measures of central tendencymeasures of central tendency chapter 6. measures of central tendency...

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Chapter 6

Measures of Central Tendency

Chapter 6. Measures of Central Tendency

Definition of a Summary Measure (page 185)

A summary measure is a single value that we compute from a collection of measurements in order to describe one of the collection’s particular characteristics.

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Definition of Measures of Central Tendency (page 185)

A measure of central tendency is a single value that can be used to represent all the other values in the collection.

Notes:

Some people refer to this measure as the “average”. This measure tells us where the “center” of the distribution

lies. The use of this measure will also facilitate the comparison of

two or more collections of measurements.


Definition of the Arithmetic Mean (page 192)

Definition 6.2:

The arithmetic mean is the sum of all the values in the collection divided by the total number of elements in the collection.

The population mean for a finite population with N elements, denoted by the lowercase Greek letter , is

N

XN

ii

1

where Xi is the measure taken from the ith element of the population.

The sample mean for a sample with n elements, denoted by X (read as “ X -bar”), is

n

XX

n

ii

1

where Xi is the measure taken from the ith element of the sample.

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Example 6.3a (page 193)

Five judges give their scores on the performance of a gymnast as follows: 8, 9, 9, 9, and 10. Find the mean score of the gymnast.

Solution: We compute for the population mean. Let Xi be the score given by the ith judge in the population. Add the scores given by the 5 judges.

5

1i

iX

= 8 + 9 + 9 + 9 + 10 = 45

Then divide the sum of the scores by the number of judges. We have N = 5 judges and get

9545

μ

The mean score of the gymnast is 9.


Example 6.3b (page 193-194)

A cereal company selects a sample of 8 cartons of cereals for a quality control check on the specified weight. The weights in grams are as follows: 256.1, 255.2, 255.0, 255.3, 255.9, 256.2, 255.8, and 255.4. Find the mean weight for the sample of cereal cartons.

Question: Can we compute for ?

Solution:

Let Xi be the weight of the ith cereal carton in the sample. Adding the weights of the 8 cereal cartons, we have,

8

1i

iX

= 256.1+255.2+255.0+255.3+255.9+256.2+255.8+255.4

= 2044.9

We divide the sum of weights by the number of cereal cartons in the sample. We have n = 8 cereal cartons and get

X 61.2558

9.2044

The mean weight of the cereals in the sample is 255.61 grams.

Note; The computed mean is not one of the measurements in the collection.

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The Mean as a “Center” of Mass Figure 6.1 (page 194)

What would happen if the first measurement had been 7 instead of 8?

7 8 9 103

11 12To balance the seesaw, the fulcrum must be at µ= 9.

The loads that we place on the seesaw are the observations, Xi,: 8,9,9,9,10

The loads that we now place on the seesaw are the observations, Xi,: 7,9,9,9,10

To balance the seesaw, the fulcrum must be at what value of µ?To balance the seesaw, the fulcrum must

be at what value of µ? µ=8.8

What would happen this time if the last measurement had been 1000 instead of 10?

To balance the seesaw, the fulcrum must

be at what value of µ? µ=206.8


Effect of an Outlier on the Mean (pages 196-197)

Definition: Outliers are data values that are markedly different from the rest of the data items.

Since the mean is the “center of mass” then its value is gravely affected by outliers. An outlier will pull the value of the arithmetic mean in its direction and away from the location of majority of the observations.

With the presence of outliers, the mean might not be a suitable measure of central tendency because it may not be a good representative of the observations in the collection.

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Example 6.6 (page 197)

a) Let us consider the monthly salaries of 5 employees: P9,500.00, P10,200.00, P9,000.00, P10,500.00, and P11,000.00. Find the mean.

Solution: .P10,040/mo5

11,00010,5009,00010,2009,500 X

Question: Is this a good representative of the values in the collection? b) Suppose the monthly salary of the fifth employee is P60,000.00 instead of P11, 000.00.

What will be the mean salary?

Solution: 9,500 10,200 9,000 10,500 60,000X P19,840/mo.

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Question: Is this still a good representative of the values in the collection?


Properties of the Mean (pages 196-197, 215)

The mean is the “center of mass”. It uses all the observed values in the calculation. It may or may not be an actual observed value in

the data set. We may treat its formula algebraically. Its value is gravely affected by outliers. The mean of a finite collection always exists and is

unique. Data values should be measured using at least an

interval scale.

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Mathematical Properties (page 215)

1 1 1

1

1 1

( )

0.

n n n

i ii i i

n

ii

n n

i ii i

X X X X

n n

X nX

n

X X

n

Theorem 1: The first central moment about the

mean is 0, that is,

N n

i ii 1 i 1

X X X0 and 0

N n

Proof:



Theorem 2. is minimum when

Proof: This is minimum when

2

1( )

n

ii

X c

.c X

2 2 2 2 2

1 1 1 1 1

2 2 2 2

1

2 2 2 2

1

2 2 2

1

( ) ( 2 ) 2

2 ( )

( 2 )

( )

n n n n n

i i i i ii i i i i

n

iin

iin

ii

X c X X c c X c X c

X cnX nc nX nX

X nX n X cX c

X nX n X c

2( ) 0, , .n X c that is c X

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Theorem 3: If we add a constant c to all original observations, then the mean of the new observations will increase by the same amount constant c. That is,

Proof: Original Data = {X1, X2, …, Xn} Transformed Data = {Y1, Y2, …, Yn} where Yi = Xi + c

1 1 1 1

1 1

( )

.

n n n n

i i ii i i i

n n

i ii i

Y X c X cY

n n n

X nc Xc X c

n n

.new originalX X c



Theorem 4: If we subtract a constant c to all original observations, then the mean of the new observations will decrease by the same amount constant c. That is,

.new originalX X c

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Theorem 5: If we multiply a constant c to all original observations, then the mean of the new observations is the original mean multiplied by the constant c. That is,

Proof: Original Data = {X1, X2, …, Xn} Transformed Data = {Y1, Y2, …, Yn} where Yi = cXi

1 1 1

( )

.

n n n

i i ii i i

Y cX c XY

n n ncX

.new originalX X xc



Theorem 5: If we divide all the original observations by a constant c, then the mean of the new observations is the original mean divided by the constant c. That is,

.new originalX X c

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Example

In a sample of 4 days, the temperature (in centigrade) 28.5, 27.0, 31.2, 29.4

28.5 27.0 31.2 29.4 29.025 .4

X C

If observations were converted to Fahrenheit (F= (C x 9/5) + 32)): 83.30, 80.60, 88.16, 84.92

83.30 80.60 88.16 84.92 84.245 .4

Y F

OR

9 932 29.025 32 84.245 .5 5Y X x x F


Approximating the Mean for Grouped Data (FDT) page 195

Population Mean: N

Xfμ

k

1iii

Sample Mean: n

XfX

k

1iii

where fi = the frequency of the ith class iX = the class mark of the ith class k = total number of classes

N or n = total number of observations =

k

iif

1

10


Example 6.4 (page 195)The table below gives us the weight in pounds of a sample of 75 pieces of luggage. Approximate the sample mean weight of the luggage.

Weight No. of Luggage Class Mark

(in pounds) if iX ii Xf 31.5 - 41.4 …… 9 …………36.45 …… 328.05 41.5 - 51.4 …… 8 …………46.45 …… 371.60 51.5 - 61.4 …… 4 …….….. 56.45 …… 225.80 61.5 - 71.4 …… 32 …………66.45 …… 2126.40 71.5 - 81.4 …… 14 …………76.45 …… 1070.30 81.5 - 91.4 …… 5 …………86.45 …… 432.25 91.5 -101.4 …… 3 …………96.45 …… 289.35 75 if 75.4843 ii Xf

7

74843.75 64.58

75

i ii 1

ii 1

f XX pounds

f


Modifications of the mean:Weighted Mean (page 200)

Used when observations are not of equal importance

If we assign a weight to each observation, where i = 1, 2,…, n, and n is the number of observations in the sample, then the weighted sample mean is given by

n21

nn2211n

1ii

n

1iii

w W...WWXW...XWXW

W

XWX

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Example 6.9 (pages 200 – 201)

Suppose a government agency gives scholarship grants to employees taking graduate studies. Courses in graduate studies earn credits of 1, 2, 3, 4, or 5 units. They can get a partial scholarship for the next semester if they get a weighted average of 1.5 to 1.75 and a full scholarship if the average is better than 1.5, which means an average of 1.0 to 1.49. What kind of scholarship will the 2 employees get given their grades for the previous semester? Employee A Employee B Subjects Units Grade Subjects Units Grade A 1 1.0 A 1 2.0 B 2 1.25 B 2 1.75 C 3 1.5 C 3 1.5 D 4 1.75 D 4 1.25 E 5 2.0 E 5 1.0 Solution: We let the units be the weights Wi and the grades be the Xi.

Weighted average of employee A: (1)(1) (2)(1.25) (3)(1.5) (4)(1.75) (5)(2) 25 1.67

1 2 3 4 5 15WX

Weighted average of employee B: (1)(2) (2)(1.75) (3)(1.5) (4)(1.25) (5)(1) 20 1.33

1 2 3 4 5 15WX


Modifications of the Mean:Combined Mean or the Mean of Means(page 201 – 202)

Suppose that k finite populations having k21 .,N..,,NNmeasurements, respectively, have means k21 ,μ...,,μμ . The combined population mean, c , if we combine the measurements of all the populations is

k21

kk2211k

1ii

k

1iii

c N...NNμN...μNμN

N

μNμ

If samples of size n1, n2, . . . , nk, selected from these k populations, have the sample means, iX , respectively, thecombined sample mean, cX , if we combine the measurements in all the samples is

k

kkk

ii

k

iii

c nnnXnXnXn

n

XnX

......

21

2211

1

1

12



Three sections of a statistics class containing 28, 32, and 35 students averaged 83, 80, and 76, respectively, on the same final examination. What is the combined population mean for all 3 sections? Solution: We let N1 = 28, N2 = 32, N3 = 35, 1 = 83, 2 = 80, 3 = 76

(28)(83) (32)(80) (35)(76) 79.428 32 35C

Thus, the mean grade of the 3 sections is 79.4.


Modifications of the Mean: Trimmed Mean (page 202)

Objective: remove the influence of possible outliers

Choose (Greek letter alpha: the proportion of observations that will be deleted), 0 < < 1.

To find the (/2)(100)%-trimmed mean for a given data set, we first order the data according to magnitude, say, lowest to highest. Then, we remove (/2)(100)% of the observations in both the lower and upper ends of the array. We then calculate the arithmetic mean for the remaining observations.

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Example 6.11 (pages 202-203)

Compute for the arithmetic mean and 5% trimmed mean for the given data set:

10 10 11 11 11 12 12 12 12 12 12 13 13 14 14 14 14 14 15 15 15 15 16 16 17 17 18 18 18 18 18 18 19 20 20 20 20 20 500 524

Solution:

The arithmetic mean is

40

1 1598 39.9540 40

iiX

. We will notice that the

values 500 and 524 are outliers. These outliers pulled the value of the mean in their direction. As a result, the computed mean of 39.95 is not a good representative of the observations in the data set because it is so much higher than the values of majority of the observations.


Example 6.11 cont’d (pages 202-203)To compute for the 5% trimmed mean, we delete the bottom 5% and the top 5% of the observations in the array. Since there are 40 observations and 5% of 40 is 2 then we would have to delete the first two observations in the array and the last two observations. This will leave us with 36 observations. These are:

11 11 11 12 12 12 12 12 12 13 13 14 14 14 14 14 15 15 15 15 16 16 17 17 18 18 18 18 18 18 19 20 20 20 20 20

The 5% trimmed mean is the mean of these remaining 36

observations. Thus, we have

36

1 554 15.3936 36

iiX

. This value is a better

summary measure to represent the observations in the original data set.

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Assignment (pages 204-206)

Exercise 3. (No need to interpret.) Exercise 5. Exercise 6. Exercise 8. (No need to compare.) If

you were asked to compute for the weighted mean, what do you suggest that we use as weights?


Definition of the Median (page 206)

Definition 6.3.

The median divides the array into two equal parts.

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Finding the Median (page 206-207)

Step 1: Arrange the observations in an array. We let X(i) be the ith

observation in the array, where i = 1,2, . . . , n. Thus, X(1) is the smallest observation while X(n) is the largest observation.

Step 2: Determine the median, Md.

Case 1: If the number of observations n is odd,

12nMd X

Case 2: If the number of observations n is even,

1

2 2

2

n nX XMd


Example 6.13a (page 207)

The following are the total receipts of 7 mining companies (in million pesos): 1.3 6.6 10.5 12.6 50.7 4.7 7.3

Solution: Array: 1.3 4.7 6.6 7.3 10.5 12.6 50.7 Notation: X(1) X(2) X( 3) X(4) X( 5) X(6) X(7)

Since n=7 is odd, Md = 12nX

= 7 1

2

X

= X( 8/2) = X( 4 )= 7.3 million pesos

How many observations are to the left of the median? How many observations are to the right of the median? A median of 7.3 million pesos indicates that companies with total receipt of less than 7.3 million pesos belong in the lower half of the array; whereas, companies with total receipt greater than 7.3 million pesos belong in the upper half of the array.

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Example 6.13b (pages 207-208)The following are the number of years of operation of 8 mining companies: 9 11 16 12 17 20 18 19 Solution: Arrange the observations from lowest to highest. Array: 9 11 12 16 17 18 19 20

Notation: X(1) X(2) X(3) X( 4) X(5) X( 6) X(7) X( 8) Two middle observations

Since n=8 is even, Md = 1

2 2

2

n nX X

=

8 8 12 2

2

X X

= 16.5

21716

2

(5)4 XX

How many observations are to the left of the median? How many observations are to the right of the median? A median of 16.5 indicates that companies operating for less than 16.5 years belong in the lower half of the array while companies operating for more than 16.5 years belong in the upper half of the array.


Interpretation of the Median (page 208)

The exact interpretation of the median is as follows:

“At least half of the observations are less than or equal to the median and at the same time at least half of the observations are greater than or equal to the median. ”

This general interpretation can handle all types of data sets, including those with tied values in the middle of the array.

Example: n=12

Array: 3, 4, 4, 4, 4, 5, 5, 5, 5, 7, 8, 9Median = 5

How many are less than 5? How many are greater than 5?

A median of 5 means that at least half of the observations are less than or equal to 5 and at the same time at least half are greater than or equal to 5.

The interpretation will simplify to “half of the observations are less than the median and half are greater than the median” if the median is not one of the observed values, that is, n is even and there are no ties (see previous example).

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Effect of Outliers on MedianExample 6.16 (page 211)

Let us consider the data set in Example 6.6 on the monthly salaries of 5 employees (in pesos): 9,500 10,200 9,000 10,500 60,000

The mean monthly salary is P19,840. Let us now determine the median.

Array: 9000 9500 10200 10500 60000 Notation: X(1) X(2) X(3) X(4) X(5)

(3)1 5 12 2

10,200nMd X X X

The median monthly salary is P10,200. Which is a better measure of central tendency?


Characteristics of the Median (pages 210, 215)

The median is the “center” of the array. The median is also a measure of position/location. An observation

whose value is smaller than the median belongs in the lower half of the array while an observation whose value is higher than the median belongs in the upper half of the array.

Unlike the mean, it uses only the middle value/s in the array for its computation.

Unlike the mean, the median is not affected by outliers (observations whose values are extremely different from the others in the data set).

Unlike the mean, the median is not amenable to algebraic manipulation.

Unlike the mean, the median is still interpretable when the level of measurement is as low as ordinal.

The median will always exist and is unique.

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Approximating the Median for Grouped Data (FDT) (page 209)

Step 1. Calculate n/2, where n is the number of observations=1

k

iif

Step 2. Construct the < cumulative frequency distribution (< CFD). Step 3. Starting from the top, locate the value in the <CFD column that is

greater than or equal to n/2 for the first time. The class interval corresponding to that value is the median class.

Step 4. Approximate the median using the formula given below.

Md = LCBMd + C

Md

Md

fCFDn 12/

where LCBMd is the lower class boundary of the median class C is the class size of the median class n is the number of observations <CFMd-1 is the less than cumulative frequency of the class

preceding the median class fMd is the frequency of the median class


Example 2 (page 210)

The table below gives us the weight in pounds of a sample of 75 pieces of luggage. Approximate the median weight of luggage.

Weight Class Boundaries No. of Luggage < CF (in pounds) LCB - UCB fi 31.5 - 41.4 31.45 - 41.45 9 9 41.5 - 51.4 41.45 – 54.45 8 17 51.5 - 61.4 51.45 – 61.45 4 21 61.5 - 71.4 61.45 – 71.45 32 53 71.5 - 81.4 71.45 – 81.45 14 67 81.5 - 91.4 81.45 – 91.45 5 72 91.5-101.4 91.45 – 101.45 3 75 75 The median class is 61.5 – 71.4 since n/2= 75/2 =37.5 and the <CF of the interval, 61.5 – 71.4, is 53 which is the value that is greater than or equal to 37.5 for the first time from the top.

Md LCBMd + C

Md

Md

fCFDn 12/

= 61.45 + 10 61.6632

215.37

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RationaleConsider the points A, B and C. Point A: (UCBMd-1, <CFDMd-1) Point B: (Md, n/2) Point C: (UCBMd, <CFDMd) These three points all belong on the same line and the slope of this line can be determined by any two of these three points. Thus, equating the formula for the slope using Points A and B and the formula for the slope using Points A and C, we have:

1 1

1 1

/ 2 Md Md Md

Md Md Md

n CFD CFD CFDMd UCB UCB UCB

1/ 2 Md Md

Md

n CFD fMd LCB C

1/ 2 MdMd

Md

n CFDMd LCB Cf

Less than Ogive

n/2

Md

A

B C


Definition of the Mode (page 211)

The mode is the observed value that occurs with the greatest frequency in a data set.

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Example 6.17 (pages 211-212)

Determine the mode. b) Given the number of dogs owned by 23 students: 0, 0, 0, 1, 1, 1, 1, 2, 2, 2 , 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3 d) Consider the scores of 15 students in a quiz: 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20 c) Consider the shoe size of 24 faculty members: 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 11


Multimodality Detection of multimodality can provide insights into the

underlying nature of the data set. For example: A metal alloy manufacturer was concerned with customer complaints about the lack of uniformity in the melting points of one of the firm's alloy filaments. Sixty filaments were selected from the production process and their melting points determined. The resulting frequency polygon showed that the distribution is bimodal. Closer examination of the data revealed that 26 of the filaments were produced by the first shift and the other 34 by the second shift. The frequency polygons for each shift were constructed. This revealed that the bimodal nature of the original frequency distribution is the result of a difference in the positions of the distributions for the two shifts. They then discovered that the second shift was using the wrong alloy mixture specification. Correction of the mixture resulted in production of filaments with more uniform melting points.

Entire Data Set

0

5

10

15

20

25

295 305 315 325 335 345 355

Melting Point (in Celcius)

Num

ber o

f Fila

men

ts

Separation by Shift

0

5

10

15

20

295 305 315 325 335 345 355

Melting Point (in Celcius)

Num

ber o

f Fila

men

ts

First shift

Second shift

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Given the frequency distribution of the civil status data of 24 employees,

what is the mode?

Civil Status Number of Faculty

Single …………………………….. 15

Married …………………………… 8

Widowed ……………………….. 0

Separated ……………………….. 1

Total ………………………. 24


Characteristics of the Mode (pages 219,221)

1. The mode is the “center” in the sense that it is the most typical value in a set of observations.

2. Outliers do not affect the mode. 3. The mode will not always exist; and if it does, it may not be unique. A data

set is said to be unimodal if there is only one mode, bimodal if there are two modes, trimodal if there are three modes, and so on.

4. The value of the mode is always one of the observed values in the data set. 5. We can get the mode for both quantitative and qualitative types of data; that is,

the mode is interpretable even if the level of measurement is as low as nominal. 6. The mode is generally not as useful a measure of central tendency as the

mean and the median when the data consist of only a few numbers. For example, for the numbers 7, 12, 18, 22, 31, 31, the mode is 31 since it appears twice and all other numbers only once. But 31 cannot be considered a good measure of central tendency for these data since it is in fact at the extreme high end of the values and its frequency exceeds the frequency of the other values by only 1.

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Approximating the Mode for Grouped Data (page 212)

Step 1. Locate the modal class. For frequency distribution with equal class sizes, the modal class is the class interval with the highest frequency. Otherwise, compute for adjusted frequencies first.

Step 2. Approximate the mode using the following formula:

Mo = LCBMo + C

21

1

2 fffff

Mo

Mo

where LCBMo is the lower class boundary of modal class C is the class size f Mo is the frequency of the modal class f 1 is the frequency of the class preceding the

modal class f 2 is the frequency of the class following the

modal class



The table below gives us the weight in pounds of a sample of 75 pieces of luggage. Approximate the modal weight of the luggage.

Weight Class Boundaries No. of Luggage (in pounds) LCB - UCB fi 31.5 - 41.4 31.45 - 41.45 9 41.5 - 51.4 41.45 – 54.45 8 51.5 - 61.4 51.45 – 61.45 4 Modal class 61.5 - 71.4 61.45 – 71.45 32 71.5 - 81.4 71.45 – 81.45 14 81.5 - 91.4 81.45 – 91.45 5 91.5-101.4 91.45 – 101.45 3

Mo = LCBMo + C

21

1

2 fffff

Mo

Mo

= 61.45 + 10 54.67144)32(2

432

23


Rationale

A B A

C

D

E

Mo

G A

F A

AEC and BED are similar triangles. Thus, AC EFBD EG

1

2 2

Mo Mo

Mo

f f Mo LCBf f LCB Mo

1 2 2( )( ) ( )( )Mo Mo Mof f LCB Mo Mo LCB f f

2 1 2 2 1( )( ) ( )( ) (( ) ( ))Mo Mo Mo Mo MoLCB f f LCB f f Mo f f f f

1 2 1 2( )( ) ( )( ) (2 )Mo Mo Mo Mo MoLCB C f f LCB f f Mo f f f

1 1 2 1 2( ) ( )(2 ) (2 )Mo Mo Mo MoC f f LCB f f f Mo f f f

1

1 2

( )(2 )

MoMo

Mo

C f f LCB Mof f f


Assignment

1. Using the data on height (in meters) of a sample of 50 trees in page 183, no. 5, determine the mean, median and mode.

2. Using the distribution of scores in page 205, no. 4, approximate the median and the mode for the sample of girls and the sample of boys.

measures of central tendencymeasures of central tendency chapter 6. measures of central tendency...

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