Lectures 1-13 +Outlines-6381-2016-Work Book Version.pdf

7/25/2019 Lectures 1-13 +Outlines-6381-2016-Work Book Version.pdf

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Management 6381: Managerial Statistics

Lectures and Outlines

2016 Bruce Cooil


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2

See the Bottom Right Corner of Each Page for the Document Page Numbers Listed Here.

TABLE OF CONTENTS

Lecture 1

Descriptive Statistics

(Including Stem &

Leaf Plots, Box Plots,

Regression Example) 1Stem & Leaf Di splay 1

Descri ptive Statistics: Means,

Median, Std.Dev., IQR 2

Box Plot 3

Regression 10

Lecture 2

Central Limit

Theorem & CIs 12Statement of Theorem 12

Simulations 13

Practical I ssues & Examples 15

Tail Probabil iti es & Z-values 16

Z-Value Notation 17

Picture of CLT 19

Everythi ng There I s to Know 20

Summary & 3 Types of CI s 21

Glossary 22

Lecture 3CIs & Introduction to

Hypothesis Testing 23Examples of Two Main Types

of CI s 23

Hypothesis Testing 25

Type I & Type I I Er ror 27

Pictures of the Right and Left

Tai l P-Values 29

Big Pictur e Recap 30

Glossary 31

Lecture 4

One- & Two-Tailed

Tests, Tests on

Proportion, & Two

Sample Test 32When to Use t-Values (Case 2) 34

Test on Sample Proporti on

(Case 3) 34

Means from Two Samples

(Case 4) 35

About t-Distri bution 38

Lecture 5

More Tests on Means

from Two Samples 39Tests on Two Proportions 39

Odds, Odds Ratio, & Relati ve 44

Tests on Pair ed Samples 45

F inding the Right Case 47

Lecture 6

Simple Linear

Regression 48Purposes 48

The Thr ee Assumptions,

Terminology, & Notation 49

Modeli ng Cost in Terms of

Units 50

Estimation & I nterpretation of

Coefficients 51

Decompositi on of SS(Total ) 52

Main Regression Output 53

Measures of F it 54Correlation 56

Di scussion Questions 57I nterpretation of Plots

59How to Do Regression in

Minitab 61

How to Do Regression i n Excel 62


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TABLE OF CONTENTS

Lecture 6 Addendum

Terminology, Examples

&Notation 63

Synonym Groups 63

Main I deas 63Examples of Corr elati on 64

Notation for Types of Var iation

and R2 66

Lecture 7

Inferences About

Regression Coefficients

& Confidence/Prediction

Intervals for Y/Y 67

Modeli ng Home Pri ces Using 68

Regression Output 72

Two Basic Tests 73

Test for Lack- of-F it 74

Test on Coeff icients 75

Prediction I ntervals & Confi dence

I ntervals 76

How to Generate These Intervals

in M ini tab 17 77

Lecture 8

Introduction to Multiple

Regression 80

Application to Predicting Product

Share (Super Bowl Broadcast) 81

3-D Scatterplot 82

Regression Output 84

Sequenti al Sums of Squares 85

Squared Coeff icient t-Ratio

Measures Marginal Value 86

Di scussion Questions on

I nterpreting Output 88

Lecture 9

More Multiple

Regression Examples 90Modeli ng Salaries (NF L

Coaches

2015) 90Modeli ng Home Pri ces 93

Regression Di alog Boxes 99

Lecture 10

Strategies for Finding the

Best Model 102Stepwise Approach 102

Best Subsets Appr oach 103

Procedure for F inding Best Model 104

Studying Successfu l Products (TV

Shows) 105

Best Subsets Output 108

Stepwise Options 109

Stepwise Output 110

Best Predictive Model 111

Regression on All Candidate

Predictors to Find Redundant

Predictors 113

Other Cr iteri a for Selecting Models 114

Discoverers 115

Lecture 11

1-Way Analysis of

Variance (ANOVA) as a

Multiple Regression 116Comparing Dif ferent Types of

Mutual Funds116

Meaning of the Coeff icients 118

Purpose of Overall F -test and

Coeff icient t-Tests 120

Comparing Network Share by

Location of Super Bowl 122

Standard Formulation of 1-WayANOVA 125

Analysis of Covariance 126

Looking Up F Critical Values 128


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TABLE OF CONTENTS

Lecture 12

Chi-Square Tests for

Goodness-of-Fit &

Independence 129Goodness-of -F it Test 129

Test for I ndependence 130

Using M in itab 132

Lecture 13

Executive Summary &

Notes for Final Exam,

Outline of the Course &

Review Questions 133Executive Summary & Notes for

Final 133

Outli ne of Course 135

Review Questions with Answers 140

Appendix for Review Questions 145

The Outlines

Tests Concerning Means

and Proportions &

Outline of Methods for

Regression 149Tests Concerni ng M eans and

Proportions151

Conf idence I ntervals for the Seven

Cases 153

Outl ine of M ethods for Regression 154


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Lecture 1: Descriptive StatisticsManagerial Statistics

Reference: Ch. 2: 2.4, 2.6pp. 56-59, 67-68; Ch. 3: 3.1-3.4 --pp. 98-105, 108-116, 118-129Outline:!Stem and Leaf displays

!Descriptive StatisticsMeasures of the Center: mean, quartiles, trimmed mean, median

Measures of Dispersion: standard deviation, interquartile range!Box plots & Regression as Descriptive Tools

Stem and Leaf Displays The rules:1) List extremes separately;

2) Divide the remaining observations into from 5 to 15 intervals;

3) The stem represents the first part of each observation and is used to label the interval, while the

leaf represents the next digit of each observation;4) Dont hesitate to bend or break these rules.

Famous Ancient Example (modified slightly): Salaries of 10 college graduates in thousands (1950s):

2.1, 2.9, 3.2, 3.3, 3.5, 4.6, 4.8 , 5.5, 7.9, 50.

Stem and Leaf

(With trimming)

Units:0.10 Thousand$

2|19

3|235

4|68

5|5

6|

7|9

High: 500MINITABs Version:This is an option in the Graph Menu, Or you can give the commands shown.

Stem-and-Leaf Displays

Stem and Leaf Display When Numbers Above are

in Units of $100,000 (i.e., same data X 100)

UNITS: 0.1 *100 = 10 Thousand $

Same Display

High: 500

No Trimming! (Here the extreme observations are

included in the main part of the display.)

MTB> Stem c1

Leaf Unit = 1.0

(9) 0 223334457

1 1

1 2

1 3

1 4

1 5 0

With Trimming

MTB > Stem c1;

SUBC> trim.

Leaf Unit = 0.10

2 2 19

5 3 235

5 4 68

3 5 5

2 6

2 7 9

HI 500

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Another Example: Make S&L of 11 customer expenditures at an electronics store(dollars): 235, 403, 539, 705, 248, 350, 909, 506, 911, 418, 283.

Units: $10

3 2|348

4 3|5(2) 4|01

5 5|30

3 6|

3 7|0

2 8|

2 9|01

Now reconsider the first example with the 10 salaries!I put those 10 observations into the first column of a Minitab spreadsheet (orworksheet) and then asked for descriptive statistics.

MTB > desc c1

Descriptive Statistics

Variable N Mean Median TrMean StDev SE Mean

Salaries 10 8.78 4.05 4.46 14.58 4.61

Variable Minimum Maximum Q1 Q3

Salaries 2.10 50.00 3.12 6.10

What do the Mean, TrMmean, Median, Q1" and Q3" represent?

Mean: Average of Sample

TrMean (5% Trimmed Mean): Average of middle 90% of sample

Median: Middle Observation (when n is even: average of middle two obs.)

OR 50th

Percentile

Q1 & Q3 (1stand 3

rdquartiles):

25thand 75thPercentiles

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Note how the median is much better measure of a typical central value in this case.

Recall how standard deviation is calculated.

First the sample variance is calculated:

S2= estimate of average squared distance from the mean

= {sum of squared differences (Obs-Mean)2}/(n-1)

={2.1 -8.78)2+(2.9 -8.78)

2+...+ (50 -8.78)

2}/9= 212.6.

Then the sample standard deviation is calculated as the square root of the

sample variance:

s = (212.6)= 14.58 .

As a descriptive statistic, s is usually interpreted as the typical distance of anobservation from the mean. But what does s actually measure?

Square root of average squared distance from mean

Whats the disadvantage ofSas a measure of dispersion (or spread)?

Sensitive to extreme observations (large and small)

Whats an alternative measure of dispersion that is insensitive to extremes?

0.75 * (Q3 - Q1)

[Q3-Q1] is referred to as the interquartile range (IQR). If the distribution is

approximately normal, then

(0.75)(Q3 - Q1) (0.75) IQR

provides an estimate of the population standard deviation ().

For sample of 10 salaries:(0.75) IQR =0.75(6.10 - 3.12) =2.2.

(Compare with s= 14.58.)The Boxplot

Elements: Q1, median, and Q3 are represented as a box, and 2 sets of fences

(inner and outer) are graphed at intervals of 1.5 IQR below Q1 and above Q3.

The figures on pages 122-125 (in our text by Bowerman et al.) provide goodillustrations.

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Inner Fences

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MINITAB Boxplot of the 10 Salaries

Result of Menu Commands: GraphBoxplot

50

40

30

20

10

0

Salaries

Boxplot of Salaries

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More Examples with Another Data Set Where We Compare Distributions

These data are fromhttp://www.google.com/finance and consist of daily closing prices and

returns (in %) for Google stock and the S&P500 index (see the variables Google_Return and

S&P500_Return below), and a column of standard normal observations.

. . .

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http://www.google.com/financehttp://www.google.com/financehttp://www.google.com/financehttp://www.google.com/finance


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(Recall what the Standard Normal distribution looks like, e.g.http://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svg.)

MTB > describe c3 c5 c6

(Or to do same analysis from menus: start from Statmenu, got to Basic Statistics & then to Display

Descriptive Statistics, then in the dialog box select c3, c5,and c6 as the variables.)

Descriptive Statistics: Google_Return, S&P_Return, Standard_Normal

Descriptive Statistics: Google_Return, S&P_Return, Standard_Normal

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum

Google_Return 29 1 -0.207 0.260 1.401 -5.414 -0.772 -0.086 0.771 1.674

S&P_Return 29 1 0.026 0.116 0.624 -1.198 -0.414 0.017 0.534 1.051

Standard_Normal 30 0 -0.134 0.170 0.931 -1.778 -0.813 -0.184 0.598 1.871

Standard_NormalS&P_ReturnGoogle_Return

2

1

0

-1

-2

-3

-4

-5

-6

Data

22-Apr-16

Boxplot of Google_Return, S&P_Return, Standard_Normal

quarterly results

Apparently due to announcement of disappointing

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http://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svghttp://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svghttp://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svghttp://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svg


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In contrast to the boxplots on the previous page, many business distributions are

positively skewed. For example, here is a comparison of the revenue distribution

for the largest firms in three health care industries.

Pharmacy & Other ServicesMedical FacilitiesInsurance & Managed Care

1 20

1 00

80

60

40

20

0

R

evenue(Billions)

Express_Scripts_Holdin

HCA_Holdings

United_Health_ Group

Boxplot of 2014 Revenues in Three Health Care Industries

(12 Firms) (13 Firms) (13 Firms)

(for Firms That Are Among the Largest 1000 in U.S.)

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11

The regression equation is approximately:

Google_Return = - 0.2317 + 0.9408 [S&P_Return]

This equation describes the relationship between Google returns and

returns on the S&P500. If we assume the S&P500 represents the market

as a whole, then this regression model is a form of the market model (or

security characteristic line). It summarizes the relationship betweencontemporaneous returns, since the Google returns and S&P returns

occur at the same time. Consequently, it has no direct predictive value

(i.e., it does not allow me to predict future Google returns because that

would require me to know future S&P returns). Nevertheless, it allows

me to study the relationship between Google returns and the market as a

whole. For example, the expected return on Google stock, when the

S&P500 return is 0, is:

Google_Return = - 0.2317 + 0.9408 [0] = -0.23% (approximately)

The regression analysis also shows that there is a relatively weak

relationship between the two types of returns: the R-squared (adjusted)

value (to the right of the plot) indicates that only 14.5% of the variance

in Google returns is explained by the market return.

This type of regression is generally done with weekly or monthly

returns, rather than daily returns (as was done here).

1 .00.50.0-0.5-1.0-1.5

2

1

0

-1

-2

-3

-4

-5

-6

S 1 .29557

R-Sq 1 7.5%

R-Sq(adj) 1 4.5%

S& P_Return

Google_

Return

Fitted Line PlotGoogle_Return = - 0.231 7 + 0.9408 S&P_Return

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Lecture 2: The Central Limit Theorem & Confidence Intervals

Outline (Reference: Ch. 7-8: 7.1-7.3, App 7.3, 8.1-8.5, App 8.3): The Central Limit Theorem (CLT):

Sample Means Have Approximately a Normal Distribution (given asufficiently large sample, from virtually any parent distribution)

Illustration of How Sample Proportion is a Sample Mean (with simulations) Two Examples: Example 1: sample mean; Example 2: sample proportion Introduction to Confidence Intervals (CLT application): Z-values, Picture of

CLT, a Quiz, & Major Types of CIs

Assume you have a "large" sample size n, and that you find the sample mean,, as the average of n observations, each of which is from a parentdistribution (or population) with mean and standard deviation .

Statement of the Centr al L imit Theorem:

The sample mean has approximately a normal distribution with mean and standard deviation /n.

Note: A sample proportion(

) is simply the mean of n independentobservations where each observation takes on the value "1" with probability p,and takes on the value 0" with probability 1-p.

For example, suppose a company studies the probability (p) that an individualcustomer complains. Think of each customer response as a random variableX, which takes on the value "1" if they complain, and "0" otherwise. Imaginethat each customer complains with a probability p = 0.1.

What i s the name of the parent distr ibution of X ?

X has a Bernoulli distribution, which is also referred to as a binomial

distribution with 1 trial (n=1, p=0.1). The mean () is n*p =1*0.1=0.1,

and the standard deviation () is [np(1-p)]=[1*0.1*0.9]1/2= 0.3.

By Central Limit Theorem: if I sample 100 customers, then the proportion whocomplain (p) will have a normal distribution (approximately), with the same mean

(0.1) but a much smaller standard deviation of 0.3/n = 0.3/100 = 0.03.

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Here is a picture of the parent distribution.

10

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Value of O bservation (1 : Complaint; 0 : No Complaint)

Frequency

Parent Distribution: Binomial (n=1, p=0.1)

In a simulation, I repeatedly took a random sample of 100observations from the parent distribution above, and calculatedthe mean of each sample of 100 observations. I did this a 1000times. Here is a histogram of those 1000 means.

0.210.180.150.120.090.060.03

160

140

120

100

80

60

40

20

0

Value of the Mean

Frequency

Mean 0.09962

StDev 0.02918N 1000

Histogram of 1000 Means (Each is the Average of 100 Observations)(and Comparison with Normal Distribution)

As predicted by the Central Limit Theorem: this distribution isapproximately normal and the sample mean (the mean of themeans) is approximately 0.1 (same as the parent), and thesample standard deviation (of the means) is approximately 0.03(= [parent distributions std.dev.]/ n = 0.3/100).

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Note: this is

approximately 0.03


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Another simulation: suppose we toss one fair die. Here is theprobability distribution of the outcome.

654321

0.18

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

Outcome of Tossed Die

Frequency

Parent Distribution: Integers 1-6 Are Equally Probable

I repeatedly take a random sample of 2 observations from theparent distribution above, and calculate the mean of each sampleof 2 observations. I do this a 1000 times. Here is a histogram ofthose 1000 means (each mean is of only 2 observations).

6.45.64.84.03.22.41.60.8

180

160

140

120

100

80

60

40

20

0

Value of the mean

Frequency

M e an 3. 53 9

S tD ev 1 .18 4

N 1000

Histogram of 1000 Means (Each is the Average of 2 Observations)

As predicted by the Central Limit Theorem: this distribution isapproximately normal and the sample mean (the mean of themeans) is approximately 3.5 (same as the parent), and thesample standard deviation (of the means) is approximately 1.2

(1.2 = [parent distributions std. dev.]/n = 1.7/2). Page 14 of 156

Note: This is

approximately 1.2


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Practical Issues & Two Examples

How large should n be? Here are two guides:

1)for a typical sample mean, : n > 30 (this is a conservative rule);2)for a sample proportion: n large enough so that & ( ) .

Example 1

Here are descriptive statistics for 40 annual returns on the S&P500(these returns are simple annual percent gain or loss on index, withoutcompounding or inclusion of dividends), 1975-2014.

MTB > desc 'S&P_Return' StDev/N = [16.57/40

Descriptive Statistics: S&P_Return

Variable N N* Mean SE Mean StDev Minimum

S&P_Return 40 0 13.41 2.62 16.57 -36.55

Variable Q1 Median Q3 MaximumS&P_Return 4.99 15.75 27.74 37.20

This summary shows that:

n = 40, = 13.41 (this is an estimate of ), s = 16.57 (an estimate of );and s/n = 16.57/(40) = 2.62

Describe the distr ibution of (the sample mean), assuming the actualdistr ibution of S&P_Retur n remains unchanged dur ing 1975-2014:The distribution is approximately Normal with mean of approximately

13.41 and std. dev. of approx. 2.62 .

Example 2

Suppose I interview 100 people and 20 prefer a new product (to

competing brands). I want to estimate: p proportion of population thatprefer the new brand. (Each customer preference is a Bernoulli

observation, with an approx. mean of 0.20 and approx. variance of[0.20 0.8]=0.16.)

I n summary, the sample proportion,p , is: 20/100 = 0.2.

p behaves as though it has a normal distribution,

with a mean of approximately 0.2 (this is our estimate) and a standard

deviation of approximately[0.2*0.8/100]1/2 = 0.04 .

Recall that forBernoulli Distribut = p, 2=p(1-p).Consequently:

/n= [p(1-p)/n]1/2

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Tail-Probabilities & the Corresponding Normal Values (Z-values)

0.4

0.3

0.2

0.1

0.0

Frequency

General Normal Distribution

Tail Probabilities

0.10 >

0.05 >

0.025 >

Value of Normal Random Variable

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6

Z-Value Notation

z is used to represent the standard normal value above which there is a tail

probability of .Tail probability is

z

Verify that z0.10= 1.28, z0.05=1.645, and that z0.025= 1.96. (Use normaltable, e.g.,http://www2.owen.vanderbilt.edu/bruce.cooil/cumulative_standard_normal.pdf.)

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To verify that Z = 1.28:

0.10

0.10

Z

0.90

Tail probability is 0.10,

So find Z-value that

corresponds

to cumulative prob. of 0.9 .

=> It's 1.28

To verify that Z = 1.645:0.05

Z0.05

Tail probability is 0.05,

So find Z-value that

corresponds

to cumulative prob. of 0.95.

=> It's 1.645

Verify that Z = 1.96 !0.025

Z

0.025


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0z

Cumulative probabilities for POSITIVE z-values are in the following table:

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

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7

Picture of the Central Limit Theorem

Acknowledgment: This picture of the Central Limit Theorem is based on a much prettier graph made for this course by Tim Keiningham,Global Chief Strategy Officer and Executive Vice President,Ipsos Loyalty (also a student in an earlier version of this course).

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8

Everything There is to Know About the Normal Distribution,

The Central Limit Theorem, and Confidence Intervals

The Central Limit Theorem states that the distribution of (the

distribution of sample means) is approximately normal with mean and

variance 2/n, abbreviated:

is approximately N(,2/n), where:

is the mean of the distribution of ( is also the mean of the

population from which the observations were sampled).

is the sample mean. (The sample is taken from a population with

mean and variance 2. Think ofas an "estimate" of .)

2/n is the variance of the distribution of . Also referred to as the

variance of.

/n is the standard error of (the sample mean). It is also sometimes

called the SE mean or standard deviation of.

The figure on the top of the previous page indicates:

is within 1.28 standard errors*of with probability 80% .



*Remember that the standard error of is /n.

Another Way of Saying the Same Thing

(1.28) (/n) is an __80__% confidence interval for .

(1.645)(/n) is a __90__% confidence interval for .

(1.96) (/n) is a __95__% confidence interval for .

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9

Br ief Summary of Chapter 8

Three Types of Confidence Intervals Are Introduced in Chapter 8

1) 100(1-)% confidence Interval for when n > 30:

/(/ ).2) 100(1-)% confidence Interval for p:

/ ( )/.This assumes n is large enough so that & ( ) .

3) 100(1-)% confidence Interval for when n< 30:

/()(/ ).

This is the same as confidence interval (1), except that a t-value is now used in place of the z-

value.

NOTE: The text refers to "/()

" as "t/2".

General Form of Confidence Intervals:Estimate [ (t- or z-value) x (Standard Dev.of Estimate)]

Examp le 1:Consider the return data: n=40, =13.41, s/n=2.62.Find a 90% confidence interval for :

13.41 1.645 (2.62)

Find a 95% C.I. for :

13.41 1.96 (2.62)

Examp le 2: Consider the product preference example above.

Here = 0.2 and n=100.Find a 90% confidence interval for p(the actual proportion):

0.2 (1.645)[.2(.8)/100]

Find an 80% C.I. for p:

0.2 1.28(0.04)

Examp le 3:Suppose we consider only the last 16 changes in S&P:

n=16, = 6.93, s/n= 4.69. Must use t-value because n


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10

Glossary

Reference: Chapter 5 (pp. 188, 190) versus Chapter 3 (pp. 100, 110)

The Mean of a Distribution: () (). (1)

The mean of a distribution (or a random variable X) is simply the weighted average of its

realizable outcomes, where each realizable value is weighted by its probability, P(x).

Contrast this definition with the definition of a sample mean:

x x n ( / )1 (= ). (2)The only difference is that (1/n), the frequency with which each observation occurs in the

sample, replaces P(x) in equation (1).

The Variance of a Distribution: ( ) ( )(). (3)

The variance of a distribution (of a random variable X) may also be calculated as = 2() 2. Note the first term in this last expression is just the expectation or averagevalue of X2.

Standard Deviation of a Distribution:

= ( )()/ = / . (4)Compare this with the definition of the sample standard deviation:

= ( ) ()/

= ( )/ ( )/.

(The sample variance is: = [ ( )/ ( )].)

_________________________________________________________________

ANSWERS to Examples (on Bottom of Previous Page)

Example 1

1) 90% CI: Z0.05 (s/n) = 13.41 1.645 (2.62) = 13.41 4.31OR: (9.1, 17.7)

2) 95% CI: Z0.025(s/n) = 13.41 1.96 (2.62) = 13.41 5.13OR: (8.3, 18.5)

Example 2

1) 90% CI: 0.05 (1 )/ = 0.2 (1.645)[.2(.8)/100]= 0.2 0.066

OR: (13%, 27%)

2) 80% CI: 0.0 (1 )/ = 0.2 1.28(0.04) = 0.2 0.051OR: (15%, 25%)

Example 399% CI:

/()

(/ ) = 6.93 2.947 (4.69) = 6.93 13.82 OR: (-6.9, 20.8)Page 22 of 156


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Lecture 3

Confidence Intervals for Means and Proportions& Introduction to Hypothesis Testing (Large Sample Mean)

Outline (Ch.9: 9.1-9.2) Recap of C.I.s for Means and Proportions

One-tailed tests on sample meanWhat is type I error? Type II error? Power?

Everything to Know About the Test Statistic and P-value

Recap of Confidence IntervalsExample 1I have just designed a new type of mid-size car with a hybrid engine.To determine its average fuel efficiency (mpg), I sample 30 mile per

gallon measurements from 30 different cars (city driving).MTB > print c1

MPG

70.4 70.5 70.8 71.2 72.5 73.5 75.1 77.0 77.2 77.4 77.9 78.3

80.3 80.9 81.1 81.4 84.2 84.2 84.3 85.4 85.6 86.3 86.3 86.7

89.3 89.7 89.9 90.6 91.0 92.1

MTB > stem c1;

SUBC> trim.

Stem-and-Leaf Display: MPGStem-and-leaf of MPG N = 30; Leaf Unit = 1.0

4 7 0001

6 7 23

7 7 5

11 7 7777

12 7 8

(4) 8 0011

14 8

14 8 44455

9 8 666

6 8 999

3 9 01

1 9 2

MTB > desc c1

Descriptive Statistics: MPGVariable N Mean SE Mean TrMean StDev Minimum Q1

MPG 30 81.37 1.24 81.43 6.80 70.40 76.53

Variable Median Q3 Maximum

MPG 81.25 86.40 92.10

Note: SE Mean (or 1.24) = / = (6.8/30)

MPG

95

90

85

80

75

70

Boxplot of MPG

Page 23 of 156


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2

Find a 95% confidence interval for the real mean mpg () andinterpret it.

C.I.: ./ = 81.37 1.96 (1.24)

= 81.37 2.43 or (78.9, 83.8)

Interpretation: This covers the real mean () with 95%

probability

Would an 80% confidence interval be longer or shorter?

Shorter!

(Use Z0.10= 1.28, and interval becomes (79.8, 83.0).)

(The convention is : Use t-values when n


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3

Hypothesis Testing

Reconsider the new hybrid car example (example 1). Suppose that I want

to show that my new car has an average mpg () that is better than that of

the best performing competitor, for which the average mpg is 78. Formally, I want

to "disprove" a null hypothesis

H0: =78 (or sometimes written as 78)

in favor of the alternative hypothesis:

H1: > 78.

Note that: n=30,

=81.37, s= 6.8, s/n = 1.24. (For n < 30, the procedure isidentical except when we find the critical value. That case will also be discussed.)

To build a case for H1, I follow 3 logical steps (typical of all hypothesis testing).

1) Assume H0is true.

2) Construct a test statisticwith a known distribution (using H0).

In this case I use the test statisti c,z [ - 78]/(s/n)

which should have approximately a standard normal

distribution if H0is true. (WHY? CLT, since n is large)

3) Reject H0in favor of H1if the value of z supports H1.

("Large" values of z support H1in this case.)

Regarding step 3, if H0is true, I would see values of z greater than z0.05= 1.645

only 5% of the time. This seems improbable and it supports H1and so a reasonable

decision rule is to: reject H0in favor of H1if z is greater than 1.645. This assumes

that I am wil l ing to make a mistake 5% of the time.

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4

In this sample,

z = [ - 78]/(s/n) = [81.37-78]/1.24 = 2.72 > 1.645.

Therefore, I reject H0in favor of H1.

SUMMARY: to test H0: = 78 versus H1: > 78

we use the decision rule: reject H0if

z = [ - 78]/(s/n) > z

or equivalently if: > 78 + z(s/n).

Otherwise we accept H0.

In this case z= 2.72, so I reject the null hypothesis H0at the 0.05 level, and

conclude in favor of the alternative hypothesis H1. That is, I conclude that

the average mpg of the new hybrid automobile is significantly greater than

78, but using this decision rule (i.e., rejecting H0whenever z>z0.05) there is

a 5% chance that I have erroneously rejected H0and that the real average

mpg () really is only 78 (or less).

Above we chose = 0.05, so that z0.05= 1.645. This probability isreferred

to as the significance level, and it is the maximum probability of making a

type I error: type I error refers to the error we make if we reject H0when H0

is in fact true. Typically we use = 0.001, 0.010, 0.025, 0.05, 0.1, or 0.2

so that z= 3.09, 2.33, 1.96, 1.645, 1.28, or 0.84, respectively

(the corresponding t-values are very similar for moderate values of n:

for n=20: t

19

= 3.6, 2.5, 2.1, 1.7, 1.3, or 0.86;

for n=30: t( )29

= 3.4, 2.5, 2.0, 1.7, 1.3, or 0.85 ).

Suppose that I had chosen = 0.001, then since z0.001 = 3.09, and z = 2.72,

I would accept H0because z =2.72 >/ z0.001=3.09. In this case, I would be

concerned that I made a type II error. Type II error refers to the case where

the null hypothesis H0is really false but I fail to reject it! The following

figure summarizes the situation with type I and II errors.

Page 26 of 156

Z0.05


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5

DECISION WHAT IS REALLY TRUE

H0IS TRUE H1IS TRUE

REJECT H0 Type I Error Correct

Decision

ACCEPT H0 Correct

Decision

Type II Error

Good Lingo: Cannot Reject H0 can be used for Accept H0.

Bad Lingo: Accept H1 should not be used for Reject H0.

How do we protect against:

Type 1 Error? Small Type II Error? Large n

Note that to make a decision on whether to reject or accept H0: =78, we simply

need to compare the test statistic z = [ - 78]/(s/n) with an appropriate normalvalue, z, that corresponds to the significance level that is chosen beforehand. Ifz > z, we reject H0(otherwise accept H0).

Distribution of Test Statistic (Z) When H0Is True

z0.05 z z0.001

1.645 2.72 3.09

Alternatively, we could simply look up the tail probability that corresponds to

the test statistic z (this is called the p-value) and compare it to the

significance level . If thep-value is less than (p-value < ), wereject H0

(otherwise accept H0).

In this case z = 2.72, and the p-value for H0: = 78 versus H1: > 78, is theright tail-probability (because this is a one-tailed test where the alternative

hypothesis goes to the right-side). What is the p-value in this case?

P-value (probability to the right of 2.72) = 1 - [Cumulative probability at 2.72]

=1 - 0.9967 0.0033

Can we reject H0at the 0.05 level? YES At the 0.001 level? NO!Page 27 of 156

P-Value: the

probability to right

of z


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6

Another One-Tailed Test (in the other direction):

Suppose I make the claim that my cars average mpg is 80 (theobservations on page 1 were really drawn from a normal distribution

with = 80). My competitor might be interested in testing:

H0: = 80 (sometimes written > 80) versus H1: < 80.

And suppose my competitor chooses a significance level of = 0.10.

In this case the test statistic is:

z = [ - 80]/(s/n) = [81.37 - 80]/1.24 = 1.10

andsmall values of z support the alternative hypothesis H1: z 0.05= 1.645 z < - z0.1= -1.28

Page 28 of 156


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7

Alternatively, we can find the p-value that corresponds to the test

statistic, z, for this hypothesis test and compare it with , and (as always)

we only reject H0if the p-value is less than . Remember that when

the alternative hypothesis goes to the left side, the p-value refers to the

tail probability to the left of the test statistic z. Given the way the p-value

is calculated, we always reject H0when p-value < , and accept H0

otherwise.

Given the test statistic z =1.10 for H0: = 80 versus H1: 78, and the test statistic was z= 2.72?

This was calculated on page 5 as 0.0033.

43210-1-2-3-4

2.5

2.0

1.5

1.0

0.5

0.0

43210-1-2-3-4

2.5

2.0

1.5

1.0

0.5

0.0

Page 29 of 156

1.10

2.72

Here P-value is a left tail

probability because H1goe

to the left !

Here P-value is a rig

tail probability beca

H1goes to the right


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8

Big Picture RecapLet 0represent the constant benchmark to which we wish to compare , &

consider three scenarios.

1)H0: = 0 2)H0: = 0 3)H0: = 0

H0

also written as: 0 0 No Other WayAlternative

Hypothesis H1H1: > 0 H1: < 0 H1: 0

Critical Value z -z z/2

Decision Rule

Reject H0 if: z > z z < z |z| > z/2(Note that z is the test statistic )

Definition ofp-value Tail prob. > z Tail prob. < z Tail prob. > |z|

Example Example 1

(see bottom p.6)

Example 2

(see bottom p.6)

Example 3

(new example)

Picture of

test statistic and

p-value (shaded area)relative to standard

normal distribution

Null Hypothesis H0 : = 78 H0 : = 80 H0 : = 80Alternative H1: > 78 H1: < 80 H1: 80Significance Level = 0.05 = 0.10 = 0.10

Test Statisticz = = . = . | | = .

Critical Value Z0.05= 1.645 -Z0.10= -1.28 Z0.10/2=Z0.05=1.6

Decision Reject H0 Accept H0Accept H0Becau

|1.10| 1.645

Page 30 of 156

P-Value=0.0033

P-value=0.86

P-Value=0.27


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9

Glossary

= significance level = maximum probability of making a type I error.

p-value = tail-probability that corresponds to test statistic, that is calculated for

specific alternative hypothesis H1.

= probability of making a type II error (not rejecting H0when H1is true).

Power = 1- = probability of making correct decision when H1is true.

How does power change with sample size?

Power increases as sample size increases (ceteris paribus).

Because as n increases, the test statistic becomes larger in absolute

value, and is more likely to exceed the critical value in the appropriate

direction. See the 3rd-to-last row of the table on the last page (i.e., the

test statistic formulas). Another way to think about it: as the test

statistic becomes larger in absolute value in the direction supporting H1,the p-value decreases.

How does power change with ?

Power increases as increases (ceteris paribus).

Because as increases, the critical value decreases in absolute value,

and is more likely to be exceeded by the test statistic, see the

penultimate row of last page (i.e., the critical values and how theychange with ).

Bruce Cooil, 2016

Page 31 of 156


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Lecture 4

One and Two-Tailed Tests, Tests on a Sample

Proportion, & Introduction to Tests on Two Samples

Main References

(1) Ch.9: 9.3-9.4, Summary, Glossary, App. 9.3;Ch.10: 10.1

(2) The Outline "Tests on Means and

Proportions"(referred to as "The Outline")

TopicsI. Tests on Means and Propor tions from One Sample (Reference:

9.3-9.4)

Example of a two-tailed test (Case 1) When to use t-values (Case 2) Tests on a sample proportion (Case 3)

II. Tests on M eans from Two Samples (Ref: 10.1)

Tests on means from two large samples (Case 4) Tests on means when it is appropriate to assume variances are

equal(Case 5)

I. Tests on Means & Proportions from One Sample

Summary of Last Time (1-Tailed Versions of Case 1)

Last time we first considered the one-tailed hypothesis test:

H0: = 78 versus H1: > 78.

(OR H0: 78 )In this case we use the decision rule: reject H0if:

z = [ - 78 ] /(s/n) > z,

or equivalently if > 78 + z(s/n). Otherwise we

accept H0.

Page 32 of 156


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2

Then we considered the one-tailed test going the other way (

still represents the mean mpg of my new hybrid). I make the

claim that the average mpg is 80, and so my competitor wants to

test:

H0: = 80 versus H1: < 80 .

The decision rule will be to reject H0in favor of H1if:

=

( : =.

.= . )

supports H1. If is calculated using observations from adistribution where < 80 (as my competitor believes is the

case), then we will tend to get small values of z. So the decision

rule would be, reject H0in favor of H1if

=

<

(or equivalently if: < (/ ).

[Note that this is just Case 1 in the outline: 0refers to the constant used

in the null hypothesis, which is "80" in this last case.]

Example of a 2-tai led TestA two-tailed test would be:

H0: = 80 versus H1: 80.So, for example if =0.05, we would reject H0in favor of H1if

|z| > z0.025(because /2 = 0.025).

What do we conclude if we do this 2-tailed test?

(Recall that: n= , = . , / =1.24.)Test Stat:Z =(81.37 - 80)/1.24 = 1.10 (SAME as above )

Critical Value:Z0.025= 1.96

Conclusion:AcceptH0 .

(is not significantly different from 80.) Page 33 of 156


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3

When to Use T-values

Case 2 (p.1 of outline) is identical to Case 1, except that the

critical values in Case 2 are based on the t-statistic. When

should we use Case 2?

A good conservative approach is to always use t-values when

you have to estimate (which isalways) but it does not make

much of a difference if n >30.

Lets dothe two-tailed test above, using Case 2 (=0.05):

H0: = 80 versus H1: 80 (SAME)Test Statistic: t = 1.10 (Same)

Critical Value: ./ = . = .Conclusion: Accept H0.

Tests on a Sample Proport ion

Example: Let p = proportion of customers in the population that

prefer my new product. Suppose I need to test ( = 0.05):

H0: p = 0.1 versus H1: p > 0.1.

This is case 3 in the outline (with p0= 0.1).The details are just

like case 1 except we use a different standard error of the mean.

I f 30 of 100( =0.3 )randomly selected customers prefer myproduct, can I show that more than 10% of the population of

all customers prefer my product at the 0.05 level?

Test Stat ist ic :

= ()

= ....

= .. = . > . =.

Crit ical Value:Z0.05= 1.645

Conclus ion :Reject H0(YES!).

Only

difference

from Case 1

Page 34 of 156


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4

II. Means from Two Samp les

Case 4: What To Do When Both Samples Are Large

Example:

The owner of two fast-food restaurants wants to compare the

average drive-thru experience times for lunch customers at eachrestaurant (experience time is the time from when vehicle

entered the line to when the entire order was received). There is

reason to believe that Restaurant 1 has lower average experience

times than Restaurant 2 because its staff has more training.

Suppose n1experience times during lunch are randomly selected

for Restaurant 1, n2from Restaurant 2 with following results(units: minutes): n

1

= 100 = s1= 0.7n

2

= 50 = . s2= 0.5 .Why do we use Case 4 on page 1 of the outline?

Both Samples 30 (& Independent).

If we want to show Restaurant 1 has a lower average experience

time, what are the appropriate hypotheses and what can we

conclude (at the 0.1 level)?H0: 1- 2= 0 (OR: 0 ) In Outline: D0= 0.

H1: 1< 2 OR 1- 2< 0

Test Statistic:

=

+

= .(.) +

(.)

= ..

= .

Critical Value:-Z0.10= -1.28 Conclusion: Reject H0.

(YES!)

What would happen if we test at the 0.01 level?

NewCritical Value:-Z0.01= -2.33 (Still Reject H0)

Is there any reason to pick in advance?

Yes, its more objective!

Page 35 of 156


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5

Would Welchs t-test (p. 376) make a difference?In this case we use the same test statistic but compare it with a

critical value from the t-distribution with degrees of freedom,

So for the =0.1 and =0.01, the criticalvalues are:

0.()

= 1.29, 0.0()

= 2.35, respectively, and

the conclusions are the same in each case!

Case 5: What If We Are Willing To Assume Equal

Variances?

Example : I'm comparing weekly returns on the same stock

over two different periods. The average sample return is larger

during period 2. Can one show that the return during period 2

is significantly higher than during period 1 at the 0.01 level?The data are: n1 = 21, = . %,

= .

n2 = 11, = . %, = . .

What are the appropriate hypotheses?

H0: 1- 2= 0

H1: 1< 2 1- 2< 0.

It may be risky to rely only on the CLT. (Why?)

Technically I make 3 additional assumptions if I use Case 5:

(1) observations are approximately normal,

(2) the two populations have equal variances and

(3) samples are independent.

.133

150

)50/5.0(

1100

)100/7.0(

)100/1(

1

)/(

1

)/(

)//(2222

2

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

n

ns

n

ns

nsnsdf

Page 36 of 156


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6

The test statistic in Case 5 allows us to use a pooled estimate of

the variance:

. .

.

. . The test statistic is:

t

. ..

..Suppose I do this test at the 0.01 significance level. What would

be the critical value for the test statistic "t" and what would be

the conclusion?

Critical Value: .? . .-2.457Conclusion: Reject H0 . (YES!)

What would be the two-tailed test in this case? (Specify H0&

H1.) Also give the critical value and conclusion if testing at the

0.01 level?

H0: 1- 2= 0 versus H1: 1- 20

Test Statistic: t = -2.6 (Same as for one-tailed test)

Critical Value: ./ . = 2.75Conclusion: Accept H0 .(No!)

(Because |t|=2.6 < 2.75 .)

JustlikeCase4withspused

inplaceofs1"ands2

Page 37 of 156


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7

About the t-Distribution (Reference: Bowerman, et al., pp. 344-346)

According to the Central Limit Theorem, (the sample mean of n observations),

has approximately a normal distribution with mean , and standard deviation /n .

Also, this approximation improves as the sample size, n, increases. Consequently,by the Central Limit Theorem, the standardized mean,

=

,

has approximately a standard normal distribution. We have been using this single

resul t to justi fy the construction of confidence intervals and hypothesis tests.

When using this result, we have generally been approximating by substituting

the sample standard deviation, s,for it. If the sample is large enough, thisdoesnt imposemuch additional error. But when samples are smaller (e.g., n < 30),

the convention is to accommodate the additional error (caused when using s for )

by using the fact that i f theoriginal distributionwas normal, then the t-statistic,

=

,

really has what is referred to as at-distr ibution wi th n-1 degrees of freedom. The

degrees offreedom number, n-1, refers to the amount of information that thesample standard deviation, s, contains about the true standard deviation . If we

have only 1 observation, we have no information about (n-1= 1-1 = 0), if we

have 2 observations we have essentially 1 piece of information about , and so on.

This is the reason we divide by the degrees of freedom n-1, when calculating s,

= [ ( )/ ( )].

The real question becomes: why should we use the t-distribution when i t rel ies

on the strong assumption that the ori ginal distribution is normal, which is

exactly the type of assumption we were trying to avoid by using the Central Limit

Theorem?! The answer is essentially this: by using t-values in place of z-values

we are doing something that accommodates the additional inaccuracy we generate

by using s to estimate , and inpractice it works quite well even when the parent

distri bution is not normal! Of course, t-values converge to z-values as the sample

size increases: see the t-table.

Page 38 of 156


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Lecture 5: More Tests on Means from Two Samples

Outline: (Reference: Bowerman et al., 10.2-10.3, Appendix 10.3; the Outline

Tests Concerning Mean and Proportions)

Tests on Two Proportions (Case 6, Ch. 10.3) Everything to Know About Odds, Odds Ratios and Relative Risk

Tests on Paired Samples (Case 7, Ch. 10.2)

Tests on Two Proportions(Case 6: Large Samples)

This example comes from an article, 10 Most Popular

Franchises published in the Small Business section of

CNN.com (April, 2010):http://money.cnn.com/galleries/2010/smallbusiness/1004/gallery.Franchise_failure_rates/index.html.

(More recent data through early 2016 consist primarily of a

smaller sample of settled loans from the same period:

http://fitsmallbusiness.com/best-franchises-sba-default-rates/# . )

It provides franchise failure rates based on loan data from theSmall Business Administration (October, 2000 through

September, 2009) and it illustrates all of the issues one will

typically face when comparing rates (expressed as proportions).

The 10 most popular franchises are: 1)Subway, 2)Quiznos,

3)The UPS Store, 4)Cold Stone Creamery, 5)Dairy Queen,

6)Dunkin Donuts, 7)Super 8 Motel, 8)Days Inn, 9)Curves for

Women, and 10)Matco Tools. Super 8 Motel and Days Inn have

the highest start-up costs (average SBA loan sizes are 0.91 and

1.02 million dollars, respectively), and nominally Super 8Motels seem to have a lower failure rate. Here are the data.

SBA Loans Failures*

Super 8 Motel 456 18

Days Inn 390 23

*Failures are loans in liquidation or charged off.

Page 39 of 156
http://money.cnn.com/galleries/2010/smallbusiness/1004/gallery.Franchise_failure_rates/index.htmlhttp://money.cnn.com/galleries/2010/smallbusiness/1004/gallery.Franchise_failure_rates/index.htmlhttp://money.cnn.com/galleries/2010/smallbusiness/1004/gallery.Franchise_failure_rates/index.htmlhttp://money.cnn.com/galleries/2010/smallbusiness/1004/gallery.Franchise_failure_rates/index.htmlhttp://money.cnn.com/galleries/2010/smallbusiness/1004/gallery.Franchise_failure_rates/index.html


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2

Is there a higher failure rate for SBA loans to Days Inn than for

Super 8 Motel at the 0.05 level?

H0:

p1- p2= 0 (Or 0)

H1: p1- p2> 0

(Where p1= proportion of Days Inn failures;

p2= proportion of Super 8 Motel failures.)

Are the sample sizes sufficiently large to use the normal

approximation in CASE 6?(In Case 6, the relevant sample sizes are the number of successes and failures ineach sample; each must be at least 5, i.e., )p-(1n,pn),p-(1n,pn 22221111 .)

YES, all 4 groups 5.

The sample estimates of p1and p2are:

= 2= 0.0590; 2= 846= 0.0395.Consequently, the test statistic is:

=

(

) +

(

)

= . . .(.) + .(.)

= 0.01950.0151= 1.30.

Page 40 of 156

D0= 0


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3

OR, following the texts approach (which is appropriate only

when the null hypothesis states that the proportions are equal),

we could also use the overall rate of failure to calculate the

standard error of the test statistic. Since,

=

=23 +1

39 + = 0.0485 (see data on p.1),the test statistic becomes:

= .9 .39 .(.) + .(.)

=.19.1 = 1.32 .

With either test statistic we get essentially the same result:

Critical Value: Conclusion:

Z0.05= 1.645 Accept H0(No,the rate at Days Inn is not significantly higher.)

Which approach does MINITAB take?

Page 41 of 156

Case 1

Case 2

Cases 4 & 5

Case 7

Case 3

Case 6


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4

If We Do NOT Pool (which is the default unless we click on the usedpooled estimate... option ):

Test and CI for Two Proportions

Sample X N Sample p

1 23 390 0.058974

2 18 456 0.039474

Difference = p (1) - p (2)

Estimate for difference: 0.0195007

95% CI for difference: (-0.00992794, 0.0489293)

Test for difference = 0 (vs not = 0): Z = 1.30

P-Value = 0.194

Wait!! This p-value is for a two-sided test!

We need the p-value forH1: p1>p2 , which is: 0.194/2 =0.097

=> Accept H0.

Three options are provided here

1)Both samples in one column

2)Each sample in its own colum

3)Summarized data.

I could have selected the

appropriate one-sided alternati

here but instead used the defaul

option (the two-sided test).

3210-1-2-3

0.4

0.3

0.2

0.1

0.0

Page 42 of 156

D0= 0

The default setting is to notpool !

Sum of two tail probabilities

tail prob. is

0.194/2 =0.097


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5

If We Pool:

Test and CI for Two Proportions

Sample X N Sample p

1 23 390 0.058974

2 18 456 0.039474

Difference = p (1) - p (2)

Estimate for difference: 0.0195007

95% CI for difference: (-0.00992794, 0.0489293)

Test for difference = 0 (vs not = 0):

Z = 1.32 P-Value = 0.188 1-sided p-value=0.094 .

Other Caveats and Notes

1)1& 2 may seriously underestimate actual rates offailure, since the study includes recent loans to franchises

that probably will fail within 5 years (but had not yetfailed during the study period). To get better estimates,

each loan should be observed over a period of equal

duration. For example, we might observe each over a 5

year period (from the time of the loan is granted), and

1& 2would then be legitimate estimates of the failurerate of SBA loans to each franchise.

2) Sometimes data of this type are summarized in terms of

odds and odds ratios, especially in health/medical care

applications.

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6

Odds, Odds Ratio and Relative Risk

DefinitionIf an event occurs with probability p, then the odds of it occurring is

defined as p/(1-p).

I n the ExampleIf we use 1 = 6%as an estimate of the failure rate at Days Innfranchises and 2 = 4%as the corresponding estimate for Super 8Motel franchises, then the odds of failure are:

Days Inn franchises: 0.06/(1-0.06)= 0.0638;

Super 8 franchises: 0.04/(1-0.04)= 0.0417.

And the odds ratio (or ratio of odds for failure for Days Inn versus

Super 8 Motels) is:

Odds Ratio/() /() =

.. = . , (1)

indicating that the odds of failure is 1.5 times higher for the Days Innfranchises. (To turn this into a health care example: imagine

companies are people, and that failure is a disease to which certain

people are more susceptible.)

Alternatively, since this is a prospective study, sometimes the results

are summarized in terms of the relative riskof failure (for small versus

large), which is simply the ratio of

1 2:

Relative Risk = .. = . , (2)indicating that failure is 1.5 times more likely for the Days Inn. Of

course, remember that 1& 2are not good estimates, which is acommon problem in health/medical applications. Also, 1is not evensignificantly larger than 2at the 0.05 level!

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8

If I expect the CGM Focus Fund to outperform Fidelity Growth

Strategies Fund during 2010-2015, I might ask the following research

question: Does the CGM Focus have an average return that is

significantly more than 0.5% higher than the average annual

return of the Fidelity Growth Strategies during 2010-2014 (=0.1)?

Then: H0:CGM- Fidelity =0.5(OR < 0.5) H1:CGM- Fidelity > 0.5 .

The actual data are below.

Year CGM Focus

Fund

F idel ity Growth

Strategies Fund

Differences:

=CGMFidelity2010 16.94 25.63 - 8.692011 - 26.29 - 8.95 - 17.34

2012 14.23 11.78 2.452013 37.61 37.87 - 0.262014 1.39 13.69 - 12.302015 - 4.11 3.17 - 7.28Mean 6.63 13.87 - 7.24

We cant apply cases 4 or5 to this problem because the annual returnsare from the same years and are affected by the same market forces.

Consequently,

the two samples are not independent!But we can take differences (CGM minus Fidelitysee the last columnin the table above) and apply Case 2 to the single sample of differences.

The following hypotheses are equivalent to the ones above but arewritten in terms of the differences:

H0: Differences=0.5 (OR < 0.5) H1: Differences> 0.5.The mean and standard deviation of the five differences are:

=7.24; = ( ) = 7.38. Thus, the standard error of themean is:

=

.6 =3.01. Here are the details of the case 2 test.

Test statistic: = =7.240.5

3.01 = 2.57Critical Value: () = .() =. Conclusion:Accept H0 (No!)

The averagedifference makeclear we cannotreject H0 (Fidelitoutperforms CGBut we formallyapply the testanyway (as anillustration).

Because t is not greater than 1.48Page 46 of 156


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9

Addendum: Finding the Right Case

Large Sample (Case 1)

Mean

Small Sample (Case 2)

1 Sample

Proportion (Case 3)

Large Samples (Case 4)

When Either Sample Is Not Large, Use Welchs t

Means (OR: always useWelchs t)

Equal Variances (Case 5)

2 samples

Proportions (Case 6)

Paired Samples (Case 7)

Bruce Cooil, 2016

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Lectu re 6: Simp le Lin ear Regression

Outline: Main reference is Ch. 13, especially 13.1-13.2, 13.5, 13.8

The Why, When ,What and How of Regression

Purposes of Regression Three Basic Assumptions:

Linearity, Homoscedasticity, Random Error Estimation and Interpretation of the Coefficients

Decomposition of SS(Total) = ( )=1(See third equation on page 492:

SS(Total) is referred to there as Total Variation.) Measures of fit: MS(Error) (the variance of error), R2(Adjusted)

Purposes

1. To predict values of one variable (Y) given the values of

another (X). This is important because the value of X may

be easier to obtain, or may be known earlier.

2. To study the strength or nature of the relationship between

two variables.

3. To study the variable Y by controlling for the effects (orremoving the effects) of another variable X.

4. To provide a descriptive summary of the relationship

between X and Y.

Assumpt ions

The basic model is of the form:

(1) 0 + 1 + ,where 0, and 1are called coefficients, and represent unknownconstants (that will be estimated in the regression analysis), and

is used to represent random error. The error,, is assumed

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What

How

Why

When


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2

to come from a distribution with mean 0 and constant variance

2. The main result of the regression analysis is to provide

estimates of the coefficients so that we can use the estimatedregression equation,

(2) 0 + 1to predict Y.

Notes on Terminology and Notation

yis the predicted value and is referred to as the "fit" or the

"fitted value."

The residuals, ei,(the observed errors) are defined as thedifference between the actual and the predicted value of Y,

i.e.,

ei= [residual for observation i]= . Note that the theoretical error term, i, from equation (1), is

slightly different from the residuals:

i yi (0+ 1xi) versus ei yi (b0+ b1xi).

Formal ly the model makes the assumption that the errors (the

i)are a random sample from a distr ibution with mean 0 and

variance 2. This one assumption is sometimes referred to in 3

parts.

1. Linearity: there is a basic linear relationship between y andx as shown in (1), which is equivalent to saying that the

real mean of the errors (the i) is 0.

2. Homoscedasticity: the variance of the errors iis constantfor all yi.

3. Random Error: the errors iare independent from one

another.

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3

Two plots provide a way of checking these assumptions:

To check linearity: the plot of y versus x;

To check linearity, homoscedasticity and randomness: the

plot of the residuals, ( ), versus the fit values,.Plots of standardized residuals versus fit are especiallyuseful.

Imagine I have developed a special new product and that Idevelop a model to estimate the cost of producing it using data

from the first 5 orders.

Order Numberof

Units(x)

Cost(y)($1000)

Predicted Cost(or fit)

Residual

( )1 1 6 5 12 3 14 11 3

3 4 10 14 -4

4 5 14 17 -3

5 7 26 23 3

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4

Estimation and Interpretat ion o f Coefficients

In the plot above, the open circles are the actual observations of

y & x (cost & units), and the solid circles are the values of y & x(predicted, or fitted cost, & units). The vertical distances

between open circles and solid circles represent the observederrors or residuals of the regression model. The estimated

regression line is:

0 + 1(3) 2 + 3.The two estimated coefficients,

0 2 and 1 3,are chosen to minimize the sum of squared residuals or errors

that are made when we use the estimated regression equation to

predict cost (y). Note that in this case the sum of squared

residuals (or errors) is (see last column of table on previouspage):

SS(Error) = ( )=1= 12+32+(-4)2 +(-3)2+32= 44.

This is apparently the smallest value of sum of squared error

obtainable among all possible choices of b0and b1.

Please interpret these coefficients.

b0: predicted (or average) value of Y when X=0

(in this application it is the fixed cost) ;

b1: average change in Y per unit change in X

(in this application it is the variable cost).

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5

Decompo sit ion o f SS(Total)

Without this regression model, we might be forced to use the

average, , to predict future values of y. To get an indication ofhow we1l would do as a prediction, we can find the sum ofsquared differences between each yi& :

( )

=1=(6-14)2+(14-14)2+(10-14)2+(14-14)2+(26-14)2=224

(see the 3rdcolumn of the table on the next page). This sum ofsquares is referred to as SS(Total),i.e.,

SS(Total) = ( )= = 224 .

The regression model succeeds in reducing the uncertainty about

yif SS(Error)is significantly less than SS(Total). Also,

regression models actually allow us to decompose SS(Total)into two parts, SS(Error)and SS(Regression):

SS(Total) = SS(Regression) + SS(Error);

where: SS(Regression)= ( ) =1= the sum of squares of the fitted values around

their mean (the mean of the values is ).=(5-14)2+(11-14)2+(14-14)2+(17-14)2+(23-14)2

= 180

(see the 4th column of the table on the next page).

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6

So in this case, the decomposition of SS(Total) works out as

follows: SS(Total) = SS(Regression) + SS(Error)

224 = 180 + 44.

Summary of the Decomposition of SS(Total)

Units(x) Cost(y) ( ) ( ) ( )1 6 (6-14)2 (5-14)2 12

3 14 (14-14)2 (11-14)2 32

4 10 (10-14)2 (14-14)2 (-4)2

5 14 (14-14)2 (17-14)2 (-3)2

7 26 (26-14)2 (23-14)2 32

TOTALS: 224 = 180 + 44Name of SS: SS(Total)= SS(Regress.)+ SS(Error)

Minitab Summary: Main Regression Output of Version 17

(See Page 11 for a Compariso n w ith Excel)

Regression Analysis: Cost(y) versus Units(x)

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-ValueRegression 1 180.00 180.00 12.27 0.039

Units(x) 1 180.00 180.00 12.27 0.039

Error 3 44.00 14.67

Total 4 224.00

Model SummaryS R-sq R-sq(adj) R-sq(pred)

3.82971 80.36% 73.81% 38.11%

CoefficientsText Notat ion:

s

Term Coef SE Coef T-Value P-Value VIFConstant 2.00 3.83 0.52 0.638Units(x) 3.000 0.856 3.50 0.039 1.00

Regression EquationCost(y) = 2.00 + 3.000 Units(x)

224/4

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"MS" refers to "Mean Square" which is always t

corresponding SS (Sum of Squares) divided byDF (degrees of freedom): MS=SS/DF.

Variance of Error

Variance of Y

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7

Measures of Fit (Mod el Summary)

Note that on the line just below the Analysis of Variance table in

the MINITAB output, there are 4 primary measures of fit:

s=3.83, R-sq=80.4%, R-Sq(adj)=73.8%, R-sq(pred)=38.1 .

The first three can be calculated using the information in the

Analysis of Variance table. The standard deviation srepresents the estimated standard deviation of the residuals, or

observed errors, also written as s,s= [Variance of observed errors]

1/2

= [SS(Error)/(n-[# parameters in model])]1/2

= [44/(5-2)]1/2= [14.67]1/2= 3.83.

[The text calls "s" the "standard error," and writes it as simply s.See the shaded expressions on page 479.]

Note that the variance of the errors, 14.67, is also provided in

the Analysis of Variance table, and is often referred to there as

"MS(Error)" or Adj MS(Error)which is shorthand for"adjusted mean squared error." "Mean Square" is generally used

as a synonym for "variance." For these data:

s= [MS(Error)]1/2 = [14.67]1/2= 3.83 .

Another obvious overall measure of how well the model

performs would be:

()() ()()

which is the proportion of SS(Total) generated or "explained" by

the model. R2is often referred to as the "coefficient of

determination."

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8

In this example:

()() . . %

(OR: ()() )where 180," 44,and 224" are all shown in the Analysis of

Variance table.

A better measure of fit is found by adjusting R2so that it

estimates the proportion of the varianceof y that is explained

by the fitted values from the model. This proportion is referredto as "R2(Adjusted),"

() ()() .In this case:

()

( )

[ ( ) ] .

. . %

Note that 14.67" is shown in the Analysis of Variance

table.

R2(Adj) represents the proportion of the variance of Y that is"explained" (or generated) by the regression equation, while s

represents the estimated standard deviation of the residuals. I n

this example, 73.8% of the variance in cost (Y) is" explained"

by the model that uses units (X) as a predictor and the

standard deviation of the errors made by this model is 3.8

thousand doll ars.

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9

In simple linear regression, R2(unadjusted) is generally written

as "r2" and it represents the squared correlation coefficient (alsosee page 494 of the text). The estimated correlation between

cost (Y) and units (X) is 0.896. See the correlation matrixbelow.

MTB > corr c1 c2 c3Correlations (Pearson)

Units(x) Cost(y)Cost(y) 0.896

0.039

FITS1 1.000 0.896* 0.039

Formulas for Correlation (Pearson Correlation) :

.

)1(

1*

)1(

1

)1/(

2/1

n

1i

2n

1i

2

n

1i

yyn

xxn

nyyxx

ii

ii

(See pp. 125-127, 492-495 of the text for more examples and discussion.)

Alternatively:r = (sign of b1) * [square root of R2(f rom simple regression)].

I n the example: r = + 896.0804.0 .

Note that in general, r is between -1 and +1.

Cell Contents: Correlation

P-value

In spreadsheet:c1: Units(x); c2: Cost(y); and c3: Fits1

r

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10

Discussion Questions

(The Regression Output Is Redisplayed on the Next Page.)

1. Use the regression equation to predict the cost(Y) when number of units(X) is 4.

y = b0+ b1X = 2 + 3(4) = 14 thousand dollars

2. What was the actual cost for an order when units =4? (What is theresidual or error at that point?)

From page 3: When Units (X) is 4, Y is 10,

thus: residual = y - y = 10 - 14 = -4.

3. What is the sample variance of cost (Y)? (See next page.)

SY2= SS(Total)/4 = 224/4 = 56( SY= 56 = 7.5 thousand )

4. What is the estimated variance of the residuals (or errors) of theregression?

S2 = MS(ERROR) = 14.67 (Find this in Analysis of Variance Table!)

5. How good is the fit?There are two ways of measuring fit (see the Model Summary):

S= 3.83 thousand dollars

R2(Adjusted) = 73.8%

(74% of the Variance in cost(Y) is explained by the model.)

6. Show how R2(Adjusted) is related to the variance of cost and the varianceof the residuals?

= 1 (

)

= 1 14.6756 7. Show how R2(unadjusted) is related to the correlation between cost (Y) and units (X).

R2 = r2 (r represents the sample correlation; this only works insimple linear regression!!)

(On next page: R2= 0.804; on page 9: r = 0.896.)

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11

Appendix

Comparison of MINITAB Output (Versions 14-17) with Excel

MINITAB 17 Output

Regression Analysis: Cost(y) versus Units(x)

Analysis of VarianceSource DF Adj SS Adj MS F-Value P-ValueRegression 1 180.00 180.00 12.27 0.039Units(x) 1 180.00 180.00 12.27 0.039

Error 3 44.00 14.67 #4Total 4 224.00

Model Summary

224/4 = Var(Y) for #3S R-sq R-sq(adj) R-sq(pred)

3.82971 80.36% 73.81% 38.11%#5

CoefficientsTerm Coef SE Coef T-Value P-Value VIFConstant 2.00 3.83 0.52 0.638Units(x) 3.000 0.856 3.50 0.039 1.00

Regression EquationCost(y) = 2.00 + 3.000 Units(x)

Excel OutputSUMMARY OUTPUT

Regression Statistics

Multiple R 0.896421

R Square 0.803571

Adjusted R

Square 0.738095

Standard

Error 3.829708

Observations 5

ANOVA

Df SS MS F

Significance

FRegression 1 180 180 12.27273 0.039389

Residual 3 44 14.66667

Total 4 224

Coefficients

Standard

Error t Stat P-value Lower 95%

Upper

95%

Lower

95.0%

Upper

95.0%

Intercept 2 3.829708 0.522233 0.637618 -10.18784 14.18784

-

10.18784 14.18784

X Variable 1 3 0.856349 3.503245 0.039389 0.274716 5.725284 0.274716 5.725284

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12

Interpreting the Plot of Residuals versus Fit

One way to check on the three assumptions (linearity,

homoscedasticity and random error) is to plot the residuals

(errors) against the predicted (or fitted) values .

There are hardly enough observations here to be very confident

in the assumptions. But in general we look for symmetry around

the horizontal line through zero as an indication that the

assumptions of linearity and randomness are met. To confirmhomoscedasticity, we look for roughly constant vertical

dispersion around the horizontal line through zero.

The ideal situation generally looks something like the followingplot.

-10 0 10 20

-3

-2

-1

0

1

2

Fitted Value

Residual

Residuals Versus the Fitted Values

(response is C3)

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13

Here is a situation where the linearity assumption is violated.

Here is a common situation (below) where homoscedasticity is

violated: notice how the residuals show increasing vertical

dispersion around the horizontal line through zero as the fitted

values increase.

5 10 15

-10

0

10

20

30

Fitted Value

Residual

Residuals Versus the Fitted Values

(response is C3)

-10 0 10 20 30

-10

0

10

Fitted Value

Residual

Residuals Versus the Fitted Values(response is C3)

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14

How to Do This Regression Analysis in MINITABMinitab 17

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15

How to do a Regression Analysis in Excel

Click into the Data menu and check for the Data Analysis

option (far right).

If the Data Analysis option is not there :

Start from File menuClick on OptionsClick onAdd-InsSelect Analysis ToolPak & hit Go near

the bottom of the dialog box.

Otherwise start from the Data menu:Click on Data Analysis (far right), Select Regression and

then specify the Y- and X-range in the dialog box.

(You can simply click into each range box and then move themouse directly into the spreadsheet to select the numerical

data cells from the appropriate column(s) of the spreadsheet.

The appropriate range of cells should then appear in the range

box. The Input X Range may consist of several columns,

each column for a different predictor.)

Other Good References

See page 519 of the text for an example with great screenpictures. Another good reference is:

www.wikihow.com/Run-Regression-Analysis-in-Microsoft-Excel.

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Lecture 6 Addendum: Terminology, Examples, and Notation

Regression Terminology

Synonym Groups

1) Y, Dependent Variable, Response Variable

2) X, Predictor Variable, Independent Variable

3) , Prediction, Predicted Value, Fit, Fitted Value4) Variance of Y, MS(Total), Adj MS(Total)

5) Variance of Error, MS(Error), Adj MS(Error)

6) , Error, Residual7) Coefficientsare sometimes referred to using the more general term parameters.

Coefficientsare the parametersthat are used in linear models.

Main Ideas

Simple linear regression refers to a regression model with only one predictor. The underlying

theoretical model is:

= 0 + 1 + ,where yrepresents a value of the dependent variable, xis a value of the predictor, representsrandom error and and 1represent unknown constants.The corresponding estimate regression equation is:

= 0 + 1.The regression coefficient b0and b1 refer to sample estimates of the true coefficients 0and 1,

respectively.

The sample correlation coefficient, r, estimates the true (or population) value of the correlation,

, which is a measure of the degree to which two variables (Y and X) are linearly related.

Of course, the sample correlation (r) and the slope coefficient (b1)are closely related:

1 = = (,) () , (*)where and are the sample standard deviations of Y and X respectively.The corresponding relationship between the true values, 1 and ,is:

1 = =(,)

=

(,)2

.

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Examples of Correlation

Correlation: r= 0.725 (R2(unadjusted) = r2x100% = 52.6%); y= -756.6 + 12.25 x.

Change in GDP (Y): change in Annual U.S. GDP in billions of dollars

Consumer Sentiment (X): Index of financial well-being and the degree to which consumer

expectations are positive (based on five questions on a survey conducted by the University of

Michigan. (https://data.sca.isr.umich.edu/fetchdoc.php?docid=24770)

Correlation: r= 0.725 (R2(unadjusted) = r2x100% = 51.3%); y= 44.45 +0.3045 x.

Data are from the World Health Organization (Life Expectancy (Y) as of 2015, Literacy Rate (X)

for 2007-2012).

11010090807060

500

250

0

-250

-500

Consumer Sentiment

ChangeinGDP

Change in GDP vs Consumer Sentiment (1995-2015)

2009

1999

2015

1009080706050403020

85

80

75

70

65

60

55

50

Literacy Rate (for People >=15 years-old)

LifeExpe

ctancy(BothSexesinyears)

Life Expectancy(Both_Sexes) vs Literacy Rate (>=15 years) for 112 Nations

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Correlation: r = -0.891 (R2(unadjusted) = r2x100% = 79.4%);

MPG = 41.710.006263 Weight.

Data are for 14 automobiles (2005) from www.chegg.com.

Correlation: r= 0.018 (R2(unadjusted) = r2x100% = 0.1%);

Random Y = 0.06153 + 0.01688 Random X.

Y and X are two sets of 1000 standard normal random numbers.

4000375035003250300027502500

28

26

24

22

20

18

16

Weight

MPG

MPG (City) vs Weight (Lbs)

43210-1-2-3-4

4

3

2

1

0

-1

-2

-3

-4

Random X

RandomY

Random Y vs Random X

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Notation for Types of Variation and R2

For linear regression models (with one or more predictor variables), the basic types of sums of

squares represent three types of variation.

1. Total Variation = ( ) ()The sum of squares of the observations of Y around their mean.

2. Explained Variation = ( ) ()The sum of squares of the predicted values of Y ( ) around their mean (which is also ).

3. Unexplained Variation = ( ) ()The sum of squares of the differences between each observation and the corresponding

predicted value.

Note:

Total Variation = Explained Variation + Unexplained Variation

Or: SS(Total) = SS(Regression) + SS(Error)

The R2 (Unadjusted) is sometimes called the simple coefficient of determination and it is

the square of the correlation:

() = = =()

()= ()()

R2 (Adjusted) is a more accurate assessment of the strength of the relationship between Y

and X. In general:

R2 (Adjusted) = ()()

=

() ( [# ])

() ( )

For simple linear regression, which includes a constant and a slope coefficient: [# of

parameters in model] = 2.

[Reference: pp. 493-495, Essentials of Business Statistics (2015), 5thEdition, Bowerman

et al.]

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Lecture 7Inferences About Regression Coefficients &

Confidence/Prediction Intervals for Y/Y

Outline:(Ref: Ch. 13: 13.3-13.4, 13.6-13.7) Recap of Main Ideas from Lecture 6

Testing Lack-of-Fit

Inferences Based on Regression Coefficients (Ch. 13.3)

Prediction Intervals versus Confidence Intervals for Y (Ch. 13.4)

(Please read pp. 486-489, not for details on how PIs and

CIs are calculated but for the main idea of what they tell

us about Y!)

Summary of Ideas from Lecture 6

3 Assumptions:

The basic relationship between Y and X is linear up to a

random error term that has mean 0 (linearity) and constant

variance (homoscedasticity). Errors are random in the sense

that they are independent of each other and do not depend on

the value of Y.

One way to check these assumptions is to plot residuals versus

fitted values.

The coefficients estimates b0, and b1, are chosen to minimize the

sum of squared errors (or residuals).

b1represents the average change in Y that is associated with aone unit change in X.

Regression is useful because it allows us to reduce the

uncertainty regarding Y. We can think about this is in terms of

the decomposition of SS(Total) (NOTE: SS is used in

regression to refer to Sums of Squares):

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2

SS(Total) = SS(Regression) + SS(Error)

( )2= = ( ) 2= + ( ) 2=

= +

This decomposition of total uncertainty (SS(Total)) suggests two

useful summaries of how well the model fits:

1) 2() 1 ()/([# ])()/() 1 ()() .(Recall that:

2() 1 ())() . )2) ()

{()/( # )}.

Appl icat ion

Data are available from Blackboard (with this lecture note).These data are from 43 urban communities (2012)

Citation: Cost of Living Index, Council for Community and

Economic Research, January, 2013.

MTB > info c1-c3

Information on the WorksheetColumn Count Name

T C1 43 URBAN AREA AND STATE

C2 43 HOME PRICEAvg for 2400 sq. ft. new home, 4 bed, 2 bath on 8000 sq.ft. lot

C3 43 Apt RentAvg for 950 sq. ft. unfurnished apt., 2 bed, 1.5-2bathexcluding all utilities except water.

Other interesting data sets on home prices and rental rates by city:

https://smartasset.com/mortgage/price-to-rent-ratio-in-us-cities

https://smartasset.com/mortgage/rent-vs-buy#map.

SS of observed

y-valuesaround mean

SS of fitted

values aroundthe mean

SS of errors (errors

are the actual y minusfitted or predicted y)

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Variance of Error

Variance of Y


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3

Regression for All 43 Cities

Regression Using 38 Cities Where Rent < $1500

4000350030002500200015001000

1400000

1200000

1000000

800000

600000

400000

200000

S 70561.6

R-Sq 88.2%

R-Sq(adj) 87.9%

Apt Rent

HOMEPRICE

Fitted Line PlotHOME PRICE = - 61366 + 339.3 Apt Rent

New York (Manhattan) NY

New York (Brooklyn) NY

San Francisco CA

Honolulu HI

New York (Queens) NY

150014001300120011001000

500000

450000

400000

350000

300000

250000

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7/25/2019 Lectures 1-13 +Ou

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