Measures of Variability - University of Utahlzhang/teaching/3070spring2009/Daily Update… ·...

Measures of Variability

Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70Sample II: 30, 41, 48, 49, 50, 51, 52, 59, 70Sample III: 41, 45, 48, 49, 50, 51, 52, 55, 59


Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70Sample II: 30, 41, 48, 49, 50, 51, 52, 59, 70Sample III: 41, 45, 48, 49, 50, 51, 52, 55, 59


I Sample Range: the difference between the largest and thesmallest sample values.

e.g. for Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70the sample range is 40(= 70− 30).

I Deviation from the Sample Mean: the diffenence betweenthe individual sample value and the sample mean.e.g. for Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70the sample mean is 50 and thus the deviation from the samplemean for each data is -20, -15, -10, -5, 0, 5, 10, 15, 20.


I Sample Range: the difference between the largest and thesmallest sample values.e.g. for Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70the sample range is 40(= 70− 30).




I Deviation from the Sample Mean: the diffenence betweenthe individual sample value and the sample mean.

e.g. for Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70the sample mean is 50 and thus the deviation from the samplemean for each data is -20, -15, -10, -5, 0, 5, 10, 15, 20.





I Sample Variance: the mean (or average) of the sum ofsquares of the deviations from the sample mean for eachindividual data.

If our sample size is n, and we use x̄ to denote the samplemean, then the sample variance s2 is given by:

s2 =

∑ni=1(xi − x̄)2

n − 1=

Sxx

n − 1

I Sample Standard Deviation: the square root of the samplevariance

s =√

s2


I Sample Variance: the mean (or average) of the sum ofsquares of the deviations from the sample mean for eachindividual data.If our sample size is n, and we use x̄ to denote the samplemean, then the sample variance s2 is given by:

s2 =

∑ni=1(xi − x̄)2

n − 1=

Sxx

n − 1


s =√

s2


I Sample Variance: the mean (or average) of the sum ofsquares of the deviations from the sample mean for eachindividual data.If our sample size is n, and we use x̄ to denote the samplemean, then the sample variance s2 is given by:

s2 =

∑ni=1(xi − x̄)2

n − 1=

Sxx

n − 1


s =√

s2


e.g. for Sample I: 30, 35, 40, 45, 50, 55, 60, 65, 70, the mean is50 and we havexi 30 35 40 45 50 55 60 65 70xi − x̄ -20 -15 -10 -5 0 5 10 15 20(xi − x̄)2 400 225 100 25 0 25 100 225 400

Therefore the sample variance is(400 + 225 + 100 + 25 + 0 + 25 + 100 + 225 + 400)/(9− 1) = 187.5and the standard deviation is

√187.5 = 13.7.


e.g. for Sample II: 30, 41, 48, 49, 50, 51, 52, 59, 70, the mean isalso 50 and we have

xi 30 41 48 49 50 51 52 59 70xi − x̄ -20 -9 -2 -1 0 1 2 9 20

(xi − x̄)2 400 81 4 1 0 1 4 81 400


√121.5 = 11.0.


e.g. for Sample III: 41, 45, 48, 49, 50, 51, 52, 55, 59, the mean isalso 50 and we have

xi 41 45 48 49 50 51 52 55 59xi − x̄ -9 -5 -2 -1 0 1 2 5 9

(xi − x̄)2 81 25 4 1 0 1 4 25 81


√27.75 = 4.9.


sample variance for Sample I is 187.5, for Sample II is 121.5 andfor Sample III is 27.75.


Remark: 1. Why use the sum of squares of the deviations? Whynot sum the deviations?

Because the sum of the deviations from the sample mean EQUALTO 0!

n∑i=1

(xi − x̄) =n∑

i=1

xi −n∑

i=1

x̄

=n∑

i=1

xi − nx̄

=n∑

i=1

xi − n(1

n

n∑i=1

xi )

= 0


Remark: 1. Why use the sum of squares of the deviations? Whynot sum the deviations?Because the sum of the deviations from the sample mean EQUALTO 0!

n∑i=1

(xi − x̄) =n∑

i=1

xi −n∑

i=1

x̄

=n∑

i=1

xi − nx̄

=n∑

i=1

xi − n(1

n

n∑i=1

xi )

= 0


Remark: 1. Why use the sum of squares of the deviations? Whynot sum the deviations?Because the sum of the deviations from the sample mean EQUALTO 0!

n∑i=1

(xi − x̄) =n∑

i=1

xi −n∑

i=1

x̄

=n∑

i=1

xi − nx̄

=n∑

i=1

xi − n(1

n

n∑i=1

xi )

= 0


Remark:2. Why do we use divisor n − 1 in the calculation of samplevariance while we use use divisor N in the calculation of thepopulation variance?

The variance is a measure about the deviation from the“center”. However, the “center” for sample and population aredifferent, namely sample mean and population mean.If we use µ instead of x̄ in the definition of s2, thens2 =

∑(xi − µ)/n.

But generally, population mean is unavailable to us. So ourchoice is the sample mean. In that case, the observations x ′i s tendto be closer to their average x̄ then to the population averageµ. So to compensate, we use divisor n − 1.


Remark:2. Why do we use divisor n − 1 in the calculation of samplevariance while we use use divisor N in the calculation of thepopulation variance?The variance is a measure about the deviation from the“center”. However, the “center” for sample and population aredifferent, namely sample mean and population mean.

If we use µ instead of x̄ in the definition of s2, thens2 =

∑(xi − µ)/n.



Remark:2. Why do we use divisor n − 1 in the calculation of samplevariance while we use use divisor N in the calculation of thepopulation variance?The variance is a measure about the deviation from the“center”. However, the “center” for sample and population aredifferent, namely sample mean and population mean.If we use µ instead of x̄ in the definition of s2, thens2 =

∑(xi − µ)/n.



Remark:2. Why do we use divisor n − 1 in the calculation of samplevariance while we use use divisor N in the calculation of thepopulation variance?The variance is a measure about the deviation from the“center”. However, the “center” for sample and population aredifferent, namely sample mean and population mean.If we use µ instead of x̄ in the definition of s2, thens2 =

∑(xi − µ)/n.



Remark:3. It’ customary to refer to s2 as being based on n − 1 degrees offreedom (df).

s2 is the average of n quantities: (x1 − x̄)2, (x2 − x̄)2, . . . ,(xn − x̄)2. However, the sum of x1 − x̄ , x2 − x̄ , . . . , xn − x̄ is 0.Therefore if we know any n − 1 of them, we know all of them.

e.g. {x1 = 4, x2 = 7, x3 = 1, and x4 = 10}.Then the mean is x̄ = 5.5 and x1 − x̄ = −1.5, x2 − x̄ = 1.5 andx3 − x̄ = −4.5. From that, we know directly that x4 − x̄ = 4.5since their sum is 0.


Remark:3. It’ customary to refer to s2 as being based on n − 1 degrees offreedom (df).s2 is the average of n quantities: (x1 − x̄)2, (x2 − x̄)2, . . . ,(xn − x̄)2. However, the sum of x1 − x̄ , x2 − x̄ , . . . , xn − x̄ is 0.Therefore if we know any n − 1 of them, we know all of them.




e.g. {x1 = 4, x2 = 7, x3 = 1, and x4 = 10}.

Then the mean is x̄ = 5.5 and x1 − x̄ = −1.5, x2 − x̄ = 1.5 andx3 − x̄ = −4.5. From that, we know directly that x4 − x̄ = 4.5since their sum is 0.





Some mathematical results for s2:

I s2 = Sxxn−1 where Sxx =

∑(xi − x̄)2 =

∑x2

i −(∑

xi )2

n ;

I If y1 = x1 + c, y2 = x2 + c , . . . , yn = xn + c , then s2y = s2

x ;

I If y1 = cx1, y2 = cx2, . . . , yn = cxn, then sy =| c | sx .Here s2

x is the sample variance of the x ’s and s2y is the sample

variance of the y ’s. c is any nonzero constant.




∑(xi − x̄)2 =

∑x2

i −(∑

xi )2

n ;


x ;







∑(xi − x̄)2 =

∑x2

i −(∑

xi )2

n ;


x ;







∑(xi − x̄)2 =

∑x2

i −(∑

xi )2

n ;


x ;





e.g. in the previous example, Sample III is {41, 45, 48, 49, 50, 51,52, 55, 59} then we can calculate the sample variance as following

xi 41 45 48 49 50 51 52 55 59x2

i 1681 2025 2304 2401 2500 2601 2704 3025 3481∑xi 450∑x2

i 22722

Therefore the sample variance is

(22722− 4502

9)/(9− 1) = 27.75


Boxplots

e.g. A recent article (“Indoor Radon and Childhood Cancer”) presented the

accompanying data on radon concentration (Bq/m2) in two different samples of

houses. The first sample consisted of houses in which a child diagnosed with cancer

had been residing. Houses in the second sample had no recorded cases of childhood

cancer. The following graph presents a stem-and-leaf display of the data.

1. Cancer 2. No cancer

9683795 0 9576839767899386071815066815233150 1 12271713114

12302731 2 994941918349 3 839

5 47 5 55

67 Stem: Tens digit8 5 Leaf: Ones digit


Boxplotse.g. A recent article (“Indoor Radon and Childhood Cancer”) presented the

accompanying data on radon concentration (Bq/m2) in two different samples of

houses. The first sample consisted of houses in which a child diagnosed with cancer

had been residing. Houses in the second sample had no recorded cases of childhood

cancer. The following graph presents a stem-and-leaf display of the data.

1. Cancer 2. No cancer

9683795 0 9576839767899386071815066815233150 1 12271713114

12302731 2 994941918349 3 839

5 47 5 55

67 Stem: Tens digit8 5 Leaf: Ones digit


The boxplot for the 1st data set is:


The boxplot for the 2nd data set is:


We can also make the boxplot for both data sets:


Some terminology:

I Lower Fourth: the median of the smallest half

I Upper Fourth: the median of the largest half

I Fourth spread: the difference between lower fourth andupper fourth

fs = upper fourth− lower fourth

I Outlier: any observation farther than 1.5fs from the closestfourthAn outlier is extreme if it is more than 3fs from the nearestfourth, and it is mild otherwise.


Some terminology:







Some terminology:







Some terminology:





I Outlier: any observation farther than 1.5fs from the closestfourth

An outlier is extreme if it is more than 3fs from the nearestfourth, and it is mild otherwise.


Some terminology:







The boxplot for the 2nd data set is:

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