Summaries

Session 0.1

26^th May 2010

Course Overview

Session 1.1

Sampling a Simple Population

We use random sampling to estimate an empirical model of a population. We check the empirical model by direct inspection of the population. 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.

Random sampling is the basis for obtaining information in statistical activities. Sampling is necessary, tedious, time consuming and expensive. Random sampling incorporates reliability, precision and uncertainty.

In this session, we begin the study of probability. We begin with a very basic example of a population, and explore the process of sampling a population.

In our first case, we begin with a fair, six-sided die. We track predicted and observed face values in six random samples of 50 tosses of the die. We then compare our samples to what is expected under the fair model.

We examine two modes of sampling a population: census (total enumeration), in which every member of the population is examined; and random sampling with replacement (SRS/WR), in which single members are repeatedly selected from the population. One practical reason why we would want a sampling process is that we wish to estimate some property of the population. Total enumeration allows a definitive settling of the question, and random sampling allows an approximate answer. In most practical settings, the populations of interest are too difficult to totally enumerate – the population is too large, or too complex, or cannot be accessed in total. In practical applications, it is sufficient (and usually necessary) to use a suitable random sample in lieu of the total population.

In our second case, we begin with a color bowl whose true color frequencies are known. We compute a population frequency model and then compute the expected structure for random samples from that bowl. We obtain six (6) random samples, each consisting of 50 draws with replacement (SRS/WR). We then compute sample color frequencies and compare them to the true structure of the bowl.

We then explore a bit of decision theory by playing with Ellsberg’s Urns.

Prediction and Probabilistic Randomness: Predicting the Behavior of a Six-sided Die

Process

We have a fair, six-sided die, with face values 1, 2, 3, 4, 5 and 6. Prior to each toss of the die, a member of the group predicts the face value that will be observed on that toss. Upon tossing the die, the group notes the observed face value, as well as the correctness (or lack thereof) of the prediction. Each group produces a sample of 50 tosses.

Sample Worksheet

Prediction and the Fair Die

Sample Grid n=50

Each cell corresponds to a single toss of the die.

X
X
X
X
X	X	X	X	X	X	X	X	X	X

Sample results are tabulated in the form below.

Face Value	Count	Prediction	Count
1		Hit(Correct)
2		Miss(Incorrrect)
3		Total
4
5
6
Total

Samples – Face Values and Predictions

Here are the results for our six samples. You should be able to begin with the counts in the table and work out the proportions and percentages.

	#1			#2			Pooled12			Model
Face Value	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent	Probability	Percent
1	8	8/50 = 0.16	100*.16=16	10	0.2	20	8+10=18	0.18	18	1/6≈0.167	16.667
2	8	8/50 =0.16	100*.16=16	17	0.34	34	8+17=25	0.25	25	1/6≈0.167	16.667
3	13	13/50 = 0.26	100*.26=26	6	0.12	12	13+6=19	0.19	19	1/6≈0.167	16.667
4	5	5/50 = 0.1	100*.10=10	3	0.06	6	5+3=8	0.08	8	1/6≈0.167	16.667
5	6	6/50 = 0.12	100*.12=12	7	0.14	14	6+7=13	0.13	13	1/6≈0.167	16.667
6	10	10/50 = 0.2	100*.20=20	7	0.14	14	10+7=17	0.17	17	1/6≈0.167	16.667
Total	50	1	100	50	1	100	100	1	100	1.000	100.000
Prediction	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent
Hit(Correct)	8	8/50 =0.16	100*.16=16	7	0.14	14	15	0.15	15	1/6≈0.167	16.667
Miss(Incorrect)	42	42/50=0.84	100*.84=84	43	0.86	86	85	0.85	85	5/6≈0.833	83.333
Total	50	1	100	50	1	100	100	1	100	1.000	100.000
	#3			#4			Pooled34			Model
Face Value	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent	Probability	Percent
1	12	0.24	24	8	0.16	16	20	0.2	20	0.167	16.667
2	7	0.14	14	8	0.16	16	15	0.15	15	0.167	16.667
3	9	0.18	18	9	0.18	18	18	0.18	18	0.167	16.667
4	4	0.08	8	11	0.22	22	15	0.15	15	0.167	16.667
5	7	0.14	14	8	0.16	16	15	0.15	15	0.167	16.667
6	11	0.22	22	6	0.12	12	17	0.17	17	0.167	16.667
Total	50	1	100	50	1	100	100	1	100	1.000	100.000
Prediction	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent
Hit	9	0.18	18	7	0.14	14	16	0.16	16	0.167	16.667
Miss	41	0.82	82	43	0.86	86	84	0.84	84	0.833	83.333
Total	50	1	100	50	1	100	100	1	100	1.000	100.000
	#5			#6			Pooled56			Model
Face Value	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent	Probability	Percent
1	10	0.2	20	9	0.18	18	19	0.19	19	0.167	16.667
2	7	0.14	14	8	0.16	16	15	0.15	15	0.167	16.667
3	8	0.16	16	11	0.22	22	19	0.19	19	0.167	16.667
4	9	0.18	18	5	0.1	10	14	0.14	14	0.167	16.667
5	9	0.18	18	9	0.18	18	18	0.18	18	0.167	16.667
6	7	0.14	14	8	0.16	16	15	0.15	15	0.167	16.667
Total	50	1	100	50	1	100	100	1	100	1.000	100.000
Prediction	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent
Hit	12	0.24	24	8	0.16	16	20	0.2	20	0.167	16.667
Miss	38	0.76	76	42	0.84	84	80	0.8	80	0.833	83.333
Total	50	1	100	50	1	100	100	1	100	1.000	100.000
	Pooled135			Pooled246			PooledAll			Model
Face Value	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent	Probability	Percent
1	8+12+10=30	30/150 =0.200	20.000	27	0.180	18.000	30+27=57	57/300=0.190	19.000	0.167	16.667
2	8+7+7=22	22/150=0.147	14.667	33	0.220	22.000	22+33=55	55/300=0.183	18.333	0.167	16.667
3	13+9+8=30	30/150=0.200	20.000	26	0.173	17.333	30+33=56	56/300=0.187	18.667	0.167	16.667
4	5+4+9=18	18/150=0.120	12.000	19	0.127	12.667	18+19=37	37/300=0.123	12.333	0.167	16.667
5	6+7+9=22	22/150=0.147	14.667	24	0.160	16.000	22+24=46	46/300=0.153	15.333	0.167	16.667
6	10+11+7=28	28/150=0.187	18.667	21	0.140	14.000	28+21=49	49/300=0.163	16.333	0.167	16.667
Total	50+50+50=150	1.000	100.000	150	1.000	100.000	150+150=300	1.000	100.000	1.000	100.000
Prediction	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion	Percent
Hit	8+9+12=29	29/150=0.193	19.333	22	0.147	14.667	29+22=51	51/300=0.170	17.000	0.167	16.667
Miss	42+41+38=121	121/150=0.807	80.667	128	0.853	85.333	121+128=249	249/300=0.830	83.000	0.833	83.333
Total	150	1.000	100.000	150	1.000	100.000	150+150=300	1.000	100.000	1.000	100.000

In the fair die model for this case, in long runs of tosses of the die: approximately 16⅔% of tosses show “1”, approximately 16⅔% of tosses show “2”, approximately 16⅔% of tosses show “3”, approximately 16⅔% of tosses show “4”, approximately 16⅔% of tosses show “5”, and approximately 16⅔% of tosses show “6.” The sample data are generally compatible with a fair die assumption (equally-likely face values) and with a baseline expected prediction success rate of (1/6), or 16⅔%. Sample performance seems to improve with increasing sample size – but the samples do not exactly fit the fair assumption.

Sample versus Fair Model

Face Value 1: 19.0% (Sample) versus 16.7% (Fair Model)

Face Value 2: 18.3% (Sample) versus 16.7% (Fair Model)

Face Value 3: 18.7% (Sample) versus 16.7% (Fair Model)

Face Value 4: 12.3% (Sample) versus 16.7% (Fair Model)

Face Value 5: 15.3% (Sample) versus 16.7% (Fair Model)

Face Value 6: 16.3% (Sample) versus 16.7% (Fair Model)

Prediction “Hit”: 17.0%(Sample) versus 16.7% (Fair Model)

Prediction “Miss”: 83.0%(Sample) versus 83.3% (Fair Model)

Case Study 1.1: A Color Bowl

In random sampling, we might get a complete list of colors - we'd need a total sample (census) for that kind of listing. The sample proportions of each listed color approximate the corresponding model proportion in the bowl itself. In census sampling, every object in the bowl is counted. The listing is complete, and the model proportions may be calculated directly.

The basic idea in case study 1.1 is that random samples give imperfect pictures of what is being sampled. However, with sufficiently large samples, these samples can reliably yield good pictures of the processes or populations being sampled. And the essence of many statistical applications is the study of selected processes or populations. For a sense of the efficiency of the samples, compare sample and true percentages.

Some Formulas – Proportions, Percentages, Counts

The class represents some property or attribute, for example, blue, or red. Each member, or unit, of a sample can be classified – the result of the classification of the unit is the unit’s class.

Sample Proportion (p)

n_class ~ number of units of sample in class

n_total ~ total number of units in sample

p_class = n_class / n_total

p_class ~ proportion of sample in class

Sample Percent (pct)

n_class ~ number of units of sample in class

n_total ~ total number of units in sample

p_class = n_class / n_total

pct_class = 100*(n_class / n_total)

pct_class = 100* p_class

pct_class ~ percent of sample in class

Population Proportion (P)

N_class ~ number of units of population in class

N_total ~ total number of units in population

P_class = N_class / N_total

P_class ~ proportion of population in class

Population Percent (PCT)

N_class ~ number of units of population in class

N_total ~ total number of units in population

P_class = N_class / N_total

PCT_class = 100*(N_class / N_total)

PCT_class = 100* P_class

PCT_class ~ percent of population in class

In this setting,

nblue ~ number of blue draws in sample ntotal ~ total number of draws per sample pblue = nblue / ntotal pblue ~ proportion of sample draws showing blue pctblue = 100*pblue pctblue ~ percent of sample draws showing blue Nblue ~ number of blue marbles in bowl Ntotal ~ total number of marbles in bowl Pblue = Nblue / Nblue Pblue ~ proportion of marbles in bowl that are blue

ngreen ~ number of green draws in sample ntotal ~ total number of draws per sample pgreen = ngreen / ngreen pgreen ~ proportion of sample draws showing green pctgreen = 100*pgreen pctgreen ~ percent of sample draws showing green Ngreen ~ number of green marbles in bowl Ntotal ~ total number of marbles in bowl Pgreen = Ngreen / Ngreen Pgreen ~ proportion of marbles in bowl that are green

nred ~ number of red draws in sample ntotal ~ total number of draws per sample pred = nred / nred pred ~ proportion of sample draws showing red pctred = 100*pred pctred ~ percent of sample draws showing red Nred ~ number of red marbles in bowl Ntotal ~ total number of marbles in bowl Pred = Nred / Nred Pred ~ proportion of marbles in bowl that are red

nyellow ~ number of yellow draws in sample ntotal ~ total number of draws per sample pyellow = nyellow / nyellow pyellow ~ proportion of sample draws showing yellow pctyellow = 100*pyellow pctyellow ~ percent of sample draws showing yellow Nyellow ~ number of yellow marbles in bowl Ntotal ~ total number of marbles in bowl Pyellow = Nyellow / Nyellow Pyellow ~ proportion of marbles in bowl that are yellow

The True State of the Bowl

Color	Count	Proportion	E50	E100	E150	E300
Blue	8	8/22=0.364	50*(8/22)=18.182	100*(8/22)=36.364	150*(8/22)=54.545	300*(8/22)=109.091
Green	6	6/22=0.273	50*(6/22)=13.636	100*(6/22)=27.273	150*(6/22)=40.909	300*(6/22)=81.818
Red	6	6/22=0.273	50*(6/22)=13.636	100*(6/22)=27.273	150*(6/22)=40.909	300*(6/22)=81.818
Yellow	2	2/22=0.091	50*(2/22)=4.545	100*(2/22)=9.091	150*(2/22)=13.636	300*(2/22)=27.273
Total	22	1	50	100	150	300

The true proportions are probabilities:

Pr{Blue Shows} = 4/11 ≈ .364

In long runs of draws with replacement from the bowl, approximately 36.4 percent of draws with replacement from the bowl show blue.

Pr{Green Shows} = 3/11 ≈ .273

In long runs of draws with replacement from the bowl, approximately 27.3 percent of draws with replacement from the bowl show green.

Pr{Red Shows} = 3/11 ≈ 273

In long runs of draws with replacement from the bowl, approximately 27.3 percent of draws with replacement from the bowl show red.

Pr{Yellow Shows} = 1/11 ≈ 091

In long runs of draws with replacement from the bowl, approximately 9.1 percent of draws with replacement from the bowl show yellow.

If we fix sample size, the probabilities imply perfect or expected counts: E=n*P:

n=50

E_Blue=50*Pr{Blue} = 50*(4/11) ≈ 18.182

E_Green=50*Pr{Green} = 50*(3/11) ≈ 13.636

E_Red=50*Pr{Red} = 50*(3/11) ≈ 13.636

E_Yellow=50*Pr{Yellow} = 50*(1/11) ≈ 4.545

In samples of 50 draws with replacement from the bowl, approximately 18 or 19 draws show blue.

In samples of 50 draws with replacement from the bowl, approximately 13 or 14 draws show green.

In samples of 50 draws with replacement from the bowl, approximately 13 or 14 draws show red.

In samples of 50 draws with replacement from the bowl, approximately 4 or 5 draws show yellow.

n=100

E_Blue=100*Pr{Blue} = 100*(4/11) ≈ 36.364

E_Green=100*Pr{Green} = 100*(3/11) ≈ 27.273

E_Red=100*Pr{Red} = 100*(3/11) ≈ 27.273

E_Yellow=100*Pr{Yellow} = 100*(1/11) ≈ 9.091

In samples of 100 draws with replacement from the bowl, approximately 36 or 37 draws show blue.

In samples of 100 draws with replacement from the bowl, approximately 27 or 28 draws show green.

In samples of 100 draws with replacement from the bowl, approximately 27 or 28 draws show red.

In samples of 100 draws with replacement from the bowl, approximately 9 or 10 draws show yellow.

n=150

E_Blue=150*Pr{Blue} = 150*(4/11) ≈ 54.545

E_Green=150*Pr{Green} = 150*(3/11) ≈ 40.909

E_Red=150*Pr{Red} = 150*(3/11) ≈ 40.909

E_Yellow=150*Pr{Yellow} = 150*(1/11) ≈ 13.636

In samples of 150 draws with replacement from the bowl, approximately 54 or 55 draws show blue.

In samples of 150 draws with replacement from the bowl, approximately 40 or 41 draws show green.

In samples of 150 draws with replacement from the bowl, approximately 40 or 41 draws show red.

In samples of 150 draws with replacement from the bowl, approximately 13 or 14 draws show yellow.

n=300

E_Blue=300*Pr{Blue} = 300*(4/11) ≈ 109.091

E_Green=300*Pr{Green} = 300*(3/11) ≈ 81.818

E_Red=300*Pr{Red} = 300*(3/11) ≈ 81.818

E_Yellow=300*Pr{Yellow} = 300*(1/11) ≈ 27.273

In samples of 300 draws with replacement from the bowl, approximately 109 or 110 draws show blue.

In samples of 300 draws with replacement from the bowl, approximately 81 or 82 draws show green.

In samples of 300 draws with replacement from the bowl, approximately 81 or 82 draws show red.

In samples of 300 draws with replacement from the bowl, approximately 27 or 28 draws show yellow.

Process

We have a four color bowl, with blue, green, red and yellow marbles. Prior to each draw from the bowl, the bowl is thoroughly mixed, giving each resident marble an approximately equal chance of selection. After mixing, a blind (made without looking into the bowl) draw of a single marble is made. The group notes the color of the marble, and the marble is returned to the bowl – this is sampling with replacement. The mixing makes the sampling random. Each group produces a sample of 50 tosses.

Each cell corresponds to a single draw with replacement from the bowl.

Sample Grid (n=50)

0

0

0

0

0

0

0

0

0

0

Sample results are tabulated in the form below.

Table – Draws with Replacement

Color	Count
Blue
Green
Red
Yellow
Total

Samples from the Color Bowl

Here are the six samples from our groups. You should be able to begin with the counts in the table and work out the proportions and percentages.

	#1			#2			Pooled12
Color	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion
Blue	19	19/50 =0.38	100*.38=38	18	0.36	36	19+18=37	37/100=0.37
Green	13	13/50 =0.26	100*.26=26	13	0.26	26	13+13=26	26/100=0.26
Red	14	14/50 =0.28	100*.28=28	14	0.28	28	14+14=28	28/100=0.28
Yellow	4	4/50 =0.08	100*.08=8	5	0.1	10	4+5=9	9/100=0.09
Total	50	1	100	50	1	100	100	1
	#3			#4			Pooled34
Color	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion
Blue	22	0.44	44	18	0.36	36	40	0.4
Green	11	0.22	22	12	0.24	24	23	0.23
Red	12	0.24	24	15	0.3	30	27	0.27
Yellow	5	0.1	10	5	0.1	10	10	0.1
Total	50	1	100	50	1	100	100	1
	#5			#6			Pooled56
Color	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion
Blue	18	0.36	36	19	0.38	38	37	0.37
Green	14	0.28	28	15	0.3	30	29	0.29
Red	11	0.22	22	11	0.22	22	22	0.22
Yellow	7	0.14	14	5	0.1	10	12	0.12
Total	50	1	100	50	1	100	100	1
	Pooled135			Pooled246			PooledAll
Color	Count	Proportion	Percent	Count	Proportion	Percent	Count	Proportion
Blue	19+22+18=59	59/150=0.393	39.333	55	0.367	36.667	59+55=114	114/300=0.380
Green	13+11+14=38	38/150=0.253	25.333	40	0.267	26.667	38+40=78	78/300=0.260
Red	14+12+11=37	37/150=0.247	24.667	40	0.267	26.667	37+40=77	77/300=0.257
Yellow	4+5+7=16	16/150=0.107	10.667	15	0.1	10	16+15=31	31/300=0.103
Total	50+50+50=150	1	100	150	1	100	150+150=300	300/300=1

Sample versus Population

38.0%(Sample) versus 36.4%(Population)

26.0%(Sample) versus 27.3%(Population)

25.7%(Sample) versus 27.3%(Population)

10.3%(Sample) versus 9.1%(Population)

We see reasonable, but not exact matches between the sample proportions and the probabilities (P). We see reasonable, but not exact matches between the sample and expected counts.

We didn’t get to these, so read up on Ellsberg I and Ellsberg II.

Regarding Ellsberg I

The 1^st Game: The first bowl is 50%/50% split between blue and green. The best we can do is break even, regardless of strategy. The simplest strategy involves picking one of the colors and always betting on that color.

The 2^nd Game: The second bowl is an unknown composite of red and yellow. We might be able to win this game if 1) there is a dominant color and 2) we can determine that dominant color. A simple strategy here is to pick one color and ride it for awhile. Then stop betting and check the number of winning bets. If the color being betted is losing on a regular basis, switch colors.

The 3^rd Game: This game only makes sense if the second bowl is dominant in red, bet on red – if red consistently shows, stay on the second bowl. Otherwise, either stop playing, or stick with the first bowl.

Regarding Ellsberg II

The 1^st Game: The first bowl is 20% red / 40% black / 40% white. The simplest strategy involves picking one of the colors and always betting on that color. Regardless of betting choice, there is a 40% chance of losing for the single bet, and 20% for getting kicked off the game.

The 2nd Game: The second bowl is 20% red / 80% black or white. The simplest strategy involves picking one of the colors and always betting on that color. If either white or black is sufficiently dominant, this game might be worth playing. The problem is that regardless of the possible advantage in the white/black part of the bowl, there is still a 20% chance of getting killed (permanently losing). But to detect this advantage, one is forced to pick a betting color (white or black) and spend some money.

The idea underlying the Ellsberg games is to illustrate the concept of making decisions about selected processes or populations by making decisions using random samples.