.5 11.2 10.4 10.8

624.0421.0411.011.1 22 rsss xyyx

94.195.11

22

cr

nrt 0 0.5 1 1.5 2 2.5 3

9.2

9.4

9.6

9.8

10

10.2

10.4

10.6

10.8

11

Pol

ymer

izat

ion

Rat

e

Hydrogen Concentration

Matlab: Correlation Analysis Example

>> h = [0 0.1 0.3 0.5 1 1.5 2 3];

>> p = [9.7 9.2 10.7 10.1 10.5 11.2 10.4 10.8];

>> R = corrcoef(h,p)R= 1.0000 0.6238 0.6238 1.0000

>> t = ttest(h,p)t=1 (reject hypothesis)