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Testing seed mixtures

Júlia Barabás PhD (Hungary)

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Introduction

In year 2000 two questions were addressed by the Purity Committee to the Statistics Committee.

1. How could we compare the test result of a sample against the labelled proportion of seed mixture-lot?(= does the lot has the labelled proportion?)2. How could we compare two test results from a seed mixture-lot?(= we have made a test on our submitted sample but we or another lab had to perform another test, “are the two results compatible?” )

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Suggested methods and their mathematical background

Subgroup to discuss the possibilities by e_mail (leadership J. Barabás).After about 6 circulars and different contributions, the mathematical background was chosen and a software was prepared to address 4 different cases

1 check against label 2 compare two testsa % in number of seeds 1 a label number 2 a compare numbersb % in weight of seeds 1 b label weight 2 b compare weights

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case 1 awhen we want to compare the labelled proportion with one test result of the sample on the basis of the count of the number of seeds of different components.

In this case we could use a well known statistical test the Chi-squared test for goodness of fit. It measures the difference between the expected and the experienced frequencies.

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H o: p pi o i if i= 1 ,2 ..k , an d w h ere , p i is th e lab e lled p ro p o rtio n o f th e i-th co m p o n en t in th e w h o le lo t.

H 1 : p pi o i ex is t even o n e co m p o n en t w h ich h as n o t th e g iven p ro p o rtio n .

T h e tes t s ta tis tic :

22

1

( )f n p

n pi i o

i oi

k

,

w h ere f i is th e o b served freq u en cy o f th e i.- th co m p o n en t.R e jec tio n reg io n :

2 2: . (e .g . fo r 5% an d k= 5 th e c ritica l va lu e= 9 ,488 )

n=number of seeds tested Pi0=number of seeds of a component

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case 1 bwhen we want to compare the labelled proportion with one test result of the sample and on the basis of the weight of the different components.(we need to know or measure the 1000 seed-weights of the components).We transform the original formula and suggest to apply in this case as follows:

Ho and H1 and rejection region are the same as inthe first case.

Ho =all frequencies are as expected

H1= at least one of the frequencies is different from expected

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T h e te s t s ta t i s t ic :

21 0 0 0 0

0

0

0

2

sm

p

mpm

sm

p

mpm

sm

i

i

i

ii

i

i

i

i

ii

i

i

i

w h e re s i i s th e m e a su re d w e ig h t o f th e i - th c o m p o n e n ta n d m i i s th e 1 0 0 0 s e e d s w e ig h t o f th e i - th c o m p o n e n t ,a n d a l l th e g o e s f ro m 1 to k , i f th e re a re k d i f f e re n tc o m p o n e n ts .

Pi0 = the labelled proportion % of the i-th component

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Case 2 awhen we want to compare two test results from the same seed mixture lot on the basis of the count of the number of seeds of different components in both tests.

The base statistic is a homogeneity Chi-squared test.

Ho: F1(x)= F2(x), that is the two distributions are the same, H1 : F1(x) not = F2(x) that is the two distributions are not equal.

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The test statistic:

2 1 2

2

1

1 2

( )h

n

ln

h ln n

i i

i ii

k

where hi is the seed-number of the i-thcomponent in the first test and li is theseed-number of the i-th component in thesecond test, and n1, n2 the total seed-numbers in each test

Rejection region: 2 2: and degree of

freedom = the number of components -1.

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2 b case when we want to compare two test results from the same seed mixture lot on the basis of the weight of different components in both tests.

The base statistic is the same but we need to transform the original formula.

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2

2

11000

vivimi

wiwimi

mi vi wii

kvimi

wimi( )

Ho and H1 and rejection region are the same as incase 2/a.

Where vi is the weight of the i-th component in

the first test and wi is the weight of the i-th

component in the second test, and mi the 1000 seed weight.

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Excel spreadsheet to compute tests in seed mixtures

• In order to allow the practical use in the many possible circumstances

• When work will be validated by Statistic and Purity committee, it will be available through ISTA

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Chi 2 t est f or seed mi xt ur es check agai nst pr opor t i ons i n number of seeds

labelled proport ion (ie 0.5 for 50%)

observed number of seeds

observed proport ion expected number of seeds Chi2 contribution

component

species 1 0.15 68 0.1700 60 1.0667species 2 0.38 160 0.4000 152 0.4211species 3 0.42 153 0.3825 168 1.3393species 4 0.02 9 0.0225 8 0.1250species 5 0.03 10 0.0250 12 0.3333

6 0.0000 0 0.00007 0.0000 0 0.00008 0.0000 0 0.00009 0.0000 0 0.0000

10 0.0000 0 0.0000

sums 1 400 1 400 3.2853

degrees of freedom 4

Observed Chi2 probability 0.0505 is 5.05% 0. 5113 critical Chi2 at 5% 9.487728465 NS

critical Chi2 at 1% 13.27669856 NScritical Chi2 at 0.1% 18.46622617 NS

how t o pr oceed:step 1 click on clear data buttonstep 2 enter data , proport ions and observed number of seedsstep 3 see results if probability is less than the crit ical probability chosen the test is signif icant < = > not conform to labelNB: you can enter the name of the components

Case where seed mixture composition is checked by numbers of seeds

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C h i 2 t e s t f o r s e e d mi x t u r e s c h e c k a g a i n s t p r o p o r t i o n s i n we i g h t o f s e e d sl a b e l l e d % o f

c o m p o n e n t s ( i e 5 0 f o r 5 0 %) P i %

W e i g h t o f c o m p o n e n t s S i

(g r a m s )

W e i g h t o f 1 0 0 0 s e e d s M i (g r a m s ) P i / M i S i / M i n u m e r a t o r d e n o m i n a t o r

n u m e r a t o r / d e n o m i n a t o r

c o m p o n e n t

sp eci es x x x 2 0 5 2 0 1 . 0 0 0 0 0 . 2 5 0 0 1 . 9 8 2 0 7 9 1 8 9 0 . 2 0 5 4 7 9 4 5 2 9 . 6 4 6 1 1 8 7 2 1

sp eci es y y y 3 0 6 2 4 1 . 2 5 0 0 0 . 2 5 0 0 0 . 0 4 6 9 1 3 1 1 7 0 . 2 5 6 8 4 9 3 1 5 0 . 1 8 2 6 4 8 4 0 2

sp eci es zz z 1 0 3 1 2 0 . 8 3 3 3 0 . 2 5 0 0 6 . 2 0 4 2 5 9 7 1 1 0 . 1 7 1 2 3 2 8 7 7 3 6 . 2 3 2 8 7 6 7 1sp eci es ssss 2 0 2 . 5 1 0 2 . 0 0 0 0 0 . 2 5 0 0 2 5 . 9 0 7 7 6 8 8 1 0 . 4 1 0 9 5 8 9 0 4 6 3 . 0 4 2 2 3 7 4 4sp eci es t t t t t 2 0 5 2 0 1 . 0 0 0 0 0 . 2 5 0 0 1 . 9 8 2 0 7 9 1 8 9 0 . 2 0 5 4 7 9 4 5 2 9 . 6 4 6 1 1 8 7 2 1

6789

1 0

s u ms 1 0 0 2 1 . 5 8 6 6 . 0 8 3 3 3 3 3 3 3 1 . 2 5 0 0 3 6 . 1 2 3 1 0 0 0 2 1 . 2 5 1 1 8 . 7 5 o b s e r v e d c h i s q u a r e

d e g r e e s o f f r e e d o m 4

O b s e r v e d C h i 2 p r o b a b i l i t y 0 . 0 5 0 5 i s 5 . 0 5 % 0 . 0 0 0 0 c r it ic a l C h i2 a t 5 % 9 .4 8 7 7 2 8 4 6 5

c r it ic a l C h i2 a t 1 % 1 3 .2 7 6 6 9 8 5 6c r it ic a l C h i2 a t 0 .1 % 1 8 .4 6 6 2 2 6 1 7

h o w t o p r o c e e d :s t e p 1 c l i c k o n c l e a r d a t a b u t t o ns t e p 2 e n t e r d a t a , i n g r e e n c o l u m n s f o r y o u r c o m p o n e n t ss t e p 3 s e e r e s u l t s i f p r o b a b i l i t y i s l e s s t h a n t h e c r i t i c a l p r o b a b i l i t y c h o s e n t h e t e s t i s s i g n i f i c a n t < = > n o t c o n f o r m t o l a b e lN B : y o u c a n e n t e r t h e n a m e o f t h e c o m p o n e n t s i n t h e c o l u m n c o m p o n e n t s

c le a r d a ta

Case where seed mixture composition is checked by weight of seeds

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computation of the expected number of seeds for a given species in a seed mixture when the expected percentage is known

400 type in the number of seeds (for instance 400)

55 type in the labelled/expected percentage (for instance 55 if 55%)confidence intervals

220 average number of seeds expected alpha risk min average max delta

2.48747 theoretical standard deviation 20% 207 220 233 12.7358

10% 204 220 236 16.3178

5% 200 220 240 19.5018

you can obtain distributions at the following web adress: 1% 198 220 242 22.2877

http://www.ruf.rice.edu/~lane/stat_sim/normal_approx/index.html 0.10% 194 220 246 25.5712

this is a hard copy of what you can get

Another question : Which number of seeds shall I expect for a given labelled % and number of seeds