Analysis of Economic Data 4. Measures of dispersionkafuwong.econ.hku.hk/teaching/econ1003/2003F/ppt/Chap002.pdf · 2 Ka-fu Wong © 2003 Chap 2-7 EXAMPLE 1 Dr. Tillman is Dean of the

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Ka-fu Wong © 2003 Chap 2-1

Dr. Ka-fu Wong

ECON1003Analysis of Economic Data


Chapter TwoDescriptive StatisticsDescriptive Statistics

GOALS1. Organize data into a frequency distribution.2. Plots:

Histogram,stem-and-leaf display.

3. Measures of central tendency4. Measures of dispersion5. Some properties of mean and standard

deviations


Frequency Distribution

A Frequency distribution is a grouping of data into mutually exclusive categories showing the number of observations in each class.


Construction of a Frequency Distribution

1. Question to be addressed.2. Collect data (raw data).3. Organize data.

Frequency distribution4. Present data.5. Draw conclusion.


Terms Related to Frequency Distribution

In constructing a frequency distribution, data are divided into exhaustive and mutually exclusive classes.Mid-point: A point that divides a class into two equal parts. This is the average of the upper and lower class limits.Class frequency: The number of observations in each class.Class interval: The class interval is obtained by subtracting the lower limit of a class from the lower limit of the next class.


Four steps to construct frequency distribution

1. Decide on the number of classes. 2. Determine the class interval or width.3. Set the individual class limits.4. Tally the data into the classes and count the

number of items in each.

For illustration, it is convenient to carry an example with us – example 1.

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EXAMPLE 1

Dr. Tillman is Dean of the School of Business Socastee University. He wishes to prepare a report showing the number of hours per week students spend studying. He selects a random sample of 30 students and determines the number of hours each student studied last week.

15.0, 23.7, 19.7, 15.4, 18.3, 23.0, 14.2, 20.8, 13.5, 20.7, 17.4, 18.6, 12.9, 20.3, 13.7, 21.4, 18.3, 29.8, 17.1, 18.9, 10.3, 26.1, 15.7, 14.0, 17.8, 33.8, 23.2, 12.9, 27.1, 16.6.

Organize the data into a frequency distribution.Ka-fu Wong © 2003 Chap 2-8

Step 1: Decide on the number of classes.

The goal is to use just enough groupings or classes to reveal the shape of the distribution.“Just enough” Recipe –“2 to the k rule”

Select the smallest number (k) for the number of classes such that 2k is greater than the number of observations (n).


2 to the k rule

Sample size (n) = 8021=2; 22=4; 23=8; 24=16; 25=32; 26=64; 27=128; …The rule suggest 7 classes.

Sample size (n) = 100021=2; 22=4; 23=8; 24=16; 25=32; 26=64; 27=128; 28=256; 29=512; 210=1024 …The rule suggest 10 classes.



2 to the k rule

Sample size (n)=1000021=2; 22=4; 23=8; 24=16; 25=32; 26=64; 27=128; …,213=8192; 214 = 16384The rule suggest 14.

Sample size (n)=10000022=4; 23=8; 24=16; 25=32; 26=64; 27=128; …



2 to the k rule

Select the smallest number (k) for the number of classes such that 2k is greater than the number of observations (n).We want to find smallest k such that 2k > n.

Smallest k such that k log 2 > log nSmallest k such that k > (log n) / (log 2)

Example: If n=10000, (log n) / (log 2) = 13.28. Hence the recipe suggest 14 classes.

Note: Same result for base 10 log and natural log.


Step 1: Decide on the number of classes.

2 to the k ruleSelect the smallest number (k) for the number of classes such that 2k is greater than the number of observations (n).

Example 1 (continued):Sample size (n) = 3021=2; 22=4; 23=8; 24=16; 25=32; 26=64; 27=128; …The rule suggest 5 classes.Alternative, by computing (Log 30/log 2) = 4.91, we get the same suggestion of 5 classes.

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Step 2: Determine the class interval or width.

Generally the class interval or width should be the same for all classes.The classes all taken together must cover at least the distance from the lowest value in the raw data to the highest value.The classes must be mutually exclusive and exhaustive.

Class interval ≥

(Highest value – lowest value) / number of classes.

kLHi −

≥

Usually we will chose some convenient number as class interval that satisfy the inequality.


Step 2: Determine the class interval or width.

Example 1 (continued):Highest value = 33.8 hoursLowest value = 10.3 hoursk=5.Hence, class interval ≥ (33.8-10.3)/5 = 4.7We choose class interval to be 5, some convenient number.

Class interval ≥

(Highest value – lowest value) / number of classes.

kLHi −

≥


Step 3: Set the individual class limits

The class limits must be set so that the classes are mutually exclusive and exhaustive.Round up to some convenient numbers.



Example 1 (continue):Highest value = 33.8 hours.Lowest value = 10.3 hours.Range = highest – lowest = 23.5.K=5; Interval = 5. With k=5 and interval = 5, the classes will cover a range of 25. Let’s split the surplus in the lower and upper tail equally. (25-23.5)/2 = 0.75. Hence, the lower limit of the first class should be around (10.3 – 0.75)=9.55 and upper limit of the last class should be (33.8 + 0.75)=34.55. 9.55 and 34.55 look odd. Some convenient and close numbers would be 10 and 35.



Example 1 (continue):We have determined K=5; Interval = 5. The lower limit of the first class = 10 and The upper limit of the last class = 35.

Hours studying Frequency, f

10 up to 15

15 up to 20

20 up to 25

25 up to 30

30 up to 35

“10 up to 15” means the interval from 10 to 15 that includes 10 but not 15.


Step 4: Tally the data into the classes and count the number of items in each

10 up to 15Hours studying

15 up to 2020 up to 2525 up to 3030 up to 35

15.0, 23.7, 19.7, 15.4, 18.3, 23.0, 14.2, 20.8, 13.5, 20.7, 17.4, 18.6, 12.9, 20.3, 13.7, 21.4, 18.3, 29.8, 17.1, 18.9, 10.3, 26.1, 15.7, 14.0, 17.8, 33.8, 23.2, 12.9, 27.1, 16.6.

7

12

731

4


EXAMPLE 1 (continued)

Hours studying Frequency, f

10 up to 15 7

15 up to 20 12

20 up to 25 7

25 up to 30 3

30 up to 35 1



A relative frequency distribution shows the percent of observations in each class.

Hours f Relative Frequency

10 up to 15 7 7/30=.233 15 up to 20 12 12/30=.400 20 up to 25 7 7/30=.233 25 up to 30 3 3/30=.100 30 up to 35 1 1/30=.033

TOTAL 30 30/30=1


Stem-and-leaf Displays

Stem-and-leaf display: A statistical technique for displaying a set of data. Each numerical value is divided into two parts: the leading digits become the stem and the trailing digits the leaf.

Note:An advantage of the stem-and-leaf display over a frequency distribution is we do not lose the identity of each observation.An disadvantage is that it is not good for large data sets.


EXAMPLE 2

Colin achieved the following scores on his twelve accounting quizzes this semester:

86, 79, 92, 84, 69, 88, 91, 83, 96, 78, 82, 85.

Construct a stem-and-leaf chart.



86, 79, 92, 84, 69, 88, 91, 83, 96, 78, 82, 85.

Stem Leaf

6

7

8

9

6

9

2

4

9

8

1

3

6

8

2 5



86, 79, 92, 84, 69, 88, 91, 83, 96, 78, 82, 85.

Stem Leaf

6

7

8

9

2

8

1

3

9

4

2

5

6

9

6 8

1

2

(6)

3

Freq

5


Graphic Presentation of a Frequency Distribution

The three commonly used graphic forms are histograms,frequency polygons, and a cumulative frequency distribution.

A Histogram is a graph in which the classes are marked on the horizontal axis and the class frequencies on the vertical axis.

The class frequencies are represented by the heights of the bars and the bars are drawn adjacent to each other.


Graphic Presentation of a Frequency Distribution

A frequency polygon consists of line segments connecting the points formed by the class midpoint and the class frequency.

A cumulative frequency distribution is used to determine how many or what proportion of the data values are below or above a certain value.


Histogram for Hours Spent Studying

02468

101214

12.5 17.5 22.5 27.5 32.5

Hours spent studying

Freq

uen

cy

Example 1 (continued):A Histogram is a graph in which the classes are marked on the horizontal axis and the class frequencies on the vertical axis.


Frequency Polygon for Hours Spent Studying

02468

101214

0 10 20 30 40

Hours spent studying

Freq

uenc

yExample 1 (continued):

A frequency polygon consists of line segments connecting the points formed by the class midpoint and the class frequency.


Cumulative Frequency Distribution For Hours Studying

05

101520253035

0 10 20 30 40Hours Spent Studying

Freq

uenc

y

Example 1 (continued):A cumulative frequency distribution is used to determine how many or what proportion of the data values are below or above a certain value.


Bar Chart

A bar chart can be used to depict any of the levels of measurement (nominal, ordinal, interval, or ratio).

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Example 3

Construct a bar chart for the number of unemployed per 100,000 population for selected cities during 2001

City Number of unemployed per 100,000 population

Atlanta, GA 7300

Boston, MA 5400

Chicago, IL 6700

Los Angeles, CA 8900

New York, NY 8200

Washington, D.C. 8900


Bar Chart for the Unemployment Data

7300

5400

6700

8900

8200

8900

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

Atlant

a

Boston

Chica

go

Los An

gles

New Yor

kD.C

.

Cities

# u

nem

ploy

ed/1

00,0

00

Example 3 (continued):


Pie Chart

A pie chart is useful for displaying a relative frequency distribution. A circle is divided proportionally to the relative frequency and portions of the circle are allocated for the different groups.


EXAMPLE 4

Type of shoe # of runners

Nike 92

Adidas 49

Reebok 37

Asics 13

Other 9

A sample of 200 runners were asked to indicate their favorite type of running shoe. Draw a pie chart based on the following information.



Type of shoe # of runners Percentage (relative frequency)

Degree

Nike 92 46%=100*92/200 165.6

Adidas 49 24.5%=100*49/200 88.2

Reebok 37 18.5%=100*37/200 66.6

Asics 13 6.5%=100*13/200 23.4

Other 9 4.5%=100*9/200 16.2

Compute the percentage and degree each type occupy out of 360o.

Degree occupied in a circle = percentage x 360


Pie Chart for Running Shoes

Adidas24%

Reebok19%

Asics7%

Other5%

Nike45%

NikeAdidasReebokAsicsOther

Example 4 (continued):

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Measures of Central Tendency

1. Mean2. Median3. Mode


Population Parameters and Sample Statistics

A population parameter is number calculated from all the population measurements that describes some aspect of the population.

The population mean, denoted µ, is a population parameter and is the average of the population measurements.

A point estimate is a one-number estimate of the value of a population parameter.A sample statistic is number calculated using sample measurements that describes some aspect of the sample.


The Mean

Population X1, X2, …, XN

µ

Population Mean

N

X

N

1=ii∑

=µ

Sample x1, x2, …, xn

Sample Mean

n

xx

n

1=ii∑

=

x


Population Mean

For ungrouped data, the population mean is the sum of all the population values divided by the total number of population values:

NX

=µ∑

where µ is the population mean.N is the total number of observations.X is a particular value.Σ indicates the operation of adding.


Sample Mean

For ungrouped data, the sample mean is the sum of all the sample values divided by the number of sample values:

Where n is the total number of values in the sample.

nΣX

=X

This sample mean is also referred as arithmetic mean or simply sample average.


EXAMPLE 5

A sample of five executives received the following bonus last year ($000):

14.0, 15.0, 17.0, 16.0, 15.0

15.4=577

=5

15.0+...+14.0=

nΣX

=X

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Population and Sample Proportions

Population X1, X2, …, XN

p

Population Proportion


Sample Proportion

n

xˆ

n

1=ii∑

=p

p̂

xi = 1 if characteristic present, 0 if not


Properties of the Arithmetic Mean

1. Every set of interval-level and ratio-level data has a mean.

2. All the values are included in computing the mean.

3. A set of data has a unique mean.4. The mean is affected by unusually large or

small data values.5. The arithmetic mean is the only measure of

central tendency where the sum of the deviations of each value from the mean is zero.



A sample of five executives received the following bonus last year ($000):

7.0, 15.0, 17.0, 16.0, 15.0Changing the first observation from 14.0 to 7.0 will change the sample mean.

14=570

=5

15.0+...+7.0=

nΣX

=X

15.4=577

=5

15.0+...+14.0=

nΣX

=X


EXAMPLE 6

Consider the set of values: 3, 8, and 4. The mean is 5. Illustrating the fifth property:

[ ] 0=5)(4+5)(8+5)(3=)X -Σ(X ---


Weighted Mean

The weighted mean of a set of numbers X1, X2, ..., Xn, with corresponding weights w1, w2, ...,wn, is computed from the following formula:

)n21

nn2211w ...w+w+(w

)Xw+...+Xw+X(w=X


EXAMPLE 7

During a one hour period on a hot Saturday afternoon cabana boy Chris served fifty drinks. He sold five drinks for $0.50, fifteen for $0.75, fifteen for $0.90, and fifteen for $1.10. Compute the weighted mean of the price of the drinks.

$0.89=50

$44.50=

15+15+15+515($1.15)+15($0.90)+15($0.75)+5($0.50)

=X w

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The Median

The Median is the midpoint of the values after they have been ordered from the smallest to the largest.

There are as many values above the median as below it in the data array.

For an even set of values, the median will be the arithmetic average of the two middle numbers.


EXAMPLE 8

The ages for a sample of five college students are:21, 25, 19, 20, 22

Arranging the data in ascending order gives: 19, 20, 21, 22, 25. Thus the median is 21.

The heights of four basketball players, in inches, are: 76, 73, 80, 75Arranging the data in ascending order gives: 73, 75, 76, 80. Thus the median is 75.5


Properties of the Median

1. There is a unique median for each data set.

2. It is not affected by extremely large or small values and is therefore a valuable measure of central tendency when such values occur.

3. It can be computed for ratio-level, interval-level, and ordinal-level data.

4. It can be computed for an open-ended frequency distribution if the median does not lie in an open-ended class.


The Mode

The mode is the value of the observation that appears most frequently.

EXAMPLE 6: The exam scores for ten students are: 81, 93, 84, 75, 68, 87, 81, 75, 81, 87.

Because the score of 81 occurs the most often, it is the mode.


Geometric Mean

The geometric mean (GM) of a set of n numbers is defined as the nth root of the product of the nnumbers. The formula is:

The geometric mean is used to average percents, indexes, and relatives.

The geometric mean is not applicable when some numbers are negative.

n n321 ))...(X)(X)(X(X=GM


EXAMPLE 9

The interest rate on three bonds were 5, 21, and 4 percent.The geometric mean is

The arithmetic mean is (5+21+4)/3 =10.0 The GM gives a more conservative profit figure because it is not heavily weighted by the rate of 21 percent.

7.49=(5)(21)(4)=GM 3

10


Geometric Mean continued

Another use of the geometric mean is to determine the percent increase in sales, production or other business or economic series from one time period to another.

1period) of beginningat (Value

period) of endat (Value=GM n -


EXAMPLE 10

The total number of females enrolled in American colleges increased from 755,000 in 1992 to 835,000 in 2000. That is, the geometric mean rate of increase is 1.27%.

.0127=1 - 755,000835,000

=GM 8


EXAMPLE 11

A banker wants to get an annual return of 100% on its loan in credit card business. What monthly interest rate should he charge?

A monthly interest rate of 5.9%.

.059=1 - 100200

=GM 12


EXAMPLE 12

The Chinese government claimed in 1990 that their GDP will double in 20 years. What must the annual GDP growth rate be for this dream to come true?

A annual GDP growth of 3.5%.

.035=1 - 100200

=GM 20


EXAMPLE 13

If we want to double our wealth in k years. What should the annual return rate be?

1 - 100200

=GM k

3.5%20

4.7%15

7.17%10

14.8%5

18.9%4

26.0%3

41.4%2

100%1

Annual return ratek


The Mean of Grouped Data

The mean of a sample of data organized in a frequency distribution is computed by the following formula:

nΣXf

=X

11


EXAMPLE 14

A sample of ten movie theaters in a large metropolitan area tallied the total number of movies showing last week. Compute the mean number of movies showing.

Movies showing

frequency f

1 up to 3 1

3 up to 5 2

5 up to 7 3

7 up to 9 1

9 up to 11 3

Total 10


EXAMPLE 14 continued

Movies showing

frequency f

class midpoint

X

(f)(X)

1 up to 3 1 2 2

3 up to 5 2 4 8 5 up to 7 3 6 18 7 up to 9 1 8 8

9 up to 11 3 10 30 Total 10 66

6.6=1066

=nΣX

=X


The Median of Grouped Data

The median of a sample of data organized in a frequency distribution is computed by:

(i)f

CF2n

+L=Median -

where L is the lower limit of the median class, CFis the cumulative frequency preceding the median class, f is the frequency of the median class, and i is the median class interval.


Finding the Median Class

To determine the median class for grouped data:Construct a cumulative frequency distribution.Divide the total number of data values by 2.Determine which class will contain this value. For example, if n=50, 50/2 = 25, then determine which class will contain the 25th

value.


EXAMPLE 15

Movies showing

Frequency Cumulative Frequency

1 up to 3 1 1

3 up to 5 2 3

5 up to 7 3 6

7 up to 9 1 7

9 up to 11 3 10


The idea behind the formula

x

Cum

ulat

ive

freq

uenc

y

1-3 3-5 5-7 7-9 9-11

1

3

6

7

5

12


The idea behind the formula

x

Cum

ulat

ive

freq

uenc

y

5 7

3

6

5

(5 – 3)/(6-3) = k/(7-5)(n/2 – CF)/f = k/i

k =(n/2 – CF)/f (i)

k

Median = 5+kMedian = L+k


EXAMPLE 15 continued

From the table, L=5, n=10, f=3, i=2, CF=3

6.33=(2)3

32

10

+5=(i)f

CF2n

+L=Median - -


The Mode of Grouped Data

The mode for grouped data is approximated by the midpoint of the class with the largest class frequency.

Example 13 (continued): The modes in EXAMPLE 13are 6 and 10.

Movies showing

Frequency Cumulative Frequency

1 up to 3 1 1

3 up to 5 2 3

5 up to 7 3 6

7 up to 9 1 7

9 up to 11 3 10

When two values occur a large number of times, the distribution is called bimodal, as in Example 13.


Symmetric Distribution

zero skewness: mode = median = mean

Density Distribution(the height may be interpreted as relative frequency)

The area under the density distribution is 1. The sum of relative frequency is 1.Thus median always splits the density distribution into two equal areas.


Right Skewed Distribution

Positively skewed: (Skew to the right)

Mean and Median are to the right of the Mode.

Mode<Median<Mean


Left Skewed Distribution

Negatively Skewed:(skew to the left)

Mean and Median are to the left of the Mode.

Mean<Median<Mode

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Measures of dispersionMeasures of dispersion

l

1. Range2. Mean Deviation3. Variance and standard deviation4. Coefficient of variation5. Box plots.


Range

The range is the difference between the largest and the smallest value.

Only two values are used in its calculation.It is influenced by an extreme value.It is easy to compute and understand.


Mean Deviation

The Mean Deviation is the arithmetic mean of the absolute values of the deviations from the arithmetic mean.

All values are used in the calculation.It is not influenced too much by large or small values.The absolute values are difficult to manipulate.

Mean deviation is also known as Mean Absolute Deviation (MAD).

nX-X Σ

=MD


EXAMPLE 16

The weights of a sample of crates containing books for the bookstore (in pounds ) are:

103, 97, 101, 106, 103Find the range and the mean deviation.

Range = 106 – 97 = 9


Example 16

The first step is to find the mean weight.

The mean deviation is:

102=5

510=

nΣX

=X

2.4=5

5+4+1+5+1=

5102-103+...+102-103

=n

X-X Σ=MD


Population Variance

The population variance is the arithmetic mean of the squared deviations from the population mean.

All values are used in the calculation.More likely to be influenced by extreme values than mean deviation.The units are awkward, the square of the original units.

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The VariancePopulation X1, X2, …, XN

Population Variance

(X - )

N2

i2

i=1

N

σµ

=∑

σ2


Sample Variance

1-n

)x - (x =s

n

1=i

2i

2∑

s2

Note in the sample variance formula the sum of deviation is divided by (n-1) instead of n in order to yield an unbiased estimator of the population variance.


EXAMPLE 17

The ages of the Dunn family are:2, 18, 34, 42

What is the population variance?

24=4

96=

nΣX

=µ

( ) ( )

236=4

944=

424-42+...+24-2

=N

µ)-Σ(X=σ

2222


The Population Standard Deviation

The population standard deviation (σ) is the square root of the population variance.

For EXAMPLE 2, the population standard deviation is 15.36, found by

15.36=236=σ=σ 2


EXAMPLE 18

The hourly wages earned by a sample of five students are:

$7, $5, $11, $8, $6. Find the variance.

7.40=537

=nΣX

=X

( ) ( ) ( )

5.30=1-5

21.2=

1-57.4-6+...+7.4-7

=1-nX-XΣ

=s222

2


Sample Standard Deviation

The sample standard deviation is the square root of the sample variance.

In EXAMPLE 3, the sample standard deviation is 2.30

2.30=5.29=s=s 2


Sample Variance For Grouped Data

The formula for the sample variance for grouped data is:

1-nxn-Σfx

1-nxnx2n-Σfx

1-nxΣfΣfxx2-Σfx

1-n)xxx2-Σf(x

1-Σf)x-Σf(x=s

22

222

22

2222

=

+=

+=

+=

15


Interpretation and Uses of the Standard Deviation

Chebyshev’s theorem: For any set of observations, the minimum proportion of the values that lie within k standard deviations of the mean is at least:

where k2 is any constant greater than 1.

2k1

-1


Chebyshev’s theorem

97.22%6

96.00%593.75%488.89%375.00%20%1CoverageK

Chebyshev’s theorem: For any set of observations, the minimum proportion of the values that lie within kstandard deviations of the mean is at least 1- 1/k2


Interpretation and Uses of the Standard Deviation

Empirical Rule: For any symmetrical, bell-shaped distribution:

About 68% of the observations will lie within 1sthe mean, About 95% of the observations will lie within 2s of the meanVirtually all the observations will be within 3s of the mean

Empirical rule is also known as normal rule.


µ−3σ

µ−2σ µ−1σ µ µ+1σ µ+2σ µ+ 3σ

Bell-shaped Curve showing the relationship between σ and µ


Why are we concern about dispersion?

Dispersion is used as a measure of risk.Consider two assets of the same expected (mean) returns.

-2%, 0%,+2%-4%, 0%,+4%

The dispersion of returns of the second asset is larger then the first. Thus, the second asset is more risky. Thus, the knowledge of dispersion is essential for investment decision. And so is the knowledge of expected (mean) returns.


Relative Dispersion

The coefficient of variation is the ratio of the standard deviation to the arithmetic mean, expressed as a percentage:

(100%)Xs

=CV

16


Sharpe Ratio and Relative Dispersion

Sharpe Ratio is often used to measure the performance of investment strategies, with an adjustment for risk.If X is the return of an investment strategy in excess of the market portfolio, the inverse of the CV is the Sharpe Ratio.An investment strategy of a higher Sharpe Ratio is preferred.

http://www.stanford.edu/~wfsharpe/art/sr/sr.htm


Skewness

Skewness is the measurement of the lack of symmetry of the distribution.The coefficient of skewness can range from 3.00 up to 3.00.A value of 0 indicates a symmetric distribution. It is computed as follows:

Smedian)-x3(

=sk

−∑3

sxx

2)-1)(n-(nn

=skOr


Why are we concerned about skewness?

Skewness measures the degree of asymmetry in risk.

Upside riskDownside risk

Consider the distribution of asset returns:Right skewed implies higher upside risk than downside risk.Left skewed implies higher downside risk than upside risk.


Percentiles and Quartiles

For a set of measurements arranged in increasing order, the pth percentile is a value such that p percent of the measurements fall at or below the value and (100-p) percent of the measurements fall at or above the value.The first quartile Q1 is the 25th percentile The second quartile (or median) Md is the 50th percentileThe third quartile Q3 is the 75th percentile.The interquartile range IQR is Q3 - Q1


Example: Quartiles

20 customer satisfaction ratings:

1 3 5 5 7 8 8 8 8 8 8 9 9 9 9 9 10 10 10 10

Md = (8+8)/2 = 8

Q1 = (7+8)/2 = 7.5 Q3 = (9+9)/2 = 9

IRQ = Q3 - Q1 = 9 - 7.5 = 1.5


Box Plots

A box plot is a graphical display, based on quartiles, that helps to picture a set of data.

Five pieces of data are needed to construct a box plot: the Minimum Value, the First Quartile, the Median, the Third Quartile, and the Maximum Value.

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EXAMPLE 19

Based on a sample of 20 deliveries, Buddy’s Pizza determined the following information. The minimum delivery time was 13 minutes and the maximum 30 minutes. The first quartile was 15 minutes, the median 18 minutes, and the third quartile 22 minutes. Develop a box plot for the delivery times.

12 14 16 18 20 22 24 26 28 30 32

min maxmedianQ1 Q3


Box and Whiskers Plots


Pareto Chart

Pareto Chart of Labeling Defects


Scatter Plots

Restaurant Ratings: Mean Preference vs. Mean Taste


Misleading Graphs and Charts:Scale Break

Mean Salaries at a Major University, 1999 - 2002


Misleading Graphs and Charts:Horizontal Scale Effects

Mean Salary Increases at a Major University, 1999 - 2002

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Working with mean and Standard Deviation

18.04 35.56 725645353227212012(10)

4.08 12.00 17127(9)

2.05 6.67 974(8)

2.05 5.33 853(7)

0.82 20.00 212120201919(6)

2.45 60.00 636057(5)

1.63 40.00 424038(4)

0.71 20.00 21202019(3)

0.82 0.00 10-1(2)

0.82 20.00 212019(1)

St DevMeanDataSet


(2) = (1) – mean(1):Mean(2)=0; Stdev(2)=Stdev(1)

(3) = (1) + mean(1)Mean(3)=Mean(1); Stdev(3)<Stdev(1).

(4) = (1)*2; (5) = (1)*3Mean(4)=mean(1)*2; mean(5)=mean(1)*3Stdev(4)=stdev(1)*2; stdev(5)=stdev(1)*3


2.45 60.00 636057(5)

1.63 40.00 424038(4)

0.71 20.00 21202019(3)

0.82 0.00 10-1(2)

0.82 20.00 212019(1)

St DevMeanDataSet



18.04 35.56 725645353227212012(10)

4.08 12.00 17127(9)

2.05 6.67 974(8)

2.05 5.33 853(7)

0.82 20.00 212120201919(6)

0.82 20.00 212019(1)

St DevMeanDataSet

(6)=(1) multiplied by some frequencyMean(6)=Mean(1); Stdev(6)=Stdev(1).

(9) = (7)+(8)Mean(9)=mean(7)+mean(8)

(10) = (7) *(8)Mean(10)=mean(7)*mean(8)


Further results about mean and variance of transformed variables

E(X) = mean or expected valuesV(X) = E[(X-E(X))2]=E(X2) – E(X)2

E(a+bX) = a+bE(X)E(X+Y) = E(X) + E(Y)V(X+Y) = V(X) + V(Y) if X and Y are independent.


Further results about mean and variance of transformed variables

E(a+bX) = a+bE(X)E(X+Y) = E(X) + E(Y)Suppose we invest $1 in two assets. $a in asset X and $(1-a) in asset Y. Their expected returns are respectively E(X) and E(Y). We will expect a return of E(aX+(1-a)Y) = aE(X) + (1-a)E(Y) for this investment portfolio. If these two assets are independent or uncorrelated so that C(X,Y) =0, then the variance is V(aX+(1-a)Y) = a2V(X) + (1-a)2V(Y)


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Chapter TwoDescriptive StatisticsDescriptive Statistics

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Analysis of Economic Data 4. Measures of dispersionkafuwong.econ.hku.hk/teaching/econ1003/2003F/ppt/Chap002.pdf · 2 Ka-fu Wong © 2003 Chap 2-7 EXAMPLE 1 Dr. Tillman is Dean of the