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Stick Tossing and Confidence Intervals

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Presentation Transcript

Stick Tossing and Confidence Intervals

David Sklar

San Francisco State University

Bruce Cohen

Lowell High School, SFUSD

http://www.cgl.ucsf.edu/home/bic

Asilomar - December 2006

Ver. 0.5

An Old Problem:

When a thin stick of unit length is “randomly” tossed onto a grid of parallel lines spaced one unit apart what is the probability that the stick lands crossing a grid line?

We would like to take a purely experimental and statistical approach to the problem of finding, or at least estimating, the desired probability.

Our experiments will consist of tossing a stick some fixed number of times, keeping track of how many times the stick lands crossing a grid line (the data), and computing the percentage of times this event occurs (a statistic).

Basic statistical theory will help us understand how to interpret these results.

Estimating a simple probability

Toss sticks, gather data

Estimating the probability

Estimating the uncertainty in

the estimate of the probability

Confidence intervals and what

they mean

Background material

The average and standard

deviation of a list of numbers

Histograms, what they are and

what they aren’t

The average and standard

deviation of a histogram

The normal curve

Box models and histograms

for the sum of the draws

The Central Limit Theorem

Where does the procedure for finding confidence intervals come from? Why does it work?

A mathematical model for the

data

The mathematics of the model

Sketch of a proof of a special case of the Central Limit Theorem

Conclusions: Based on this data an approximate 68% confidence interval for the

probability that the stick lands crossing a line is

47.3%

63.9%

an approximate 95% confidence interval is

39.0%

72.2%

Estimating the Probability: A Sample Calculation

Result: 20 line crossings in 36 tosses

63.9% confidence interval for the

47.3%

74.6%

58.8%

61.1%

44.5%

28 77.8% 6.9%

84.7%

70.9%

23 63.9% 8.0%

71.2%

55.9%

25 69.4% 7.7%

77.1%

61.7%

24 66.7% 7.9%

74.6%

58.8%

27 75.0% 7.2%

82.2%

67.8%

23 63.9% 8.0%

71.2%

55.9%

21 58.3% 8.2%

66.5%

50.1%

50%

70%

40%

60%

80%

68% Confidence Intervals for 10 Experiments

(36 tosses per experiment)

estimated

cross prob SE

20 55.6% 8.3%

24 66.7% 7.9%

19 52.8% 8.3%

62.5% confidence interval for the

67.5%

an approximate 95% confidence interval is

60.0%

70.0%

Pooling the data

Result: 234 line crossings in 360 (independent) tosses

Conclusions: Based on this data an approximate 68% confidence interval for the

probability that the stick lands crossing a line is

62.5% confidence interval for the

67.5%

68% Confidence Intervals for 10 Experiments

(36 tosses per experiment)

estimated

cross prob error

20 55.6% 8.3%

63.9%

47.3%

24 66.7% 7.9%

74.6%

58.8%

19 52.8% 8.3%

61.1%

44.5%

28 77.8% 6.9%

84.7%

70.9%

23 63.9% 8.0%

71.2%

55.9%

25 69.4% 7.7%

77.1%

61.7%

24 66.7% 7.9%

74.6%

58.8%

27 75.0% 7.2%

82.2%

67.8%

23 63.9% 8.0%

71.2%

55.9%

21 58.3% 8.2%

66.5%

50.1%

234 65.0% 2.5%

50%

70%

40%

60%

80%

Some 95% Confidence Intervals confidence interval for the

0 confidence interval for the

?

??

1

Where Does the Procedure for Finding

Confidence Intervals Come From?

As with all “real world” applications of mathematics we begin with a Mathematical Model.

Box Model

The number of line crossings in n tosses of the stick is like the Sum of values of n draws at random with replacement from a box with two kinds of numbered tickets. Those numbered 1 correspond to the stick landing crossing a line, and those numbered 0 to not crossing. The percentage of tickets numbered 1 in the box is not known.

The set oftickets in the box is called the population, and the (unknown) % of 1’s in the population is a parameter.

The n drawn tickets are a sample, and the % of 1’s in the sample is a statistic.

This unknown percentage corresponds to the probability that a stick lands crossing a line.

Note: this kind of box is called a zero–one box.

The Mathematics of the Model confidence interval for the

The goal for the rest of the talk is to develop the mathematics of the box model.

We first review some basic background material which we then use to

understand the behavior of the sum of the draws from a box of known

composition. Finally we use this understanding to see why the confidence

levels come from areas under the normal curve.

25 confidence interval for the

35

The Average and Standard Deviation of a List of Numbers

Example List: 21, 28, 30, 30, 34, 37

The SD measures the spread.

The mean measures the “center” of the list.

The average is the balance point.

The SD measures the spread of the list about the mean. It has the same units as the values in the list. It is a natural scale for the list: we are often more interested in how many SD’s a value is from the mean than in the value itself.

The Average and Standard Deviation of a List of Numbers confidence interval for the

For a list consisting of just 0’s and 1’s we have:

and with some algebra we can show that

We can now re-interpret the procedure for estimating our probability

Properties of The Average and Standard Deviation confidence interval for the

- If we add a constant, B, to each element of a
- list the average of the new list is the old average + B.

- If we multiply each element of a list by a constant,
- A, the average of the new list is A times the old average.

- If we add a constant, B, to each element of a
- list the SD of the new list is the old SD.

- If we multiply each element of a list by a constant,
- A, the SD of the new list is |A| times the old SD.

Standard Units confidence interval for the

We are often more interested in how many SD’s a value is from the mean than in the value itself. For example: 37 is 1.4 SD’s above the average or 28 is 0.4 SD’s below the average.

The value of an element in Standard Units is the the number of SD’s it is above (positive), or below (negative) the mean. To convert a value to standard units use

value in standard units

Example

List: 21, 28, 30, 30, 34, 37 with average 30 and SD 5

In Standard Units: -1.8, -0.4, 0, 0, 0.8, 1.4

A list in standard units will have mean 0 and SD 1.

Adding a constant to each element of a list or multiplying each element by a constant will not change the values of the elements in standard units.

For many lists roughly 68% of the values lie

within 1 SD of the mean and 95% lie within 2 SD’s.

density confidence interval for the

class intervals

#

%

(% /point)

2.0

20 - 38

5

13.9

0.8

38 - 50

50 - 74

16

44.4

1.9

1.5

74 - 90

6

16.7

1.1

90 - 100

3

8.3

0.8

1.0

Density (% per point)

0.5

0.0

scores

20

40

60

80

100

13.9%

16.7%

16.7%

44.4%

8.3%

(1.4)

(0.8)

(1.0)

(0.8)

(1.9)

From Lists to Histograms

Example: 36 Exam Scores

23, 29, 30, 31, 35, 38, 40, 41, 42, 45, 46, 51, 52, 54, 55, 55, 57, 58, 59, 60, 61, 63, 69, 70, 70, 71, 71, 74, 75, 75, 82, 85, 86, 91, 91, 93. Note:

Av = 59.1, SD = 18.9

6

16.7

1.4

Endpoint convention: class intervals contain left endpoints, but not right endpoints

A Histogram represents the percentages by areas (not by heights).

A histogram is not a bar chart.

Bar Chart of Scores confidence interval for the

Histogram of Scores

44.4%

2.0

40

1.5

30

1.0

Density (% per point)

20

% of total papers

16.7%

16.7%

13.9%

0.5

10

8.3%

8.3%

0

0.0

20

38

50

74

90

100

scores

scores

20

40

60

80

100

16.7%

13.9%

16.7%

44.4%

(1.9)

(1.4)

(1.0)

(0.8)

(0.8)

A Histogram is Not A Bar Chart

A Histogram represents the percentages by areas (not by heights).

A histogram is not a bar chart.

2.0 confidence interval for the

(1.9)

1.5

(1.4)

(1.0)

1.0

Density (% per point)

(0.8)

(0.8)

0.5

8.3%

0.0

scores

Av = 60.5

SD = 19

For many histograms roughly 68% of the area lies within 1 SD of the mean and 95% lies within 2 SD’s.

20

40

60

80

100

16.7%

13.9%

16.7%

44.4%

The Average and Standard Deviation of a Histogram

To find the mean or average of a histogram first list the center of each class interval then multiply each by the area of the block above it and finally sum.

Class intervals: 20 to 38, 38 to 50, 50 to 74, 74 to 90, 90 to 100

List of midpoints: 29, 44, 62, 82, 95

To find the standard deviation of a histogram find the squared deviations of the center of each class interval, then multiply each by the area of its corresponding block, then sum, and finally take the square root.

[Note for the original data: Av = 59.1, SD = 18.9]

2.0 confidence interval for the

1.5

Density (% per point)

1.0

Av = 60.5

0.5

SD = 19

0.0

scores

-3

-2

-1

1

2

3

0

Standard Units

Histograms and Standard Units

Area confidence interval for the

(percent)

Height

(% per Std.U.)

The Normal Curve

The normal curve was discovered by Abraham De Moivre around 1720. Around 1870 Adolph Quetelet had the idea of using it as an ideal histogram to which histograms for data could be compared. Many histograms follow the normal curve and many do not.

From: Freedman, Pisani, and Purves, Statistics, 3rd Ed.

2.0 confidence interval for the

1.5

Density (% per point)

1.0

Av = 60.5

0.5

SD = 19

0.0

scores

-3

-2

-1

1

2

3

0

Standard Units

Histograms, Standard Units, and the Normal curve

Data Histograms and Probability Histograms confidence interval for the

Discrete data convention

From: Freedman, Pisani, and Purves, Statistics, 3rd ed.

Data Histograms and Probability Histograms for confidence interval for the

the Sum of the Draws

The Central Limit Theorem confidence interval for the

There are many Central Limit Theorems. We state two in terms of box models. The second is a special case of the first and it covers the model we are dealing with in our stick tossing problem. It goes back to the early eighteenth century.

When drawing at random with replacement from a box of numbered tickets (with bounded range), the probability histogram for the sum of the draws will follow the standard normal curve, even if the the contents of the box do not. The histogram must be put into standard units, and the number of draws must be reasonably large.

De Moivre – La Place version: When drawing at random with replacement from a zero-one box, the probability histogram for the sum of the draws will follow the standard normal curve, even if the the contents of the box do not. The histogram must be put into standard units, and the number of draws must be reasonably large.

provides a box model for counting the number of heads in confidence interval for then tosses of a fair coin.

1

0

Histogram for the box

100

50

0

0

1

The Normal Curve and Probability Histograms for

the Sum of the Draws

From: Freedman, Pisani, and Purves

The Normal Curve and Probability Histograms for confidence interval for the

the Sum of the Draws

From: Freedman, …

Histogram for the box confidence interval for the

1

2

9

The Normal Curve and Probability Histograms for

the Sum of the Draws

From: Freedman, …

The Central Limit Theorems confidence interval for the

When drawing at random with replacement from a box of numbered tickets (with bounded range), the probability histogram for the sum (and average) of the draws will follow the standard normal curve, even if the the contents of the box do not. The histogram must be put into standard units, and the number of draws must be reasonably large.

De Moivre – La Place version: When drawing at random with replacement from a zero-one box, the probability histogram for the sum (and average) of the draws will follow the standard normal curve, even if the the contents of the box do not. The histogram must be put into standard units, and the number of draws must be reasonably large.

The probability histogram for the average of the draws, when put in standard units is the same as for the sum because multiplying each value of the sum by 1/(# of draws) won’t change the corresponding values in standard units.

Estimated SE for the average confidence interval for the

True SE for the average of the draws

1

True Population Average

Standard units

Where Does the 68 % Confidence Level Come From?

Sample Average

Since the estimated SE for the average computed from sample is, on average, about equal to the true SE a 68% confidence interval will cover the true population mean whenever the sample mean is within 1 SE of the true mean. The probability of this happening is, by the central limit theorem, the area within 1 standard unit of 0 under the normal curve, and this area is about 68%.

How to Prove The De Moivre – La Place Version of The Central Limit Theorem

Show that the probability that the sum of n draws at random with replacement from a zero-one box is exactly k given by the binomial formula

Then using “Stirling’s Formula”

Use the series for the log to show that, Central Limit Theorem

Which implies

How to Prove The De Moivre – La Place Version of The Central Limit Theorem -- continued

The limiting processes in these steps require some care. Both k and n must go to infinity together in a fixed relationship to each other, and we need to understand why values of x for which |x|>npq are unimportant.

Bibliography Central Limit Theorem

- Freedman, Pisani, & Purves, Statistics, 3rd Ed., W.W. Norton, New York,
- 1998

- W. Feller, An Introduction to Probability Theory and Its Applications, Volume I, 2nd Ed., John Wiley & Sons, New York, London, Sydney, 1957

3. F. Mosteller, Fifty Challenging Problems in Probability with Solutions, Addison-Wesley, Palo Alto, 1965.

4. http://www-history.mcs.st-andrews.ac.uk/Biographies/De_Moivre.html

5. R Development Core Team, R: A language and environment

for statistical computing, R Foundation for Statistical

Computing, Vienna, Austria, 2006, <http://www.R-project.org>

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