Text Book : Moore, McCabe and Craig, Introduction …€¦ · Text Book : Moore, McCabe and Craig,...

� Text Book : Moore, McCabe and Craig, Introduction to the Practice of Statistics, 6th ed.

� Class Web Page: http://www.stt.msu.edu/academics/classpages/

Choose STT421.

READ YOUR SYLLABUS WHICH IS ALSO POSTED ON THE CLASS PAGE!!!!!!!

Why are you taking this course?

a) It is required.

b) I am so bored that I enrolled the first course that I came across.

c) I heard that statistics can lie and I just wanted to check if it was true.

d) The name is funny and interesting.

e) Is it the statistics class that I am in? I will definitely drop this course as soon as this boring class ends.

f) I am aware of the fact that the science of statistics would really help me in my daily life let alone for my profession.

“It is easy to lie with statistics. It is hard to

tell the truth without it.”

Andrejs Dunkels

Looking at Data - DistributionsDisplaying Distributions with Graphs

IPS Chapter 1.1

© 2009 W.H. Freeman and Company

Objectives (IPS Chapter 1.1)

Displaying distributions with graphs

� Variables

� Types of variables

� Graphs for categorical variables

� Bar graphs

� Pie charts

� Graphs for quantitative variables

� Histograms

� Stemplots

� Stemplots versus histograms

� Interpreting histograms

� Time plots

What Is Statistics?

Why?1. Collecting Data

e.g., Survey

2. Presenting Data

e.g., Charts & Tables

3. Characterizing Data

e.g., Average

Data

Analysis

Decision-

Making

© 1984-1994 T/Maker Co.

© 1984-1994 T/Maker Co.

What Is Statistics?

Statistics is the science of data. It involves collecting, classifying, summarizing, organizing, analyzing, and interpreting numerical information.

Application Areas

�Economics

� Forecasting

� Demographics

�Sports

� Individual & Team

Performance

�Engineering

� Construction

� Materials

�Business

� Consumer Preferences

� Financial Trends

Statistical Methods

Statistical

Methods

Descriptive

Statistics

Inferential

Statistics

Descriptive Statistics

1. Involves

• Collecting Data

• Presenting Data

• Characterizing Data

2. Purpose

• Describe Data

X = 30.5 S2 = 113

0

25

50

Q1 Q2 Q3 Q4

$

1. Involves

• Estimation

• Hypothesis Testing

2. Purpose

• Make decisions about population characteristics

Inferential Statistics

Population?

What is a Statistical Inference?

� It is an estimate or prediction or some generalization about a population based on information contained in a sample.

Fundamental Elements

� Experimental unit(cases or individuals)

� Object upon which we collect data

� Example: people, animals, plants, or any object of interest.

� Population

� All items of interest

� Variable

� Characteristic of an individual experimental unit

� A variable varies among individuals.

� Example: age, height, blood pressure, ethnicity, leaf length, first language

� The distribution of a variable tells us what values the variable takes and how often it takes these values.

� Sample

� Subset of the units of a population

Two types of variables

� Variables can be either quantitative…

� Something that takes numerical values for which arithmetic operations,

such as adding and averaging, make sense.

� Example: How tall you are, your age, your blood cholesterol level, the

number of credit cards you own.

� … or categorical (qualitative).

� Something that falls into one of several categories. What can be counted

is the count or proportion of individuals in each category.

� Example: Your blood type (A, B, AB, O), your hair color, your ethnicity,

whether you paid income tax last tax year or not.

How do you know if a variable is categorical or quantitative?

Ask:

� What are the n individuals/units in the sample (of size “n”)?

� What is being recorded about those n individuals/units?

� Is that a number (� quantitative) or a statement (� categorical)?

Individualsin sample

DIAGNOSIS AGE AT DEATH

Patient A Heart disease 56

Patient B Stroke 70

Patient C Stroke 75

Patient D Lung cancer 60

Patient E Heart disease 80

Patient F Accident 73

Patient G Diabetes 69

QuantitativeEach individual is

attributed a numerical value.

CategoricalEach individual is assigned to one of several categories.

Ways to chart categorical data

� Summary Table

� Bar graphs

� Pie chart

� Lists categories & number of elements in category

� Obtained by tallying responses in category

� May show the class frequencies and/or relative

frequencies

Summary table

Example: Top 10 causes of death in the United States 2001

Rank Causes of death Counts% of top

10s% of total

deaths

1 Heart disease 700,142 37% 29%

2 Cancer 553,768 29% 23%

3 Cerebrovascular 163,538 9% 7%

4 Chronic respiratory 123,013 6% 5%

5 Accidents 101,537 5% 4%

6 Diabetes mellitus 71,372 4% 3%

7 Flu and pneumonia 62,034 3% 3%

8 Alzheimer’s disease 53,852 3% 2%

9 Kidney disorders 39,480 2% 2%

10 Septicemia 32,238 2% 1%

All other causes 629,967 21%

For each individual who died in the United States in 2001, we record what was

the cause of death. The table above is a summary of that information.

0

100

200

300

400

500

600

700

800

Hear

t dis

ease

s

Canc

ersC

erebr

ovas

cula

r

Chro

nic

resp

irato

ryA

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ents

Dia

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ellit

usFlu

& p

neum

onia

Alz

heim

er's d

isea

seK

idne

y dis

orde

rsS

eptic

emia

Counts

(x1000)

Top 10 causes of deaths in the United States 2001

Bar graphs

Each category is represented by one bar. The bar’s height shows the count (or

sometimes the percentage) for that particular category.

The number of individuals who died of an accident in

2001 is approximately 100,000.

0

100

200

300

400

500

600

700

800

Hear

t dis

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s

Canc

ersC

erebr

ovas

cula

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Chro

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resp

irato

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ccid

ents

Dia

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ellit

usFlu

& p

neum

onia

Alz

heim

er's d

isea

seK

idne

y dis

orde

rsS

eptic

emia

Counts

(x1000)

Bar graph sorted by rank� Easy to analyze

Top 10 causes of deaths in the United States 2001

0

100

200

300

400

500

600

700

800

Acc

iden

ts

Alz

heim

er's d

isease

Canc

ersC

erebr

ovas

cula

r

Chroni

c re

spira

tory

Dia

bete

s m

ellit

usFlu

& p

neum

onia

Hear

t dis

ease

sK

idne

y diso

rder

sS

eptic

emia

Counts

(x1000) Sorted alphabetically

� Much less useful

Percent of people dying fromtop 10 causes of death in the United States in 2000

Pie chartsEach slice represents a piece of one whole. The size of a slice depends on what

percent of the whole this category represents. Angle size:(360°) (9%) = 32.4°

Make sure all percents add up to 100.

Bar Chart vs. Pie Chart

� Bar chart is used more often to represent the actual values while pie chart is used to represent relative proportions (in %).

� When comparison of relative proportion is important, pie chart is more appropriate.

� When the absolute counts or values are more important, a bar chart should be used.

Ways to chart quantitative data

� Histograms and stem-and-leaf plots (or stemplots)

These are summary graphs for a single variable. They are very useful to

understand the pattern of variability in the data.

� Line graphs: time plots

Use when there is a meaningful sequence, like time. The line connecting

the points helps emphasize any change over time.

Histograms

The range of values that a

variable can take is divided

into equal size intervals.

The histogram shows the

number of individual data

points that fall in each

interval.

The first column represents all states with a Hispanic percent in their

population between 0% and 4.99%. The height of the column shows how

many states (27) have a percent in this range.

The last column represents all states with a Hispanic percent in their

population between 40% and 44.99%. There is only one such state: New

Mexico, at 42.1% Hispanics.

Stem plots

How to make a stemplot:

1) Separate each observation into a stem, consisting of

all but the final (rightmost) digit, and a leaf, which is

that remaining final digit. Stems may have as many

digits as needed, but each leaf contains only a single

digit.

2) Write the stems in a vertical column with the smallest

value at the top, and draw a vertical line at the right

of this column.

3) Write each leaf in the row to the right of its stem, in

increasing order out from the stem.

STEM LEAVES

Example

Suppose we have the following data on weights (in lb) of 17 school-kids:

88 47 68 76 46 106 49 63 72 64 84 66

68 75 72 81 44

How do they work?

Sorted data:

44 46 47 49 63 64 66 68 68 72

72 75 76 81 84 88 106

Stem Leaf

4 4 6 7 9

5

6 3 4 6 8 8

7 2 2 5 6

8 1 4 8

9

10 6

key: 6|3 = 63

leaf unit: 1.0

stem unit: 10.0

S tate Percent

A labam a 1.5A laska 4.1A rizona 25.3A rkansas 2.8C alifo rn ia 32.4

C olorado 17.1C onnec ticut 9.4D elaw are 4.8Florida 16.8G eorg ia 5.3H aw aii 7.2

Idaho 7.9Illino is 10.7Ind iana 3.5Iow a 2.8K ansas 7

K entucky 1.5Louis iana 2.4M aine 0.7M aryland 4.3M assachusetts 6.8

M ichigan 3.3M innesota 2.9M iss iss ipp i 1.3M issouri 2.1M ontana 2N ebraska 5.5

N evada 19.7N ew H am psh ire 1.7N ew Jersey 13.3N ew M exico 42.1N ew York 15.1

N orthC aro lina 4.7N orthD akota 1.2O hio 1.9O klahom a 5.2O regon 8

P ennsylvan ia 3.2R hodeIs land 8.7S outhC aro lina 2.4S outhD akota 1.4Tennessee 2

Texas 32U tah 9V erm ont 0.9V irg in ia 4.7W ash ington 7.2W estV irg in ia 0.7

W iscons in 3.6W yom ing 6.4

Percent of Hispanic residents

in each of the 50 states

Step 2:

Assign the values to

stems and leaves

Step 1:

Sort the data

S ta te P e rc ent

M a in e 0 .7W es tV irg in ia 0 .7V erm on t 0 .9

N orth D ak o ta 1 .2M is s is s ip p i 1 .3

S ou th D ak o ta 1 .4A la bam a 1 .5K en tu c k y 1 .5

N ew H a m ps h ire 1 .7O h io 1 .9M o n ta na 2

Te nne s se e 2M is s ou ri 2 .1

Lo u is ia na 2 .4S ou th C aro lina 2 .4A rk an s as 2 .8

Iow a 2 .8M inn es o ta 2 .9P enn s ylvan ia 3 .2

M ic h iga n 3 .3Ind ian a 3 .5W is c o ns in 3 .6

A la s ka 4 .1M a ryland 4 .3

N orth C aro lina 4 .7V irg in ia 4 .7D e law are 4 .8

O k lah om a 5 .2G e org ia 5 .3N eb ra s k a 5 .5

W yom in g 6 .4M a s s ac hu se tts 6 .8K ans a s 7

H aw a ii 7 .2W as h ing ton 7 .2

Idah o 7 .9O reg on 8R ho de Is lan d 8 .7

U ta h 9C on nec tic u t 9 .4Ill ino is 10 .7

N ew J ers ey 13 .3N ew Yo rk 15 .1F lo rida 16 .8

C o lo rad o 17 .1N eva da 19 .7

A rizo na 25 .3Te xas 32C a lifo rn ia 32 .4

N ew M exic o 42 .1

Stem Plot

� To compare two related distributions, a back-to-back stem plot with common stems is useful.

� Stem plots do not work well for large datasets.

� When the observed values have too many digits, trim the numbers before making a stem plot.

� When plotting a moderate number of observations, you can spliteach stem.

� Unlike histograms, stem-and-leaf displays retain the original data.

� Stemplots are quick and dirty histograms that can easily be done by

hand.

� However, they are rarely found in scientific or laymen publications.

Stemplots versus histograms

Interpreting histograms

When describing the distribution of a quantitative variable, we look for the

overall pattern and for striking deviations from that pattern. We can describe

the overall pattern of a histogram by its shape, center, and spread.

Histogram with a line connecting

each column � too detailed

Histogram with a smoothed curve

highlighting the overall pattern of

the distribution

Most common distribution shapes

� A distribution is symmetric if the right and left

sides of the histogram are approximately mirror

images of each other.

Symmetric distribution

Skewed distribution

� A distribution is skewed to the right if the right

side of the histogram (side with larger values)

extends much farther out than the left side. It is

skewed to the left if the left side of the histogram

extends much farther out than the right side.

Alaska Florida

Outliers

An important kind of deviation is an outlier. Outliers are observations

that lie outside the overall pattern of a distribution. Always look for

outliers and try to explain them.

The overall pattern is fairly

symmetrical except for 2

states that clearly do not

belong to the main trend.

Alaska and Florida have

unusual representation of

the elderly in their

population.

A large gap in the

distribution is typically a

sign of an outlier.

A Few Questions

� How to choose the bin size? Let the computer decide it for

you. ☺

Line graphs: time plots

A trend is a rise or fall that

persists over time, despite

small irregularities.

In a time plot, time always goes on the horizontal, x axis.

We describe time series by looking for an overall pattern and for striking

deviations from that pattern. In a time series:

A pattern that repeats itself

at regular intervals of time is

called seasonal variation.

Retail price of fresh oranges

over time

This time plot shows a regular pattern of yearly variations. These are seasonal

variations in fresh orange pricing most likely due to similar seasonal variations in

the production of fresh oranges.

There is also an overall upward trend in pricing over time. It could simply be

reflecting inflation trends or a more fundamental change in this industry.

Time is on the horizontal, x axis.

The variable of interest—here

“retail price of fresh oranges”—

goes on the vertical, y axis.

1918 influenza epidemic

Date # Cases # Deaths

week 1 36 0week 2 531 0week 3 4233 130week 4 8682 552week 5 7164 738week 6 2229 414week 7 600 198week 8 164 90week 9 57 56week 10 722 50week 11 1517 71week 12 1828 137week 13 1539 178week 14 2416 194week 15 3148 290week 16 3465 310week 17 1440 149

01000

2000300040005000

600070008000

900010000

wee

k 1

wee

k 3

wee

k 5

wee

k 7

wee

k 9

wee

k 11

wee

k 13

wee

k 15

wee

k 17

# c

ase

s d

iag

no

se

d

0

100

200

300

400

500

600

700

800

# d

ea

ths r

ep

ort

ed

# Cases # Deaths

A time plot can be used to compare two or more

data sets covering the same time period.

The pattern over time for the number of flu diagnoses closely resembles that for the

number of deaths from the flu, indicating that about 8% to 10% of the people

diagnosed that year died shortly afterward, from complications of the flu.

Death rates f rom cancer (US, 1945-95)

0

50

100

150

200

250

1940 1950 1960 1970 1980 1990 2000

Years

Death

rate

(per

thousand)

Death rates from cancer (US, 1945-95)

0

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250

1940 1960 1980 2000

Years

Death

rate

(per

thousand)


0

50

100

150

200

250

1940 1960 1980 2000

Years

Death

rate

(per

thousand)

A picture is worth a thousand words,

BUT

There is nothing like hard numbers.

�� Look at the scales.

Scales matter

How you stretch the axes and choose your scales can give a different impression.


120

140

160

180

200

220

1940 1960 1980 2000

Years

Death

rate

(per

thousand)

Looking at Data - DistributionsDescribing distributions with numbers

IPS Chapter 1.2



Describing distributions with numbers

� Measures of center: mean, median, mode

� Mean versus median

� Measures of spread: quartiles, standard deviation

� Five-number summary and boxplot

� Choosing among summary statistics

� Changing the unit of measurement

Two Characteristics

The central tendency of the set of measurements–that is, the tendency of the data to cluster, or center, about certain numerical values.

Central Tendency

(Location)

Center

Two Characteristics

The variability (spread) of the set of measurements–that is, the spread of the data.

Variation

(Dispersion)

Sample A

Sample B

Variation of Sample B

Variation of Sample A

The mean or arithmetic average

To calculate the average, or mean, add

all values, then divide by the number of

individuals. It is the “center of mass.”

Sum of heights is 1598.3

divided by 25 women = 63.9 inches

58.2 64.0

59.5 64.5

60.7 64.1

60.9 64.8

61.9 65.2

61.9 65.7

62.2 66.2

62.2 66.7

62.4 67.1

62.9 67.8

63.9 68.9

63.1 69.6

63.9

Measure of center: the mean

x =1598.3

25= 63.9

Mathematical notation:

∑=

=n

iix

nx

1

1

woman(i)

height(x)

woman(i)

height(x)

i = 1 x1= 58.2 i = 14 x14= 64.0

i = 2 x2= 59.5 i = 15 x15= 64.5

i = 3 x3= 60.7 i = 16 x16= 64.1

i = 4 x4= 60.9 i = 17 x17= 64.8

i = 5 x5= 61.9 i = 18 x18= 65.2

i = 6 x6= 61.9 i = 19 x19= 65.7

i = 7 x7= 62.2 i = 20 x20= 66.2

i = 8 x8= 62.2 i = 21 x21= 66.7

i = 9 x9= 62.4 i = 22 x22= 67.1

i = 10 x10= 62.9 i = 23 x23= 67.8

i = 11 x11= 63.9 i = 24 x24= 68.9

i = 12 x12= 63.1 i = 25 x25= 69.6

i = 13 x13= 63.9 n= 25 ΣΣΣΣ=1598.3

Learn right away how to get the mean using your calculators.

n

xxxx n+++

=...21

Your numerical summary must be meaningful.

63.9x =

The distribution of women’s heights appears coherent and

symmetrical. The mean is a good numerical summary.

Measure of center: the medianThe median is the midpoint of a distribution—the number such

that half of the observations are smaller and half are larger.

1. Sort observations by size.n = number of observations

______________________________

1 1 0.6

2 2 1.2

3 3 1.6

4 4 1.9

5 5 1.5

6 6 2.1

7 7 2.3

8 8 2.3

9 9 2.5

10 10 2.8

11 11 2.9

12 3.3

13 3.4

14 1 3.6

15 2 3.7

16 3 3.8

17 4 3.9

18 5 4.1

19 6 4.2

20 7 4.5

21 8 4.7

22 9 4.9

23 10 5.3

24 11 5.6

n = 24 �n/2 = 12

Median = (3.3+3.4) /2 = 3.35

2.b. If n is even, the median is the mean of the two middle observations.

1 1 0.6

2 2 1.2

3 3 1.6

4 4 1.9

5 5 1.5

6 6 2.1

7 7 2.3

8 8 2.3

9 9 2.5

10 10 2.8

11 11 2.9

12 12 3.3

13 3.4

14 1 3.6

15 2 3.7

16 3 3.8

17 4 3.9

18 5 4.1

19 6 4.2

20 7 4.5

21 8 4.7

22 9 4.9

23 10 5.3

24 11 5.6

25 12 6.1

n = 25 (n+1)/2 = 26/2 = 13 Median = 3.4

2.a. If n is odd, the median is observation (n+1)/2 down the list

Mode

1. Measure of central tendency

2. Value that occurs most often

3. Not affected by extreme values

4. May be no mode or several modes

5. May be used for quantitative or qualitative data

Mode Example

� No ModeRaw Data: 10.3 4.9 8.9 11.7 6.3 7.7

� One ModeRaw Data: 6.3 4.9 8.9 6.3 4.9 4.9

� More Than 1 ModeRaw Data: 21 28 28 41 43 43

Modes

The peaks of a histogram are called modes.

A distribution is

� unimodal if it has one mode,

� bimodal if it has two modes,

� multimodal if it has three or more modes.

Unimodal, Bimodal or Multimodal?

Unimodal Bimodal Multimodal

Uniform Histogram

� A histogram that doesn’t appear to have any mode.

� All the bars are approximately the same.

Right-SkewedLeft-Skewed Symmetric

Mean = MedianMean Median Median Mean

Mode

Mode= Mode

The median, on the other hand,

is only slightly pulled to the right

by the outliers (from 3.4 to 3.6).

The mean is pulled to the

right a lot by the outliers

(from 3.4 to 4.2).

Pe

rce

nt o

f p

eo

ple

dyi

ng

Mean and median of a distribution with outliers

4.3=x

Without the outliers

2.4=x

With the outliers

Disease X:

Mean and median are the same.

Symmetric distribution…

4.3

4.3

=

=

M

x

Multiple myeloma:

5.2

4.3

=

=

M

x

… and a right-skewed distribution

The mean is pulled toward the skew.

Impact of skewed data

Spread of a Distribution

Are the values concentrated around the center of the distribution or they are spread out?

� Range,

� Interquartile Range,

� Variance,

� Standard Deviation.

Note: Variance and standard deviation are more appropriate when the distribution is symmetric.

Range

� Range of the data is defined as the difference between the maximum and the minimum values.

� Data: 23, 21, 67, 44, 51, 12, 35.Range = maximum – minimum = 67 – 12 = 55.

� Disadvantage: A single extreme value can make it very large, giving a value that does not really represent the data overall. On the other hand, it is not affected at all if some observation changes in the middle.

Interquartile Range (IQR)

� What is IQR?

IQR = Third Quartile (Q3) – First Quartile (Q1).

� What are quartiles?

Recall: Median divides the data into 2 equal halves.

The first quartile, median and the third quartile divide the data into 4 roughly equal parts.

M = median = 3.4

Q1= first quartile = 2.2

Q3= third quartile = 4.35

1 1 0.6

2 2 1.2

3 3 1.6

4 4 1.9

5 5 1.5

6 6 2.1

7 7 2.3

8 1 2.3

9 2 2.5

10 3 2.8

11 4 2.9

12 5 3.3

13 3.4

14 1 3.6

15 2 3.7

16 3 3.8

17 4 3.9

18 5 4.1

19 6 4.2

20 7 4.5

21 1 4.7

22 2 4.9

23 3 5.3

24 4 5.6

25 5 6.1

The quartiles

The first quartile, Q1, is the value in the

sample that has 25% of the data at or

below it (� it is the median of the lower

half of the sorted data, excluding M).

The third quartile, Q3, is the value in the

sample that has 75% of the data at or

below it (� it is the median of the upper

half of the sorted data, excluding M).

Outlier-sensitivity

� Data: 10, 13, 17, 21, 28, 32

Without the outlier� IQR = 15 Range = 22

� Data: 10, 13, 17, 21, 28, 32, 59

With the outlier� IQR = 19 Range = 49

Conclusion: IQR is less outlier-sensitive than range.

M = median = 3.4

Q3= third quartile = 4.35

Q1= first quartile = 2.2

25 6 6.1

24 5 5.6

23 4 5.3

22 3 4.9

21 2 4.7

20 1 4.5

19 6 4.2

18 5 4.1

17 4 3.9

16 3 3.8

15 2 3.7

14 1 3.6

13 3.4

12 6 3.3

11 5 2.9

10 4 2.8

9 3 2.5

8 2 2.3

7 1 2.3

6 6 2.1

5 5 1.5

4 4 1.9

3 3 1.6

2 2 1.2

1 1 0.6

Upper whisker=Largest = max. = 6.1

Lower whisker = Smallest = min = 0.6

Disease X

0

1

2

3

4

5

6

7

Ye

ars

un

til d

ea

th

Five-number summary:

min Q1 M Q3 max

Five-number summary and boxplot

BOXPLOT

� Data must be ordered from lowest value to highest value before finding the 5 number summary.

� Upper whisker of the box-plot extends to the largest value within the upper inner fence (Q3+1.5*IQR)

� Lower whisker of the box-plot extends to the smallest value within the lower inner fence (Q1-1.5*IQR)

0123456789

101112131415

Disease X Multiple Myeloma

Ye

ars

un

til d

ea

th

Comparing box plots for a normal and a right-skewed distribution

Boxplots for skewed data

Boxplots remain

true to the data and

depict clearly

symmetry or skew.

Suspected outliers

Outliers are troublesome data points, and it is important to be able to

identify them.

One way to raise the flag for a suspected outlier is to compare the

distance from the suspicious data point to the nearest quartile (Q1 or

Q3). We then compare this distance to the interquartile range

(distance between Q1 and Q3).

We call an observation a suspected outlier if it falls more than 1.5

times the size of the interquartile range (IQR) above the first quartile or

below the third quartile. This is called the “1.5 * IQR rule for outliers.”

Q3 = 4.35

Q1 = 2.2

25 6 7.9

24 5 6.1

23 4 5.3

22 3 4.9

21 2 4.7

20 1 4.5

19 6 4.2

18 5 4.1

17 4 3.9

16 3 3.8

15 2 3.7

14 1 3.6

13 3.4

12 6 3.3

11 5 2.9

10 4 2.8

9 3 2.5

8 2 2.3

7 1 2.3

6 6 2.1

5 5 1.5

4 4 1.9

3 3 1.6

2 2 1.2

1 1 0.6

Disease X

0

1

2

3

4

5

6

7

Ye

ars

un

til d

ea

th

8

Interquartile rangeQ3 – Q1

4.35 − 2.2 = 2.15

Distance to Q3

7.9 − 4.35 = 3.55

Individual #25 has a value of 7.9 years, which is 3.55 years above

the third quartile. This is more than 3.225 years, 1.5 * IQR. Thus,

individual #25 is a suspected outlier.

Histogram vs. Box plot

� Both histogram and box plot capture the symmetry or skewness of distributions.

� Box plot cannot indicate the modality of the data.

� Box plot is much better in finding outliers.

� The shape of histogram depends to some extent on the choice of bins.

Which type of car has the largest median Time to

accelerate?

A. upscale

B. sports

C. small

D. large

E. family

Which type of car always take less than 3.6

seconds to accelerate?

A. upscale

B. sports

C. small

D. Large

E. Luxury

Which type of car has the smallest IQR for

Time to accelerate?

A. upscale

B. sports

C. small

D. Large

E. Luxury

What is the shape of the distribution of

acceleration times for luxury cars?

A. Left skewed

B. Right skewed

C. Roughly symmetric

D. Cannot be determined from the information given.

What percent of luxury cars accelerate to 30 mph

in less than 3.5 seconds?

A. Roughly 25%

B. Exactly 37.5%

C. Roughly 50%

D. Roughly 75%

E. Cannot be determined from the information given

What percent of family cars accelerate to 30

mph in less than 3.5 seconds?

A. Less than 25%

B. More than 50%

C. Less than 50%

D. Exactly 75%

E. None of the above

The standard deviation “s” is used to describe the variation around the

mean. Like the mean, it is not resistant to skew or outliers.

2

1

2 )(1

1xx

ns

n

i −−

= ∑

1. First calculate the variance s2.

2

1

)(1

1xx

ns

n

i −−

= ∑

2. Then take the square root to get

the standard deviation s.

Measure of spread: the standard deviation

Mean± 1 s.d.

x

Calculations …

We’ll never calculate these by hand, so make sure to know how to get the standard deviation using your calculator or software.

2

1

)(1

xxdf

sn

i −= ∑

Mean = 63.4

Sum of squared deviations from mean = 85.2

Degrees freedom (df) = (n − 1) = 13

s2 = variance = 85.2/13 = 6.55 inches squared

s = standard deviation = √6.55 = 2.56 inches

Women height (inches)

i xi x (xi-x) (xi-x)2

1 59 63.4 -4.4 19.0

2 60 63.4 -3.4 11.3

3 61 63.4 -2.4 5.6

4 62 63.4 -1.4 1.8

5 62 63.4 -1.4 1.8

6 63 63.4 -0.4 0.1

7 63 63.4 -0.4 0.1

8 63 63.4 -0.4 0.1

9 64 63.4 0.6 0.4

10 64 63.4 0.6 0.4

11 65 63.4 1.6 2.7

12 66 63.4 2.6 7.0

13 67 63.4 3.6 13.3

14 68 63.4 4.6 21.6

Mean 63.4

Sum 0.0

Sum 85.2

Properties of Standard Deviation

� s measures spread about the mean and should be used only when the mean is the measure of center.

� s = 0 only when all observations have the same value and there is no spread. Otherwise, s > 0.

� s is not resistant to outliers.

� s has the same units of measurement as the original observations.

Choosing among summary statistics

� Because the mean is not

resistant to outliers or skew, use

it to describe distributions that are

fairly symmetrical and don’t have

outliers.

� Plot the mean and use the

standard deviation for error bars.

� Otherwise use the median in the

five number summary which can

be plotted as a boxplot.

Height of 30 Women

58

59

60

61

62

63

64

65

66

67

68

69

Box Plot Mean +/- SD

He

igh

t in

In

ch

es

Boxplot Mean ± SD

Changing the unit of measurement

Variables can be recorded in different units of measurement. Most

often, one measurement unit is a linear transformation of another

measurement unit: xnew = a + bx.

Temperatures can be expressed in degrees Fahrenheit or degrees Celsius.

TemperatureFahrenheit = 32 + (9/5)* TemperatureCelsius � a + bx.

Linear transformations do not change the basic shape of a distribution

(skew, symmetry, multimodal). But they do change the measures of

center and spread:

� Multiplying each observation by a positive number b multiplies both

measures of center (mean, median) and spread (IQR, s) by b.

� Adding the same number a (positive or negative) to each observation

adds a to measures of center and to quartiles but it does not change

measures of spread (IQR, s).

� Temperature data (in F): 10, 13, 18, 22, 29� Range = 19 F, IQR =14 F, s = 7.5 F,

s2 = 56.25 F2.� Suppose transformed data = (-3)*original data.� So transformed data (in F): -30, -39, -54, -66, -87� Range = |-3|*19 = 57 F, IQR = |-3|*14 = 42 F, s = |-3|* 7.5 = 22.50 F, s2 = (-3)2*56.25 = 506.25 F2.

� Temperature data (in F): 10, 13, 18, 22, 29

� Range = 19 F, IQR =14 F, s = 7.5 F, s2 = 56.25 F2.

� Suppose transformed data = original data + 2.5.

� Hence transformed data (in F): 12.5, 15.5, 20.5, 24.5, 31.5

� Range = 19 F, IQR =14 F, s = 7.5 F, s2 = 56.25 F2.

Software output for summary statistics:

Excel - From Menu:

Tools → Data Analysis

→Descriptive Statistics

Give common

statistics of your

sample data.

Minitab

MINITAB

For graphs

From the Menu

�Graphs→Histogram

�Graphs → Stem-and-Leaf

�Graphs → Box-Plot

For measures of central tendency and spread

� Stat → Basic Statistics → Display Descriptive

Statistics

Comparing Distributions

Use of Stem-and-Leaf Display

and

5 number summary

Comparing Distributions: Use of Stem-and-Leaf Plot

East Lansing Gas Prices

2.53

2.59

2.69

2.61

2.49

2.49

2.56

2.74

2.74

2.76

2.69

2.53

2.49

2.68

Ann Arbor Gas Prices

2.69

2.69

2.72

2.94

2.69

2.95

2.69

2.72

2.63

2.89

2.75

2.96

2.93

2.69

2.89

Ann Arbor

Gas Prices

East Lansing

Gas Prices

24|9 = $2.49

Gas prices in Ann

Arbor are unimodal

with right skew.

They are between

$2.63 and $2.96 per

gallon, centered at

about $2.75 per

gallon

Gas prices in

East Lansing

are roughly

uniform

between $2.49

and $2.76 per

gallon,

centered at

$2.60 per

gallon

Gas prices tend to be

higher in Ann Arbor and

have more variability. The

range of prices in Ann

Arbor is 33¢ and in East

Lansing is only 27¢

Ann Arbor

Gas Prices

East Lansing

Gas Prices

24|9 = $2.49

Min Q1 Median Q3 Max

Ann Arbor Gas Prices $2.63 $2.69 $2.72 $2.93 $2.96

East Lansing Gas Prices $2.49 $2.53 $2.60 $2.69 $2.76

5-Number Summary

Comparing Distributions

Use of Histograms

Which data have more variability?

A. Graph A

B. Graph B

C. Both have the same variability

0 2 4 6 8 10

1

2

3

4

5

12

6

SCORE

1

2

3

4

5

6

FR

EQ

UE

NC

Y

0 2 4 6 8 10 12SCORE

AB


A. Graph A

B. Graph B

C. Both have the same variability

AB

0 2 4 6 8 10

1

2

3

4

5

12

6

SCORE

1

2

3

4

5

6

0 2 4 6 8 10 12SCORE

Which data have a higher median?

A. Graph A

B. Graph B

C. Both have the same median

AB

0 2 4 6 8 10

1

2

3

4

5

12

6

SCORE

1

2

3

4

5

6

0 2 4 6 8 10 12SCORE

88


A. Graph A

B. Graph B

C. Roughly, both have the same variability

AB

0 2 4 6 8 10

1

2

3

4

5

12

6

SCORE

1

2

3

4

5

6

FR

EQ

UE

NC

Y

0 2 4 6 8 10 12SCORE

89

Looking at Data - DistributionsDensity Curves and Normal Distributions

IPS Chapter 1.3



Density curves and Normal distributions

� Density curves

� Measuring center and spread for density curves

� Normal distributions

� The 68-95-99.7 rule

� Standardizing observations

� Using the standard Normal Table

� Inverse Normal calculations

� Normal quantile plots

Density curvesA density curve is a mathematical model of a distribution.

The total area under the curve, by definition, is equal to 1, or 100%.

The area under the curve for a range of values is the proportion of all observations for that range.

Histogram of a sample with the

smoothed, density curve

describing theoretically the

population.

Density curves come in any

imaginable shape.

Some are well known

mathematically and others aren’t.

Median and mean of a density curve

The median of a density curve is the equal-areas point: the point that

divides the area under the curve in half.

The mean of a density curve is the balance point, at which the curve

would balance if it were made of solid material.

The median and mean are the same for a symmetric density curve.

The mean of a skewed curve is pulled in the direction of the long tail.

Normal distributions

e = 2.71828… The base of the natural logarithm

π = pi = 3.14159…

Normal – or Gaussian – distributions are a family of symmetrical, bell-

shaped density curves defined by a mean µ (mu) and a standard

deviation σ (sigma) : N(µ,σ).

21

21( )

(2 )

x

f x e

µσ

σ π

− − =

xxxxxxxx

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

A family of density curves

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

Here, means are different

(µ = 10, 15, and 20) while standard

deviations are the same (σ = 3)

Here, means are the same (µ = 15)

while standard deviations are

different (σ = 2, 4, and 6).

mean µ = 64.5 standard deviation σσσσ = 2.5

N(µ, σσσσ) = N(64.5, 2.5)

Reminder: µ (mu) is the mean of the idealized curve, while is the mean of a sample.

σ (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample.

The 68-95-99.7% Rule for Normal Distributions

� About 68% of all observations

are within 1 standard deviation

(σ) of the mean (µ).

� About 95% of all observations

are within 2 σ of the mean µ.

� Almost all (99.7%) observations

are within 3 σ of the mean.

Inflection point

x

Because all Normal distributions share the same properties, we can

standardize our data to transform any Normal curve N(µ,σ) into the

standard Normal curve N(0,1).

The standard Normal distribution

For each x we calculate a new value, z (called a z-score).

N(0,1)

=>

zx

N(64.5, 2.5)

Standardized height (no units)

How to compare apples with oranges?

� A college admissions committee is looking at the files of two candidates, one with a total SAT score of 1500 and another with an ACT score of 22. Which candidate scored better?

� How do we compare things when they are measured on different scales?

� We need to standardize the values so that they will be free of units.

z =(x − µ )

σ

A z-score measures the number of standard deviations that a data

value x is from the mean µ.

Standardizing: calculating z-scores

� When x is larger than the mean, z is positive.

� When x is smaller than the mean, z is negative.

1 , ==−+

=+=σσ

σµσµ

σµ zxfor

When x is 1 standard deviation larger

than the mean, then z = 1.

222

,2 ==−+

=+=σσ

σµσµ

σµ zxfor

When x is 2 standard deviations larger

than the mean, then z = 2.

Interpretation of z-scores

� The z-scores measure the distance of the data values from the mean in the standard deviation scale.

� A z-score of 1 means that data value is 1 standard deviation above the mean.

� A z-score of -1.2 means that data value is 1.2 standard deviations below the mean.

� Regardless of the direction, the further a data value is from the mean, the more unusual it is.

� A z-score of -1.3 is more unusual than a z-score of 1.2.

z-scores: An Example

Data: 4, 3, 10, 12, 8, 9, 3 (n=7 in this case)Mean = (4+3+10+12+8+9+3)/7 = 49/7 =7.Standard Deviation = 3.65.

Original Value z-score --------------------------------------------------------------

4 (4 – 7)/3.65 = -0.823 (3 – 7)/3.65 = -1.10

10 (10 – 7)/3.65 = 0.82 12 (12 – 7)/3.65 = 1.37 8 (8 – 7)/3.65 = 0.279 (9 – 7)/3.65 = 0.553 (3 – 7)/3.65 = -1.10

--------------------------------------------------------------

How to use z-scores?

� A college admissions committee is looking at the files of two candidates, one with a total SAT score of 1500 and another with an ACT score of 22. Which candidate scored better?

� SAT score mean = 1600, std dev = 500.

� ACT score mean = 23, std dev = 6.

� SAT score 1500 has z-score = (1500-1600)/500 = -0.2.

� ACT score 22 has z-score = (22-23)/6 = -0.17.

� ACT score 22 is better than SAT score 1500.

Which is more unusual?A. A 58 in tall woman

z-score = (58-63.6)/2.5 = -2.24.B. A 64 in tall man

z-score = (64-69)/2.8 = -1.79.C. They are the same.

Heights of adult men have

mean of 69.0 in.

std. dev. of 2.8 in.

Heights of adult women have

mean of 63.6 in.

std. dev. of 2.5 in.

Using z-scores to solve problems

An example using height data and U.S. Marine and Army height requirements

Question: Are the height restrictions set up by the

U.S. Army and U.S. Marine more restrictive for

men or women or are they roughly the same?

Heights of adult women have � mean of 63.6 in.

� standard deviation of 2.5 in.

Data from a National Health Survey

Heights of adult men have– mean of 69.0 in. – standard deviation of 2.8 in.

Men Minimum

Women Minimum

U.S. Army 60 in 58 in

U.S. Marine Corps 64 in 58 in

Height Restrictions

Heights of adult men have � mean of 69.0 in.


Heights of adult women have – mean of 63.6 in.– standard deviation of 2.5 in.

Men Minimum Women minimum

U.S.

Army

U.S.

Marine

60 in

z-score = -3.21

Less restrictive

58 in

z-score = -2.24

More restrictive

64 in

z-score = -1.79

More restrictive

58 in

z-score = -2.24

Less restrictive

Effect of Standardization

� Standardization into z-scores does not change the shape of the histogram.

� Standardization into z-scores changes the center of the distribution by making the mean 0.

� Standardization into z-scores changes the spread of the distribution by making the standard deviation 1.

mean µ = 64.5"

standard deviation σ = 2.5"

x (height) = 67"

We calculate z, the standardized value of x:

mean from dev. stand. 1 15.2

5.2

5.2

)5.6467( ,

)(=>==

−=

−= z

xz

σµ

Because of the 68-95-99.7 rule, we can conclude that the percent of women

shorter than 67” should be, approximately, .68 + half of (1 - .68) = .84 or 84%.

Area= ???

Area = ???

N(µ, σ) = N(64.5, 2.5)

µµµµ = 64.5” x = 67”

z = 0 z = 1

Ex. Women heights

Women’s heights follow the N(64.5”,2.5”)

distribution. What percent of women are

shorter than 67 inches tall (that’s 5’6”)?

Using the standard Normal table

(…)

Table A gives the area under the standard Normal curve to the left of any z value.

.0082 is the

area under

N(0,1) left

of z = -

2.40

.0080 is the area

under N(0,1) left

of z = -2.41

0.0069 is the area

under N(0,1) left

of z = -2.46

Area ≈ 0.84

Area ≈ 0.16

N(µ, σ) =

N(64.5”, 2.5”)

µ = 64.5” x = 67” z = 1

Conclusion:

84.13% of women are shorter than 67”.

By subtraction, 1 - 0.8413, or 15.87% of

women are taller than 67".

For z = 1.00, the area under

the standard Normal curve

to the left of z is 0.8413.

Percent of women shorter than 67”

Tips on using Table A

Because the Normal distribution is

symmetrical, there are 2 ways

that you can calculate the area

under the standard Normal curve

to the right of a z value.

area right of z = 1 - area left of z

Area = 0.9901

Area = 0.0099

z = -2.33

area right of z = area left of -z

Tips on using Table A

To calculate the area between 2 z- values, first get the area under N(0,1)

to the left for each z-value from Table A.

area between z1 and z2 =

area left of z1 – area left of z2

A common mistake made by

students is to subtract both z

values. But the Normal curve

is not uniform.

Then subtract the

smaller area from the

larger area.

�� The area under N(0,1) for a single value of z is zero.

(Try calculating the area to the left of z minus that same area!)

Calculating the area for normal

distribution by TI calculators

Use TI 83/84 Plus.

� Press [2nd] & [VARS] (i.e. [DISTR])

� Select 2: normalcdf

� Format of command:

normalcdf(lower bound, upper bound, mean, std.dev.)

� For percentage between a and b

normalcdf(a, b, mean, sigma)

� For area (percentage) below a specific value b

normalcdf(-1000, b, mean, sigma)

� For area (percentage) above any specific value a

normalcdf( a, 1000, mean, sigma)

Approximately what percent of U.S. women do you expect to be

less than 64 in tall?

� Heights of adult women are normally distributed with

� mean of 63.6 in,


� Note that here upper bound is 64, but there is no mention of lower bound.

� So take a very small value for lower bound, say -1000.

For this problem

normalcdf(-1000, 64, 63.6, 2.5) = 0.5636.

i.e. about 56.4% of adult U.S. women have heights less than 64 in.

The National Collegiate Athletic Association (NCAA) requires Division I athletes to

score at least 820 on the combined math and verbal SAT exam to compete in their

first college year. The SAT scores of 2003 were approximately normal with mean

1026 and standard deviation 209.

What proportion of all students would be NCAA qualifiers (SAT ≥ 820)?

820

1026

209

( )

(820 1026)

209

2060.99

209

Table A: area under

N(0,1) to the left of

z = -0.99 is 0.1611

or approx. 16%.

x

xz

z

z

µ

σ

µσ

=

=

=

−=

−=

−= ≈ −

area right of 820 = total area - area left of 820= 1 - 0.1611

≈ 84%

normalcdf (820,1000,1026,209)

The NCAA defines a “partial qualifier” eligible to practice and receive an athletic

scholarship, but not to compete, with a combined SAT score of at least 720.

What proportion of all students who take the SAT would be partial qualifiers?

That is, what proportion have scores between 720 and 820?

720

1026

209

( )

(720 1026)

209

3061.46

209

Table A: area under

N(0,1) to the left of

z = -.99 is -1.46

or approx. 7%.

x

xz

z

z

µ

σ

µσ

=

=

=

−=

−=

−= ≈ −

About 9% of all students who take the SAT have scores

between 720 and 820.

area between = area left of 820 - area left of 720

720 and 820 = 0.1611 - 0.0721

≈ 9%

normalcdf (720,820,1026,209)

N(0,1)

z =(x − µ )

σ

The cool thing about working with normally distributed data is that we can manipulate it, and then find answers to questions that involve comparing seemingly non-comparable distributions.

We do this by “standardizing” the data. All this involves is changing the scale so that the mean now = 0 and the standard deviation =1. If you do this to different distributions it makes them comparable.

180 200 220 240 260 280 300 320

Gestation time (days)

Vitamins only Vitamins and better food

What is the effect of better maternal care on gestation time and preemies?

The goal is to obtain pregnancies 240 days (8 months) or longer.

µ 250σ 20

µ 266σ 15

Ex. Gestation time in malnourished mothers

What improvement did we get

by adding better food?

240

250

20

( )

(240 250)

20

100.5

20

(half a standard deviation)

Table A: area under N(0,1) to

the left of z = -0.5 is 0.3085.

x

xz

z

z

µ

σ

µσ

=

=

=

−=

−=

−= = −

Vitamins Only

Under each treatment, what percent of mothers failed to carry their babies at

least 240 days?

190 210 230 250 270 290 310


Vitamins only: 30.85% of women

would be expected to have gestation

times shorter than 240 days.

µ=250, σ=20, x=240

normalcdf(-1000,240,250,20)

240

266

15

( )

(240 266)

15

261.73

15

(almost 2 sd from mean)

Table A: area under N(0,1) to

the left of z = -1.73 is 0.0418.

x

xz

z

z

µ

σ

µσ

=

=

=

−=

−=

−= = −

Vitamins and better food

221 236 251 266 281 296 311


Vitamins and better food: 4.18% of women

would be expected to have gestation

times shorter than 240 days.

µ=266, σ=15, x=240

Compared to vitamin supplements alone, vitamins and better food resulted in a much

smaller percentage of women with pregnancy terms below 8 months (4% vs. 31%).

normalcdf(-1000,240,266,15)

Inverse normal calculations

We may also want to find the observed range of values that correspond

to a given proportion/ area under the curve.

For that, we use Table A backward:

� we first find the desired

area/ proportion in the

body of the table,

� we then read the

corresponding z-value

from the left column and

top row.

For an area to the left of 1.25 % (0.0125), the z-value is -2.24

Use TI 83/84 Plus.

� Press [2nd] & [VARS] (i.e. [DISTR])

� Select 3: invNorm

� Format of command:

invNorm(fraction, mean, std.dev.)

� invNorm only considers percentage or fraction in the lower tail of normal distribution.

25695.255

)15*67.0(266

)*()(

0.67.-about is N(0,1)

under 25% arealower

for the valuez :A Table

?

%25arealower

%75areaupper

15

266

≈=

−+=

+=⇔−

=

=

=

=

=

=

x

x

zxx

z

x

σµσ

µ

σ

µ

Vitamins and better food

221 236 251 266 281 296 311


µ=266, σ=15, upper area 75%

How long are the longest 75% of pregnancies when mothers with malnutrition are

given vitamins and better food?

?

upper 75%

� The 75% longest pregnancies in this group are about 256 days or longer.

Remember that Table A gives the area to the

left of z. Thus, we need to search for the lower

25% in Table A in order to get z. invnorm(0.25, 266,15)=256

One way to assess if a distribution is indeed approximately normal is to

plot the data on a normal quantile plot.

The data points are ranked and the percentile ranks are converted to z-

scores with Table A. The z-scores are then used for the x axis against

which the data are plotted on the y axis of the normal quantile plot.

� If the distribution is indeed normal the plot will show a straight line,

indicating a good match between the data and a normal distribution.

� Systematic deviations from a straight line indicate a non-normal

distribution. Outliers appear as points that are far away from the overall

pattern of the plot.

Normal quantile plots

Normal quantile plots are complex to do by hand, but they are standard

features in most statistical software.

Good fit to a straight line: the

distribution of rainwater pH

values is close to normal.

Curved pattern: the data are not

normally distributed. Instead, it shows

a right skew: a few individuals have

particularly long survival times.

By Minitab

� Graph-Probability plot-Single (For one sample)

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