ƒ(µ1¸σ1)

μ1 μ2

ƒ(µ2¸σ2)

Biometrics for Plant ScientistsPl. Sci 547 ~ Spring 2009

StatisticsConvert information from events, occurrences, situations, or some process into numerical form.Analyze or summarize the numbers collected.Make inference or interpret statistics and make decisions or recommendations.

Why do Experiments fail?Proposing a wrong hypothesis or asking the wrong question.Incomplete consideration of the requirements needed to interpret the data.Ignorance of methods used in analyses.

Potato Yield Trial(2 irrigations & four plant spacing)

Irrigation Spacing Site A Site B Site CHigh 50 271 253 177High 40 248 251 169High 30 199 241 208High 20 192 220 237Low 50 227 207 121Low 40 191 182 144Low 30 179 179 162Low 20 141 164 182

0

50

100

150

200

250

High Low

Yie

ld

Irrigation

Think Question

175

180

185

190

195

200

205

210

215

20 30 40 50

Yie

ld

Row spacing

Think Question

0

50

100

150

200

250

A B C

Yie

ld

Sites

Think Question

0

50

100

150

200

250

High Low

Yie

ld

Irrigation

Think Question

0

50

100

150

200

250

300

High Low

Yie

ld

Irrigation

Think Question

0

50

100

150

200

250

300

Site A Site B Site C

Yie

ld

Sites

Think Question

Probability and

Statistical Parameters

Chevalier de Méré

Chevalier de Méré

Probability of a six in one throw = 1/6Probability of a six in four throws = 4x1/6

= 2/3Probability of 2x6’s in one throw = 1/36

Probability of 2x6’s in 24 throws = 24x1/36= 2/3

Chevalier de MéréWhy am I still loosing?

In 1654 de Méré posed this question to a famous mathematician called BlaisePascal.He began a series of letter communications with another mathematician called Pierre de Fermat.Between them theory formed the theory of probability.

Chevalier de Méré

Probability

Rule 1

If a trial has k likely outcomes, or possibilities, of which one, and only one, can occur, then the probability of any single individual outcome is 1/k.

Rule 1

The probability of rolling a six in a single roll of an unbiased dice is 1/6 as there are six sides on a dice.

Rule 2

If an event is satisfied by any one of a group of possible outcomes, then the probability of the event is the sum of the probabilities of each individual event.

Rule 2

Consider again rolling an unbiased dice, and wanting the probability of rolling a 1 OR a 2.

This would be 1/6 + 1/6 = 2/6 or 1/3, as the probability of a 1 is 1/6 and a 2 is 1/6.

Rule 2

The probability of a two children family having at least one boy is

P[1st a B] + P[2nd a B] = ½ + ½ = 1

Oops!

Rule 2

The probability of a two children family having at least one boy is

P[BG] + P[GB] – P[BB]

½ + ½ - ¼ = 3/4

Rule 3

In a series of independent trials, the probability that each of a specific series of events happens is the product of the probabilities of the individual events.

Rule 3

The probability of throwing two sixes on each of two unbiased dices is 1/6 x 1/6 = 1/36.

Rule 3

The probability of a two children family having at least one boy is

1 - P[GG]

1 – (½) x (½) = 3/4

Chevalier de Méré

Chevalier de Méré

Prob(not a 6) = 5/6

Prob(not a 6 in 4 throws) = (5/6)4 = 0.482

Prob(at least one 6 in 4 throws) = 1- (5/6)4 = 0.518

Chevalier de Méré

Prob(not a double 6) = 35/36

Prob(not 2x6 in 24 throws) = (35/36)24 = 0.509

Prob(at least one 2x6 in 24 throws) = 1- (35/36)24 = 0.491

Binomial Distribution

Binomial DistributionPopulations with only two possibilities (i.e. male:female, odd:even, infected:healthy).So if 52% of the worlds population are women, then 48% (by default) are men.In a random sample of n humans, what is the probability that the sample will contain 1, 2, 3, …. , n women? Often called the Bernoulli Distribution after Bernelli first published it in 1713.

Sex of children in family of size n

Consider a simple example of a childssex in a family of two children

1BB : 1BG : 1GB : 1GG

1BB : 2BG : 1GG

¼ BB : ½ BG : ¼ GG

(½)2 BB : 2(½)(½) BG : (½)2 GG

Sex of children in family of size 3GBB BGG

BBB BGB GBG GGG

BBG GGB

1 3 3 1

(½)3 3(½)2(½) 3(½)(½)2 (½)3

3 Boys 2xB & 1xG 2xG & 1xB 3 Girls

Binomial expansion

(a+b)2 = a2 + 2ab + b2

(a+b)3 = a3 + 3a2b + 3ab2 + b3

(a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Binomial expansion

(a+b)5 = a5 +5a4b +10a3b2 +10a2b3 +5ab4 +b5

What is that probability that a family of 5 children will have 2

boys and 3 girls?

10(½)2(½)3 = 10/32 = 5/16

Regularities: Pascal’s Triangle

Power Coefficients

1 = 1 1

2 = 1 2 1

3 = 1 3 3 1

4 = 1 4 6 4 1

5 = 1 5 10 10 5 1

Regularities ~ Powers

(a+b)5 = a5 +5a4b +10a3b2 +10a2b3 +5ab4 +b5

General FormIf the probability of a dwarf (short) F2plant is p and the probability of a tall

plant is q: the in a sample of n plants the probability of s short plants and t tall

plants is:

(n!)/(s!t!)psqt

(n!)/(s!t!)

[n x (n-1)x(n-2)x…x(1)] [s x (s-1) x … t x (t-1) x …][8 x 7 x 6 x 5 x 4 x 3 x 2 x 1] [6 x 5 x 4 x 3 x 2 x 1][2 x 1]

[8 x 7][2 x 1]

General Form

General Form ~ Example

If the probability of a dwarf (short) F2 plant is ¼ and the probability of a tall plant is ¾ : then in a sample of 8 plants the probability

of 2 short plants and 6 tall plants is:

(8!)/(2!6!)p2q6

28 (¼)2 (¾)6

0.31

Think Question

Test #1 ~ Question #3

Pig mastitis is a serious problem

P[Pig mastitis] = 37.35/1000 = 0.03735

Must treat whole herd if > 8% infected

Farmer Scratchings has 50 pigs

8% = more than 4 pigs

Probability of failure due to mastitis.

Prob[0] = [50!/0!50!](0.037)0(0.963)50 = 0.1492 Prob[1] = [50!/1!49!](0.037)1(0.963)49 = 0.2892 Prob[2] = [50!/2!48!](0.037)2(0.963)48 = 0.2749 Prob[3] = [50!/3!47!](0.037)3(0.963)47 = 0.1707 Prob[4] = [50!/4!46!](0.037)4(0.963)46 = 0.0778

Total 0.9617

Prob[failure] = 1 - 0.9617 = 0.0383

Test #1 ~ Question #3a

p = 0.2

p = 0.8

p = 0.5

9 classes

0

200

400

600

800

1000

1200

Yield

11 Classes

0

100

200

300

400

500

600

700

800

900

1000

Yield

n classes

Normal Distribution

Statistical Parameters

Mean

μ = x = (x1 + x2 + x3 + …. + xn) n

μ = x = ∑in (xi)/n

Standard Deviation

σ = √[∑in (xi - μ)2/n]

^ S = √[∑in (xi - x)2/(n-1)]

σ = √[∑in xi

2 – [(∑in xi)2/n]

^ S = √[∑in xi

2 – [(∑in xi)2/(n-1)]

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Distribution

μ

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Distribution

μ

σ16% 16%

. .

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Distribution

μ

2σ 2.5%2.5%

..

95%

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Distribution

μ

3σ 0.5%0.5%

.

.

99%

Variance

σ2 = ∑in (xi - μ)2/n

^ S2 = ∑in (xi - x)2/(n-1)

σ2 = ∑in xi

2 – (∑in xi)2/n

^ S2 = ∑in xi

2 – (∑in xi)2/(n-1)

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Variance Partition

σ2P

..

.

.

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Variance Partition

σ2P

σ2G

σ2GE

σ2e

..

.

.

Think Question

Think Question: Ch 1, Q1Cultivar Weed Density Non-irrigated IrrigatedStevens 1:25 2035 3610

1:20 1945 37561:15 1932 38161:10 1865 3919

Steptoe 1:25 1321 16611:20 1050 15621:15 923 15111:10 600 1423

Lambert 1:25 2678 41451:20 2421 35011:15 2220 22681:10 1967 1940

Think Question: Ch 1, Q1

0

500

1000

1500

2000

2500

3000

3500

Stevens Steptoe Lambert

Yie

ld

Think Question: Ch 1, Q1

0500

1000150020002500300035004000

Non-irrigated Irrigated

Yie

ld

Stevens Steptoe Lambert

1

1

2

2

33

Think Question: Ch 1, Q1

0

500

1000

1500

2000

2500

3000

3500

1:25 1:20 1:15 1:10

Yie

ld

Non-irrigated Irrigated

Think Question: Ch 1, Q1

0

500

1000

1500

2000

2500

3000

3500

4000

1:25 1:20 1:15 1:10

Yie

ld

Stevens Steptoe Lambert

Characteristics of the Normal Distribution