TABLES,

FORMULAS AND

EXERCISES WITH KEY FOR

BIOMETRICS





assist. prof. Tadeja Kraner Šumenjak, Ph.D. and assist. Vilma Sem, Ph.D.





Maribor, June 2018





Title:

Tables, formulas and exercises with key for biometrics





Authors:

assist. prof. Tadeja Kraner Šumenjak, Ph.D.

(University of Maribor, Faculty of Agriculture and Life Sciences)



assist. Vilma Sem, Ph.D.

(University of Maribor, Faculty of Agriculture and Life Sciences)





Review:

assoc. prof. Jože Nemec, Ph.D.

(University of Maribor, Faculty of Agriculture and Life Sciences)



assist. prof. Aleksandra Tepeh, Ph.D.

(University of Maribor, Faculty of Electrical Engineering and Computer Science)





Proofreading:

Katja Težak (University of Maribor, Faculty of Agriculture and Life Sciences)





Technical editor:

sen. lect. Peter Berk, M.S. (University of Maribor, Faculty of Agriculture and Life Sciences) and Jan Perša, M.D.

(University of Maribor Press).





Cover designer:

Jan Perša, M.D. (University of Maribor Press)





Other Graphics


Authors.



Co-published by / Izdajateljica:

University of Maribor, Faculty of Agriculture and Life Sciences

Pivola 10,2311 Hoče, Slovenia

http://fkbv.um.si, fkbv@um.si



Published by / Založnik:

University of Maribor Press

Slomškov trg 15, 2000 Maribor, Slovenia

http://press.um.si, zalozba@um.si



Edition:

1st





Publication type:

e-publication





Available at:

http://press.um.si/index.php/ump/catalog/book/336





Published:

Maribor, June 2018



© University of Maribor Press

All rights reserved. No part of this book may be reprinted or reproduced or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publisher.



CIP - Kataložni zapis o publikaciji

Univerzitetna knjižnica Maribor



57.087.1:631.421(075.8)



KRANER Šumenjak, Tadeja

Tables, formulas and exercises with key for biometrics [Elektronski vir] : for biometrics /

Tadeja Kraner Šumenjak and Vilma Sem. - 1st ed. - Maribor : University of Maribor Press, 2018



Način dostopa (URL): http://press.um.si/index.php/ump/catalog/book/336



ISBN 978-961-286-166-7



doi: 10.18690/978-961-286-166-7

1. Sem, Vilma

COBISS.SI-ID 94606337



ISBN:

978-961-286-166-7 (PDF)





DOI:

https://doi.org/10.18690/978-961-286-166-7





Price:

Free copy





For publisher:

full prof. dr. Žan Jan Oplotnik, Vice-rector (University of Maribor)





TABLES, FORMULAS AND EXERCISES WITH KEY FOR BIOMETRICS

T. Kraner Šumenjak & V. Sem





Tables, formulas and exercises with key for biometrics



TADEJA KRANER ŠUMENJAK & VILMA SEM1



Abstract This publication includes some materials we use in the one semester

bachelor course entitled Biometrics, at the Faculty of Agriculture and Life

Sciences.



In the last few years, several foreign students came to study at the University

of Maribor. A certain number of them came to study at the Faculty of

Agriculture and Life Sciences, and because the foreign students are not able to

follow the lectures in Slovene, we organized lectures in English. Since there was

no learning material in English that would deal with the basics of

statistics through solving real problems from agriculture using the statistical

program SPSS, we decided to translate and publish these materials, and thus

make them more available.



Keywords: • biometrics • statistical tables • statistical formulas • SPSS •

exercises •





CORRESPONDENCE ADDRESS: Tadeja Kraner Šumenjak, Ph.D., Assistant Professor, University of Maribor, Faculty of Agriculture and Life Sciences, Pivola 10,2311 Hoče, Slovenia, e-mail: tadeja.kraner@um.si. Vilma Sem, Ph.D., Assistant, University of Maribor, Faculty of Agriculture and Life Sciences, Pivola 10,2311 Hoče, Slovenia, e-mail: vilma.sem@um.si.



DOI https://doi.org/10.18690/978-961-286-166-7





ISBN 978-961-286-166-7

© 2018 University of Maribor Press

Available at: http://press.um.si.





TABLE OF CONTENTS


1

FORMULAS ................................................................................................................................................................ 1

1.1

RELATIVE NUMBERS .................................................................................................................................................. 1

1.1.1 STRUCTURES ......................................................................................................................................................... 1

1.1.2 INDICES ................................................................................................................................................................. 1

1.2

DESCRIPTIVE STATISTICS ........................................................................................................................................... 2

1.2.1 FREQUENCY DISTRIBUTIONS ................................................................................................................................. 2

1.2.2 QUANTILES ............................................................................................................................................................ 2

1.2.3 MEASURES OF CENTRAL TENDENCY ..................................................................................................................... 2

1.2.4 MEASURES OF VARIABILITY .................................................................................................................................. 3

1.3

INFERENTIAL STATISTICS ........................................................................................................................................... 4

1.3.1 ONE POPULATION ................................................................................................................................................. 4

1.3.2 TWO POPULATIONS .............................................................................................................................................. 4

1.3.3 MORE POPULATIONS – ANALYSIS OF VARIANCE ................................................................................................. 5

1.3.4 CORRELATION ....................................................................................................................................................... 6

1.3.5 LINEAR REGRESSION ............................................................................................................................................. 8

2

TABLES ....................................................................................................................................................................... 9

2.1

STANDARD NORMAL DISTRIBUTION TABLE .............................................................................................................. 9

2.2

CRITICAL VALUES OF STUDENTS’ 𝑇-DISTRIBUTION ................................................................................................. 10

2.3

CRITICAL VALUES OF THE F-DISTRIBUTION (FISHER’S DISTRIBUTION) ..................................................................... 11

2.4

CRITICAL VALUES OF THE CHI-SQUARE DISTRIBUTION ............................................................................................ 12

2.5

CRITICAL VALUES OF PEARSON'S CORRELATION COEFFICIENT ............................................................................... 13

2.6

CRITICAL VALUES OF SPEARMAN'S CORRELATION COEFFICIENT ............................................................................ 14

3

EXERCISES ................................................................................................................................................................ 15

3.1

DESCRIPTIVE STATISTICS ......................................................................................................................................... 15

3.2

NORMAL DISTRIBUTION .......................................................................................................................................... 16

3.3

CONFIDENCE INTERVAL ........................................................................................................................................... 17

3.4

ONE SAMPLE T-TEST ................................................................................................................................................ 17

3.5

INDEPENDENT SAMPLES T-TEST .............................................................................................................................. 18

3.6

PAIRED SAMPLES T-TEST ......................................................................................................................................... 19

3.7

ANALYSIS OF VARIANCE .......................................................................................................................................... 20

3.8

PEARSON' S CORRELATION COEFFICIENT ................................................................................................................ 21

3.9

SPEARMAN' S CORRELATION COEFFICIENT ............................................................................................................. 22

3.10

CHI-SQUARED TEST ................................................................................................................................................. 23

3.11

LINEAR REGRESSION................................................................................................................................................ 24



i



4

ANSWERS TO EXERCISES .......................................................................................................................................... 26

4.1

DESCRIPTIVE STATISTICS ......................................................................................................................................... 26

4.2

NORMAL DISTRIBUTION .......................................................................................................................................... 27

4.3

CONFIDENCE INTERVAL ........................................................................................................................................... 28

4.4

ONE SAMPLE T-TEST ................................................................................................................................................ 28

4.5

INDEPENDENT SAMPLES T-TEST .............................................................................................................................. 29

4.6

PAIRED SAMPLES T-TEST ......................................................................................................................................... 29

4.7

ANALYSIS OF VARIANCE .......................................................................................................................................... 30

4.8

PEARSON' S CORRELATION COEFFICIENT ................................................................................................................ 30

4.9

SPEARMAN' S CORRELATION COEFFICIENT ............................................................................................................. 31

4.10

CHI-SQUARED TEST ................................................................................................................................................. 31

4.11

LINEAR REGRESSION................................................................................................................................................ 32

5

REFERENCES ............................................................................................................................................................ 34

6

INDEX ...................................................................................................................................................................... 36





ii





1 FORMULAS

1.1 RELATIVE NUMBERS

1.1.1 STRUCTURES

PROPORTION

PERCENTAGE





ANGLE


of units in the ith group

of units in the ith group

which corresponds to the ith group

f

f

f

f 0

i

i

i

i =



f

⋅ 100

φ

⋅ 360°

∑K f

i% = ∑K

i = ∑K

i=1 i

f

i=1 i

f

i=1 i





where K is the number of groups

where K is the number of groups

where K is the number of groups

1.1.2 INDICES

FIXED BASE INDEX NUMBER





COMPUTING


Yj


fixed base index number from the chain index number

Ij =

⋅ 100

Y



0





CHAIN INDEX NUMBER


V


After base period:

I

j⋅Ij−1

j =



100

Y

I

j

j

V

Before base period:

Ij−1 =

⋅ 100

j =

⋅ 100

Y

Vj

j−1

NOTATIONS





AVERAGE OF RELATIVES


I


k

j

fixed base index number for the jth year

I = √V1V2 … Vk, k is the number of chain indices

Vj

chain index number for the jth year

or

Yj

data for the jth year

k−1 Y

I = 100 ⋅

√ k , k is the number of years in the time

Y0

data for the base period

Y1

Yj−1

data for the ( j − 1) th year

series

Ij−1

fixed base index number for the ( j − 1) th year





GROWTH RATE


Sj = Vj − 100





1





1.2 DESCRIPTIVE STATISTICS

1.2.1 FREQUENCY DISTRIBUTIONS

COMULATIVE FREQUENCY





NOTATIONS


F𝑘+1 = Fk + f𝑘

xk,min lower bound of the kth class

x





FREQUENCY DENSITY


k,max upper bound of the kth class

fk

frequency of the kth class

𝒇

𝒈

𝒌

𝒌 =



xk

midpoint of the kth class

𝒊𝒌

𝑖𝑘

width of the kth class





CLASS WIDTH


i𝑘 = xk,max − xk,min

1.2.2 QUANTILES

RELATIVE RANK





RANK


r−0.5

p =



r = n ⋅ p + 0.5

n

LINEAR INTERPOLATION





NOTATIONS


x−x0

x

= 1−x0

n

number of units

rx−r0

r1−r0

x

value, for which the rank is computed

QUANTILE 𝒙, WHICH CORRESPONDS TO 𝒓𝒙

x0

value in the array one place before x

x1

value in the array one place after x

x = x0 + (rx − r0)(x1 − x0)

rx

rank, which corresponds to x

RANK 𝒓𝒙, WHICH CORRESPONDS TO 𝒙

r0

rank, which corresponds to x0

x−x

r

0 + r

r1

rank, which corresponds to x1

x = x

0

1−x0

1.2.3 MEASURES OF CENTRAL TENDENCY

SAMPLE MEAN





NOTATIONS


1


1

x̅ = (x

∑n x

𝑥𝑖

ith observation of variable 𝑋

n

1 + x2 + ⋯ + xn) = n i=1 i



𝑓𝑖

frequency of the ith group

1

∑k

f

𝑛

sample size

Grouped data: x̅ = ∑k f

= i=1 ixi

n

i=1 ixi

∑k

f

i=1 i

𝑘

number of groups





SAMPLE GEOMETRIC MEAN


SAMPLE MODE (𝒎𝒐)

g = √

n x

is the value that occurs most frequently in a data set.

1 ⋅ x2 … xn





SAMPLE HARMONIC MEAN


SAMPLE MEDIAN (𝒎𝒆)

𝑛

ℎ =



is a quantile with a relative rank of 𝟎. 𝟓 or central value

1

1

1

+

+⋯+

𝑥1 𝑥2

𝑥𝑛

in the array.

Note: For computation of population parameters the same expressions can be used, only the mean, geometric mean and harmonic mean are computed over all members of the population.





2





1.2.4 MEASURES OF VARIABILITY

SAMPLE RANGE





NOTATIONS


vr = x𝑚𝑎𝑥 − x𝑚𝑖𝑛

𝑥max

observed maximum

𝑥𝑚𝑖𝑛

observed minimum

𝑞

SAMPLE QUARTILE DEVIATION OR SEMI-

1

first quartile

𝑞





INTERQUARTILE RANGE


3


third quartile

x̅

sample mean

1

𝑞0 = (𝑞3 − 𝑞1)

me

sample median

2

𝑥





SAMPLE INTERQUARTILE RANGE


𝑖

ith observation of variable 𝑋

𝑓𝑖

frequency of the ith class

𝑞𝑟 = 𝑞3 − 𝑞1

N

population size





OUTLIERS


M


population mean

n

sample size

are observations that are out of interval:

k

number of groups

(𝑞1 − 1.5𝑞𝑟, 𝑞1 + 1.5𝑞𝑟)





VARIANCE


MEAN ABSOLUTE DEVIATION AROUND A CENTRAL

1





POINT


Population:


σ2 = ∑N x2 − M2

N

i=1 i

1

1

∑

Sample:

s2 =

∑n (x

=

Around mean: ADx̅ =

|x

n

i − x

̅|

n−1

i=1

i − x

̅)

1

1

n

∑|



=

∑n x2 −

x̅2

Around median: 𝐴𝐷𝑚𝑒 =

𝑥

𝑛

𝑖 − 𝑚𝑒|

n−1

i=1 i

n−1

Grouped data:





STANDARD DEVIATION


1


𝜎2 = ∑𝑁 𝑓 2 − 𝑀2

Population:

σ = √σ2

𝑁

𝑖=1 𝑖𝑥𝑖

𝑛

1

𝑛

Sample:

𝑠 = √𝑠2

𝑠2 =

∑ 𝑓 2 −

𝑥̅2



𝑛 − 1

𝑖𝑥𝑖

𝑛 − 1

𝑖=1





3





1.3 INFERENTIAL STATISTICS

1.3.1 ONE POPULATION

CONFIDENCE INTERVAL FOR THE POPULATION MEAN





NOTATIONS


s


s

x̅ − t

M

population mean

α

≤ M ≤ x̅ + tα



2 √n

2 √n

MH

hypothetical mean

Degrees of freedom for t-distribution (Table 2.2): df = x̅

sample mean

s2

sample variance

n  1.

s

sample standard deviation

ONE SAMPLE T-TEST

tα

critical t-value for a two-tailed area

2

a) Hypotheses

n

sample size

𝐻0: 𝑀𝐻 = 𝑀

𝐻1: 𝑀𝐻 ≠ 𝑀

b) Sample test statistic

(𝑥̅ − 𝑀

𝑡 =

𝐻)

𝑠



√𝑛

This test statistic follows a t-distribution (Table 2.2) with

degrees of freedom: df = n − 1.

1.3.2 TWO POPULATIONS

INDEPENDENT SAMPLES T-TEST

PAIRED SAMPLES t-TEST

a) Hypotheses

a) Hypotheses

𝐻0: 𝑀1 = 𝑀2

𝐻0: 𝑀1 = 𝑀2 or 𝑀1 − 𝑀2 = 0

𝐻1: 𝑀1 ≠ 𝑀2

𝐻1: 𝑀1 ≠ 𝑀2 or 𝑀1 − 𝑀2 ≠ 0



b) Sample test statistic

b) Sample test statistic

𝑛

𝑠2(𝑛

2(𝑛

1

𝑛

𝑠2 = 1

1 − 1) + 𝑠2

2 − 1)

𝑠2 =

∑ 𝑑2 −

𝑑̅2

𝑛

𝑑

𝑖

1 + 𝑛2 − 2

𝑛 − 1

𝑛 − 1

𝑥̅

𝑖=1

𝑡 =

1 − 𝑥̅2

,

𝑑̅

𝑛

𝑡 =



√𝑠2 ⋅ 1 + 𝑛2

𝑠𝑑

𝑛1𝑛2

√𝑛

This statistic follows a t-distribution (Table 2.2) with This statistic follows a t-distribution (Table 2.2) with degrees of freedom: df = n1 + n2  2.

degrees of freedom: df = n − 1.





NOTATIONS


s


pooled standard deviation

M1

mean of population 1

s2

sample variance from population 1

M

1

2

mean of population 2

s2

x̅

2

sample variance from population 2

1

sample mean from population 1

d

n

i

pair difference

1

sample size from population 1

x̅

d̅

mean of the differences di

2

sample mean from population 2

2

n

sd

sample variance of the differences di

2

sample size from population 2

s2

pooled variance

n

number of data pairs





4





1.3.3 MORE POPULATIONS – ANALYSIS OF VARIANCE

a) HYPOTHESES

H0: M1 = M2 = ⋯ = Mk

H1: The means are not all equal.



b) ANOVA TABLE

Source of

Sums of

Degrees of

Mean square

F-statistic

variation

squares

freedom

Between groups

SSB

df

SSB

MSB

1 = k – 1

MSB =



F =



df1

MSW

SSW

This statistic follows an F-distribution (Tab. 2.3 )

Within groups

SSW

df2 = N − k

MSW =



df2

F(k − 1, N − k)

Total

SST

df = N – 1





c) NOTATIONS

Mi

mean of the ith population

ni

number of observations in the ith group

SST

sum of squares total

xij

jth observation at ith group

SSB

sum of squares between

N

total number of observations

𝑆𝑆𝑊 sum of squares within

MSB

mean square between groups

k

number of groups (treatments)

MSW mean square within groups



d) COMPUTING SUMS OF SQUARES AND DEGREES OF FREEDOM FOR GROUPS WITH EQUAL SIZES

𝑛1 = 𝑛2 = ⋯ = 𝑛𝑘 = 𝑛



2

k

n



1

C =

(∑ ∑ x )

2

N

ij

𝑘

𝑛

1

i=1 j=1

𝑆𝑆𝐵 =

∑ (∑ 𝑥𝑖𝑗) − 𝐶

𝑘

𝑛

𝑛 𝑖=1 𝑗=1

𝑆𝑆𝑇 = ∑ ∑ 𝑥2𝑖𝑗 − 𝐶

𝑆𝑆𝑊 = 𝑆𝑆𝑇  𝑆𝑆𝐵

𝑖=1 𝑗=1





5





1.3.4 CORRELATION

PEARSON'S CORRELATION COEFFICIENT





NOTATIONS




a) Hypotheses

n

number of pairs

H

x

0: ρ = 0 (no linear correlation between X and Y)

i

ith observation of variable 𝑋

H

x̅

sample mean of variable X

1: ρ ≠ 0 (linear correlation between X and Y)

y



i

ith observation of variable 𝑌

y̅

sample mean of variable Y

b) Sample Pearson's correlation coefficient



∑n (x

r

i=1

i − x

̅)(yi − y̅)

xy =

=

√∑n (x

n

i=1

i − x

̅)2 ∑

(y

i=1

i − y

̅)2

n ∑n x

(∑n x )(∑n y )

=

i=1 iyi −

i=1 i

i=1 i



√n ∑n (x

n

n

n

)2

i=1

i)2 − (∑

x )2

i=1 i

√n ∑

(y

i=1

i)2 − (∑

y

i=1 i



c) Compare your obtained correlation coefficient to the critical

values in the table 2.5 or use the approximation

rxy√n − 2

t =

,

√1 − r2xy

that follows a t-distribution (Table 2.2) with the following

degrees of freedom:

df = n − 2.

SPEARMAN'S CORRELATION COEFFICIENT





NOTATIONS


a) Hypotheses

n

number of pairs

H0: ρS = 0 (no monotone correlation between X and Y)

di

difference in ranks

H1: ρS ≠ 0 (monotone correlation between X and Y)

x̅

sample mean of ranked variable X



y̅

sample mean of ranked variable Y

b) Sample Spearman's correlation coefficient



Convert the raw data on each variable into ranks.

If there are no ties in the ranks:

6 ∑n d 2

r

i=1 i

S = 1 −



n(n2 − 1)

otherwise:

∑n (x

r

i=1

i − x

̅)(yi − y̅)

S =



√∑n (x

n

i=1

i − x

̅)2 ∑

(y

i=1

i − y

̅)2

Compare your obtained correlation coefficient’s 𝑟𝑆 against the

critical values in the table 2.6 (Critical values of the Spearman's

correlation coefficient).

6



PEARSON'S CHI-SQUARE TEST





NOTATIONS


a) Hypotheses

𝑋 variable with categories 𝐴1, 𝐴2, … , 𝐴𝑟

H0: Variable 𝑋 and variable 𝑌 are independent.

𝑌 variable with categories 𝐵1, 𝐵2, … , 𝐵𝑐

H1: Variable 𝑋 and variable 𝑌 are not independent.

Contingency table of observed frequencies:





B1

B2

…

Bc

Sum

b) Sample test statistic

A1

O11

O12

…

O1c

O1.

𝑟

𝑐

A

(𝑂

2

O21

O22

…

O2c

O2.

𝜒2 = ∑ ∑

𝑖𝑗 − 𝐸𝑖𝑗 )2

⋮

⋮

⋮

⋱

⋮

⋮

𝐸𝑖𝑗

A

𝑖=1 𝑗=1

r

Or1

Or2

…

Orc

Or.



Sum

O.1

O.2

…

O.c

N



c) This statistic follows a chi-square distribution (Table 2.4) with 𝑂

the following degrees of freedom: 𝑑𝑓 = (𝑟 − 1)(𝑐 − 1).

𝑖𝑗



observed frequency

𝑂

𝐸

𝑖.⋅𝑂.𝑗

𝑖𝑗 =



expected frequency

𝑁

N



total number of observations

YATES CORRECTION





NOTATIONS


a) Hypotheses

Contingency table for

H0: Variable 𝑋 and variable 𝑌 are independent.

variable X (with categories A1 and A2) and

H1: Variable 𝑋 and variable 𝑌 are not independent.

variable Y (with categories B1 and B2 ).





b) Sample test statistic



B1

B2

Sum

𝑛 2

A1

N11

N12

L1

𝑛 (|𝑁11𝑁22 − 𝑁21𝑁12| −

A

𝜒2 =

2)

2

N21

N22

L2

𝐿1𝐿2𝑆1𝑆2

Sum

S1

S2

N





c) This statistic follows a chi-square distribution (Table 2.4) with

the following degrees of freedom: 𝑑𝑓 = 1.





7





1.3.5 LINEAR REGRESSION

LEAST-SQUARES LINE





NOTATIONS


Population:



Y = α + βX

n

number of pairs

Sample estimate:

Y

̂ = a + bX

xi

ith observation of variable 𝑋



x̅

sample mean of variable X

𝑛 ∑𝑛 𝑥

(∑𝑛 𝑥

∑𝑛 𝑦

y

𝑏 =

𝑖=1 𝑖𝑦𝑖 −

𝑖=1 𝑖)(

𝑖=1 𝑖)

i

ith observation of variable 𝑌



2

𝑛 ∑𝑛 𝑥2 − (∑𝑛 𝑥 )

y̅

sample mean of variable Y

𝑖=1 𝑖

𝑖=1 𝑖

𝑎 = 𝑦̅ − 𝑏𝑥̅

b

slope of the line (sample)

n ∑n x

(∑n x

∑n y

r

i=1 iyi −

i=1 i)(

i=1 i)

a

intercept of the line (sample)

xy =



√n ∑n (x

n

2 √

n

n

2

β

slope of the line (population)

i=1

i)2 − (∑

x

i=1 i)

n ∑

(y

i=1

i)2 − (∑

y

i=1 i)

α

intercept of the line (population)

∑𝑛 (𝑦 − 𝑦̂)2

∑𝑛 𝑦2 − 𝑎 ∑𝑛 𝑦 − 𝑏 ∑𝑛 𝑥

r

𝑠

𝑖=1

𝑖=1 𝑖

𝑖=1 𝑖

𝑖=1 𝑖𝑦𝑖

xy

correlation coefficient

𝑒 = √

= √



𝑛 − 2

𝑛 − 2

r2xy

coefficient of determination

se

standard error of estimate

TESTING 𝜷





NOTATIONS


a) Hypotheses



H0: β = 0 (variable X has no effect on Y)

β

slope of the line (population)

H1: β ≠ 0 (variable X has linear effect on Y)

se

standard error of estimate



b

slope of the line (sample)

b) Sample test statistic

n

number of pairs

n

n

2

𝑏

1

𝑡 =

√∑ x2 − (∑ x )

𝑠

i

i

𝑒

n

i=1

i=1



c) This statistic follows t-distribution (Table 2.2) with degrees

of freedom: 𝑑𝑓 = 𝑛 − 2.





8





2 TABLES

2.1 STANDARD NORMAL DISTRIBUTION TABLE

This table presents the area under the standard normal curve

between the z score and infinity: p = P(Z ≥ z).



Example:

Area under the standard normal curve for z = 0.75 is p = 0.2266.



Z

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681

0.4641

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.4286

0.4247

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3897

0.3859

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.3557

0.3520

0.3483

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3228

0.3192

0.3156

0.3121

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.2877

0.2843

0.2810

0.2776

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

0.2546

0.2514

0.2483

0.2451

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

0.9

0.1841

0.1814

0.1788

0.1762

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

1.0

0.1587

0.1562

0.1539

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1379

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1170

1.2

0.1151

0.1131

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

0.0985

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

0.0838

0.0823

1.4

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0.0708

0.0694

0.0681

1.5

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0594

0.0582

0.0571

0.0559

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

0.0475

0.0465

0.0455

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

0.0392

0.0384

0.0375

0.0367

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0322

0.0314

0.0307

0.0301

0.0294

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

0.0256

0.0250

0.0244

0.0239

0.0233

2.0

0.0228

0.0222

0.0217

0.0212

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

2.1

0.0179

0.0174

0.0170

0.0166

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

2.2

0.0139

0.0136

0.0132

0.0129

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

2.3

0.0107

0.0104

0.0102

0.0099

0.0096

0.0094

0.0091

0.0089

0.0087

0.0084

2.4

0.0082

0.0080

0.0078

0.0075

0.0073

0.0071

0.0069

0.0068

0.0066

0.0064

2.5

0.0062

0.0060

0.0059

0.0057

0.0055

0.0054

0.0052

0.0051

0.0049

0.0048

2.6

0.0047

0.0045

0.0044

0.0043

0.0041

0.0040

0.0039

0.0038

0.0037

0.0036

2.7

0.0035

0.0034

0.0033

0.0032

0.0031

0.0030

0.0029

0.0028

0.0027

0.0026

2.8

0.0026

0.0025

0.0024

0.0023

0.0023

0.0022

0.0021

0.0021

0.0020

0.0019

2.9

0.0019

0.0018

0.0018

0.0017

0.0016

0.0016

0.0015

0.0015

0.0014

0.0014

3.0

0.0013

0.0013

0.0013

0.0012

0.0012

0.0011

0.0011

0.0011

0.0010

0.0010

3.1

0.0010

0.0009

0.0009

0.0009

0.0008

0.0008

0.0008

0.0008

0.0007

0.0007

3.2

0.0007

0.0007

0.0006

0.0006

0.0006

0.0006

0.0006

0.0005

0.0005

0.0005

3.3

0.0005

0.0005

0.0005

0.0004

0.0004

0.0004

0.0004

0.0004

0.0004

0.0003

3.4

0.0003

0.0003

0.0003

0.0003

0.0003

0.0003

0.0003

0.0003

0.0003

0.0002

3.5

0.0002

0.0002

0.0002

0.0002

0.0002

0.0002

0.0002

0.0002

0.0002

0.0002

3.6

0.0002

0.0002

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

3.7

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

3.8

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

0.0001

3.9

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

4.0

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

9





2.2 CRITICAL VALUES OF STUDENTS’ 𝑇-DISTRIBUTION

α

The table entry is the value of tα, having an area to the right of

2

2

under t-distribution with df degrees of freedom (two-sided test):

α

p = P (T ≥ tα) = P (T ≤ −tα) = .

2

2

2



Example:

α

α = 0.05 ⟹ = 0.025 and df = 10

2

tα(10) = 2.228

2

df

0.1

0.05

0.025

0.01

0.005

0.001

0.0005

1

3.078

6.314

12.706 31.821 63.656 318.289 636.578

2

1.886

2.920

4.303

6.965

9.925

22.328

31.600

3

1.638

2.353

3.182

4.541

5.841

10.214

12.924

4

1.533

2.132

2.776

3.747

4.604

7.173

8.610

5

1.476

2.015

2.571

3.365

4.032

5.894

6.869

6

1.440

1.943

2.447

3.143

3.707

5.208

5.959

7

1.415

1.895

2.365

2.998

3.499

4.785

5.408

8

1.397

1.860

2.306

2.896

3.355

4.501

5.041

9

1.383

1.833

2.262

2.821

3.250

4.297

4.781

10

1.372

1.812

2.228

2.764

3.169

4.144

4.587

11

1.363

1.796

2.201

2.718

3.106

4.025

4.437

12

1.356

1.782

2.179

2.681

3.055

3.930

4.318

13

1.350

1.771

2.160

2.650

3.012

3.852

4.221

14

1.345

1.761

2.145

2.624

2.977

3.787

4.140

15

1.341

1.753

2.131

2.602

2.947

3.733

4.073

16

1.337

1.746

2.120

2.583

2.921

3.686

4.015

17

1.333

1.740

2.110

2.567

2.898

3.646

3.965

18

1.330

1.734

2.101

2.552

2.878

3.610

3.922

19

1.328

1.729

2.093

2.539

2.861

3.579

3.883

20

1.325

1.725

2.086

2.528

2.845

3.552

3.850

21

1.323

1.721

2.080

2.518

2.831

3.527

3.819

22

1.321

1.717

2.074

2.508

2.819

3.505

3.792

23

1.319

1.714

2.069

2.500

2.807

3.485

3.768

24

1.318

1.711

2.064

2.492

2.797

3.467

3.745

25

1.316

1.708

2.060

2.485

2.787

3.450

3.725

26

1.315

1.706

2.056

2.479

2.779

3.435

3.707

27

1.314

1.703

2.052

2.473

2.771

3.421

3.689

28

1.313

1.701

2.048

2.467

2.763

3.408

3.674

29

1.311

1.699

2.045

2.462

2.756

3.396

3.660

30

1.310

1.697

2.042

2.457

2.750

3.385

3.646

40

1.303

1.684

2.021

2.423

2.704

3.307

3.551

60

1.296

1.671

2.000

2.390

2.660

3.232

3.460

120

1.289

1.658

1.980

2.358

2.617

3.160

3.373

∞

1.282

1.645

1.960

2.326

2.576

3.090

3.291





10





2.3 CRITICAL VALUES OF THE F-DISTRIBUTION (FISHER’S DISTRIBUTION)

The table entry is the value of Fα, having an area to the right of α

under F-distribution with df1and df2 degrees of freedom:

p = P(F ≥ Fα) = α.



Example:

α = 0.05, df1 = 2, df2 = 10

F0.05(2, 10) = 4.103





df

df

1

2

1

2

3

4

5

6

7

8

10

12

20

25

∞

2

18.51 19.00

19.16

19.25 19.30 19.33 19.35 19.37 19.40 19.41 19.45 19.46 19.50

3

10.13 9.552

9.277

9.117 9.013 8.941 8.887 8.845 8.785 8.745 8.660 8.634 8.527

4

7.709 6.944

6.591

6.388 6.256 6.163 6.094 6.041 5.964 5.912 5.803 5.769 5.628

5

6.608 5.786

5.409

5.192 5.050 4.950 4.876 4.818 4.735 4.678 4.558 4.521 4.365

6

5.987 5.143

4.757

4.534 4.387 4.284 4.207 4.147 4.060 4.000 3.874 3.835 3.669

7

5.591 4.737

4.347

4.120 3.972 3.866 3.787 3.726 3.637 3.575 3.445 3.404 3.230

8

5.318 4.459

4.066

3.838 3.688 3.581 3.500 3.438 3.347 3.284 3.150 3.108 2.928

9

5.117 4.256

3.863

3.633 3.482 3.374 3.293 3.230 3.137 3.073 2.936 2.893 2.707

10

4.965 4.103

3.708

3.478 3.326 3.217 3.135 3.072 2.978 2.913 2.774 2.730 2.538

11

4.844 3.982

3.587

3.357 3.204 3.095 3.012 2.948 2.854 2.788 2.646 2.601 2.405

12

4.747 3.885

3.490

3.259 3.106 2.996 2.913 2.849 2.753 2.687 2.544 2.498 2.297

13

4.667 3.806

3.411

3.179 3.025 2.915 2.832 2.767 2.671 2.604 2.459 2.412 2.207

14

4.600 3.739

3.344

3.112 2.958 2.848 2.764 2.699 2.602 2.534 2.388 2.341 2.131

15

4.543 3.682

3.287

3.056 2.901 2.790 2.707 2.641 2.544 2.475 2.328 2.280 2.066

16

4.494 3.634

3.239

3.007 2.852 2.741 2.657 2.591 2.494 2.425 2.276 2.227 2.010

17

4.451 3.592

3.197

2.965 2.810 2.699 2.614 2.548 2.450 2.381 2.230 2.181 1.961

18

4.414 3.555

3.160

2.928 2.773 2.661 2.577 2.510 2.412 2.342 2.191 2.141 1.917

19

4.381 3.522

3.127

2.895 2.740 2.628 2.544 2.477 2.378 2.308 2.155 2.106 1.879

20

4.351 3.493

3.098

2.866 2.711 2.599 2.514 2.447 2.348 2.278 2.124 2.074 1.844

21

4.325 3.467

3.072

2.840 2.685 2.573 2.488 2.420 2.321 2.250 2.096 2.045 1.812

22

4.301 3.443

3.049

2.817 2.661 2.549 2.464 2.397 2.297 2.226 2.071 2.020 1.784

23

4.279 3.422

3.028

2.796 2.640 2.528 2.442 2.375 2.275 2.204 2.048 1.996 1.758

24

4.260 3.403

3.009

2.776 2.621 2.508 2.423 2.355 2.255 2.183 2.027 1.975 1.734

25

4.242 3.385

2.991

2.759 2.603 2.490 2.405 2.337 2.236 2.165 2.007 1.955 1.712

26

4.225 3.369

2.975

2.743 2.587 2.474 2.388 2.321 2.220 2.148 1.990 1.938 1.691

27

4.210 3.354

2.960

2.728 2.572 2.459 2.373 2.305 2.204 2.132 1.974 1.921 1.672

28

4.196 3.340

2.947

2.714 2.558 2.445 2.359 2.291 2.190 2.118 1.959 1.906 1.655

29

4.183 3.328

2.934

2.701 2.545 2.432 2.346 2.278 2.177 2.104 1.945 1.891 1.638

30

4.171 3.316

2.922

2.690 2.534 2.421 2.334 2.266 2.165 2.092 1.932 1.878 1.623

40

4.085 3.232

2.839

2.606 2.449 2.336 2.249 2.180 2.077 2.003 1.839 1.783 1.510

60

4.001 3.150

2.758

2.525 2.368 2.254 2.167 2.097 1.993 1.917 1.748 1.690 1.390

120

3.920 3.072

2.680

2.447 2.290 2.175 2.087 2.016 1.910 1.834 1.659 1.598 1.255

∞

3.842 2.996

2.605

2.372 2.214 2.099 2.010 1.939 1.831 1.752 1.571 1.506 1.025

11





2.4 CRITICAL VALUES OF THE CHI-SQUARE DISTRIBUTION

The table entry is the value of 𝜒2

𝛼, having an area to the right of α

under chi-square distribution with df degrees of freedom:

p = P(χ2 ≥ χ2α) = α.



Example:

α = 0.05, df = 3

𝜒2

(9) = 7.815

0,05



df

0.995

0.975

0.20

0.10

0.05

0.025

0.02

0.01

0.005

0.002

0.001

1

0.000

0.001

1.642

2.706

3.841

5.024

5.412

6.635

7.879

9.550

10.828

2

0.010

0.051

3.219

4.605

5.991

7.378

7.824

9.210

10.597

12.429

13.816

3

0.072

0.216

4.642

6.251

7.815

9.348

9.837

11.345

12.838

14.796

16.266

4

0.207

0.484

5.989

7.779

9.488

11.143

11.668

13.277

14.860

16.924

18.467

5

0.412

0.831

7.289

9.236

11.070

12.833

13.388

15.086

16.750

18.907

20.515

6

0.676

1.237

8.558

10.645 12.592

14.449

15.033

16.812

18.548

20.791

22.458

7

0.989

1.690

9.803

12.017 14.067

16.013

16.622

18.475

20.278

22.601

24.322

8

1.344

2.180

11.030

13.362 15.507

17.535

18.168

20.090

21.955

24.352

26.124

9

1.735

2.700

12.242

14.684 16.919

19.023

19.679

21.666

23.589

26.056

27.877

10

2.156

3.247

13.442

15.987 18.307

20.483

21.161

23.209

25.188

27.722

29.588

11

2.603

3.816

14.631

17.275 19.675

21.920

22.618

24.725

26.757

29.354

31.264

12

3.074

4.404

15.812

18.549 21.026

23.337

24.054

26.217

28.300

30.957

32.909

13

3.565

5.009

16.985

19.812 22.362

24.736

25.472

27.688

29.819

32.535

34.528

14

4.075

5.629

18.151

21.064 23.685

26.119

26.873

29.141

31.319

34.091

36.123

15

4.601

6.262

19.311

22.307 24.996

27.488

28.259

30.578

32.801

35.628

37.697

16

5.142

6.908

20.465

23.542 26.296

28.845

29.633

32.000

34.267

37.146

39.252

17

5.697

7.564

21.615

24.769 27.587

30.191

30.995

33.409

35.718

38.648

40.790

18

6.265

8.231

22.760

25.989 28.869

31.526

32.346

34.805

37.156

40.136

42.312

19

6.844

8.907

23.900

27.204 30.144

32.852

33.687

36.191

38.582

41.610

43.820

20

7.434

9.591

25.038

28.412 31.410

34.170

35.020

37.566

39.997

43.072

45.315

21

8.034 10.283

26.171

29.615 32.671

35.479

36.343

38.932

41.401

44.522

46.797

22

8.643 10.982

27.301

30.813 33.924

36.781

37.659

40.289

42.796

45.962

48.268

23

9.260 11.689

28.429

32.007 35.172

38.076

38.968

41.638

44.181

47.391

49.728

24

9.886 12.401

29.553

33.196 36.415

39.364

40.270

42.980

45.559

48.812

51.179

25 10.520 13.120

30.675

34.382 37.652

40.646

41.566

44.314

46.928

50.223

52.620

26 11.160 13.844

31.795

35.563 38.885

41.923

42.856

45.642

48.290

51.627

54.052

27 11.808 14.573

32.912

36.741 40.113

43.195

44.140

46.963

49.645

53.023

55.476

28 12.461 15.308

34.027

37.916 41.337

44.461

45.419

48.278

50.993

54.411

56.892

29 13.121 16.047

35.139

39.087 42.557

45.722

46.693

49.588

52.336

55.792

58.301

30 13.787 16.791

36.250

40.256 43.773

46.979

47.962

50.892

53.672

57.167

59.703

31 14.458 17.539

37.359

41.422 44.985

48.232

49.226

52.191

55.003

58.536

61.098

32 15.134 18.291

38.466

42.585 46.194

49.480

50.487

53.486

56.328

59.899

62.487

33 15.815 19.047

39.572

43.745 47.400

50.725

51.743

54.776

57.648

61.256

63.870

34 16.501 19.806

40.676

44.903 48.602

51.966

52.995

56.061

58.964

62.608

65.247

35 17.192 20.569

41.778

46.059 49.802

53.203

54.244

57.342

60.275

63.955

66.619

12





2.5 CRITICAL VALUES OF PEARSON'S CORRELATION COEFFICIENT

The results are significant if the calculated value of rxy is greater or equal than the table value.

n

0.1

0.05

0.01

3

0.988

0.997

0.999

4

0.900

0.950

0.990

5

0.805

0.878

0.959

6

0.729

0.811

0.917

7

0.669

0.754

0.875

8

0.621

0.707

0.834

9

0.582

0.666

0.798

10

0.549

0.632

0.765

11

0.521

0.602

0.735

12

0.497

0.576

0.708

13

0.476

0.553

0.684

14

0.458

0.532

0.661

15

0.441

0.514

0.641

16

0.426

0.497

0.623

17

0.412

0.482

0.606

18

0.400

0.468

0.590

19

0.389

0.456

0.575

20

0.378

0.444

0.561

21

0.369

0.433

0.549

22

0.360

0.423

0.537

23

0.352

0.413

0.526

24

0.344

0.404

0.515

25

0.337

0.396

0.505

26

0.330

0.388

0.496

27

0.323

0.381

0.487

28

0.317

0.374

0.479

29

0.311

0.367

0.471

30

0.306

0.361

0.463

40

0.264

0.312

0.403

50

0.235

0.279

0.361

60

0.214

0.254

0.330

70

0.198

0.235

0.306

80

0.185

0.220

0.286

90

0.174

0.207

0.270

100

0.165

0.197

0.256





13





2.6 CRITICAL VALUES OF SPEARMAN'S CORRELATION COEFFICIENT

The results are significant if the calculated value of rs is greater or equal than the table value.

𝑛

0.1

0.05

0.01

1





2





3





4

1.000





5

0.900

1.000



6

0.829

0.886

1.000

7

0.714

0.786

0.929

8

0.643

0.738

0.881

9

0.600

0.700

0.833

10

0.564

0.648

0.794

11

0.536

0.618

0.755

12

0.503

0.587

0.727

13

0.484

0.560

0.703

14

0.464

0.538

0.679

15

0.446

0.521

0.654

16

0.429

0.503

0.635

17

0.414

0.485

0.615

18

0.401

0.472

0.600

19

0.391

0.460

0.584

20

0.380

0.447

0.570

21

0.370

0.435

0.556

22

0.361

0.425

0.544

23

0.353

0.415

0.532

24

0.344

0.406

0.521

25

0.337

0.398

0.511

26

0.331

0.390

0.501

27

0.324

0.382

0.491

28

0.317

0.375

0.483

29

0.312

0.368

0.475

30

0.306

0.362

0.467

31

0.301

0.356

0.459

32

0.296

0.350

0.452

33

0.291

0.345

0.446

34

0.287

0.340

0.439

35

0.283

0.335

0.433

36

0.279

0.330

0.427

37

0.275

0.325

0.421

38

0.271

0.321

0.415

39

0.267

0.317

0.410

40

0.264

0.313

0.405





14





3 EXERCISES

3.1 DESCRIPTIVE STATISTICS

1) The given frequency table describes the weights of bulls (hypothetical data):



Weights in kg

Number of bulls

Above 170 to 190

18

Above 190 to 210

38

Above 210 to 230

68

Above 230 to 250

76

Above 250 to 270

30

Above 270 to 290

16

Above 290 to 310

4

Total

250



a) Express this frequency distribution with a histogram.

b) Compute the sample mean, sample variance and sample standard deviation.

2) The average hourly earnings of production workers for selected periods are given below (hypothetical data).



Year

2007

2008

2009

2010

2011

2012

2013

Average hourly earnings (EUR)

6.32

6.57

6.83

7.43

8.28

8.74

9.34



a) From the following data, compute the fixed base index number for 2013 taking 2010 as the base year and interpret the result.

b) Calculate the chain index numbers and the average of relatives. Interpret the chain index number for 2010.

3) The table shows chained index numbers for production of wine (hypothetical data).



Year

2011

2012

2013

2014

2015

Chained index number

–

117.2

113.6

78.8

146.6



Calculate the fixed base index numbers taking 2014 as the base year.

4) A man travels from city A to city B. The first half of the way, he drove at the constant speed of 90 km per hour. Then he (instantaneously) increased his speed and travelled the remaining distance at 130 km per hour.

Find the average speed of the motion.

5) A man was travelling from city A to city B. For the first hour, he drove at the constant speed of 90 km per hour. Then he (instantaneously) increased his speed and, for the next hour, kept it at 130 km per hour. Find the average speed of the motion.



15





6) Lentil seed yields (kg/plot) were as shown in the following table:



52.2

34.9

29.7

28.4

28.0

22.8

18.5

19.4

18.8

19.5

13.1

10.1

41.5

36.3

31.7

31.0

28.2

33.0

26.0

30.6



a) From the data compute the sample mean, the sample standard deviation.

b) Compute the median, the first quartile, the third quartile and the interquartile range.

c) Make a box plot of the data.

7) The following data are the average annual temperatures from 2001-2010 for meteorological station Maribor (Source: ARSO, 2018).

Year

Average annual temperatures

2001

10.983

2002

11.799

2003

11.249

2004

10.406

2005

10.099

2006

10.781

2007

11.711

2008

11.510

2009

11.277

2010

10.470



a) From the data compute the sample mean, the sample standard deviation and the median.

b) Construct a box-plot of the data.

c) Find the 35th percentiles.

3.2 NORMAL DISTRIBUTION

8) Assume 𝑋 is normally distributed with a mean of 10 and a standard deviation of 2. Determine the following: a) 𝑃(𝑋 ≤ 13),

b) 𝑃(−2 ≤ 𝑋 ≤ 8),

c) 𝑃(𝑋 ≥ 11.5).

9) The heights of students (variable X) at the Faculty of Agriculture and Life Sciences in Hoče are normally distributed with a mean of 174 cm and a standard deviation of 8 cm (hypothetical data).

a) What percentage of the students are between 165 and 185 cm tall?

b) Find 𝑃(𝑋 ≤ 150).

c) What is the value that separates the top 1 % of heights from the rest of the population?

d) Find 𝑥1 and 𝑥2 for the middle 95 % of the area under the standard normal curve.



16





10) The plant height in a rice crop is normally distributed with a mean of 75 cm and standard deviation 5 cm.

Find the probability that a random sample of 𝑛 = 25 rice plants will have sample mean

a) less than 73.5 cm.

b) between 74 and 75.4 cm.

c) more than 77 cm.

3.3 CONFIDENCE INTERVAL

11) A study aims to estimate the mean systolic blood pressure of Slovenian adults by randomly sampling and measuring the blood pressure of 100 adults from the population. From their sample, they estimate the sample mean to be 70mmHg and the sample standard deviation to be 8mmHg (hypothetical data). Find the 95% (99 %) confidence interval.



12) Over the past several months, an adult patient has been treated for tetany. This condition is associated with an average total calcium level. Recently, the patients total calcium tests gave the following readings (Source: Brase and Brase, 2006):



9.3

8.8

10.1

8.9

9.4

9.8

10.0

9.9

11.2

12.1



Assume that the population of values has an approximately normal distribution. Find a 99% confidence interval for the population mean of the total calcium value in the patient’s blood.

3.4 ONE SAMPLE T-TEST

13) A sample of 12 eggs are randomly selected and their weights (in g) recorded, which are as follows: 60.2

53.8

67.2

56.9

58.6

60.0

66.3

50.7

56.0

63.3

58.2

64.1



Can we say that the mean egg weight in the population is different from 62.0 g (α=0.05)? Assume the variable is normally distributed.

14) A grain yield of standard variety of wheat is around 150 kg/plot. A new variety is planted on twenty-five randomly selected plots. The observed sample average for the new variety is 159.4 kg of grain per plot with a standard deviation of 14.1 kg. Should the new variety be used instead of the standard one, by using a 5%

level of significance by assuming normal distribution of selected variable?





17





15) The mean length of apple tree roots of apple tree varieties obtained from previous researches is 57.62 cm.

Interpret the SPSS output for observed sample which contains 18 trees.

a) Write 𝐻0.

b) Write the test statistic.

c) Determine the level of significance.

d) Write the p-value.

e) Compare p-value with level of significance and interpret the result.



One-Sample Statistics



N

Mean

Std. Deviation

Std. Error

Mean

Length

18

54.8333

19.30331

4.54983





One-Sample Test



Test Value = 57.62

t

Df

Sig. (2-tailed) Mean Difference 95% Confidence Interval of the

Difference

Lower

Upper

Length -.612

17

.548

-2.78667

-12.3860

6.8126

3.5 INDEPENDENT SAMPLES T-TEST

16) The experiment was made for comparing grain yields (kg/parcel) for two varieties of corn. The grain yields were as follows:

Parcel

Variety A

Variety B

1

33

27

2

25

43

3

20

36

4

19

20

5

42

22



Test whether these two varieties differ significantly at a 5 percent level of probability with respect to the grain yield (assume that population variances are equal and variables are normally distributed).

17) The difference between two types of fertilizers (organic and chemical) are being observed in order to compare the yield of grapevine. The organic fertilizer was applied on 44 vines with an average grape yield of 3.5 kg per vine and a standard deviation of 1.5 kg. The chemical fertilizer was used on 47 vines with an average grape yield of 3.9 kg per vine and standard deviation of 1.1 kg. Can we say that the chemical fertilizer produces a higher yield than the organic (𝛼 = 0.05)?





18





18) A newly developed variety of alfalfa was investigated in two different climatic zones to test the significance of the difference in the yield (kg/plot). According to the SPSS output, do the following tasks:

a) Write 𝐻0.

b) Write the test statistic.

c) Determine the level of significance.

d) Write the p-value.

e) Compare the p-value with the level of significance and interpret the result.





Group Statistics




N

Mean





Std. Deviation Std. Error Mean


1


10

24.7000

4.27005

1.35031

Yield

2

10

28.5000

2.75882

0.87242





Independent Samples Test


Levene's Test


t-test for Equality of Means

for Equality

of Variances

F

Sig.

t

df

Sig. (2-

Mean

Std.

95% Confidence

tailed)

Differe

Error

Interval of the

nce

Differ

Difference

ence

Lower

Upper

Yield

Equal

1.271 0.274

-2.364

18

0.030

-3.800

1.608

-7.177

-0.423

variances

assumed

Equal





-2.364

15.399

0.032

-3.800

1.608

-7.219

-0.381

variances

not

assumed



3.6 PAIRED SAMPLES T-TEST

19) Two laboratories carry out independent estimates of protein content in rice. Eight estimates were made by each laboratory. Each time one sample was taken. Half the quantity was sent to Lab 1 and the other half to Lab 2. We assume that the differences follow from a normally distributed population. The results are as follows (𝛼 = 0.05).



Lab 1

8

10

8

10

10

9

11

9

Lab 2

9

11

9

11

10

8

10

8



Do the two laboratories report the same results?

19





20) Consider the following study in which standing and supine systolic blood pressures were compared. This study was performed on 8 patients. Their blood pressures were measured in both positions. According to the SPSS output, do the following tasks:

a) Write 𝐻0.

b) Write the test statistic.

c) Determine the level of significance.

d) Write the p-value.

e) Compare the p-value with level of significance and interpret the result.





Paired Samples Test


Paired Differences


t

df Sig. (2-

Mean

Std.

Std. Error 95% Confidence Interval of the

tailed)

Deviation

Mean

Difference

Lower

Upper

Standing -

-3.25000

3.49489

1.23563

-6.17180

-0.32820

-2.630 7

0.034

Supine



3.7 ANALYSIS OF VARIANCE

21) The plot grain yields (kg) of three varieties of oats in an experiment are:



Plot

Variety A

Variety B

Variety C

1

22

24

19

2

18

39

22

3

35

19

24

4

17

25

17

5

42

23

28



Assume that ANOVA assumptions are met. Is there any difference in the grain yield produced by these varieties (0.05)?

22) Four diets were formulated, and their effect on weight loss was investigated on people of the same age, sex, and other activities. After a certain period, the weight loss (kg) was recorded and results were analyzed via SPSS.

a) Write 𝐻0.

b) Write the test statistic.

c) Determine the level of significance.

d) Write the p-value.

e) Compare the p-value with the level of significance and interpret the result.



20





ANOVA


weightloss




Sum of Squares

df

Mean Square

F

Sig.

Between Groups

70.245

3

23.415

14.218

0.000

Within Groups

44.464

27

1.647





Total

114.710

30





3.8 PEARSON' S CORRELATION COEFFICIENT

23) Consider the following study in which systolic and diastolic blood pressures were compared. This study was performed on 7 patients. Make a scatter diagram. Can we say, at a 0.05 level of significance, that there is a linear correlation between systolic and diastolic blood pressures? Assumed that the assumptions are met.



Systolic blood pressure Diastolic blood pressure

210

130

169

122

187

124

160

104

167

112

176

101

185

121



24) The calculated Pearson's correlation coefficient from a sample of 28 observations between two variables equals 0,5. Is the correlation significant at level 0,01?



25) The ear length and number of grains per ear of maize cultivar MASTER have been investigated in order to study the relationship between variables. On a base of the SPSS output, do the following tasks:

a) Write 𝐻0.

b) Write the test statistic.

c) Determine the level of significance.

d) Write the p-value.

e) Compare p-value with level of significance and interpret the result.





Correlations


length


number

length

Pearson Correlation

1

0.896**

Sig. (2-tailed)



0.000

N

12

12

number Pearson Correlation

0.896**

1

Sig. (2-tailed)

0.000



N

12

12

**. Correlation is significant at the 0.01 level (2-tailed).

21





3.9 SPEARMAN' S CORRELATION COEFFICIENT

26) The table shows a maximum monthly averaged temperature and monthly averaged sunshine hours for the climate station Oxford, covering the period from 1981 to 2010.

(Source: Metoffice. gov.uk, 2018 ). Find out if there is a correlation between both variables.



Month Max. temp. (°C) Sunshine (hours)

Jan

7.60

62.50

Feb

8.00

78.90

Mar

10.90

111.20

Apr

13.60

160.90

May

17.10

192.90

Jun

20.30

191.00

Jul

22.70

207.00

Aug

22.30

196.50

Sep

19.10

141.20

Oct

14.80

111.30

Nov

10.50

70.70

Dec

7.70

53.80

27) Ten commercial samples (ten different brands) of yoghurt were analyzed. In order to determine whether the two evaluators agree with one another in their evaluation of ten samples (1: dislike extremely to 10: like extremely). Measure the agreement between the evaluators via the Spearman rank correlation coefficient.



Commercial samples Evaluator 1 Evaluator 2

1

6

5

2

4

6

3

9

10

4

1

2

5

2

3

6

7

8

7

3

1

8

8

7

9

5

4

10

10

9

28) With a data base for year the 2011 provided by the network of the Farm Accountancy Data (Eng. Abbreviation FADN), researchers studied the correlation between variables of the gross value added and agricultural land in use (Source: Trpin Švikart, 2016). According to SPPS Output, do the following activities:

a) Write 𝐻0.

b) Write the test statistic.

c) Determine the level of significance.

d) Write the p-value.

e) Compare the p-value with level of significance and interpret the result.





22





Correlationsa


land in use


gross value added

Spearman's rho

land in use

Correlation

1.000

0.672**

Coefficient

Sig. (2-tailed)

.

0.000

N

2818

2818

gross value added

Correlation

0.672**

1.000

Coefficient

Sig. (2-tailed)

0.000

.

N

2818

2821

**Correlation is significant at the 0.01 level (2-tailed).

Year = 2011

3.10 CHI-SQUARED TEST

29) Researchers interviewed consumers of different varieties of apples, asking them why they buy a specific variety of apples (random sample). The following results were obtained.





Variety A

Variety B

Variety C

Total

Colour

320

18

62

400

Taste

42

13

45

100

Size

110

57

83

250

Durability

28

12

10

50

Total

500

100

200

800



Find if the sale of different apple varieties and different apple properties are independent.

30) Entomologists have investigated yellow, short-leaved and spruce pines in a certain forest to see how many were being seriously attacked by insects. An investigation of 250 randomly selected trees of each species gave the following results (Source: Palaniswamy and Palaniswamy, 2006 ):



Species

Seriously damaged Not damaged Total

Yellow

58

192

250

Short leaved

80

170

250

Spruce

78

172

250



Are insects attacking one of the species more than the others?

31) In an experiment, the following contingency table was obtained. We want to test the association between row and column classification using Yates' correction for continuity.



Variable 1 \ Variable 2 Column 1 Column 2

Row 1

7

34

Row 2

16

30



23





3.11 LINEAR REGRESSION

32) In an investigation into the interdependence of water uptake and food intake in egg production, the following records were obtained from a ten day period of observation of 12 birds (hypothetical data). Assume that regression analysis assumptions are met.



Water uptake and food intake in (g) per bird/day

Egg production (eggs/10days)

265

2

492

5

366

6

514

9

406

6

336

7

343

4

377

6

143

0

204

0

360

2

292

0



a) Express this relation in a scatter diagram and add the best fitting line.

b) What is the equation of the least-squares line?

c) Calculate the predicted egg production when x=450.

d) Compute the coefficient of determination.

e) What percentage of the variation in y is explained by the regression line?

f) What percentage of the variation in y is not explained by the regression line?

g) Use a 1 % level of significance to test the claim that : 𝜌 ≠ 0.

h) Use a 1 % level of significance to test the claim that : 𝛽 ≠ 0.

33) In an investigation of the relationship between the amount of addition and chickens’ weight, the following results were obtained using SPSS.

a) Express the equation of the line of regression.

b) Find the estimated 𝑦 for 𝑥 = 21.

c) Test the hypothesis 𝛽 = 0 against : 𝛽 ≠ 0, use ; 𝛼 = 0.01.

d) What percentage of the variation in y is explained by the regression line?





24





Model Summary


Model


R

R Square

Adjusted R

Std. Error of

Square

the Estimate

1

0.937a

0.878

0.848

9.12871

a. Predictors: (Constant), food amount





Coefficientsa


Model


Unstandardized Coefficients

Standardized

t

Sig.

Coefficients

B

Std. Error

Beta

1

(Constant)

487.333

29.307



16.629

0.000

food amount

8.000

1.491

0.937

5.367

0.006

a. Dependent Variable: chicken weight





25





4 ANSWERS TO EXERCISES



4.1 DESCRIPTIVE STATISTICS

1)

a) Histogram:



b) By using the formulas given in sections 1.2.3 and 1.2.4, the following can be obtained:

𝑥̅ = 230.08; 𝑠2 = 723.69; 𝑠 = 26.9.

2) By using the formulas given in section 1.1.2, the following can be obtained: a) 125.7; from 2010 to 2013 there is 25.7 percent increase.

b) V2008 = 104; V2009 = 104; V2010 = 108.8; V2011 = 111.4; V2012 = 105.6; V2013 = 106.9; I = 106.7; this means that from 2009 to 2010 there is an 8.8 percent increase.

3) By using the formulas given in section 1.1.2, the following can be obtained: Year

2011

2012

2013

2014

2015

Fixed base index numbers (2014)

95.3

111.7

126.9

100

146.6

4) By using the formula given in section 1.2.3, it follows that:

ℎ = 106.4.

5) By using the formula given in section 1.2.3, it follows that:

𝑥̅ = 110.

6) Using the formulas given in sections 1.2.2, 1.2.3 and 1.2.4, we get: a) 𝑥̅ = 27.685; 𝑠 = 9.826.

b) 𝑚𝑒 = 28.3; 𝑞1 = 19.45; 𝑞3 = 32.35; 𝑞𝑟 = 12.90.

c) Box-plot:



26





7) Using the formulas given in sections 1.2.2, 1.2.3 and 1.2.4, we obtain:

a) 𝑥̅ = 11.0285; 𝑠 = 0.5801; 𝑚𝑒 = 11.115.

b) Box-plot:





c) 10.78.

4.2 NORMAL DISTRIBUTION

𝑿−𝑴

8) If we define a new continuous random variable 𝒁, in terms of 𝑿 as 𝒁 =

and used that 𝑷(𝑿

𝝈

𝟏 ≤ 𝑿 ≤

𝑿𝟐) = 𝑷(𝒁𝟏 ≤ 𝒁 ≤ 𝒁𝟐), then, by considering the standard normal distribution table 2.1 and symmetry of the normal distribution, we obtain the following areas under the normal curve:

a) 𝑧 =1.5; 𝑃(𝑋 ≤ 13) = 𝑃(𝑍 ≤ 1.5) = 0.9332.

b) 𝑧1 = −6; 𝑧2 = −1; 𝑃(−2 ≤ 𝑋 ≤ 8) = 𝑃(−6 ≤ 𝑍 ≤ −1) = 0.1587.

c) 𝑧 = 0.75; 𝑃(𝑋 ≥ 11.5) = 𝑃(𝑍 ≥ 0.75) = 0.2266.

You can easily verify the above results by using the normal distribution calculator.

27





9) You can solve a) and b) by following all hints given in exercise 8):

a) 𝑧1 = −1.125; 𝑧2 = 1.375; 78. 5 %.

b) 𝑧 = −3; 𝑃(𝑋 ≤ 150) = 0.0013.

c) First convert the area of the standard normal curve to 0.01, find the closest value to 0.01 which gives you z-score of 2.33 and then insert it into the equation 𝑋 = 𝑍𝜎 + 𝑀 which equals 𝑥 = 192.64.

d) You can get the solution by following hints given in case c) and by considering the symmetry of the normal distribution: 𝑥1 = 158.32; 𝑥2 = 189.68.

10) When we sample from a normal population, then sampling distribution of 𝑋̅ is a normal distribution and the

𝑋̅−𝑀

variable 𝑍 = 𝜎 follows the standard normal curve which, by considering the standard normal distribution

√𝑛

table 2.1 and symmetry, gives :

a) 𝑧 = −1.5; 0.0668.

b) 𝑧1 = −1; 𝑧2 = 0.4; 0.4967.

c) 𝑧 = 2; 0.0228.

4.3 CONFIDENCE INTERVAL

11) By using the formulas given in section 1.3.1, it follows that:

68.42 ≤ M ≤ 71.58; 67.91 ≤ M ≤ 72.09.

You can verify the above results by using the confidence interval calculator.

12) First calculate the sample mean and sample standard deviation, then, by using the formulas given in section

1.3.1, you get:

8.9 ≤ M ≤ 11.

4.4 ONE SAMPLE T-TEST

13) The solution is obtained by calculating the sample mean and sample standard deviation and then using the formulas given in section 1.3.1:

𝐻0: 𝑀 = 62.0; 𝐻1: 𝑀 ≠ 62.0; 𝛼 = 0.05; 𝑡𝑐𝑜𝑚𝑝 = −1.663; 𝑑𝑓 = 11; 𝑡𝑐𝑟𝑖𝑡 = 2.201; the null hypothesis is retained, thus at the 5% level of significance, the population mean egg weight is not significantly different from 62.0. You can verify the obtained results by using the one sample t-test calculator.

14) By using the formulas given in the section 1.3.1 it follows that:

𝐻0: 𝑀 = 150.0; 𝐻1: 𝑀 ≠ 150.0; 𝛼 = 0.05; 𝑡𝑐𝑜𝑚𝑝 = 3.33; 𝑑𝑓 = 24; 𝑡𝑐𝑟𝑖𝑡 = 2.064; the null hypothesis is rejected, thus at the 5% level of significance, we can claim that the new variety has the higher grain yield.





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15)

a) 𝐻0: 𝑀 = 57.62.

b) 𝑡 = −0.612.

c) 𝛼 = 0.05.

d) 𝑝 = 0.548.

e) At a 5% level of significance the population mean does not differ significantly from the 57.62 mean.

4.5 INDEPENDENT SAMPLES T-TEST

16) The solution is obtained by calculating the sample mean and sample standard deviation for both varieties and then using the formulas given in the section 1.3.2:

𝐻0: 𝑀1 = 𝑀2; 𝐻1: 𝑀1 ≠ 𝑀2; 𝛼 = 0.05; 𝑡𝑐𝑜𝑚𝑝 = −0.294; df = 8; 𝑡𝑐𝑟𝑖𝑡 = 2.306; we cannot reject the null hypothesis, thus we conclude that the varieties do no differ significantly with respect to grain yield. You can check the obtained results by using the t-test calculator (choose a unpaired t-test).

17) Apply the formulas given in section 1.3.2 and you get the following:

𝐻0: 𝑀1 = 𝑀2; 𝐻1: 𝑀1 ≠ 𝑀2; 𝛼 = 0.05; 𝑡𝑐𝑜𝑚𝑝 = 1.457; df = 89; 𝑡𝑐𝑟𝑖𝑡 = 1.99; the null hypothesis is retained, we cannot say that the chemical fertilizer produces a statistically significant higher yield.

18)

a) 𝐻0: 𝑀1=𝑀2.

b) 𝑡 = −2.364.

c) 𝛼 = 0.05.

d) 𝑝 = 0.030.

e) The yields differ significantly.

4.6 PAIRED SAMPLES T-TEST

19) First calculate the differences and then use the formulas from 1.3.2:

𝐻0: 𝑀1 − 𝑀2 = 0; 𝐻1: 𝑀1 − 𝑀2 ≠ 0; 𝛼 = 0.05; 𝑡𝑐𝑜𝑚𝑝 = −0.357; df = 7; 𝑡𝑐𝑟𝑖𝑡 = 2.365; we

cannot

conclude that a significant difference exists. You can verify the obtained results by using the t-test calculator

(choose a paired t-test).

20)

a) 𝐻0: 𝑀1 − 𝑀2 = 0

b) 𝑡 = −2.63.

c) 𝛼 = 0.05.

d) 𝑝 = 0.034.

e) There is a statistical difference between standing and supine blood pressures



29





4.7 ANALYSIS OF VARIANCE

21) Apply the formulas given in section 1.3.3, and you get the following:

𝐻0: 𝑀1 = 𝑀2 = 𝑀3; 𝐻1: 𝑀𝑖 ≠ 𝑀𝑗 for some 𝑖, 𝑗 ∈ {1,2,3}; 𝛼 = 0.05; 𝐹𝑐𝑜𝑚𝑝 = 0.496; 𝑑𝑓1 = 2; 𝑑𝑓2 =

12; 𝐹𝑐𝑟𝑖𝑡 = 3.885; we cannot reject the null hypothesis, thus we conclude that the varieties do not differ significantly with respect to grain yield. You can verify the obtained results by using the one-way ANOVA

calculator.





22)

a) 𝐻0: 𝑀1 = 𝑀2 = 𝑀3 = 𝑀4.

b) 𝐹 = 14.218.

c) 𝛼 = 0.05.

d) 𝑝 = 0.000.

e) We have evidence at 𝛼 = 0.05 that at least two means for weight loss differ significantly.

4.8 PEARSON' S CORRELATION COEFFICIENT

23) Apply the formula given in section 1.3.4, and you get the following:

𝐻0: 𝜌 = 0; 𝐻1: 𝜌 ≠ 0; 𝛼 = 0.05; 𝑟𝑥𝑦 = 0.726; 𝑟𝑐𝑟𝑖𝑡 = 0.754; we cannot reject the null hypothesis, at a 5%

level of significance, the evidence is not strong enough to indicate any correlation between the systolic and diastolic blood pressures. The above can be checked by using the Pearson correlation coefficient calculator.



Scatter diagram:



24) 𝐻0: 𝜌 = 0; 𝐻1: 𝜌 ≠ 0; 𝛼 = 0.01; 𝑟𝑥𝑦 = 0.5; 𝑟𝑐𝑟𝑖𝑡 = 0.479; at 1% level of significance we can reject the null hypothesis and conclude that a statistically significant positive linear correlation exists in the population.

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25)

a) 𝐻0: 𝜌 = 0.

b) 𝑟𝑥𝑦 = 0.896.

c) 𝛼 = 0.05.

d) 𝑝 = 0.000.

e) At 𝛼 = 0.05 (and also 𝛼 = 0.001) we can claim that a significant positive linear correlation between the variables exists.

4.9 SPEARMAN' S CORRELATION COEFFICIENT

26) Convert the raw data in each variable into ranks and apply the formula given in section 1.3.4 and you get the following:

𝐻0: 𝜌𝑆 = 0 ; 𝐻1: 𝜌𝑆 ≠ 0; 𝛼 = 0.05; 𝑟𝑠 = 0.937 𝑟𝑐𝑟𝑖𝑡 = 0.587; at 5 % level of significance we can reject the null hypothesis and conclude that a statistically significant positive correlation exists between the maximum monthly averaged daily temperature and monthly averaged sunshine hours. The above can be checked by using the Spearman rank-order correlation coefficient calculator.

27) By applying the formula given in section 1.3.4 , you get the following:

𝐻0: 𝜌𝑆 = 0 ; 𝐻1: 𝜌𝑆 ≠ 0; 𝛼 = 0.05; 𝑟𝑠 = 0.903; 𝑟𝑐𝑟𝑖𝑡 = 0.648; at 5 % level of significance we can reject the null hypothesis and conclude that there is convincing evidence of agreement.

28)

a) 𝐻0: 𝜌𝑆 = 0.

b) 𝑟𝑆 = 0.672.

c) 𝛼 = 0.01.

d) 𝑝 = 0.000.

e) At 𝛼 = 0.01 (also 𝛼 = 0.001) we can claim that there is a significant positive correlation between the variables. It turns out that an increase of agricultural land in use affects the increase in gross value added.

4.10 CHI-SQUARED TEST

29) By applying the formulas from 1.3.4, you obtain:

𝐻0: In this population, the two categorical variables are independent; 𝐻1: In this population, the two categorical variables are dependent;𝛼 = 0.05; χ2

2

𝑐𝑜𝑚𝑝 = 125.024; 𝑑𝑓 = 6; χ𝑐𝑟𝑖𝑡 = 12.592; a sale of different

varieties of apples and different properties are dependent at level 0.05. You can use chi-squared test

calculator to check the answer.





31





30) By applying the formulas from section 1.3.4, you obtain:

𝐻0: In the population, the two categorical variables are independent; 𝐻1: In this population, the two categorical variables are dependent; 𝛼 = 0.05; χ2

2

𝑐𝑜𝑚𝑝 = 5.774; 𝑑𝑓 = 2; χ𝑐𝑟𝑖𝑡 = 5.991; at a 0.05 level of

significance we cannot claim that insects are attacking one of the species more than the others.

31) By applying the formula from section 1.3.4, you obtain:

𝐻0: In the population, the two categorical variables are independent; 𝐻1: In this population, the two categorical variables are dependent; 𝛼 = 0.05; χ2

2

𝑐𝑜𝑚𝑝 = 2.645; 𝑑𝑓 = 1; χ𝑐𝑟𝑖𝑡 = 3.841; at a 0.05 level of

significance we cannot claim that variables are dependent. You can verify the results by applying a 2x2

contingency table calculator.



4.11 LINEAR REGRESSION

32)

a) Scatter diagram:



By applying the formulas from section 1.3.5, the following can be obtained:

b) 𝑌̂ = 0.023 ∙ 𝑋 − 3.990.

c) 𝑥 = 450; 𝑦̂ = 6.36.

d) 𝑟2

𝑥𝑦 = 0.653.

e) 65.3 %.

f) 34.7 %.

g) 𝐻1: 𝜌 ≠ 0; 𝛼 = 0.01; 𝐹𝑐𝑜𝑚𝑝 = 18.831; 𝑑𝑓1 = 1; 𝑑𝑓2 = 10; 𝐹𝑐𝑟𝑖𝑡 = 10.04; we can accept the hypothesis that 𝜌 ≠ 0, thus, water intake linearly increases the eggs production.

h) 𝐻1: 𝛽 ≠ 0; 𝛼 = 0.01; 𝑡𝑐𝑜𝑚𝑝 = 4.34; 𝑑𝑓 = 10; 𝑡𝑐𝑟𝑖𝑡 = 3.169; we can accept the hypothesis that 𝛽 ≠ 0, thus, water intake linearly increases the eggs production.

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33)

a) 𝑌̂ = 8.00 ∙ 𝑋 + 487.33.

b) 𝑥 = 21; 𝑦̂ = 655.33.

c) 𝐻1: 𝛽 ≠ 0; 𝛼 = 0.01; 𝑡𝑐𝑜𝑚𝑝 = 5.367; we can accept the hypothesis that 𝛽 ≠ 0, we can thus conclude that the amount of addition linearly increases the chickens’ weight.

d) 87.8 %.





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5 REFERENCES



Brase, C.H., Brase, C.P. Understandable statistics (8th ed.). Boston, Houghton Mifflin, 2006.



Graphpad.com. (2018). Analyze a 2x2 contingency table. [online] Available at:

https://www.graphpad.com/quickcalcs/contingency1.cfm [Accessed 8 May 2018].



Graphpad.com. (2018). One sample t test. [online] Available at:

https://www.graphpad.com/quickcalcs/oneSampleT1/ [Accessed 8 May 2018].



Graphpad.com. (2018). t test calculator. [online] Available at:

https://www.graphpad.com/quickcalcs/ttest1.cfm [Accessed 8 May 2018].



Mathportal.org. (2018). Normal distribution calculator. [online] Available at:

https://www.mathportal.org/calculators/statistics-calculator/normal-distribution-calculator.php [Accessed 8 May 2018].



Mathsisfun.com. (2018). Confidence interval calculator. [online] Available at:

https://www.mathsisfun.com/data/confidence-interval-calculator.html [Accessed 8 May 2018].



Metoffice. gov.uk. (2018) [online] Available at:

https://www.metoffice.gov.uk/public/weather/climate/gcpn7mp10 [Accessed 8 May 2018].



Palaniswamy, U.R., Palaniswamy, K.M. Handbook of statistics for teaching and research in plant and crop science.

Binghamton, NY: The Haworth Press, Inc, 2006.



Shayib, M.A. Apllied Statistics 1 st edition. (2013). [online] Available at:

http://thuvienso.bvu.edu.vn/bitstream/TVDHBRVT/15780/1/Applied-Statistics.pdf [Accessed 8 May 2018].



Sheskin, D.J. Handbook of parametric and nonparametric statistical procedures (2nd ed.). Boca Raton, Chapman & Hall /CRC, 2000.



Slovenian Environment Agency – ARSO. (2018) [online] Available at:

http://www.arso.gov.si/en/ [Accessed 8 May 2018].



Socscistatistics.com. (2018). Chi-square test calculator. [online] Available at:

http://www.socscistatistics.com/tests/chisquare2/Default2.aspx [Accessed 8 May 2018].



Socscistatistics.com. (2018). One-way ANOVA calculator. [online] Available at:

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http://www.socscistatistics.com/tests/anova/default2.Aspx [Accessed 8 May 2018].



Socscistatistics.com. (2018). Pearson correlation coefficient calculator. [online] Available at:

http://www.socscistatistics.com/tests/pearson/Default2.aspx [Accessed 8 May 2018].



Trpin Šikart, D. Analysis of the results of agricultural holdings in Slovenia processed according to FADN methodology. Master thesis, Faculty of agriculture and life sciences, 2016.



Vassarstats.net. (2018). Spearman rank-order correlation coefficient. [online] Available at:

http://vassarstats.net/corr_rank.html [Accessed 8 May 2018].





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6 INDEX



analysis of variance, 5, 20, 29

Pearson's chi-square test, 7

angle, 1

Pearson's correlation coefficient, 6, 21, 30

average of relatives, 1

percentage, 1

chain index, 1

proportion, 1

chi-square distribution, 12

quantiles, 2

chi-squared test, 23, 31

rank, 2

class width, 2

relative rank, 2

comulative frequency, 2

sample geometric mean, 2

confidence interval, 4, 17, 28

sample harmonic mean, 2

F-distribution, 11

sample interquartile range, 3

fixed base index, 1

sample mean, 2

frequency density, 2

sample median, 2

growth rate, 1

sample mode, 2

independent samples t-test, 4, 18, 29

sample quartile deviation, 3

linear regression, 8, 24, 32

sample range, 3

mean absolute deviation around a central point, 3

Spearman's correlation coefficient, 6, 21, 31

normal distribution, 9, 16, 27

standard deviation, 3

one sample t-test, 4, 17, 28

t-distribution, 10

outliers, 3

variance, 3

paired samples t-test, 4, 19, 29

Yates correction, 7

36