Examples to be solved in exercises · x1=sum(fix(5rand(1,1,'u'))+1); x2=sum(fix(5rand(1,1,'u'))+1); x=x1+x2; if x/2~=fix(x/2) s=s+1; end end Pc=s/nd // True prob. //Pa=1/2 //Pb=1/2

Examples to be solved in exercises

1 Exercise - probability

1.1 Example

// Two dice

// ------------------------------------------

[u,t,n]=file(); chdir(dirname(n(2)));

clc, clear, close, getd func, mode(0)

//What is the probability that sum of the the dashed dices

//will be greater than 3.

nd=10000; // number of experiments

x2=fix(6*rand(2,nd,'u'))+1; // nd throws with two dice

x=sum(x2,1); // sums on dice

s=0;

for i=1:nd // loop for counting positive res.

if x(i)>3

s=s+1;

end

end

P_stat=s/nd // statistical probability

P_class=1-3/36 // classical probability

1.2 Example

// Balls - realization with a function

// ------------------------------------------



function k=balls(v) // function definition

// drawing a ball

k=(rand(1,1,'u')>(v(1)/sum(v)))+1;

endfunction

//In a box we have 3w and 5b balls. We draw subsequently 2 balls without

//returning the first one. What is the probability that the second drawn

//ball will be white?

nd=10000; // numb. of experiments

nw=3; nb=5; // numb. of baals

s=0;

1

for i=1:nd // loop for experiments

n=[nw nb];

j=balls(n); // first draw

// j=(rand(1,1,'u')>(n(1)/sum(n)))+1;

n(j)=n(j)-1;

j=balls(n); // second draw

// j=(rand(1,1,'u')>(n(1)/sum(n)))+1;

if j==1 // count of positive res.

s=s+1;

end

end

P_stat=s/nd // statistical probability

P_class=3/8 // classical probability

2 Exercise - characteristics

2.1 Example

// Characteristics of r.v.

// ------------------------------------------



nd=100;

// continuous data

x1=.16*rand(nd,1,'n')+5; // variance = 0.4

x2=.64*rand(nd,1,'n')-3; // variance = 0.8

m1=mean(x1)

m2=mean(x2)

v1=variance(x1)

v2=variance(x2)

cv=covariance(x1,x2) // covariance

cr=corrcoef(x1,x2) // correlation coefficient

// binomial data

y=(rand(nd,1,'u')<.3)+0;

p=sum(y)/nd // ratio

// discrete data

z=fix(10*rand(nd,1,'u'))+fix(5*rand(nd,1,'u'))+1;

me=median(z)

2

[f,s]=histplot(min(z):max(z),z);

[xxx,j]=max(f);

mo=s(j)

2.2 Example

// One dice - conditional prob.

// ------------------------------------------



//a) What is the probability, that after a roll of a dice we obtain

// odd number?

//b) What is the above probability if we know, that the number

// is less than 5.

//c) What is the probability according to a) if we know, that the number

// is less than 4.

nd=1000;

// a)

s=0;

for i=1:nd

x=fix(6*rand(1,1,'u'))+1;

if x/2~=fix(x/2)

s=s+1;

end

end

Pa=s/nd

// b)

s=0;

for i=1:nd

x=fix(4*rand(1,1,'u'))+1;

if x/2~=fix(x/2)

s=s+1;

end

end

Pb=s/nd

// c)

s=0;

for i=1:nd

x=fix(3*rand(1,1,'u'))+1;

if x/2~=fix(x/2)

s=s+1;

end

end

Pc=s/nd

3

// True prob.

//Pa=1/2

//Pb=1/2

//Pc=2/3

2.3 Example

// Two dice - conditional prob.

// ------------------------------------------



//a) What is the probability, that after a roll of two dice the sum

// will be odd number?

//b) What is the above probability if we know, that the number on

// the first dice was not six.

//c) What is the probability according to a) if we know, that neither

// on the first nor on the second dice the result was six.

nd=10000;

// a)

s=0;

for i=1:nd

x1=sum(fix(6*rand(1,1,'u'))+1);

x2=sum(fix(6*rand(1,1,'u'))+1);

x=x1+x2;

if x/2~=fix(x/2)

s=s+1;

end

end

Pa=s/nd

// b)

s=0;

for i=1:nd

x1=sum(fix(5*rand(1,1,'u'))+1);

x2=sum(fix(6*rand(1,1,'u'))+1);

x=x1+x2;

if x/2~=fix(x/2)

s=s+1;

end

end

Pb=s/nd

// c)

s=0;

for i=1:nd

4

x1=sum(fix(5*rand(1,1,'u'))+1);

x2=sum(fix(5*rand(1,1,'u'))+1);

x=x1+x2;

if x/2~=fix(x/2)

s=s+1;

end

end

Pc=s/nd

// True prob.

//Pa=1/2

//Pb=1/2

//Pc=12/25=0.48

3 Exercise - generation

3.1 Example

// Generation

// ------------------------------------------



m=1;

n=1000;

set(scf(),'position',[600 100 800 600]);

x=grand(m,n,'nor',2,.09);

histplot(20,x);

mx=mean(x)

vx=variance(x)

return

// and try others

x = grand(m, n, "bin", N, p);

x = grand(m, n, "geom", p);

x = grand(m, n, "poi", mu);

x = grand(m, n, "nor", Av, Sd);

x = grand(m, n, "exp", Av);

x = grand(m, n, "unf", Low, High);

x = grand(m, n, "gam", shape, rate);

x = grand(m, n, "bet", A, B);

x = grand(m, n, "chi", Df);

x = grand(m, n, "f", Dfn, Dfd);

5

3.2 Example

// Výb¥r 5ti ze 100, kde jsou 3 vadné

// ------------------------------------------



//In the store, there are 100 bulbs. Three of them are bad.

//Five bulbs are randomly chosen. What is the probability, that at

//least one of the chosen bulbs will be bad?


s=0;

for i=1:nd

x=fix(100*rand(1,5,'u'))+1; // 5 chosen bulbs

z=sum(x<4);

if z>0 // count of positive

s=s+1;

end

end

P_stat=1-s/nd

P_class=0.856

// P_class=exp(combLn(95,3)-combLn(100,3))

3.3 Example

// Watch randomly stopped

// ------------------------------------------



//A watch, whose battery was weak, randomly stopped. What is

//the probability, that the big hand was stopped between

//9 and 12 o'clock? What is the probability for the small hand?


s=0;

for i=1:nd

x=12*rand(1,1,'u'); // watch stopped

if (x>9) & (x<12) // count of positive

s=s+1;

end

end

P_stat=s/nd

P_class=1/4

6

4 Exercise - sample

4.1 Example

// Limit theorems - central L.T.

// - down: dice

// - right: repeated experiments

// a chooses from {dice, coin}

// ------------------------------------------



nd=10000;

nc=50;

a=6; // 6 - dice, 2 - coin

x=fix(a*rand(nc,nd,'u')+1);

s=sum(x,1);

histplot(20,s);

4.2 Example

// Limit theorems - law of large numbers

// ------------------------------------------



nd=1000; // numb. of visible steps

h=10; // number of steps within

nc=5; // number of coins

a=6; // 6 - dice, 2 - coin

x=fix(a*rand(nc,nd*h,'u')+1);

s=sum(x,1);

for i=1:nd

c=s(1:i*h);

mc(i)=mean(c); // estimate of mean

vc(i)=variance(c)/(h*i); // variance of estimate

end

vp=mc+sqrt(vc); // upper border

vm=mc-sqrt(vc); // lower border

plot(1:nd,mc,1:nd,vp,':',1:nd,vm,':')

// variance of estimate is var(x)/n

4.3 Example

// Limit theorems - variance of estimate

7

// ------------------------------------------



ni=10000; // numb. iterations

nd=10;

nc=5; // number of coins

a=6; // 6 - dice, 2 - coin

for i=1:ni

x=fix(a*rand(nc,nd,'u')+1);

s=sum(x,1);

mc(i)=mean(s); // estimate of mean

vc(i)=variance(s);

end

varEst=variance(mc) // variance from repeated experiments

avVar=mean(vc)/nd // theoretical var. - var(x)/n

4.4 Example

// Population and Sample (repeated sampling from population)

// ------------------------------------------



P=[27 29 42 35 25 33 56 37 21 44 59 43 38 36 28];

k=length(P);

m=100;

n=5;

X=samwr(n,m,P); // m samples of the length n

mP=mean(P); // population mean

vP=variance(P) // population variance

mX=mean(X,1); // sample means (from all samples)

vX=variance(mX) // variance of sample means

vT=vP/n // theoretical var. of samp.means

set(scf(),'position',[400 100 500 400])

plot(1:length(P),P,'rx')

plot([1 k],[mP mP],'r')

plot((1:m)+20,mX,'.')

plot([1 m]+20,[mP mP],'b--')

legend('population','mean estimates','expectation','expectation');

8

5 Exercise - regression

5.1 Example

// Regression - prediction

// ------------------------------------------


clc,clear,close,mode(0),getd func

// Example 1

// In a factory, dependence of the overall costs 'n'

//( in thousands of $) on the production 'p' has been

// investigated. The following data have been measured

prod = [532 297 378 121 519 613 592 497]; // x

cost = [48 32 42 27 45 51 53 48]; // y

// a) Using linear regression estimate the costs for

// the production 1000 products

// b) For which produciton the costs would be equal

// to $ 100 000.

// estimation of regression line

param=lin_reg(prod,cost) // par=[b1,b0]

// prediction

prod0=1000;

cost_1000=lin_pred(prod0,param)

// back prediction

costs0=100; // in thousands of $

prod_100=(costs0-param(2))/param(1)

//printf '\nVerification by formulas\n'

//printf ------------------------\n\n

//

//mp=mean(prod);

//mc=mean(cost);

//vp2=var(prod);

//vc2=var(cost);

//vpc=cvar(prod,cost);

//

//b1=vpc/vp2

//b0=mc-b1*mp

//

//ypp=b1*prod0+b0

//xpp=(costs0-b0)/b1

//

9

5.2 Example

// Regression analysis

// ------------------------------------------



// Example 3

// A harmful substance leaked into the container with water.

// Neutralizing agent has been applied and the concentration of the

// harmful substance has been measured at time instants 'x'. The

// measured concentrations 'y' are

xi = [5 12 20 26 29 38 65 126];

yi = [19 17 18 17 17 15 14 7];

// Compute the correlation coefficient of linear regression

// and conclude about its suitability. If suitable, compute

// the parameters of linear regression and estimate when

// the concentration will be zero.

disp 'Correletion coefficient'

r=covariance(xi,yi)/sqrt(variance(xi)*variance(yi))

p=lin_reg(xi,yi); // p=[b1 b0]

yp=p(1)*xi+p(2);

scf(); plot(xi,yi,'x',xi,yp,':o')

disp 'Parameters'

p

disp 'Zero concentration at x0'

y0=0;

x_zero=(y0-p(2))/p(1)

5.3 Example

// Polynomial and exponential regression

// ------------------------------------------



// Example 5

// At certain process we have measured the data

xi = [5 12 20 26 29 38 40 45];

yi = [9 7 12 12 27 35 44 76];

// Perform the polynomial regression of the order 'k' and

// the exponential regression. Using prediction errors

// decide which type of regression is better.

k = 3;

xx=[min(xi):.1:max(xi)];

10

pp=pol_reg(xi,yi,k);

ypp=pol_pred(xx,pp);

scf(); plot(xi,yi,'x',xx,ypp,'markersize',4)

title 'polynomial'

pe=exp_reg(xi,yi);

yep=exp_pred(xx,pe);

scf(); plot(xi,yi,'x',xx,yep,'markersize',4)

title 'exponential'

ep=yi-pol_pred(xi,pp);

SEp=variance(ep)/variance(yi)

ee=yi-exp_pred(xi,pe);

SEe=variance(ee)/variance(yi)

scf();plot(xi,ep,':.',xi,ee,':.',[min(xi) max(xi)],[0 0],':')

title 'prediction errors'

legend('polynomial','exponential');

5.4 Example

// Multivariate regression

// ------------------------------------------



// Example 6

// At certain process we have measured the data

x1i = [15 12 11 9 9 8 5 3];

x2i = [3 9 5 11 28 14 32 58];

yi = [9 7 22 12 27 31 44 36];

// Perform multivariate linear regression and test its

// suitability. Use standard error SE.

xi=[x1i' x2i']; // all independent variables

p=lin_reg_n(xi,yi);

yp=lin_pred_n(xi,p);

SE=variance(yi-yp)/variance(yi)

scf();plot(1:length(yi),yi,'.',1:length(yi),yp,'x')

legend('output','prediction');

11

6 Exercise - con�dence intervals

6.1 Example

// Conf. interval for expectation (known variance)

// ------------------------------------------



// Example 7

// Assume, that the height of children in the age 10 has normal distribution

// with the variance "sig2". Determine the interval al-I, in which the true

// height will be if we have measured the data sample of the length "n" with

// the average "mx".

// The interval determine as a) both-sided, b) left and c) rihgt-sided.

sig2 = 38;

n = 12;

mx = 127.3;

al = .01;

// a) both sided

[lb, ub]=z_int(mx,sig2,n,'b',al);

CI_B=[lb ub]

// b) left sided

[lb, ub]=z_int(mx,sig2,n,'l',al);

CI_L=[lb ub]

// c) right sided

[lb, ub]=z_int(mx,sig2,n,'r',al);

CI_R=[lb ub]

6.2 Example

// Conf. interval for expectation (unknown variance)

// ------------------------------------------



// Example 8

// Assume, that the height of children in the age 10 has normal distribution.

// Determine the interval al-I, in which the true height will be it we have

// measured the data sample of the length "n" with the average "mx".

// The interval determine as a) both-sided, b) rihgt-sided.

n = 12;

mx = 127.3;

s2x = 38;

al = .01;

12

// both sided

[lb, ub]=t_int(mx,s2x,n,'b',al);

CI_B=[lb ub]

// left sided

[lb, ub]=t_int(mx,s2x,n,'l',al);

CI_L=[lb ub]

// right sided

[lb, ub]=t_int(mx,s2x,n,'r',al);

CI_R=[lb ub]

6.3 Example

// Conf. interval for variance

// ------------------------------------------



// Example 9

// To learn the accuracy of a method for measuring the volume

// of manganese in the steel we performed independent measurements

// of several samples. We would like to know the border for which it holds

// that only 5% of possibly measured variances will be greater than

// the value of the true variance of the method.

// The measured values are

x = [4.3 2.9 5.1 3.3 2.7 4.8 3.6];

vr=variance(x);

n=length(x);

side='r'; // it is (-inf, ub) - ub is the

al=.05;

[lb, ub]=var_int(vr,n,side,al);

printf('\nThe border is %g\n',ub)

6.4 Example

// CI for variance

// ------------------------------------------



// Example 10

// At the motorway with recommended speed 80 km/h we monitored

// the speeds of passing cars and obtained data 'xi' and 'ni'

// (values and frequencies). Determine both-sided 'al'-interval

13

// for variance of the speeds.

xi = [60 70 75 80 85 90 110]; // values

ni = [ 3 27 36 29 25 31 8]; // frequencies

al = .05;

pi=ni/sum(ni);

mx=sum(xi.*pi);

vr=sum((xi-mx)^2.*pi);

vr=variancef(xi,ni);

n=sum(ni);

[lb, ub]=var_int(vr,n,'b',al);

CI=[lb ub]

6.5 Example

// Confidence interval for proportion

// ------------------------------------------



// Example 11


// the speeds of passing cars and obtained data 'x'

// Determine both-sided 'al'-interval for the ratio of drivers

// that exceed the recommended speed by more than 'r' km/h.

x = [78 86 65 92 83 92 85 66 42 82 ...

99 92 75 81 66 76 89 76 97 76 ...

75 56 76 78 96 77 86 79 86 93];

al= .05;

r = 3;

n=length(x);

n1=sum(x>(80+r));

p=n1/n;

[lb ub]=prop_int(p,n,'b',al);

CI_B=[lb ub]

[lb ub]=prop_int(p,n,'r',al);

CI_R=[lb ub]

7 Exercise - tests with one variable

7.1 Example

// Test of expectation (known variance)

14

// ------------------------------------------



// Example 12

// From a set of steel rods with equal nominal length we have chosen

// random choice with the lengths 'x'. The producer guarantees that

// the variance of the lengths is 'sig2'. At the

// significance level 'al' test the assertion of the producer

// that the nominal length of the rods id 'd'.

x = [6.2 7.5 6.9 8.9 6.4 7.1];

d = 6.5;

sig2 = .8;

al = .05;

mu0=d;

av=mean(x);

vr=sig2;

n=length(x);

side='b';

pv=z_test(mu0,av,vr,n,side)

7.2 Example

// Test of expectation (unknown variance)

// ------------------------------------------



// Example 13

// From a set of steel rods with equal nominal length we have chosen

// random choice with the lengths 'x'. At the

// significance level 'al' test the assertion of the producer

// that the nominal length of the rods id 'd'.

x = [6.2 7.5 6.9 8.9 6.4 7.1];

d = 6.5;

al = .05;

mu0=d;

av=mean(x);

vr=variance(x);

n=length(x);

side='b';

pv=t_test(mu0,av,vr,n,side)

7.3 Example

// Test of variance

15

// ------------------------------------------



// Example 15

// The accuracy of setting of certain machine can be verified

// according to the variance of the products. If the variance is

// greater then the level 'sig2', it is necessary to perform new setting.

// A data sample has been meaasured with values 'mx' and

// frequencies of values 'nx'. On the level 'al' test if it is necessary

// to set the machine.

nx = [ 5 12 32 11 8 3];

mx = [95 100 105 110 115 120];

sig2 = 28;

al = .05;

vr0=sig2;

vr=variancef(mx,nx);

n=sum(nx);

side='r';

pv=var_test(vr0,vr,n,side)

// Note

// H0: var<var0; HA: var>var0

7.4 Example

// Test of proportion

// ------------------------------------------



// Example 18


// the speeds of passing cars and obtained data 'x'.

// At the level 'al' test the hypothesis: The ratio of drivers

// that exceed the recommended speed by more than 'r' km/h is not

// greater than 'P'%

x = [78 86 65 92 83 92 85 66 42 82 ...

99 92 75 81 66 76 89 76 97 76 ...

75 56 76 78 96 77 86 79 86 93];

al= .05;

r = 3;

P = 20;

n=length(x);

m=sum(x>80+r);

p=m/n;

p0=P/100;

16

side='r';

pv=prop_test(p0,p,n,side)

8 Exercise - tests with two variables

8.1 Example

// Test of two expectations (equal variances)

// ------------------------------------------



// Example 16

// Solidity of materials is verified by two methods A and B.

// The same material has been subdued testing by both methods.

// The results are 'xA' a 'xB'. On the level 'al' test equality of

// both methods. The variability of methods is assumed to be equal.

xA = [20.1 19.6 20.0 19.9 20.1];

xB = [20.9 20.1 20.6 20.5 20.7 20.5];

al = .05;

av1=mean(xA);

vr1=variance(xA);

n1=length(xA);

av2=mean(xB);

vr2=variance(xB);

n2=length(xB);

side='b';

pv=t_test_2s(av1,vr1,n1,av2,vr2, n2,side)

8.2 Example

// Test of two expectations (paired)

// ------------------------------------------



// Example 17

// We are going to test if the tire removal on left and right sides

// of the front wheels of cars is equal. The measured values are 'xL' a 'xP'.

// Test at the level 'al'.

xL = [1.8 1.0 2.2 0.9 1.5];

xP = [1.5 1.1 2.0 1.1 1.4];

al = .05;

sample1=xL;

sample2=xP;

17

side='b';

pv=t_test_2p(sample1,sample2,side)

8.3 Example

// Test of two expectations (independent)

// ------------------------------------------



// Example 19


// the speeds of cars going into the town and from the town. We

// obtained data 'rD' (speeds into), 'nD' (frequencies into) and

// 'rZ' (speeds from) 'nZ' (frequencies from). At the level 'al'

// test the hypothesis: From the town the cars go more quickly.

nD = [ 5 11 17 65 98 73 79 63 3];

rD = [65 70 75 80 85 90 95 100 110];

nZ = [ 8 22 13 71 48 64 89 24 5];

rZ = [65 70 75 80 85 90 95 100 110];

al = .01;

av1=meanf(rD,nD);

av2=meanf(rZ,nZ);

vr1=variancef(rD,nD);

vr2=variancef(rZ,nZ);

n1=sum(nD);

n2=sum(nZ);

side='r';

pv=t_test_2n(av1,vr1,n1,av2,vr2, n2,side)

// H0: from > into

// HA: from < into --> into - from > 0 -- 'r'

8.4 Example


// ------------------------------------------



// Example 20

// During a check of the front lights of cars we have measured

// the data 'xL' (left light) and 'xP' (right light). The values

// are distances (in cm) above (positive) and below (negative) of the

// real level with respect to the optimal level. At the level 'al'

// test if the light levels at each car are the same.

xP = [-3 5 16 9 -8 -2 23 5 -6 -3];

18

xL = [-5 -12 22 -3 -9 1 -1 2 -13 -5];

al = .1;

sample1=xP;

sample2=xL;

side='b';


8.5 Example


// ------------------------------------------



// Example 21

// During a check of the front lights of cars we have measured

// the data 'xL' (left light) and 'xR' (right light). The values

// are distances (in cm) above (positive) and below (negative) of the

// real level with respect to the optimal level. At the level 'al'

// test the hypothesis: Left lights are higher than right.

xR = [-3 5 16 9 -8 -2 23 5 -6 -3];

xL = [-5 -12 22 -3 -9 1 -1 2 -13 -5];

al = .1;

sample1=xR;

sample2=xL;

side='r';


8.6 Example

// Test of two proportions

// ------------------------------------------



// Example 22

// At a crossroads we have written down numbers of cars going

// stright (S) turning to left (L) and right (R). The measured

// data are 'xS', 'xL' a 'xR'.

// On the level 'al' test assertion that the ratio of cars

// going straight is equal to those that are turning .

xS = 62;

xL = 39;

xR = 46;

al = .1;

19

n=xS+xL+xR;

p1=xS/n; // ratio of going stright

n1=xS;

p2=(xL+xR)/n; // ratio of turning

n2=xL+xR;

side='b';

pv=prop_test_2(p1,n1,p2,n2,side)

8.7 Example

// Test of two expectations

// ------------------------------------------



// Example 23

// At a crossroads during 6 days we have written down numbers

// of cars going stright (R) turning to left (L) anf right (P).

// The measured data are 'xR', 'xL' a 'xP'.

// On the level 'al' test assertion that the number of cars

// going straight is smaller than that of cars turning .

xS = [82 78 92 83 99 97];

xL = [29 42 34 38 45 34];

xR = [31 44 36 54 31 24];

al = .05;

av1=mean(xS);

av2=mean(xL+xR);

vr1=variance(xS);

vr2=variance(xL+xR);

n1=sum(xS);

n2=sum(xL+xR);

side='r';

pv=t_test_2n(av1,vr1,n1,av2,vr2, n2,side)

9 Exercise - testes with many variables

9.1 Example

// Anova 1

// ------------------------------------------



// Example 24

// We monitor three machines. Randomly, we measure their productions

// per hour 'x1', 'x2' and 'x3. At the level 'al', test the equality

20

// of their production.

al = 0.05;

x1 = [53 55 49 58 52 61 56 55];

x2 = [49 56 52 45 51 56 44 51];

x3 = [52 53 52 54 55 53 53 52];

x=[x1' x2' x3'];

pv=anova_1(x)

9.2 Example

// Anova 1

// ------------------------------------------



// Example 25

// For one month, we monitor number of accidents at five crossroads.

// The results are in the following table.

// ---------------------------------

// year: 1999 2000 2001 2002 2003

num_1 =[3 5 2 1 3];

num_2 =[6 2 5 3 4];

num_3 =[3 2 1 1 2];

num_4 =[4 1 1 2 2];

num_5 =[4 2 5 5 6];

// --------------------------------

// At the level 'al' test hypothesis: The average number of accidents

// is equal at all monitored crossroads.

al = 0.01;

x=[num_1' num_2' num_3' num_4' num_5'];

disp 'One-way anova'

pv=anova_1(x)

disp 'Two-ways anova'

[pv_col, pv_row]=anova_2(x);

disp(pv_col,'Equality in crossoads')

disp(pv_row,'Equality in time')

9.3 Example

// Chi-square test of uniformity

// ------------------------------------------



// Example 26

21

// At a crossroads we have written down numbers of passing cars.

// The measured data are 'd' - length of monitoring and

// 'x' - number of cars

d = [15 10 20 35 10 50];

x = [71 56 98 121 44 271];

// At the level 'al' test the hypothesis that the cars go uniformly

// (the same number per time unit).

al = .05;

O=x;

E=sum(x)*d/sum(d);

pv=chisquare_test(O,E)

9.4 Example

// Chi-square test of uniformity

// ------------------------------------------



// Example 27

// The following data are frequencies of incidents at certain

// big factory

// ---------------------------------------------------

// time interval 8-10h. 10-12h. 12-13h. 13-17h.

// number of accidents 2 7 1 16

// ---------------------------------------------------

// At the level 'al' test the hypothesis that the accidents

// occur uniformly.

al = .05;

d=[2 2 1 4]; // from time interval

O=[2 7 1 16]; // number of accidents

E=sum(O)*d/sum(d);

pv=chisquare_test(O,E)

9.5 Example

// Chi-square test of independence

// ------------------------------------------



// Example 28

// A connection between color of eyes and hair has been investigated.

// I a collected data sample we obtained the following frequencies

// ----------------------------------

22

// eyes \ hair light brown dark

// blue 90 75 55

// gray 96 136 88

// brown 108 135 119

// ---------------------------------

// At the level 'al' test the hypothesis that the color of eyes

// and hair are independent.

al = .05;

T=[90 75 55

96 136 88

108 135 119];

pv=chisquare_test_i(T)

10 Exercise - nonparametric tests

10.1 Example

// Mann Whitney test

// ------------------------------------------



// Example 30

// Two doctors recommend curing the cold with two different

// methods. The results (number of days of the treatment) are

// 'x1' and 'x2'. Test equality of the methods.

x1=[5 8 7 8 4 5 5 6 9 3 5 8 6];

x2=[3 4 9 5 4 9 9 8];

// If needed, add the additionally measured data:

//x1_add=[8 7 5 8 5 7 5 6 8 4 7 7 5 6];

//v2_add=[3 3 5 3 6 4 5 6 2 2 3 4 2 3];

pv=mannwhit_test(x1,x2)

10.2 Example

// Willcoxon test

// ------------------------------------------



// Example 31

// Eight sportsmen in certain sport club were tested with respect

// to their performance. All of them threw a javelin once and then

// they were subdued to intensive training. Then they threw once more.

// The measured lengths were x1 and x2.

23

// The hypothesis is that one day of training is not enough to

// improve their performance. Test on the level 'al'.

x1=[68 81 69 72 66 91 98 89 75 68 69 75 72 83 88 79 88 76 81 85];

x2=[79 62 70 75 68 81 85 94 71 62 81 70 74 85 82 91 85 82 83 73];

sm='l';

pv=wilcoxon_test(x1,x2,sm)

// Note

// H0: x2 is not greater than x1 ... x2 <= x1

// HA: x1 < x2 left

10.3 Example

// Willcoxon test

// ------------------------------------------



// Example 32

// Test if mice and stags have equally long front legs. The measured

// values are

x1=[135 123 3.1 2.5 98 124 131 3.4 2.8 128];

x2=[136 121 2.9 2.6 101 121 130 3.5 2.9 126];

// Additionaly measured data

//x1_add=[154 135 2.9 137 2.7 3.0 3.2 131 2.8 148];

//x2_add=[162 141 2.8 142 2.9 2.8 3.0 132 3.1 151];

sm='b';

pv=wilcoxon_test(x1,x2,sm)

10.4 Example

// Friedman test

// ------------------------------------------



// Example 33

// Tree inspectors are to evaluate functionality of five fast

// food stands. Each inspector evaluates each stand. The result

// is the table Tab: rows correspond to inspectors, columns columns

// to stands. Evaluation is 1,2,..,10. 10 is the best.

// Test if the quality of the stans is equal.

Tab=[10 8 3 9 7

8 7 5 9 10

8 9 5 7 6];

24

pv=friedman_test(Tab)

// Note

// Blocks in rows (= data from individual inspectors)

10.5 Example

// Kruskal Wallis

// ------------------------------------------



// Example 34

// A factory produces some products whose weight must

// be constant. For the production it uses four machines.

// A sample of products has been taken from all machines

// to test equality of the product weights. The measured

// values are 'x1', 'x2', 'x3' and 'x4'. Test the equality.

x1=[39.4 34.8 35.6 35.1 35.8];

x2=[34.4 34.2 35.1 31.1 32.5 33.8];

x3=[30.2 35.1 34.2 36.3 30.8 35.6 35.2];

x4=[39.1 34.3 38.6 34.5 36.4 36.1];

L=list();

L(1)=x1;

L(2)=x2;

L(3)=x3;

L(4)=x4;

pv=kruskal_test(L)

10.6 Example

// Wilcoxon for one sample (too small sample)

// ------------------------------------------



// Example 35

// Let X denote the length, in centimeters, of a certain fish species.

// We obtained the data set X. Can we conclude that the median length

// of the fish species differs significantly from X0 centimeters?

X=[5.0 3.9 8.2 7.5 2.8 4.1 5.4 2.6 1.7 4.3 7.4 4.1 5.2 ..

9.3 2.7 3.4 5.9 4.3 9.4 8.2 4.8 3.3 4.7 5.3 4.2 4.0];

X0=4.1;

pv=wilcoxon_test(X,X0,'b')

25

10.7 Example

// Wilcoxon for one sample (with more data)

// ------------------------------------------



// Example 36

// The same as 35, but with additional data.

// Let X denote the length, in centimeters, of a certain fish species.

// We obtained the data set X. Can we conclude that the median length

// of the fish species differs significantly from X0 centimeters?

X=[5.0 3.9 5.2 5.5 2.8 6.1 6.4 2.6 1.7 4.3 3.5 3.2];

X0=3.7;

Sample_median=median(X)

Expected_median=X0

pv=wilcoxon_test(X,X0,'b')

// Remark:

// 1. Try to change expected median X0

//

// 2. Try to add still more data from the set

// 5.6 6.8 3.4 8.4 6.9 7.5 9.2 9.4 8.9 7.6

11 Exercise - validation of regression

11.1 Example

// Regression analysis validation

// ------------------------------------------

[u,t,n]=file(); // find working directory

chdir(dirname(n(2))); // set working directory

clc, clear, close, getd func, mode(0), warning off

//Example 37

//Try to experiment with regression analysis. Note its validation.

nd=50;

x=1:nd;

select 4 // <--- chose the case

case 1, b1=2; b0=-1; sd=1; // good regression, growing line

case 2, b1=-2; b0=3; sd=1; // good regression, decreasing line

case 3, b1=.0001; b0=-1; sd=.1; // small b1, good noise

case 4, b1=2; b0=-1; sd=1000; // good b1, big noise

case 5, b1=2; b0=-1; sd=1; // ... try your own

end

26

y=b1*x-b0+sd*rand(1,nd,'n'); // data generation

Y=y(:);

X=[x',ones(nd,1)];

disp 'reression coefficients'

th=inv(X'*X)*X'*Y; // regression coefficients

yp=X*th; // prediction

ep=y'-yp; // prediction error

disp 'standard error of residuals'

SE=variance(ep) // standard error

disp 'correlation coefficient'

r=corrcoef(x,y) // correlation coefficent

disp 'coefficint of determination'

R=r**2 // coefficient of detrmination

disp 't-test'

pv=pearson_test(x,y) // t-test

disp 'F-test'

pv=f_test_reg(x,y) // F-test

disp 'F-test of prediction'

pv=f_test_pred(y',yp,2) // F-test of prediction

set(scf(),'position',[600 300 600 400]);

plot(x,y,'.',x,yp)

27

Documents

Examples to be solved in exercises · x1=sum(fix(5*rand(1,1,'u'))+1); x2=sum(fix(5*rand(1,1,'u'))+1); x=x1+x2; if x/2~=fix(x/2) s=s+1; end end Pc=s/nd // True prob. //Pa=1/2 //Pb=1/2

Examples to be solved in exercises · x1=sum(fix(5rand(1,1,'u'))+1); x2=sum(fix(5rand(1,1,'u'))+1); x=x1+x2; if x/2~=fix(x/2) s=s+1; end end Pc=s/nd // True prob. //Pa=1/2 //Pb=1/2