Summaries

Session 1.2

24^th August 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

Revise for Fall 2009

The Bowl

Color	Count	Proportion	Percent
Blue	8	8/20 = 0.4	100*.40 = 40
Green	3	3/20 = 0.15	100*.15 = 15
Red	2	2/20 = 0.1	100*.10 = 10
Yellow	7	7/20 = 0.35	100*.35 = 35
Total	20	20/20 = 1	100*1 = 100

Probabilities with Long Run Interpretation

P_Blue = 8/20 = .40

In long runs of draws with replacement from the bowl, approximately 40% of draws show blue.

P_Green = 3/20 = .15

In long runs of draws with replacement from the bowl, approximately 15% of draws show green.

P_Red = 2/2= .10

In long runs of draws with replacement from the bowl, approximately 10% of draws show red.

P_Yellow = 7/20 = .35

In long runs of draws with replacement from the bowl, approximately 35% of draws show yellow.

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

Color	Count	Proportion	Percent
Blue	8	8/20 = 0.4	100*.40 = 40
Green	3	3/20 = 0.15	100*.15 = 15
Red	2	2/20 = 0.1	100*.10 = 10
Yellow	7	7/20 = 0.35	100*.35 = 35
Total	20	20/20 = 1	100*1 = 100

Color	E50	E100	E150	E200	E250	E300
Blue	50*.40 = 20	100*.40 = 40	150*.40 = 60	200*.40 = 80	250*.40 = 100	300*.40 =120
Green	50*.15 = 7.5	100*.15 = 15	150*.15 = 22.5	200*.15 = 30	250*.15 = 37.5	300*.15 = 45
Red	50*.10 = 5	100*.10 =10	150*.10 = 15	200*.10 = 20	250*.10 = 25	300*.10 = 30
Yellow	50*.35 =17.5	100*.35 = 35	100*.35 = 52.5	200*.35 = 70	250*.35 = 87.5	300*.35 = 105
Total	50	100	150	200	250	300

Perfect Samples 

n=50

E50_Blue  = n*P_Blue= 50*(8/20) = 20

E50_Green  = n*P_Green = 50*(3/20) = 7.5

E50_Red = n*P_Red = 50*(2/20) = 5

E50_Yellow = n*P_Yellow = 50*(7/20) = 17.5

In samples of 50 draws with replacement from the bowl, we expect approximately 20 blue draws, 7 or 8 green draws, 5 red draws, and 17 or 18 yellow draws.

n=100

E100_Blue  = n*P_Blue= 100*(8/20) = 40

E100_Green  = n*P_Green = 100*(3/20) = 15

E100_Red = n*P_Red = 100*(2/20) = 10

E100_Yellow = n*P_Yellow = 100*(7/20) = 35

In samples of 100 draws with replacement from the bowl, we expect approximately 40 blue draws, 15 green draws, 10 red draws, and 35 yellow draws.

n=150

E150_Blue  = n*P_Blue= 150*(8/20) = 60

E150_Green  = n*P_Green = 150*(3/20) = 22.5

E150_Red = n*P_Red = 150*(2/20) = 15

E150_Yellow = n*P_Yellow = 150*(7/20) = 42.5

In samples of 150 draws with replacement from the bowl, we expect approximately 60 blue draws, 22 or 23 green draws, 15 red draws, and 42 or 43 yellow draws.

n=200

E200_Blue  = n*P_Blue= 200*(8/20) = 80

E200_Green  = n*P_Green = 200*(3/20) = 30

E200_Red = n*P_Red = 200*(2/20) = 20

E200_Yellow = n*P_Yellow = 200*(7/20) = 70

In samples of 200 draws with replacement from the bowl, we expect approximately 80 blue draws, 30 green draws, 20 red draws, and 70 yellow draws.

n=250

E250_Blue  = n*P_Blue= 250*(8/20) = 100

E250_Green  = n*P_Green = 250*(3/20) = 37.5

E250_Red = n*P_Red = 250*(2/20) = 25

E250_Yellow = n*P_Yellow = 250*(7/20) = 87.5

In samples of 250 draws with replacement from the bowl, we expect approximately 100 blue draws, 37 or 38 green draws, 25 red draws, and 87 or 88 yellow draws.

n=300

E300_Blue  = n*P_Blue= 300*(8/20) = 120

E300_Green  = n*P_Green = 300*(3/20) = 45

E300_Red = n*P_Red = 300*(2/20) = 30

E300_Yellow = n*P_Yellow = 300*(7/20) = 105

In samples of 300 draws with replacement from the bowl, we expect approximately 120 blue draws, 45 green draws, 30 red draws, and 105 yellow draws.

Samples – 6.30

	#1				#2				Pooled 12
Color	E50	n	p	%	E50	n	p	%	E100	n	p	%
Blue	20	18	0.36	36	20	22	0.44	44	40	40	0.4	40
Green	7.5	5	0.1	10	7.5	9	0.18	18	15	14	0.1	14
Red	5	11	0.22	22	5	2	0.04	4	10	13	0.1	13
Yellow	17.5	16	0.32	32	18	17	0.34	34	35	33	0.3	33
Total	50	50	1	100	50	50	1	100	100	100	1	100
	#3				#4				Pooled 34
Color	E50	n	p	%	E50	n	p	%	E100	n	p	%
Blue	20	24	0.48	48	20	23	0.46	46	40	47	0.5	47
Green	7.5	9	0.18	18	7.5	8	0.16	16	15	17	0.2	17
Red	5	6	0.12	12	5	4	0.08	8	10	10	0.1	10
Yellow	17.5	11	0.22	22	18	15	0.3	30	35	26	0.3	26
Total	50	50	1	100	50	50	1	100	100	100	1	100
	#5				#6				Pooled 56
Color	E50	n	p	%	E50	n	p	%	E100	n	p	%
Blue	20	22	0.44	44	20	19	0.38	38	40	41	0.4	41
Green	7.5	4	0.08	8	7.5	8	0.16	16	15	12	0.1	12
Red	5	6	0.12	12	5	3	0.06	6	10	9	0.1	9
Yellow	17.5	18	0.36	36	18	20	0.4	40	35	38	0.4	38
Total	50	50	1	100	50	50	1	100	100	100	1	100
	Pooled 135				Pooled 246
	E150				E150				Pooled All
Color		n	p	%		n	p	%	E300	n	p	%
Blue	60	64	0.43	42.67	60	64	0.43	42.67	120	128	0.43	42.67
Green	22.5	18	0.12	12.00	23	25	0.17	16.67	45	43	0.14	14.33
Red	15	23	0.15	15.33	15	9	0.06	6.00	30	32	0.11	10.67
Yellow	52.5	45	0.30	30.00	53	52	0.35	34.67	105	97	0.32	32.33
Total	150	150	1	100	150	150	1	100	300	300	1	100

Samples – 8.00

	#1				#2				Pooled 12
Color	E50	n	p	%	E50	n	p	%	E100	n	p	%
Blue	20	17	0.34	34	20	22	0.44	44	40	39	0.4	39
Green	7.5	8	0.16	16	7.5	8	0.16	16	15	16	0.2	16
Red	5	6	0.12	12	5	6	0.12	12	10	12	0.1	12
Yellow	17.5	19	0.38	38	18	14	0.28	28	35	33	0.3	33
Total	50	50	1	100	50	50	1	100	100	100	1	100
#3
Color	E50	n	p	%	E50	n	p	%	E100	n	p	%
Blue	20	17	0.34	34	20	17	0.34	34	40	34	0.3	34
Green	7.5	5	0.1	10	7.5	5	0.1	10	15	10	0.1	10
Red	5	6	0.12	12	5	7	0.14	14	10	13	0.1	13
Yellow	17.5	22	0.44	44	18	21	0.42	42	35	43	0.4	43
Total	50	50	1	100	50	50	1	100	100	100	1	100
#5
Color	E50	n	p	%	E50	n	p	%	E100	n	p	%
Blue	20	22	0.44	44	20	17	0.34	34	40	39	0.4	39
Green	7.5	7	0.14	14	7.5	4	0.08	8	15	11	0.1	11
Red	5	3	0.06	6	5	7	0.14	14	10	10	0.1	10
Yellow	17.5	18	0.36	36	18	22	0.44	44	35	40	0.4	40
Total	50	50	1	100	50	50	1	100	100	100	1	100
Pooled 135	E150				E150
Color		n	p	%		n	p	%	E300	n	p	%
Blue	60	56	0.37	37.33	60	56	0.37	37.33	120	112	0.37	37.33
Green	22.5	20	0.13	13.33	23	17	0.11	11.33	45	37	0.12	12.33
Red	15	15	0.10	10.00	15	20	0.13	13.33	30	35	0.12	11.67
Yellow	52.5	59	0.39	39.33	53	57	0.38	38.00	105	116	0.39	38.67
Total	150	150	1	100	150	150	1	100	300	300	1	100

Pooled across Sessions (6.30 + 8.00)

	Pooled 300 (Samples 135x2)		Pooled 300 (Samples 246x2)		Pooled 600 (All 12 Samples)
	n	p	n	p	n	p
Blue	120	0.4 (versus .40)	120	0.4 (versus .40)	240	0.4 (versus .40)
Green	38	0.1267 (versus .15)	42	0.14 (versus .15)	80	0.133 (versus .15)
Red	38	0.1267 (versus .10)	29	0.097 (versus .10)	67	0.112 (versus .10)
Yellow	104	0.3467 (versus .35)	109	0.363 (versus .35)	213	0.355 (versus .35)
Total	300	1	300	1	600	1

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.