Summaries

Session 1.2

3^rd June 2009

Predicting Sample Behavior from a Model

We use a population model to predict the behavior of random samples. We check the predictions by direct inspection of samples. We repeat sampling with replacement, obtaining multiple random samples from the same population, obtained in the same process. We combine (pool) compatible samples to form larger samples. Pooling samples of size 50, we obtain samples of size 100, 150 and 300. In general, as sample size increases, samples become more precise and reliable, provided that the sampling process is reliable.

In general, if we are working with the correct model, then the predicted sample behavior reliably describes observed samples.

Session Overview

Estimation: Previously, we saw how random samples drawn from a population could be used to estimate the structure of a population as an empirical model.

Prediction: In this session, we continue our study of probability. We begin with a very basic example of a population with known structure, and use that structure to predict the behavior of random samples from that population.

We begin by constructing a probability model for our population, and define the concept of perfect sample. We then relate the perfect sample to random samples, both observed and unobserved. We then obtain real random samples, and check them against the perfect samples.

Exclude Case Study 1.2.1

Case Study 1.2.2

In this case study the idea of a perfect sample is introduced – a perfect sample matches perfectly the population from which it is sampled. On average, real samples corresponded nicely, though not perfectly, with their corresponding perfect samples.

We begin by building a color bowl. We then compute a probability model for draws with replacement(DWR) from the bowl, and then perfect samples of size 50, 100, 150, 200, 250 and 300. We then engage six groups to generate six samples each of n=50 DWR. We then compare sample frequencies and proportions to the model and to the perfect samples.

Bowl Counts and Perfect Sample Calculations

Expected Count Blue ( for Sample Size n) = n*P_Blue

Expected Count Green ( for Sample Size n) = n*P_Green

Expected Count Red ( for Sample Size n) = n*P_Red

Expected Count Yellow ( for Sample Size n) = n*P_Yellow

The Bowl

Color	N	Probability(P)	Percent
Blue	3	3/21 ≈ .1429	14.29
Green	10	10/30 ≈ .3333	33.33
Red	3	3/21 ≈ .1429	14.29
Yellow	5	5/21 ≈ .2381	23.81
Total	21	1	100

Probabilities with Long Run Interpretation

P_Blue = 3/21 ≈ .1429

In long runs of draws with replacement from the bowl, approximately 14.3% of draws show blue.

P_Green = 10/30 ≈ .3333

In long runs of draws with replacement from the bowl, approximately 33.3% of draws show green.

P_Red = 3/21 ≈ .1429

In long runs of draws with replacement from the bowl, approximately 14.3% of draws how red.

P_Yellow = 5/21 ≈ .2381

In long runs of draws with replacement from the bowl, approximately 23.8% of draws show yellow.

Perfect Counts for the Bowl – n = 50, 100, 150, 200, 250 and 300 Draws with Replacement

Color	P	E50=50*P	E100=100*P	E150=150*P	E200=200*P	E250=250*P	E300=300*P
Blue	0.1429	7.143	14.286	21.4286	28.5714	35.7143	42.857
Green	0.4762	23.81	47.619	71.4286	95.2381	119.048	142.86
Red	0.1429	7.143	14.286	21.4286	28.5714	35.7143	42.857
Yellow	0.2381	11.9	23.81	35.7143	47.619	59.5238	71.429
Total	1	50	100	150	200	250	300

Perfect Samples 

n=50

E50_Blue  = n*P_Blue= 50*(3/21) ≈ 7.1

E50_Green  = n*P_Green = 50*(10/21) ≈ 23.8

E50_Red = n*P_Red = 50*(3/21) ≈ 7.1

E50_Yellow = n*P_Yellow = 50*(5/21) ≈ 11.9

In samples of 50 draws with replacement from the bowl, we expect approximately 7 or 8 blue draws, 23 or 24 green draws, 7 or 8 red draws, and 11 or 12 yellow draws.

n=100

E100_Blue  = n*P_Blue= 100*(3/21) ≈ 14.3

E100_Green  = n*P_Green = 100*(10/21) ≈ 47.6

E100_Red = n*P_Red = 100*(3/21) ≈ 14.3

E100_Yellow = n*P_Yellow = 100*(5/21) ≈ 23.8

In samples of 100 draws with replacement from the bowl, we expect approximately 14 or 15 blue draws, 47 or 48 green draws, 14 or 15 red draws, and 23 or 24 yellow draws.

n=150

E150_Blue  = n*P_Blue= 150*(3/21) ≈ 21.4

E150_Green  = n*P_Green = 150*(10/21) ≈ 71.4

E150_Red = n*P_Red = 150*(3/21) ≈ 21.4

E150_Yellow = n*P_Yellow = 150*(5/21) ≈ 35.7

In samples of 150 draws with replacement from the bowl, we expect approximately 21 or 22 blue draws, 71 or 72 green draws, 21 or 22 red draws, and 35 or 36 yellow draws.

n=200

E200_Blue  = n*P_Blue= 200*(3/21) ≈ 28.6

E200_Green  = n*P_Green = 200*(10/21) ≈ 95.2

E200_Red = n*P_Red = 200*(3/21) ≈ 28.6

E200_Yellow = n*P_Yellow = 200*(5/21) ≈ 47.6

In samples of 200 draws with replacement from the bowl, we expect approximately 28 or 29 blue draws, 95 or 96 green draws, 28 or 29 red draws, and 47 or 48 yellow draws.

n=250

E250_Blue  = n*P_Blue= 250*(3/21) ≈ 35.7

E250_Green  = n*P_Green = 250*(10/21) ≈ 119

E250_Red = n*P_Red = 250*(3/21) ≈ 35.7

E250_Yellow = n*P_Yellow = 250*(5/21) ≈ 59.5

In samples of 250 draws with replacement from the bowl, we expect approximately 35 or 36 blue draws, 119 green draws, 35 or 36 red draws, and 59 or 60 yellow draws.

n=300

E300_Blue  = n*P_Blue= 300*(3/21) ≈ 42.9

E300_Green  = n*P_Green = 300*(10/21) ≈ 142.9

E300_Red = n*P_Red = 300*(3/21) ≈ 42.9

E300_Yellow = n*P_Yellow = 300*(5/21) ≈ 71.4

In samples of 300 draws with replacement from the bowl, we expect approximately 42 or 43 blue draws, 142 or 143 green draws, 42 or 43 red draws, and 71 or 72 yellow draws.

Samples

#1					#2					Pooled 12
Color	E50	n	p	%	Color	E50	n	p	%	Color	E100	n	p	%
Blue	7.14	7	0.14	14	Blue	7.14	4	0.08	8	Blue	14.29	11	0.11	11
Green	23.8	21	0.42	42	Green	23.8	22	0.44	44	Green	47.62	43	0.43	43
Red	7.14	4	0.08	8	Red	7.14	7	0.14	14	Red	14.29	11	0.11	11
Yellow	11.9	18	0.36	36	Yellow	11.9	17	0.34	34	Yellow	23.81	35	0.35	35
Total	50	50	1	100	Total	50	50	1	100	Total	100	100	1	100
#3					#4					Pooled 34
Color	E50	n	p	%	Color	E50	n	p	%	Color	E100	n	p	%
Blue	7.14	8	0.16	16	Blue	7.14	6	0.12	12	Blue	14.29	14	0.14	14
Green	23.8	23	0.46	46	Green	23.8	26	0.52	52	Green	47.62	49	0.49	49
Red	7.14	7	0.14	14	Red	7.14	7	0.14	14	Red	14.29	14	0.14	14
Yellow	11.9	12	0.24	24	Yellow	11.9	11	0.22	22	Yellow	23.81	23	0.23	23
Total	50	50	1	100	Total	50	50	1	100	Total	100	100	1	100
#5					#6					Pooled 56
Color	E50	n	p	%	Color	E50	n	p	%	Color	E100	n	p	%
Blue	7.14	2	0.04	4	Blue	7.14	6	0.12	12	Blue	14.29	8	0.08	8
Green	23.8	33	0.66	66	Green	23.8	33	0.66	66	Green	47.62	66	0.66	66
Red	7.14	4	0.08	8	Red	7.14	6	0.12	12	Red	14.29	10	0.1	10
Yellow	11.9	11	0.22	22	Yellow	11.9	5	0.1	10	Yellow	23.81	16	0.16	16
Total	50	50	1	100	Total	50	50	1	100	Total	100	100	1	100
Pooled 135	E150				Pooled 246	E150				Pooled All
Color		n	p	%	Color		n	p	%	Color	E300	n	p	%
Blue	21.4	17	0.113	11.33	Blue	21.4	16	0.107	10.67	Blue	42.86	33	0.11	11
Green	71.4	77	0.513	51.33	Green	71.4	81	0.54	54	Green	142.9	158	0.527	52.67
Red	21.4	15	0.1	10	Red	21.4	20	0.133	13.33	Red	42.86	35	0.117	11.67
Yellow	35.7	41	0.273	27.33	Yellow	35.7	33	0.22	22	Yellow	71.43	74	0.247	24.67
Total	150	150	1	100	Total	150	150	1	100	Total	300	300	1	100

The structure of the bowl, expressed as color proportions, determines the basic structure of samples drawn from the bowl. The perfect sample is a blueprint for the actual samples. The actual samples show choppy resemblance to the perfect sample, with samples #3, #4, pooled 34, pooled 135, pooled 246 and pooled all showing the best overall agreement. Probability models allow the prediction of sample behavior, but said predictions are only as reliable as the validity of the original model and of the sampling procedures.

The foundation of statistical applications is the careful preparation of a study population and the random sampling procedures to go with it. Proper execution of this sampling procedure ensures a potable sample.

You are now ready to learn the Long Run Argument and Perfect Sample case types in 1^st Hourly Stuff.

Exclude Case Study 1.5

We now extend our study of probability to dice. We revisit the idea of a model or population proportion as a probability, and introduce the idea of a random variable.

Models

A Fair, Six-sided Die

Face Value, d6 (FV d6)	Probability
1	1/6
2	1/6
3	1/6
4	1/6
5	1/6
6	1/6

A Fair, Three-sided Die

Face Value, d3 (FV d3)	Probability
1	1/3
2	1/3
3	1/3

Using a Fair, Six-sided Die to Simulate A Fair, Three-sided Die

Face Value, d6 (FV d6)	Mapped Face Value, d3 (FV d3)
1	1
2
3	2
4
5	3
6

Probability Calculations (fair d6→ fair d3)

Pr{E} denotes Probability for the event E.

The Fair d6 Model

FV: Face Values: 1,2,3,4,5,6

Fair Model: Equally likely face values – 1/6 per face value

Pr{d6 Shows 1} = (1/6) @ .1667 or 16.67%

In long runs of tosses, approximately 1 toss in 6 shows “1”.

Pr{d6 Shows 2} = (1/6) @ .1667 or 16.67%

In long runs of tosses, approximately 1 toss in 6 shows “2”.

Pr{d6 Shows 3} = (1/6) @ .1667 or 16.67%

In long runs of tosses, approximately 1 toss in 6 shows “3”.

Pr{d6 Shows 4} = (1/6) @ .1667 or 16.67%

In long runs of tosses, approximately 1 toss in 6 shows “4”.

Pr{d6 Shows 5} = (1/6) @ .1667 or 16.67%

In long runs of tosses, approximately 1 toss in 6 shows “5”.

Pr{d6 Shows 6} = (1/6) @ .1667 or 16.67%

In long runs of tosses, approximately 1 toss in 6 shows “6”.

The Fair d3 Model Nested within a Fair d6 Model

FV: Face Values: 1(1,2), 2(3,4), 3(5,6)

Fair Model: Equally likely face values –  (2/6 =)1/3 per face value.

Pr{d3 shows “1”} = Pr{d6 Shows 1} + Pr{d6 Shows 2}₁ = (1/6) + (1/6) = 2/6 = 1/3 @ .3333 or 33.33%

In long runs of tosses, approximately 1 toss in 3 shows “1”.

Pr{d3 shows “2”} = Pr{d6 Shows 3} + Pr{d6 Shows 4} = (1/6) + (1/6)₂ = 2/6 = 1/3 @ .3333 or 33.33%

In long runs of tosses, approximately 1 toss in 3 shows “2”.

Pr{d3 shows “3”} = Pr{d6 Shows 5} + Pr{d6 Shows 6} = (1/6) + (1/6) = 2/6 = 1/3₃ @ .3333 or 33.33%

In long runs of tosses, approximately 1 toss in 3 shows “3”.

1. Additive Rule – Map Faces to Faces

2. Inheritance of Fair Model

3. Fair d3 Model from Fair d6 Model

In the samples, compare the sample proportions (p) to the model probabilities (P) listed above.

Samples

#1						#2
FV d6	n	p	FV d3	n	p	FV d6	n	p	FV d3	n	p
1	7	0.14				1	8	0.16
2	8	0.16	1	15	0.3	2	7	0.14	1	15	0.3
3	8	0.16				3	10	0.2
4	12	0.24	2	20	0.4	4	10	0.2	2	20	0.4
5	6	0.12				5	8	0.16
6	9	0.18	3	15	0.3	6	7	0.14	3	15	0.3
Total	50	1	Total	50	1	Total	50	1	Total	50	1
#3						#4
FV d6	n	p	FV d3	n	p	FV d6	n	p	FV d3	n	p
1	9	0.18				1	5	0.1
2	9	0.18	1	18	0.36	2	8	0.16	1	13	0.26
3	6	0.12				3	11	0.22
4	8	0.16	2	14	0.28	4	8	0.16	2	19	0.38
5	10	0.2				5	7	0.14
6	8	0.16	3	18	0.36	6	11	0.22	3	18	0.36
Total	50	1	Total	50	1	Total	50	1	Total	50	1
#5						#6
FV d6	n	p	FV d3	n	p	FV d6	n	p	FV d3	n	p
1	5	0.1				1	10	0.2
2	9	0.18	1	14	0.28	2	9	0.18	1	19	0.38
3	13	0.26				3	7	0.14
4	3	0.06	2	16	0.32	4	8	0.16	2	15	0.3
5	7	0.14				5	10	0.2
6	13	0.26	3	20	0.4	6	6	0.12	3	16	0.32
Total	50	1	Total	50	1	Total	50	1	Total	50	1
Pooled 1,3,5						Pooled 2,4,6
FV d6	n	p	FV d3	n	p	FV d6	n	p	FV d3	n	p
1	21	0.14				1	23	0.153
2	26	0.173	1	47	0.313	2	24	0.16	1	47	0.313
3	27	0.18				3	28	0.187
4	23	0.153	2	50	0.333	4	26	0.173	2	54	0.36
5	23	0.153				5	25	0.167
6	30	0.2	3	53	0.353	6	24	0.16	3	49	0.327
Total	150	1	Total	150	1	Total	150	1	Total	150	1

Pooled 1,2
FV d6	n	p	FV d3	n	p
1	15	0.15
2	15	0.15	1	30	0.3
3	18	0.18
4	22	0.22	2	40	0.4
5	14	0.14
6	16	0.16	3	30	0.3
Total	100	1	Total	100	1
Pooled 3,4
FV d6	n	p	FV d3	n	p
1	14	0.14
2	17	0.17	1	31	0.31
3	17	0.17
4	16	0.16	2	33	0.33
5	17	0.17
6	19	0.19	3	36	0.36
Total	100	1	Total	100	1
Pooled 5,6
FV d6	n	p	FV d3	n	p
1	15	0.15
2	18	0.18	1	33	0.33
3	20	0.2
4	11	0.11	2	31	0.31
5	17	0.17
6	19	0.19	3	36	0.36
Total	100	1	Total	100	1
Pooled All
FV d6	n	p	FV d3	n	p
1	44	0.147
2	50	0.167	1	94	0.313
3	55	0.183
4	49	0.163	2	104	0.347
5	48	0.16
6	54	0.18	3	102	0.34
Total	300	1	Total	300	1