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Transcript
```Numerical
Descriptive
Techniques
1
Summary Measures
Describing Data Numerically
Central Tendency
Variation
Arithmetic Mean
Range
Median
Interquartile Range
Mode
Variance
Geometric Mean
Standard Deviation
Quartiles
Coefficient of Variation
Shape
Skewness
2
Measures of Central Location
• Usually, we focus our attention on two
types of measures when describing
population characteristics:
– Central location
The measure of central location
reflects the locations of all the actual
data points.
3
Measures of Central Location
• The measure of central location reflects
the locations of all the actual data
points.
• How?
With two data points,
the central location
But
if
the
third data
With one data point
should
fall inpoint
the middle
on the leftthem
hand-side
clearly the centralappears between
(in order
of
the
midrange,
it
should
“pull”of
location is at the point to reflect the location
the central
location
to the left.
itself.
both
of them).
4
The Arithmetic Mean
• This is the most popular and useful
measure of central location
Sum of the observations
Mean =
Number of observations
5
The Arithmetic Mean
Sample mean
x
n
n
ii11xxii
nn
Sample size
Population mean

N
i1 x i
N
Population size
6
The Arithmetic Mean
• Example 1
The reported time on the Internet of 10 adults are 0, 7, 12, 5, 33,
14, 8, 0, 9, 22 hours. Find the mean time on the Internet.
x
10
 i 1 xi
10

0x1  7x2
 ...  22
x10
 11.0
10
• Example 2
Suppose the telephone bills represent the population of measurements.
The population mean is
x42.19
 x38.45
 ...  x45.77
 i200
1
2
200
1 x i



200
200
43.59
7
The Arithmetic Mean
• Drawback of the mean:
It can be influenced by unusual
observations, because it uses all the
information in the data set.
8
The Median
• The Median of a set of observations is the
value that falls in the middle when the
observations are arranged in order of
magnitude. It divides the data in half.
Example 3
Comment
Find the median of the time on the internet Suppose only 9 adults were sampled
(exclude, say, the longest time (33))
for the 10 adults of example 1
Even number of observations
0, 0, 5,
0, 7,
5, 8,
7, 8,
9, 12,
14,14,
22,22,
33 33
8.59,, 12,
Odd number of observations
0, 0, 5, 7, 8 9, 12, 14, 22
9
The Median
• Median of
8 2 9 11 1 6 3
n = 7 (odd sample size). First order the data.
1 2 3 6 8 9 11
Median
•For odd sample size, median is the {(n+1)/2}th
ordered observation.
10
The Median
• The engineering group receives e-mail
requests for technical information from
sales and services person. The daily
numbers for 6 days were
11, 9, 17, 19, 4, and 15.
What is the central location of the data?
•For even sample sizes, the median is the
average of {n/2}th and {n/2+1}th ordered observations.
11
The Mode
• The Mode of a set of observations is the value
that occurs most frequently.
• Set of data may have one mode (or modal
class), or two or more modes.
The modal class
For large data sets
the modal class is
much more relevant
than a single-value
mode.
12
The Mode
• Find the mode for the data in Example 1. Here
are the data again: 0, 7, 12, 5, 33, 14, 8, 0, 9, 22
Solution
• All observation except “0” occur once. There are two “0”.
Thus, the mode is zero.
• Is this a good measure of central location?
• The value “0” does not reside at the center of this set
(compare with the mean = 11.0 and the median = 8.5).
13
Relationship among Mean, Median,
and Mode
• If a distribution is symmetrical, the mean,
median and mode coincide
Mean = Median = Mode
• If a distribution is asymmetrical, and
skewed to the left or to the right, the three
measures differ.
A positively skewed distribution
(“skewed to the right”)
Mode < Median < Mean
Mode Mean
Median
14
Relationship among Mean, Median,
and Mode
• If a distribution is symmetrical, the mean,
median and mode coincide
• If a distribution is non symmetrical, and
skewed to the left or to the right, the three
measures differ.
A positively skewed distribution
(“skewed to the right”)
A negatively skewed distribution
(“skewed to the left”)
Mode
Mean
Median
Mean
Mode
Median
Mean < Median < Mode
15
Geometric Mean
• The arithmetic mean is the most popular measure of the
central location of the distribution of a set of
observations.
• But the arithmetic mean is not a good measure of the
average rate at which a quantity grows over time. That
quantity, whose growth rate (or rate of change) we
wish to measure, might be the total annual sales of a
firm or the market value of an investment.
• The geometric mean should be used to measure the
average growth rate of the values of a variable over
time.
16
17
Example
18
19
20
21
Measures of variability
• Measures of central location fail to tell the
• A question of interest still remains
How much are the observations spread out
around the mean value?
22
Measures of variability
Observe two hypothetical
data sets:
Small variability
The average value provides
a good representation of the
observations in the data set.
This data set is now
changing to...
23
Measures of variability
Observe two hypothetical
data sets:
Small variability
The average value provides
a good representation of the
observations in the data set.
Larger variability
The same average value does not
provide as good representation of the
observations in the data set as before.
24
The range
– The range of a set of observations is the difference
between the largest and smallest observations.
– Its major advantage is the ease with which it can be
computed.
– Its major shortcoming is its failure to provide
information on the dispersion of the observations
between the two end points.
But, how do all the observations spread out?
The range cannot assist in answering this question
? Range
? ?
Smallest
observation
Largest
observation
25
The Variance


This measure reflects the dispersion of all the
observations
The variance of a population of size N, x1, x2,…,xN
whose mean is  is defined as
2 

2
N
(
x


)
i
i 1
N
The variance of a sample of n observations
x1, x2, …,xn whose mean is x is defined as
s2 
ni1( xi  x)2
n 1
26
Why not use the sum of deviations?
Consider two small populations:
9-10= -1
11-10= +1
8-10= -2
12-10= +2
A measure of dispersion
A
Can the sum of deviations
agreesofwith
this
Be aShould
good measure
dispersion?
The sum
of deviations is
observation.
zero for both populations,
8 9 10 11 12
therefore, is not a good
…but
Themeasurements
mean of both in B
measure
of
arepopulations
moredispersion.
dispersed
is 10...
4-10 = - 6
16-10 = +6
7-10 = -3
than those in A.
B
4
Sum = 0
7
10
13
16
13-10 = +3
27
Sum = 0
The Variance
Let us calculate the variance of the two populations
2
2
2
2
2
2 (8  10)  (9  10)  (10  10)  (11  10)  (12  10)
A 
2
5
2
2
2
2
2
2 (4  10)  (7  10)  (10  10)  (13  10)  (16  10)
B 
 18
5
Why is the variance defined as
the average squared deviation?
Why not use the sum of squared
deviations as a measure of
After all, the sum of squared
deviations increases in
magnitude when the variation
of a data set increases!!
28
The Variance
Let us calculate the sum
of squared
deviations
for both data sets
Which
data set has
a larger dispersion?
Data set B
is more dispersed
around the mean
A
B
1
2 3
1
3
5
29
The Variance
SumA = (1-2)2 +…+(1-2)2 +(3-2)2 +… +(3-2)2= 10
SumB = (1-3)2 + (5-3)2 = 8
SumA > SumB. This is inconsistent with the
observation that set B is more dispersed.
A
B
1
2 3
1
3
5
30
The Variance
However, when calculated on “per observation”
basis (variance), the data set dispersions are
properly ranked.
A2 = SumA/N = 10/10 = 1
B2 = SumB/N = 8/2 = 4
A
B
1
2 3
1
3
5
31
The Variance
• Example 4
– The following sample consists of the
number of jobs six students applied for: 17,
15, 23, 7, 9, 13. Find its mean and
variance
• Solution
x
i61 xi
6
17  15  23  7  9  13 84


 14 jobs
6
6

n
2

(
x

x
)
1
2
i1 i
s 

(17  14)2  (15  14)2  ...(13  14)2
n 1
6 1
 33.2 jobs2

32
The Variance – Shortcut
method
n
2
n


1
(

x
)
2
2
i1 i
s 
 x i 

n  1  i1
n

2



1  2
17

15

...

13
2
2

 17  15  ...  13 

6  1 
6



 33.2 jobs2
33
Standard Deviation
• The standard deviation of a set of
observations is the square root of the
variance .
Sample standard dev iation: s  s
2
Population standard dev iation:   
2
34
Standard Deviation
• Example 5
– To examine the consistency of shots for a
new innovative golf club, a golfer was asked
to hit 150 shots, 75 with a currently used (7iron) club, and 75 with the new club.
– The distances were recorded.
– Which 7-iron is more consistent?
35
Standard Deviation
• Example 5 – solution
Excel printout, from the
The innovation club is
more consistent, and
because the means are
close, is considered a
better club
Current
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Innovation
150.5467
0.668815
151
150
5.792104
33.54847
0.12674
-0.42989
28
134
162
11291
75
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
150.1467
0.357011
150
149
3.091808
9.559279
-0.88542
0.177338
12
144
156
11261
75
36
Interpreting Standard Deviation
• The standard deviation can be used to
– compare the variability of several distributions
– make a statement about the general shape of a
distribution.
• The empirical rule: If a sample of
observations has a mound-shaped
distribution, the interval
( x  s, x  s) contains approximately 68% of the measuremen ts
( x  2s, x  2s) contains approximately 95% of the measuremen ts
( x  3s, x  3s) contains approximately 99.7% of the measuremen ts
37
Interpreting Standard Deviation
• Example 6
A statistics practitioner wants to
describe the way returns on investment
are distributed.
– The mean return = 10%
– The standard deviation of the return = 8%
– The histogram is bell shaped.
38
Interpreting Standard Deviation
Example 6 – solution
• The empirical rule can be applied (bell shaped
histogram)
• Describing the return distribution
– Approximately 68% of the returns lie between 2% and
18%
[10 – 1(8), 10 + 1(8)]
– Approximately 95% of the returns lie between -6% and
26%
[10 – 2(8), 10 + 2(8)]
– Approximately 99.7% of the returns lie between -14% and
34%
[10 – 3(8), 10 + 3(8)]
39
The Chebyshev’s Theorem
• For any value of k  1, greater than 100(1-1/k2)% of
the data lie within the interval from x  ks to x  ks .
• This theorem is valid for any set of measurements
(sample, population) of any shape!!
k
Interval
Chebyshev
Empirical Rule
1
2
3
x  s, x  s
x  2s, x  2s
x  3s, x  3s
at least 0% (1-1/12) approximately 68%
at least 75%(1-1/22) approximately 95%
at least 89%(1-1/32) approximately 99.7%
40
The Chebyshev’s Theorem
• Example 7
– The annual salaries of the employees of a chain of
computer stores produced a positively skewed histogram.
The mean and standard deviation are \$28,000 and
\$3,000,respectively. What can you say about the salaries
at this chain?
Solution
At least 75% of the salaries lie between \$22,000 and
\$34,000
28000 – 2(3000)
28000 + 2(3000)
At least 88.9% of the salaries lie between \$\$19,000 and
\$37,000
41
28000 – 3(3000) 28000 + 3(3000)
The Coefficient of Variation
• The coefficient of variation of a set of
measurements is the standard deviation divided
by the mean value.
Sample coefficien t of variation : cv 
s
x

Population coefficien t of variation : CV 

• This coefficient provides a proportionate
measure of variation.
A standard deviation of 10 may be perceived
large when the mean value is 100, but only
moderately large when the mean value is 500
42
Sample Percentiles and Box Plots
• Percentile
– The pth percentile of a set of measurements is
the value for which
• p percent of the observations are less than that value
• 100(1-p) percent of all the observations are greater
than that value.
– Example
• Suppose your score is the 60% percentile of a SAT
test. Then
60% of all the scores lie here
40%
43
Sample Percentiles
• To determine the sample 100p percentile of a
data set of size n, determine
a) At least np of the values are less than or equal
to it.
b) At least n(1-p) of the values are greater than or
equal to it.
•Find the 10 percentile of 6 8 3 6 2 8 1
•Order the data: 1 2 3 6 6 8
•Find np and n(1-p): 7(0.10) = 0.70 and 7(1-0.10) = 6.3
A data value such that at least 0.7 of the values are less than or equal to it
44
and at least 6.3 of the values greater than or equal to it. So, the first observation
is the 10 percentile.
Quartiles
• Commonly used percentiles
– First (lower)decile = 10th percentile
– First (lower) quartile, Q1 = 25th percentile
– Second (middle)quartile,Q2 = 50th percentile
– Third quartile, Q3 = 75th percentile
– Ninth (upper)decile = 90th percentile
45
Quartiles
• Example 8
Find the quartiles of the following set of
measurements 7, 8, 12, 17, 29, 18, 4, 27,
30, 2, 4, 10, 21, 5, 8
46
Quartiles
• Solution
Sort the observations
2, 4, 4, 5, 7, 8, 10, 12, 17, 18, 18, 21, 27, 29, 30
The first quartile
15 observations
At most (.25)(15) = 3.75 observations
should appear below the first quartile.
Check the first 3 observations on the
left hand side.
At most (.75)(15)=11.25 observations
should appear above the first quartile.
Check 11 observations on the
right hand side.
Comment:If the number of observations is even, two observations
remain unchecked. In this case choose the midpoint between these
two observations.
47
Location of Percentiles
• Find the location of any percentile using
the formula
P
LP  (n  1)
100
w hereLP is the location of the P th percentile
• Example 9
Calculate the 25th, 50th, and 75th percentile of
the data in Example 1
48
Location of Percentiles
• Example 9 – solution
– After sorting the data we have 0, 0, 5, 7, 8, 9,
12, 14, 22, 33.
25
L 25  (10  1)
 2.75
100
Values 0
0
Location 2
Location 1
3.75 5
2.75
3
Location 3
The 2.75th location
Translates to the value
(.75)(5 – 0) = 3.75
49
Location of Percentiles
• Example 9 – solution continued
50
L 50  (10  1)
 5.5
100
The 50th percentile is halfway between the
fifth and sixth observations (in the middle
between 8 and 9), that is 8.5.
50
Location of Percentiles
• Example 9 – solution continued
75
L 75  (10  1)
 8.25
100
The 75th percentile is one quarter of the
distance between the eighth and ninth
observation that is
14+.25(22 – 14) = 16.
Eighth
observation
Ninth
observation
51
Quartiles and Variability
• Quartiles can provide an idea about the
shape of a histogram
Q1 Q2
Positively skewed
histogram
Q3
Q1
Q2
Q3
Negatively skewed
histogram
52
Interquartile Range
• This is a measure of the spread of the
middle 50% of the observations
• Large value indicates a large spread of the
observations
Interquartile range = Q3 – Q1
53
Box Plot
– This is a pictorial display that provides the
main descriptive measures of the data set:
•
•
•
•
•
L - the largest observation
Q3 - The upper quartile
Q2 - The median
Q1 - The lower quartile
S - The smallest observation
1.5(Q3 – Q1)
S
Whisker
1.5(Q3 – Q1)
Q1
Q2 Q 3
Whisker
L
54
Box Plot
• Example 10
Bills
42.19
38.45
29.23
89.35
118.04
110.46
.
Smallest =. 0
.
Q1 = 9.275
Median = 26.905
Q3 = 84.9425
Largest = 119.63
IQR = 75.6675
Outliers = ()
Left hand boundary = 9.275–1.5(IQR)= -104.226
Right hand boundary=84.9425+ 1.5(IQR)=198.4438
-104.226
0
9.275
84.9425 119.63
26.905
198.4438
No outliers are found
55
Box Plot
– The following data give noise levels measured
at 36 different times directly outside of Grand
Central Station in Manhattan.
NOISE
82
89
94
110
.
.
.
Smallest = 60
Q1 = 75
Median = 90
Q3 = 107
Largest = 125
IQR = 32
Outliers =
BoxPlot
75
75-1.5(IQR)=27
60
70
107
80
90
100
110
120
130
56
107+1.5(IQR)
=155
Box Plot
NOISE - continued
Q1
75
60
25%
Q2
90
Q3
107
50%
125
25%
– Interpreting the box plot results
• The scores range from 60 to 125.
• About half the scores are smaller than 90, and about half are
larger than 90.
• About half the scores lie between 75 and 107.
• About a quarter lies below 75 and a quarter above 107. 57
Box Plot
NOISE - continued
The histogram is positively skewed
Q1
75
60
25%
Q2
90
50%
Q3
107
125
25%
50%
25%
25%
58
Distribution Shape and
Box-and-Whisker Plot
Left-Skewed
Q1
Q2 Q3
Symmetric
Q1 Q2 Q3
Right-Skewed
Q1 Q2 Q3
59
Box Plot
• Example 11
– A study was organized to compare the quality
of service in 5 drive through restaurants.
– Interpret the results
• Example 11 – solution
– Minitab box plot
60
Box Plot
Jack in the Box5
Jack in the box is the slowest in service
Hardee’s
Hardee’s service time variability is the largest
C7
McDonalds
4
3
Wendy’s
2
Popeyes
1
Wendy’s service time appears to be the
shortest and most consistent.
100
300
200
C6
61
Box Plot
Times are symmetric
Jack in the Box5
Jack in the box is the slowest in service
Hardee’s
Hardee’s service time variability is the largest
C7
McDonalds
4
3
Wendy’s
2
Popeyes
1
Wendy’s service time appears to be the
shortest and most consistent.
100
300
200
C6
Times are positively skewed
62
Paired Data Sets and the
Sample Correlation Coefficient
• The covariance and the coefficient of
correlation are used to measure the
direction and strength of the linear
relationship between two variables.
– Covariance - is there any pattern to the way
two variables move together?
– Coefficient of correlation - how strong is the
linear relationship between two variables
63
Covariance
Population covariance  COV(X, Y) 
(x i   x )(y i   y )
N
x (y) is the population mean of the variable X (Y).
N is the population size.
(xi  x)(y i  y)
Sample cov ariance cov (x y, ) 
n-1
x (y) is the sample mean of the variable X (Y).
n is the sample size.
64
Covariance
• Compare the following three sets
xi
yi
(x – x)
(y – y)
(x – x)(y – y)
2
6
7
13
20
27
-3
1
2
-7
0
7
21
0
14
x=5
y =20
Cov(x,y)=17.5
xi
yi
(x – x)
(y – y)
(x – x)(y – y)
2
6
7
27
20
13
-3
1
2
7
0
-7
-21
0
-14
x=5
y =20
Cov(x,y)=-17.5
xi
yi
2
6
7
20
27
13
Cov(x,y) = -3.5
x=5 y =20
65
Covariance
• If the two variables move in the same
direction, (both increase or both
decrease), the covariance is a large
positive number.
• If the two variables move in opposite
directions, (one increases when the other
one decreases), the covariance is a large
negative number.
• If the two variables are unrelated, the
covariance will be close to zero.
66
The coefficient of correlation
Population coefficien t of correlatio n
COV ( X, Y)

xy
Sample coefficien t of correlatio n
cov(X, Y)
r
sx sy
– This coefficient answers the question: How
strong is the association between X and Y.
67
The coefficient of correlation
+1 Strong positive linear relationship
COV(X,Y)>0
 or r =
or
0
No linear relationship
-1 Strong negative linear relationship
COV(X,Y)=0
COV(X,Y)<0
68
The coefficient of correlation
• If the two variables are very strongly
positively related, the coefficient value is
close to +1 (strong positive linear
relationship).
• If the two variables are very strongly
negatively related, the coefficient value is
close to -1 (strong negative linear
relationship).
• No straight line relationship is indicated by a
coefficient close to zero.
69
```
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