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Transcript
Measures of Variation
• Range
• Standard Deviation
• Variance
The Range
the difference between the largest
and smallest values of a
distribution
Find the range:
10, 13, 17, 17, 18
The range = largest minus smallest
= 18 minus 10 = 8
The standard deviation
a measure of the average
variation of the data entries from
the mean
Standard deviation of a sample
s
 (x  x)
n 1
n = sample size
2
mean of the
sample
•
•
•
•
•
•
To calculate standard
deviation of a sample
Calculate the mean of the sample.
Find the difference between each entry (x) and the
mean. These differences will add up to zero.
Square the deviations from the mean.
Sum the squares of the deviations from the
mean.
Divide the sum by (n  1) to get the variance.
Take the square root of the variance to get
the
standard deviation.
The Variance
the square of the standard
deviation
Variance of a Sample
(
x

x
)

2
s 
n 1
2
Find the standard deviation and
variance
x
30
26
22
78
x x
4
0
4
mean=
26
(x  x )
Sum = 0
16
0
16
___
32
2
The variance
s 
2

( x  x)
n 1
2
= 32  2 =16
The standard deviation
s=
16  4
Find the mean, the
standard deviation and
variance
mean = 5
x
xx
(x - x)
4
1
1
5
0
0
5
0
0
7
2
4
4
1
1
25
6
2
The mean, the standard
deviation and variance
Mean
=5
S tan dard deviation  1.5  1.22
Variance
6

 1 .5
4
Computation formula for
sample standard deviation:
s
SS x
n 1

 x

2
where
SS x   x
2
n
To find  x
2
Square the x values, then add.
To find (  x )
2
Sum the x values, then square.
Use the computing formulas to
find s and s2
x
x2
n=5
4
16
(Sx) 2 = 25 2 = 625
5
25
Sx2 = 131
5
25
SSx = 131 – 625/5 = 6
7
49
s2 = 6/(5 –1) = 1.5
4
25
16
131
s = 1.22
Population Mean and Standard
Deviation
x

population mean   
N
population
standard deviation   
2


x

x

N
where N  number of data values in the population
COEFFICIENT OF
VARIATION:
a measurement of the relative
variability (or consistency) of data
s

CV   100 or
 100
x

CV is used to
compare variability or
consistency
A sample of newborn infants had a mean weight of
6.2 pounds with a standard deviation of 1 pound.
A sample of three-month-old children had a mean
weight of 10.5 pounds with a standard deviation of
1.5 pounds.
Which (newborns or 3-month-olds) are more
variable in weight?
To compare variability,
compare Coefficient of Variation
For
newborns:
For 3month-olds:
CV = 16%
Higher CV:
more variable
CV = 14% Lower CV:
more consistent
Use Coefficient of Variation
To compare two groups of data,
to answer:
Which is more consistent?
Which is more variable?
CHEBYSHEV'S THEOREM
For any set of data and for any number k,
greater than one, the proportion of the
data that lies within k standard deviations
of the mean is at least:
1
1 
k
2
CHEBYSHEV'S THEOREM for k = 2
According to Chebyshev’s Theorem, at
least what fraction of the data falls
within “k” (k = 2) standard deviations of
the mean?
1
3
At least 1  2 2  4  75 %
of the data falls within 2 standard deviations of
the mean.
CHEBYSHEV'S THEOREM for k = 3
According to Chebyshev’s Theorem, at
least what fraction of the data falls
within “k” (k = 3) standard deviations of
the mean?
1
8
At least 1  3 2  9  88 . 9 %
of the data falls within 3 standard deviations of
the mean.
CHEBYSHEV'S THEOREM for k =4
According to Chebyshev’s Theorem, at
least what fraction of the data falls
within “k” (k = 4) standard deviations of
the mean?
1
15
At least 1  4 2  16  93 . 8 %
of the data falls within 4 standard deviations of
the mean.
Using Chebyshev’s Theorem
A mathematics class completes an examination
and it is found that the class mean is 77 and the
standard deviation is 6.
According to Chebyshev's Theorem, between
what two values would at least 75% of the
grades be?
Mean = 77
Standard deviation = 6
At least 75% of the grades would be in the
interval:
x  2 s to x  2 s
77 – 2(6) to 77 + 2(6)
77 – 12 to 77 + 12
65 to 89