Numerical Measures in Statistics.arithmetic mean, weighted mean, median, mode, and geometric mean.Measures of Dispersion

1. 1
McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved.
Describing Data:
Numerical Measures
Chapter 3
3-2
GOALS
1. Calculate the arithmetic mean, weighted mean,
median, mode, and geometric mean.
2. Explain the characteristics, uses, advantages, and
disadvantages of each measure of location.
3. Identify the position of the mean, median, and mode
for both symmetric and skewed distributions.
4. Compute and interpret the range, mean deviation,
variance, and standard deviation.
5. Understand the characteristics, uses, advantages,
and disadvantages of each measure of dispersion.
6. Understand Chebyshev’s theorem and the Empirical
Rule as they relate to a set of observations.
3-3
Parameter Versus Statistics
PARAMETER A measurable characteristic
of a population.
STATISTIC A measurable characteristic of a
sample.
3-44
Numerical Descriptive Measures
Measures of Location
(Central Tendency)
Arithmetic Mean
Weighted Mean
Median
Mode
Geometric Mean
Measures of Dispersion
Range
Mean Deviation
Variance
Standard Deviation
3-55
Measures of Location or Central
Tendency
Measures of Central Tendency measure
the location or the central value of a set of
data
They cannot tell you about the distribution
pattern or dispersion of data.
3-6
Mean or Arithmetic Mean
For ungrouped data, the population mean is the sum of all the population values divided by the total
number of population values.
The sample mean is the sum of all the sample values divided by the total number of sample values.
EXAMPLE:

2. 2
3-77
Properties of 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. The mean is unique.
4. The sum of the deviations of each value from the
mean is zero.
5. The mean is affected by unusually large or small
data values.
6. It can not be computed for an open-ended
frequency distribution
3-88
Calculating Sample Mean:
From Grouped Data
Class Frequency
0-8 2
8-16 6
16-24 3
24-32 5
32-40 2
40-48 2
Total 20
3-99
Calculating Sample Mean:
From Grouped Data
Class Mid-Points
(x)
f d
=(x-A)÷i
fd
0-8 4 2 -2 -4
8-16 12 6 -1 -6
16-24 20=A 3 0 0
24-32 28 5 +1 5
32-40 36 2 +2 4
40-48 44 2 +3 6
Total 20 20 ∑fd=5
22=2+20=8×
20
5
+20=i×
n
fd
+A=x
∑
n
fx
=x
∑
3-1010
Weighted Mean
The weighted mean takes into account the
weight of every item
The weighted mean of a set of numbers X1, X2,
..., Xn, with corresponding weights w1, w2, ...,wn,
is computed from the following formula:
3-1111
EXAMPLE – Weighted Mean
The Carter Construction Company pays its hourly
employees $16.50, $19.00, or $25.00 per hour. There
are 26 hourly employees, 14 of which are paid at the
$16.50 rate, 10 at the $19.00 rate, and 2 at the $25.00
rate. What is the mean hourly rate paid the 26
employees?
3-12
The Median
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 centraltendency
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.
EXAMPLES:
MEDIAN The midpoint of the values after they have been ordered from the smallest to the largest, or the largest to
the smallest.
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, 75
Arranging the data in ascending order gives:
73, 75, 76, 80.
Thus the median is 75.5

3. 3
3-1313
Calculating Median From Grouped Data
Class f CF
0-8 2 2
8-16 6 8
16-24 3 11
24-32 5 16
32-40 2 18
40-48 2 20
Total 20
m
m
L+i}
f
1)+(CF-2/)1+n(
{=m
Median class Identification:
(n+1)/2 th item is the median.
(20+1)/2=10.510.5 thth item lies in the 3rd
class
CF= cumulative frequency of the class
preceding to the median class
fm= frequency of the median class
Lm= lower limit of the median class
20=16+8}
3
9)-5.10(
{=16+8}
3
1)+(8-2/)1+20(
{=m
3-14
The Mode
MODE The value of the observation that appears most frequently.
3-1515
Properties of Mode
1.Mode occurs most times
2.It is not affected by extreme values
3.There may not be a mode or may be several
modes (bi-mode or multi-mode)
4.Mode can be used for either numerical or
categorical data
5.Mode can be found out from a frequency
distribution with open-ended class
3-1616
Calculating Mode From Grouped Data
Class f
0-8 2
8-16 6
16-24 3
24-32 5
32-40 2
40-48 2
Total 20
i)
d+d
d
(+L=M
21
1
Mo0
M0= Mode from sample
LM0= Lower limit of the modal class
d1 = Difference between the frequency of
modal class and that of pre-modal class
d1 = Difference between the frequency of
modal class and that of post-modal class
4.14=8}
3)+(2
4
{+8=M0
3-17
The Relative Positions of the Mean,
Median and the Mode
In a positively or
negatively skewed
distribution, median
is the best measure
In a symmetric
distribution, all
measures will
give the same
results
3-18
The Geometric Mean
 Useful in finding the average change of percentages, ratios, indexes, or growthrates
over time.
 It has a wide application in business and economics because we are often interested in
finding the percentage changes in sales, salaries, or economic figures, such as the
GDP, which compound or build on each other.
 The geometric mean will always be less than or equal to the arithmetic mean.
 The formula for the geometric mean is written:
EXAMPLE:
Suppose you receive a 5 percent increase in salary this year and a 15 percent
increase next year. The average annual percent increase is 9.886, not 10.0. Why is
this so? We begin by calculating the geometric mean.
098861151051 .).)(.(GM 

4. 4
3-1919
EXAMPLE – Geometric Mean
The return on investment earned by Atkins
construction Company for four successive years
was: 30%, 20%, -40%, and 200 percent. What is
the geometric mean rate of return on
investment?
..).)(.)(.)(.(GM 2941808203602131 44 
3-2020
EXAMPLE –Geometric Mean
Another use of the geometric mean is to
determine the average percent change over
a period of time.
For example, if you earned $30,000 in1997 and
$50,000 in 2007, what is your annual rate of
increase over the period
tart
5.24%=0.0524=1–0524.1=1–
000,30
000,50
=GM
1–
periodofstheatValue
periodofendtheatValue
=GM
10
n
3-2121
Dispersion
Why Study Dispersion?
A measure of location only describes the center of
the data, not the spread of the data.
– Two sets of data with the same central values may differ
in distribution pattern. [0, 10, 10, 20] and [9, 10, 10, 11]
It enables us to judge the reliability of data by
having additional information
It helps us compare the spread of two or more
distributions
Wewill be careful in using widely dispersed data
3-22
Measures of Dispersion
 RANGE
 MEAN DEVIATION
 VARIANCE AND STANDARD DEVIATION
3-2323
EXAMPLE – Range
The number of cappuccinos sold at the Starbucks
location in the Orange Country Airport between 4 and
7 p.m. for a sample of 5 days last year were 20, 40,
50, 60, and 80. Determine the range for the number of
cappuccinos sold.
Range = Largest – Smallest value
= 80 – 20 = 60
3-24
EXAMPLE – Mean Deviation
EXAMPLE:
The number of cappuccinos sold at the Starbucks location in the Orange Country
Airport between 4 and 7 p.m. for a sample of 5 days last year were 20, 40, 50, 60,
and 80. Determine the mean deviation for the number of cappuccinos sold.
Step 1: Compute the mean
Step 2: Subtract the mean (50) from each of the observations, convert to positive if difference
is negative
Step 3: Sum the absolute differences found in step 2 then divide by the number of
observations
50
5
8060504020


 
n
x
x

5. 5
3-2525
Example – Population Mean Deviation from
Grouped Data
No of colds experienced
in 12 months
No of persons
(f)
0
1
2
3
4
5
6
7
8
9
15
46
91
162
110
95
82
26
13
2
Calculate mean deviation for the following
frequency distribution:
3-2626
X f fX X-3.78 fX-3.78
0
1
2
3
4
5
6
7
8
9
15
46
91
162
110
95
82
26
13
2
0
46
182
486
440
475
492
182
104
18
3.78
2.78
1.78
0.78
0.22
1.22
2.22
3.22
4.22
5.22
56.70
127.88
161.98
126.36
24.20
115.90
182.04
83.72
54.86
10.44
N = 642 fX = 2425 941.30
78.3=
642
2425
=
N
fXΣ
=µ 47.1=
642
30.941
=
N
|µ–x|fΣ
=MD
Example – Population Mean Deviation from
Grouped Data
3-27
Variance and Standard Deviation
VARIANCE The arithmetic mean of the squared deviations from the mean.
 The variance and standard deviations are nonnegative and are zero only
if all observations are the same.
 For populations whose values are near the mean, the variance and
standard deviation will be small.
 For populations whose values are dispersed from the mean, the
population variance and standard deviation will be large.
 The variance overcomes the weakness of the range by using all the
values in the population
STANDARD DEVIATION The square root of the variance.
3-28
EXAMPLE – Population Variance and
Population Standard Deviation
The number of traffic citations issued during the last five months in Beaufort County, South Carolina, is
reported below:
What is the population variance?
Step 1: Find the mean.
Step 2: Find the difference between each observation and the mean, and square that difference.
Step 3: Sum all the squared differences found in step 3
Step 4: Divide the sum of the squared differences by the number of items in the population.
29
12
348
12
1034...1719


 
N
x

124
12
488,1)( 2
2




N
X 

3-2929
Example – Population Variance and Standard
Deviation From Grouped Data
Class Frequency X fx (x-u)² f(x-u)²
700-799
800-899
900-999
1000-1099
1100-1199
1200-1299
1300-1399
1400-1499
1500-1599
1600-1699
1700-1799
1800-1899
4
7
8
10
12
17
13
10
9
7
2
1
750
850
950
1050
1150
1250
1350
1450
1550
1650
1750
1850
3000
5950
7600
10500
13800
21250
17550
14500
13950
11550
3500
1850
250000
160000
90000
40000
10000
0
10000
40000
90000
160000
250000
360000
10,00,000
11,20,000
7,20,000
4,00,000
1,20,000
0
1,30,000
4,00,000
8,10,000
11,20,000
5,00,000
3,60,000
Total 200 ∑fx=
1,25,000
f(x-u)² =
66,80,00
Levin: Example-3-66 3-3030
Example – Population Variance and Standard
Deviation From Grouped Data
1250=
100
000,25,1
=
N
fx
=µ
∑
800,66
100
000,80,66
N
2)x(f2 


5.258800,66 

6. 6
3-31
Sample Variance and
Standard Deviation
sampletheinnsobservatioofnumbertheis
sampletheofmeantheis
sampletheinnobservatioeachofvaluetheis
variancesampletheis
:Where
2
n
X
X
s
EXAMPLE
The hourly wages for a sample of part-time
employees at Home Depot are: $12, $20,
$16, $18, and $19.
What is the sample variance?
3-32
EXAMPLE:
Determine the arithmetic mean vehicle
selling price given in the frequency
table below.
The Sample Mean and Standard
Deviation of Grouped Data
EXAMPLE
Compute the standard deviation of the vehicle
selling prices in the frequencytable below.
3-3333
Uses of Standard Deviation (SD)
SD tells us the location of values of a
frequency distribution in relation to its mean
According to Chebyshev: ‘No matter what the
shape of the distribution, at least 75% of the
values will fall within ±2σ from the mean of the
distribution and at least 89% of the values will
lie within ±3 σ from the mean.’
If the distribution is symmetric, bell-shaped, we
can measure it with more precision
3-34
Chebyshev’s Theorem and Empirical Rule
The arithmetic mean biweekly
amount contributed by the Dupree
Paint employees to the company’s
profit-sharing plan is $51.54, and
the standard deviation is $7.51. At
least what percent of the
contributions lie within plus 3.5
standard deviations and minus 3.5
standard deviations of the mean?
3-3535
Standard Score
Standard Score gives us the number of
standard deviations an observation lies
below or above the mean.
Standard Score of any data point is
represented by Z.
Z= (x-u)÷σ
3-3636
Example – Sample Variance and Standard
Deviation From Grouped Data
The administrator of a Georgia hospital
surveyed the number of days 200
randomly chosen patients stayed in the
hospital following an operation. The days
are given in the table.
(a) Calculate the standard deviation and
mean
(b) According to Chebyshev’s theorem,
how many stays should be between 0
and 17 days? How many are actually in
that interval?
c) Because the distribution is roughly
bell-shaped, how many stays can we
expect between 0 and 17 days?
Class f
1 – 3
4 – 6
7 – 9
10 – 12
13 – 15
16 – 18
19 – 21
22 – 24
18
90
44
21
9
9
4
5
Total 200
Levin: Exercise Problem-3-66

7. 7
3-3737
Solution– Sample Variance and Standard
Deviation From Grouped Data
Class f x fx ∑f(x- )²
1 – 3
4 – 6
7 – 9
10 – 12
13 – 15
16 – 18
19 – 21
22 – 24
18
90
44
21
9
9
4
5
2
5
8
11
14
17
20
23
36
450
352
231
126
153
80
115
587.90
663.41
3.57
226.62
355.51
775.90
603.68
1168.16
Total 200 1543 4384.8
x
191
182
3-3838
Solution– Sample Variance and Standard
Deviation From Grouped Data
715.7=
200
1543
=
n
fxΣ
=x
69.4=
1–200
76.4384
=
1–n
)x–x(fΣ
=S
2
(b-i) 0–17 range equals mean ±2SD (7.715 ± 2(4.69)
Thus, at least150 (75% of 200) is expected.
(b-ii) Something between 182 and 191 observed
(c) 190 items expected
3-3939
Coefficient of Variation (CV)
A relative measure of dispersion, comparable
across distributions, that expresses the SD as
percentage of mean
CV is used to compare 2 or more sets of data
measured in different units
CV is sensitive to outliers
CV= (σ÷u) ×100
3-4040
Problem of CV
Regular MBA:
Evening MBA:
23
27
29
34
27
30
22
29
24
28
21
30
25
34
26
35
27
28
24
29
Students’ ages in the regular daytime MBA
program and the evening program of Central
University are described by these two samples:
If homogeneity of the class is a positive factor in
learning, use a measure of relative variability to
suggest which of the two groups will be easier to
teach.
Levin: Exercise Problem-3-76
3-4141
Solution to the Problem of CV
Levin: Exercise Problem-3-76
Regular MBA Evening MBA
X X
23
29
27
22
24
21
25
26
27
24
– 1.8
4.2
2.2
-2.8
-0.8
-3.8
0.2
1.2
2.2
– 0.8
3.24
17.64
4.84
7.84
0.64
14.44
0.04
1.44
4.84
0.64
27
34
30
29
28
30
34
35
28
29
– 3.4
3.6
-0.4
-1.4
-2.4
-0.4
3.6
4.6
-2.4
– 1.4
11.56
12.96
0.16
1.96
5.76
0.16
12.96
21.16
5.76
1.96
x = 248 = 55.6 x = 304 = 74.4
X(
)x–x(
2
)x–x( )x–x(
2
)x–x(
3-4242
Solution to the Problem of CV
Levin: Exercise Problem-3-76
8.24=
10
248
=
n
xΣ
=X -MBAReg
4.30=
10
304
=
n
xΣ
=xEMBA
485.2=
1–10
6.55
=
1–n
)x–x(Σ
=S
2
MBAgRe 876.2=
1–10
4.74
=
1–n
)x–x(Σ
=S
2
EMBA
%02.10=100×
8.24
485.2
=100×
x
s
CV MBAgRe
%46.9=100×
4.30
876.2
=100×
x
s
CVEMBA
Since 9.46% < 10.02%, EMBA is easier to teach