STAT 3022 SLIDES UMN CHAPTER3

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Intrduction Robustness Resistance Transformation Outlier

Chapter 3

A Closer Look at Assumptions

STAT 3022School of Statistic, University of Minnesota

2013 spring

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Introduction

In Chapter 2, we discussed the mechanics of using t-proceduresto perform statistical inference. Namely t-tests and confidenceinterval.

We base these procedures on certain assumptions:

we have random samples, representative of populations

data come from Normal population

samples are drawn independently.

in pooled two-sample settings, we have equal variance(1 = 2 = )

In practice, these assumptions are usually not strictly met.When are these procedures still appropriate?

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Case Study: Making it Rain

Data collected in southern Florida between 1968 - 1972 to testhypothesis that massive injection of silver iodide (AgI) intocumulus clouds can lead to increased rainfall.

This process is called cloud seeding. Over 52 days, either

seeded a target cloud or left it unseeded (as control). Randomlyassigned treatment.

Researchers were blindto the treatment - pilots flew throughcloud every day, whether treatment or control, and mechanismin plane either seeded the cloud or left it unseeded.

Question: Did cloud seeding have an effect on rainfall? If so,how much?

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Graphical Summaries

library("Sleuth2")

boxplot(Rainfall ~ Treatment, ylab='Rainfall (acre-feet)', data=case0301)

Unseeded Seeded

0

500

1000

1500

2000

2500

Rainfall(acre

feet)

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Graphical Summaries

par(mfrow=c(2,1), mar=c(4,4,1,0.5))

hist(case0301$Rainfall[case0301$Treatment=="Seeded"], breaks=10,

main="Seeded - Rainfall", xlim=c(0,3000), col="gray", xlab="")hist(case0301$Rainfall[case0301$Treatment=="Unseeded"], breaks=8,

main="Unseeded - Rainfall", xlim=c(0,3000),col="gray", xlab="")

ee e a n a

Frequency

0 500 1000 1500 2000 2500 3000

0

2

4

6

8

10

12

nsee e a n a

Frequency

0 500 1000 1500 2000 2500 3000

0

5

10

15

20

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Numerical Summaries and Interpretations

Numerical Summaries: Do it yourself (follow the R-code onpage 42 of Chapter 2 slides)

Graphical and numerical summaries indicate that rainfall

tended to be greater on seeded days. However, there areproblems with our necessary assumptions:

both distributions are very skewed

both distributions have outliers

variability is much greater in the seeded group than in theunseeded group

Can we use our usual t-tools to analyze these data? How?

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Can we do this?

> t.test(Rainfall ~ Treatment, alternative="two.sided",

+ var.equal=TRUE, data=case0301)

Two Sample t-test

data: Rainfall by Treatment

t = -1.9982, df = 50, p-value = 0.05114

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-556.224179 1.431851

sample estimates:

mean in group Unseeded mean in group Seeded

164.5885 441.9846

How much did the violations of our assumptions affect theseresults?

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I t d ti R b t R i t T f ti O tli


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Robustness

t-tools may be used even when assumptions are violated, to acertain degree, because the t-tools are robust.

Robustness: A statistical procedure is robust to departuresfrom a particular assumption if it is valid even when theassumption is not met.

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Type 1: Robustness Against Departures fromNormality

Recall that the Central Limit Theorem (CLT) states that sample

averages have approximately Normal sampling distributions,regardless of the shape of the population distribution, for largesamples.

As long as samples are large enough, the t-ratio will follow an

approximate t-distribution even if the data is non-Normal.

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Type 1: Robustness Against Departures fromNormality

Effects of Skewness

If two populations have same standard deviations andapproximately same shapes, and ifn1 n2, then validityoft-tools is affected very little by skewness.

If two populations have same standard deviations andapproximately same shapes, but n1 = n2, then validity oft-tools is affected substantially by skewness. Larger samplesize diminish this effect.

If skewness in two populations differs considerably, tools

can be very misleading with small and moderate samplesizes.

See Display 3.4 in the textbook for simulation results.

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Type 2: Robustness Against Differing StandardDeviations

When we cannot assume 1 = 2, more serious problems mayarise:

sp no longer estimates any parameterSE(x1 x2) no longer estimates the standard deviation ofthe difference between averagesthe t-ratio no longer follows a t-distribution

What can we do:

Ifn1 n2, t-tools remain fairly valid even when 1 = 2.

When n1 and n2 are very different, we need the ratio 1/2to be between 1/2 and 2 to have reliable results.

See Display 3.5 in the textbook for simulation results.

> t.test(x1, x2, alternative = 'two.sided', var.equal = FALSE)11/27


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Resistance and Outliers

An outlier is an observation judged to be far from its groupaverage.

A statistical procedure is resistant if it does not change verymuch when a small part of the data changes, perhapsdrastically.

Whether or not we should simply remove such observationsdepend on how resistant our tools are to changes in the data.

Question: Can you tell the difference between Robustnessand Resistance?

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Example of Outlier

6 4 2 0 2 4 6

1.0

0.

5

0.

0

0.

5

1.

0

3 2 1 0 1 2 3

3

2

1

0

1

2

3

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Example of Resistance

Consider a hypothetical sample:

10, 20, 30, 50, 70

The sample mean is 36, and the sample median is 30.

Now consider the sample:

10, 20, 30, 50, 700

What happens to the sample mean? What about the sample

median?

The sample median is resistant to any change in a singleobservation, while the sample mean is not.

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Practical Strategies for the Two-Sample Problem

Our task is to size up actual conditions, using available data,and evaluate appropriateness of t-tools:

1 think about possible cluster and serial effects

2 evaluate the suitability of t-tools by examining graphicaldisplays (side-by-side histograms or box plots)

3 consider alternatives

a. Transform the data (Section 3.5) to see if the transformeddata looks nicer

b. Alternative tools that do not require model assumptions(Chapter 4)

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Transformations of Data

For positive data, the most useful transformation is the

logarithm (log), particularly the natural (base e) logarithm (e =2.71828...).

log(1) = 0log(ex) = x

0 2 4 6 8 10

2

1

0

1

2

log function

x

log

(x)

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Cloud Seeding - Transformation

Recall both groups are skewed, with the seeded days having alarger average and a greater spread.> max(case0301$Rainfall[case0301$Treatment=="Seeded"])/

+ min(case0301$Rainfall[case0301$Treatment=="Seeded"])

[1] 669.6586

> max(case0301$Rainfall[case0301$Treatment=="Unseeded"])/

+ min(case0301$Rainfall[case0301$Treatment=="Unseeded"])

[1] 1202.6

> case0301$logRain head(case0301)

Rainfall Treatment logRain

1 1202.6 Unseeded 7.092241

2 830.1 Unseeded 6.721546

3 372.4 Unseeded 5.919969

4 345.5 Unseeded 5.844993

5 321.2 Unseeded 5.772064

6 244.3 Unseeded 5.498397

Unseeded Seeded

0

500

1000

1500

2000

2500

before transformation

Unseeded Seeded

0

2

4

6

8

after transformation

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Two-Sample t-Analysis

Before:> t.test(Rainfall ~ Treatment, alternative="two.sided",

+ var.equal=TRUE, data=case0301)

Two Sample t-test

data: Rainfall by Treatment

t = -1.9982, df = 50, p-value = 0.05114

alternative hypothesis: true difference in means is not equal to 0


-556.224179 1.431851

sample estimates:mean in group Unseeded mean in group Seeded

164.5885 441.9846

After:> t.test(logRain ~ Treatment, data=case0301,

+ alternative="less", var.equal=TRUE)

Two Sample t-test

data: logRain by Treatment

t = -2.5444, df = 50, p-value = 0.007041

alternative hypothesis: true difference in means is less than 0


-Inf -0.3904045

sample estimates:

mean in group Unseeded mean in group Seeded

3.990406 5.134187

There is convincing evidence that seeding increased rainfall. 21/27


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Multiplicative Treatment Effect

Definition: Suppose Z= logY. It is estimated that the responseof an experimental unit to treatment 2 will be eZ2Z1 times aslarge as its response to treatment 1 (where Z1 = average oflog(Y1)).

> m1 m2 (diffmeans (est.mult.effect


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Confidence Interval

> (test test$conf.int

[1] -2.0466973 -0.2408651

attr(,"conf.level")

[1] 0.95

> exp(test$conf.int)

[1] 0.1291608 0.7859476

attr(,"conf.level")

[1] 0.95

A 95% confidence interval for the multiplicative effect ofunseeding/seeding is 0.129 to 0.786 times.

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R i O tli d Oth D t P i t


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Removing Outliers and Other Data Points

> library(Sleuth2); ex0327[15:17, ]

Country Life Income Type

15 Portugal 68.1 956 Industrialized16 South_Africa 68.2 NaN Industrialized

17 Sweden 74.7 5596 Industrialized

> range(ex0327$Income, na.rm=TRUE)

[1] 110 5596

> data

> d1 ### dealing with Missing data ###

> (cc data2


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Q: How many conservative economists does it take to change alight bulb?

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A: None, theyre all waiting for the unseen hand of the marketto correct the lighting disequilibrium.

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STAT 3022 SLIDES UMN CHAPTER3