Chapter 2 ~ Descriptive Analysis & Presentation of Single-Variable Data

Black Bears

20

Frequency

10 0 30 40 50 60

Length in Inches

70 80

Mean:

60.07 inches

Median:

62.50 inches

Range:

42 inches

Variance:

117.681

Standard deviation:

10.85 inches

Minimum:

36 inches

Maximum:

78 inches

First quartile:

51.63 inches

Third quartile:

67.38 inches

Count:

58 bears

Sum:

3438.1 inches

1

Chapter Goals

• Learn how to present and describe sets of data • Learn measures of central tendency, measures of dispersion (spread), measures of position, and types of distributions • Learn how to interpret findings so that we know what the data is telling us about the sampled population

2

2.1 ~ Graphic Presentation of Data

• Use initial exploratory data-analysis techniques to produce a pictorial representation of the data • Resulting displays reveal patterns of behavior of the variable being studied • The method used is determined by the type of data and the idea to be presented • No single correct answer when constructing a graphic display

3

Circle Graphs & Bar Graphs

Circle Graphs and Bar Graphs:

Graphs that are used to summarize attribute data •

Circle graphs (pie diagrams)

show the amount of data that belongs to each category as a proportional part of a circle •

Bar graphs

show the amount of data that belongs to each category as proportionally sized rectangular areas

4

Example



Example:

The table below lists the number of automobiles sold last week by day for a local dealership. Describe the data using a circle graph and a bar graph:

Day

Monday Tuesday Wednesday Thursday Friday Saturday

Number Sold

15 23 35 11 12 42

5

Circle Graph Solution

Automobiles Sold Last Week

6

Frequency

Bar Graph Solution

Automobiles Sold Last Week

7

Pareto Diagram

•

Pareto Diagram:

A bar graph with the bars arranged from the most numerous category to the least numerous category. It includes a line graph displaying the cumulative percentages and counts for the bars.

Notes

:   Used to identify the number and type of defects that happen within a product or service Separates the “vital few” from the “trivial many”  The Pareto diagram is often used in quality control applications

8

Example



Example:

The final daily inspection defect report for a cabinet manufacturer is given in the table below:

Defect

Dent Stain Blemish Chip Scratch Others

Number

5 12 43 25 40 10 1) Construct a Pareto diagram for this defect report. Management has given the cabinet production line the goal of reducing their defects by 50%.

2) What two defects should they give special attention to in working toward this goal?

9

Solutions

Daily Defect Inspection Report

1) 140 120 100 80

Count

60 40 20 0

Defect: Count Percent Cum%

Blemish 43 31.9

31.9

Scratch 40 29.6

61.5

Chip 25 18.5

80.0

Stain 12 8.9

88.9

Others 10 7.4

96.3

Dent 5 3.7

100.0

40 20 0 100 80 60

Percent

2) The production line should try to eliminate blemishes and scratches. This would cut defects by more than 50%.

10

Key Definitions

Quantitative Data:

One reason for constructing a graph of quantitative data is to examine the distribution - is the data compact, spread out, skewed, symmetric, etc.

Distribution:

The pattern of variability displayed by the data of a variable. The distribution displays the frequency of each value of the variable.

Dotplot Display:

Displays the data of a sample by representing each piece of data with a dot positioned along a scale. This scale can be either horizontal or vertical. The frequency of the values is represented along the other scale.

11

Example



Example:

A random sample of the lifetime (in years) of 50 home washing machines is given below: 2.5

16.9

4.5

0.9

1.5

17.8

8.5

8.9

2.5

6.4

14.5

0.7

7.3

1.4

12.2

3.5

2.9

4.0

3.7

6.8

7.4

4.1

0.4

3.3

0.9

4.2

3.3

4.7

18.1

2.6

4.4

7.2

6.9

7.0

0.7

1.6

2.2

9.2

5.2

15.3

4.0

10.4

12.2

4.0

4.1

1.8

21.8

18.3

3.6

.

: . . .:. .

..: :.::::::.. .::. ... . : . . . :. .

+---------+---------+---------+--- -----+---------+-------

Note:

Notice how the data is “bunched” near the lower extreme and more “spread out” near the higher extreme

12

Stem & Leaf Display

• Background: – The stem-and-leaf display has become very popular for summarizing numerical data – It is a combination of graphing and sorting – The actual data is part of the graph – Well-suited for computers

Stem-and-Leaf Display:

Pictures the data of a sample using the actual digits that make up the data values. Each numerical data is divided into two parts: The leading digit(s) becomes the

stem

, and the trailing digit(s) becomes the

leaf

. The stems are located along the main axis, and a leaf for each piece of data is located so as to display the distribution of the data.

13

Example



Example:

A city police officer, using radar, checked the speed of cars as they were traveling down the main street in town. Construct a stem-and-leaf plot for this data: 41 31 33 35 36 37 39 49 33 19 26 27 24 32 40 39 16 55 38 36

Solution:

All the speeds are in the 10s, 20s, 30s, 40s, and 50s. Use the first digit of each speed as the stem and the second digit as the leaf. Draw a vertical line and list the stems, in order to the left of the line. Place each leaf on its stem: place the trailing digit on the right side of the vertical line opposite its corresponding leading digit.

14

Example

20 Speeds

-------------------------------------- 1 | 6 9 2 | 4 6 7 3 | 1 2 3 3 5 6 6 7 8 9 9 4 | 0 1 9 5 | 5 --------------------------------------- • The speeds are centered around the 30s

Note

: The display could be constructed so that only five possible values (instead of ten) could fall in each stem. What would the stems look like? Would there be a difference in appearance?

15

Remember!

1. It is fairly typical of many variables to display a distribution that is concentrated (mounded) about a central value and then in some manner be dispersed in both directions. (Why?) 2. A display that indicates two “mounds” may really be two overlapping distributions 3. A back-to-back stem-and-leaf display makes it possible to compare two distributions graphically 4. A side-by-side dotplot is also useful for comparing two distributions

16

2.2 ~ Frequency Distributions & Histograms

• Stem-and-leaf plots often present adequate summaries, but they can get very big, very fast • Need other techniques for summarizing data • Frequency distributions and histograms are used to summarize large data sets

17

Frequency Distributions

Frequency Distribution:

A listing, often expressed in chart form, that pairs each value of a variable with its frequency

Ungrouped Frequency Distribution:

Each value of

x

in the distribution stands alone

Grouped Frequency Distribution:

Group the values into a set of classes 1. A table that summarizes data by classes, or class intervals 2. In a typical grouped frequency distribution, there are usually 5-12 classes of equal width 3. The table may contain columns for class number, class interval, tally (if constructing by hand), frequency, relative frequency, cumulative relative frequency, and class midpoint 4. In an ungrouped frequency distribution each class consists of a single value

18

Frequency Distribution

Guidelines for constructing a frequency distribution:

1. All classes should be of the same width 2. Classes should be set up so that they do not overlap and so that each piece of data belongs to exactly one class 3. For problems in the text, 5-12 classes are most desirable. The square root of

n

is a reasonable guideline for the number of classes if

n

is less than 150.

4. Use a system that takes advantage of a number pattern, to guarantee accuracy 5. If possible, an even class width is often advantageous

19

Frequency Distributions

Procedure for constructing a frequency distribution:

1. Identify the high (

H

) and low (

L

) scores. Find the range.

Range =

H

-

L

2. Select a number of classes and a class width so that the product is a bit larger than the range 3. Pick a starting point a little smaller than

L

. Count from

L

by the width to obtain the class boundaries. Observations that fall on class boundaries are placed into the class interval to the right.

20

Example



Example:

The hemoglobin test, a blood test given to diabetics during their periodic checkups, indicates the level of control of blood sugar during the past two to three months. The data in the table below was obtained for 40 different diabetics at a university clinic that treats diabetic patients: 6.5 5.0 5.6 7.6 4.8 8.0 7.5 7.9 8.0 9.2

6.4 6.0 5.6 6.0 5.7 9.2 8.1 8.0 6.5 6.6

5.0 8.0 6.5 6.1 6.4 6.6 7.2 5.9 4.0 5.7

7.9 6.0 5.6 6.0 6.2 7.7 6.7 7.7 8.2 9.0

1) Construct a grouped frequency distribution using the classes 3.7 - <4.7, 4.7 - <5.7, 5.7 - <6.7, etc.

2) Which class has the highest frequency?

21

Solutions

1) Class Frequency Relative Cumulative Class Boundaries

f

Frequency Rel. Frequency Midpoint,

x

-------------------------------------------------------------------------------------- 3.7 - <4.7

1 0.025

0.025

4.2

4.7 - <5.7

5.7 - <6.7

6.7 - <7.7

6 16 4 0.150

0.400

0.100

0.175

0.575

0.675

5.2

6.2

7.2

7.7 - <8.7

8.7 - <9.7

10 3 0.250

0.075

0.925

1.000

8.2

9.2

2) The class 5.7 - <6.7 has the highest frequency. The frequency is 16 and the relative frequency is 0.40

22

Histogram

Histogram:

A bar graph representing a frequency distribution of a quantitative variable. A histogram is made up of the following components: 1. A title, which identifies the population of interest 2. A vertical scale, which identifies the frequencies in the various classes 3. A horizontal scale, which identifies the variable

x

. Values for the class boundaries or class midpoints may be labeled along the

x

axis. Use whichever method of labeling the axis best presents the variable.

 

Notes

: The relative frequency is sometimes used on the vertical scale It is possible to create a histogram based on class midpoints

23

Example



Example:

Construct a histogram for the blood test results given in the previous example

The Hemoglobin Test

Solution:

15

Frequency

10 5 0 4.2

5.2

6.2

Blood Test

7.2

8.2

9.2

24

Example



Example:

A recent survey of Roman Catholic nuns summarized their ages in the table below. Construct a histogram for this age data: Age Frequency Class Midpoint ----------------------------------------------------------- 20 up to 30 34 25 30 up to 40 58 35 40 up to 50 50 up to 60 60 up to 70 70 up to 80 80 up to 90 76 187 254 241 147 45 55 65 75 85

25

Solution

Roman Catholic Nuns

200

Frequency

100 0 25 35 45 55

Age

65 75 85

26

Terms Used to Describe Histograms

Symmetrical:

Both sides of the distribution are identical mirror images. There is a

line

of symmetry.

Uniform (Rectangular):

Every value appears with equal frequency

Skewed:

One tail is stretched out longer than the other. The direction of skewness is on the side of the longer tail. (Positively skewed vs. negatively skewed)

J-Shaped:

There is no tail on the side of the class with the highest frequency

Bimodal:

The two largest classes are separated by one or more classes. Often implies two populations are sampled.

Normal:

A symmetrical distribution is mounded about the mean and becomes sparse at the extremes

27

Important Reminders

 The

mode

is the value that occurs with greatest frequency (discussed in Section 2.3)  The

modal class

is the class with the greatest frequency  A

bimodal distribution

has two high-frequency classes separated by classes with lower frequencies  Graphical representations of data should include a descriptive, meaningful title and proper identification of the vertical and horizontal scales

28

Cumulative Frequency Distribution

Cumulative Frequency Distribution:

A frequency distribution that pairs cumulative frequencies with values of the variable • The

cumulative frequency

for any given class is the sum of the frequency for that class and the frequencies of all classes of smaller values • The

cumulative relative frequency

for any given class is the sum of the relative frequency for that class and the relative frequencies of all classes of smaller values

29

Example



Example:

A computer science aptitude test was given to 50 students. The table below summarizes the data: Class Boundaries Frequency Relative Cumulative Frequency Frequency Cumulative Rel. Frequency ------------------------------------------------------------------------------------ 0 up to 4 4 0.08

4 0.08

4 up to 8 8 up to 12 12 up to 16 16 up to 20 20 up to 24 24 up to 28 8 8 20 6 3 1 0.16

0.16

0.40

0.12

0.06

0.02

12 20 40 46 49 50 0.24

0.40

0.80

0.92

0.98

1.00

30

Ogive

Ogive:

A line graph of a cumulative frequency or cumulative relative frequency distribution. An ogive has the following components: 1. A title, which identifies the population or sample 2. A vertical scale, which identifies either the cumulative frequencies or the cumulative relative frequencies 3. A horizontal scale, which identifies the upper class boundaries. Until the upper boundary of a class has been reached, you cannot be sure you have accumulated all the data in the class. Therefore, the horizontal scale for an ogive is always based on the upper class boundaries.

Note

: Every ogive starts on the left with a relative frequency of zero at the lower class boundary of the first class and ends on the right with a relative frequency of 100% at the upper class boundary of the last class.

31

Example



Example:

The graph below is an ogive using cumulative relative frequencies for the computer science aptitude data:

Computer Science Aptitude Test

Cumulative Relative Frequency

0.6

0.5

0.4

0.3

0.2

0.1

0.0

1.0

0.9

0.8

0.7

0 4 8 12

Test Score

16 20 24 28

32

2.3 ~ Measures of Central Tendency

• Numerical values used to locate the middle of a set of data, or where the data is clustered • The term

average

is often associated with all measures of central tendency

33

Mean

Mean:

The type of average with which you are probably most familiar. The mean is the sum of all the values divided by the total number of values,

n

:

x

=

n

1 

x i

= 1

n

(

x

1 +

x

2 + . . .

+

x n

) 

Notes

: The population mean,  , (lowercase

mu

, Greek alphabet), is the mean of all

x

values for the entire population  We usually cannot measure  but would like to estimate its value  A physical representation: the mean is the value that balances the weights on the number line

34

Example



Example:

The following data represents the number of accidents in each of the last 8 years at a dangerous intersection. Find the mean number of accidents: 8, 9, 3, 5, 2, 6, 4, 5:

Solution:

x

= 1 8 ( 8 5 ) =  In the data above, change 6 to 26:

Solution:

x

= 1 8 ( 8 26 5 ) =

Note:

The mean can be greatly influenced by outliers

35

Median

Median:

The value of the data that occupies the middle position when the data are ranked in order according to size  

Notes

: Denoted by “

x

tilde”: The population median,  (uppercase

mu

, Greek alphabet), is the data value in the middle position of the entire population To find the

median

: 1. Rank the data 2. Determine the

depth

of the median:

d x n

1 2 3. Determine the value of the median

36

Example



Example:

Find the median for the set of data: {4, 8, 3, 8, 2, 9, 2, 11, 3}

Solution:

1. Rank the data: 2, 2, 3, 3, 4, 8, 8, 9, 11 2. Find the depth: (~) ( )/ = 3. The median is the fifth number from either end in the ranked data:

x

= 4 Suppose the data set is {4, 8, 3, 8, 2, 9, 2, 11, 3, 15}: 1. Rank the data: 2, 2, 3, 3, 4, 8, 8, 9, 11, 15 2. Find the depth:

d

( ) = ( 10 + 1 ) / 2 = 5 .

5 3. The median is halfway between the fifth and sixth observations: ~ ( =

37

Mode & Midrange

Mode:

The mode is the value of

x

that occurs most frequently

Note:

If two or more values in a sample are tied for the highest frequency (number of occurrences), there is

no mode

Midrange:

The number exactly midway between a lowest value data

L

and a highest value data

H

. It is found by averaging the low and the high values: midrange = + 2

38

Example



Example:

Consider the data set {12.7, 27.1, 35.6, 44.2, 18.0} Midrange 2 = 2 =

Notes

:  When rounding off an answer, a common rule-of-thumb is to keep one more decimal place in the answer than was present in the original data  To avoid round-off buildup, round off only the final answer, not intermediate steps

39

2.4 ~ Measures of Dispersion

• Measures of central tendency alone cannot completely characterize a set of data. Two very different data sets may have similar measures of central tendency.

• Measures of dispersion are used to describe the spread, or variability, of a distribution • Common measures of dispersion: range, variance, and standard deviation

40

Range

Range:

The difference in value between the highest-valued (

H

) and the lowest-valued (

L

) pieces of data: range =  • Other measures of dispersion are based on the following quantity

Deviation from the Mean:

A deviation from the mean, , is the difference between the value of

x

and the mean

x

41

Example



Example:

Consider the sample {12, 23, 17, 15, 18}.

Find 1) the range and 2) each deviation from the mean.

Solutions:

1) range 23 12 11 2)

x

= 1 5 ( 12 23 17 15 18 ) = 17 Data

x

12 23 17 15 18 Deviation from Mean -5 6 0 -2 1

42

Mean Absolute Deviation

Note

:  (

x



x

) = 0 (Always!)

Mean Absolute Deviation:

The mean of the absolute values of the deviations from the mean: Mean absolute deviation = 1

n

 |

x



x

|

For the previous example:

1

n

 |

x



x

| = 1 5 ( 5 + 6 + 0 + 2 + 1 ) = 14 5 = 2 .

8

43

Sample Variance & Standard Deviation

Sample Variance:

The sample variance,

s

2 , is the mean of the squared deviations, calculated using

n

 1 as the divisor:

s

2 =

n

1  1  (

x



x

) 2 where

n

is the sample size

Note

: The numerator for the sample variance is called the sum of squares for

x

, denoted SS(

x

):

s

2 SS( )

n x

1 where =  (

x



x

) 2 = 

x

2  1

n

( ) 2

Standard Deviation:

The standard deviation of a sample,

s

, is the positive square root of the variance:

s

=

s

2

44

Example



Example:

Find the 1) variance and 2) standard deviation for the data {5, 7, 1, 3, 8}:

Solutions:

First:

x

= 5 1 ( 5 7 1 3 8 )

x

(



) 2 Sum:

5 7 1 3 8

24

0.2

2.2

-3.8

-1.8

3.2

0

0.04

4.84

14.44

3.24

10.24

32.08

1)

s

2 = 1 4 ( 32 .

8 ) = 8 .

2 2)

s

= 8 .

2 = 2 .

86

45

Notes

 The shortcut formula for the sample variance:

s

2 = 

x

2 

n

 1  2

n

 The unit of measure for the standard deviation is the same as the unit of measure for the data

46

2.5 ~ Mean & Standard Deviation of Frequency Distribution

• If the data is given in the form of a frequency distribution, we need to make a few changes to the formulas for the mean, variance, and standard deviation • Complete the extension table in order to find these summary statistics

47

To Calculate

• In order to calculate the mean, variance, and standard deviation for data: 1. In an

ungrouped

frequency distribution, use the frequency of occurrence,

f

, of each observation 2. In a

grouped

frequency distribution, we use the frequency of occurrence associated with each class midpoint:

x

=  

xf f s

2 =  2

x f

 

f

(    1

xf f

) 2

48

Example



Example:

A survey of students in the first grade at a local school asked for the number of brothers and/or sisters for each child. The results are summarized in the table below. Find 1) the mean, 2) the variance, and 3) the standard deviation:

Solutions:

First:

Sum:

x

0 1 2 4 5

f

15 17 23 5 2

62

xf

0 17 46 20 10

93

0 17 92 80 50

239

1)

x

= 2)

s

2 = 239  (  93 62 ) 2 = 3)

s

=

49

TI-83 Calculations

• When dealing with a grouped frequency distribution, use the following technique: Input the class midpoints or data values into L1 and the frequencies into L2; then continue with Highlight: Enter: Highlight: Enter: Highlight: L3 L3 = L1*L2 L4 L4 = L1*L3 L5(1) (first position in L5 column)

50

TI-83 Calculations

Enter: To find sum use 2 nd L5(1) = sum(L2) “List”>Math>5:sum( (Σ

f

) L5(2) = sum(L3) L5(3) = sum(L4) (Σ

xf

) (Σ

x 2 f

) L5(4) = L5(2)/L5(1) to find the mean L5(5) = (L5(3)-(L5(2)) 2 /L5(1))/(L5(1)-1) to find the variance L5(6) = 2 nd √ (L5(5) to find the standard deviation Let’s work problem 2.108 as an example!

51

Problem 2.108

• Find the mean and the variance for this grouped frequency distribution:

Class Boundaries

2 – 6 6 – 10 10 – 14

f

7 15 22 14 – 18 18 – 22 14 2 Step 1: Enter the midpoints into L1 Step 2: Enter the frequencies into L2

52

Problem 2.108

• Highlight L3 and enter L1 * L2 • Highlight L4 and enter L1 * L3 • Highlight L5(1) and enter Sum(L2)

53

Problem 2.108

• Highlight L5(2) and enter Sum(L3) • Highlight L5(3) and enter Sum(L4) • Highlight L5(4) and enter L5(2)/L5(1) • Highlight L5(5) and enter (L5(3)-(L5(2)) 2 /L5(1))/(L5(1)-1) • Finally highlight L5(6) and enter 2 nd √(L5(5))

54

Mean Median Maximum Minimum Midrange

Problem 2.73

Runs At Home 9.77

9.65

13.65

7.64

10.65

Runs Away 9.80

9.78

11.06

8.67

9.87

Difference -0.03

-0.06

4.89

-1.74

1.58

55

Problem 2.73 cont.

• Teams playing the Rockies at Coors Field generated the maximum number of runs scored (13.65). On the other hand, while playing their opponents away, the Rockies and their opponents, were able to generate only 8.76 runs, which ranked second from the bottom. Collectively, these two performances produced the greatest spread (4.89) by a considerable margin between runs scored at home and runs scored away by any stadium/team combination in the major leagues. This unusually large value inflates the midrange difference. It appears the playing conditions at Coors Field, therefore, are more responsible for producing the higher combined scores than the strength of either the Rockies’ or their opponents’ bats or any weakness of the pitching staffs.

56

a.

b.

c.

d.

e.

f.

g.

h.

Problem 2.75

∑ x need to be 500, therefore need any three numbers that total 330.

Need two numbers smaller than 70 and one larger Need multiple 87’s Need any two numbers that total 140 for the extreme values where one is 100 or larger Need two numbers smaller than 70 and one larger than 70 so their total is 330 Need two numbers of 87 and a third number large enough so that the total of all five is 500.

Mean equal to 100 requires the five data to total 500 and the midrange of 70 requires the total of L and H to be 140; 40, __, 70, ___, 100; that is a sum of 210, meaning the other two data must total 290. One of the last two numbers must be larger than 145, which would then become H and change the midrange. Impossible.

There must be two 87’s in order to have a mode, and there can only be two data larger than 70 in order for 70 to be the median. , 70, 87, 87, 100; Impossible

57

Problem 2.83

a.

b.

Range = H – L = 9 – 2 = 7 1 st : find the mean: 6 8 ∑ 30 9 2 x 4 7 x – xbar -4 -2 1 2 3 0 (x – xbar) 2 16 4 1 4 9 34 s 2 = ∑(x-xbar) 2 /(n-1) = 34/4 = 8.5

s = √s 2 = √8.5 = 2.9

58

2.6 ~ Measures of Position

• Measures of position are used to describe the relative location of an observation • Quartiles and percentiles are two of the most popular measures of position • An additional measure of central tendency, the midquartile, is defined using quartiles • Quartiles are part of the 5-number summary

59

Quartiles

Quartiles:

Values of the variable that divide the ranked data into quarters; each set of data has three quartiles 1. The first quartile,

Q

1 , is a number such that at most 25% of the data are smaller in value than

Q

1 and at most 75% are larger 2. The second quartile,

Q

2 , is the median 3. The third quartile,

Q

3 , is a number such that at most 75% of the data are smaller in value than

Q

3 and at most 25% are larger Ranked data, increasing order

L

25%

Q

1 25%

Q

2 25%

Q

3 25%

H

60

Percentiles

Percentiles:

Values of the variable that divide a set of ranked data into 100 equal subsets; each set of data has 99 percentiles. The

k

th percentile,

P k

, is a value such that at most

k

% of the data is smaller in value than

P k

and at most (100 

k

)% of the data is larger.

at most

k

% at most (100 -

k

)%

L P k H Notes:

 The 1st quartile and the 25th percentile are the same:

Q

1 =

P

25  The median, the 2nd quartile, and the 50th percentile are all the same: =

Q

2 =

P

50

61

Finding P

k

(and Quartiles)

•

Procedure for finding P k (and quartiles):

1. Rank the

n

observations, lowest to highest 2. Compute

A

= (

nk

)/100 3. If

A

is an integer: –

d

(

P k

) =

A

.5 (depth) –

P k

is halfway between the value of the data in the

A

th position and the value of the next data If

A

is a fraction: –

d

(

P k

) =

B

, the next larger integer – P

k

is the value of the data in the

B

th position

62

Example



Example:

The following data represents the pH levels of a random sample of swimming pools in a California town. Find: 1) the first quartile, 2) the third quartile, and 3) the 37th percentile: 5.6

6.0

6.7

7.0

5.6

6.1

6.8

7.3

5.8

6.2

6.8

7.4

5.9

6.3

6.8

7.4

6.0

6.4

6.9

7.5

Solutions:

1)

k

= 25: (20) (25) / 100 = 5, depth = 5.5,

Q

1 = 6 2)

k

= 75: (20) (75) / 100 = 15, depth = 15.5,

Q

3 = 6.95

3)

k

= 37: (20) (37) / 100 = 7.4, depth = 8,

P

37 = 6.2

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Midquartile

Midquartile:

The numerical value midway between the first and third quartile: midquartile = 1 + 3 2 

Example:

Find the midquartile for the 20 pH values in the previous example: midquartil e =

Q

1 +

Q

3 2 = 6 + 6 .

95 2 = 12 .

95 2 = 6 .

475

Note

: The mean, median, midrange, and midquartile are all measures of central tendency. They are

not

necessarily equal. Can you think of an example when they would be the same value?

64

5-Number Summary

5-Number Summary:

The 5-number summary is composed of: 1.

2.

3.

4.

5.

L

, the smallest value in the data set

Q

1 , the first quartile (also

P

25 ) , the median (also

P

50 and 2nd quartile)

Q

3 , the third quartile (also

P

75 )

H

, the largest value in the data set

Notes

:  The 5-number summary indicates how much the data is spread out in each quarter  The

interquartile range

is the difference between the first and third quartiles. It is the range of the middle 50% of the data

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Box-and-Whisker Display

Box-and-Whisker Display:

A graphic representation of the 5-number summary: • The five numerical values (smallest, first quartile, median, third quartile, and largest) are located on a scale, either vertical or horizontal • The box is used to depict the middle half of the data that lies between the two quartiles • The whiskers are line segments used to depict the other half of the data • One line segment represents the quarter of the data that is smaller in value than the first quartile • The second line segment represents the quarter of the data that is larger in value that the third quartile

66

Example



Example:

A random sample of students in a sixth grade class was selected. Their weights are given in the table below. Find the 5-number summary for this data and construct a boxplot: 63 64 76 76 81 83 85 86 88 89 90 91 92 93 93 93 94 97 99 99 99 101 108 109 112

Solution:

63

L

85

Q

1 92 99

Q

3 112

H

67

Boxplot for Weight Data

Weights from Sixth Grade Class

60

L

70 80 90

Weight

Q

1 100

Q

3 110

H

68

z-Score

z-Score:

The position a particular value of

x

has relative to the mean, measured in standard deviations. The

z

-score is found by the formula:

z

= value  mean = st.dev.

x



x s Notes

:  Typically, the calculated value of

z

hundredth is rounded to the nearest  The

z

-score measures the number of standard deviations above/below, or away from, the mean 

z

-scores typically range from -3.00 to +3.00



z

-scores may be used to make comparisons of raw scores

69

Example



Example:

A certain data set has mean 35.6 and standard deviation 7.1. Find the

z

-scores for 46 and 33:

Solutions:

z

=

x



s x

= 46  35 .

6 7 .

1 = 1 .

46 46 is 1.46 standard deviations

above

the mean

z x

= 0 .

37

s

33 is 0.37 standard deviations

below

the mean.

70

2.7 ~ Interpreting & Understanding Standard Deviation

• Standard deviation is a measure of variability, or spread • Two rules for describing data rely on the standard deviation: –

Empirical rule

: applies to a variable that is normally distributed –

Chebyshev’s theorem

: applies to any distribution

71

Empirical Rule

Empirical Rule:

If a variable is normally distributed, then: 1. Approximately 68% of the observations lie within 1 standard deviation of the mean 2. Approximately 95% of the observations lie within 2 standard deviations of the mean 3. Approximately 99.7% of the observations lie within 3 standard deviations of the mean   

Notes:

The empirical rule is more informative than Chebyshev’s theorem since we know more about the distribution (normally distributed) Also applies to populations Can be used to determine if a distribution is normally distributed

72

Illustration of the Empirical Rule

99.7% 95% 68%

x

 3

s x

 2

s x x

+ 2

s x

+ 3

s

73

Example



Example:

A random sample of plum tomatoes was selected from a local grocery store and their weights recorded. The mean weight was 6.5 ounces with a standard deviation of 0.4 ounces. If the weights are normally distributed: 1) What percentage of weights fall between 5.7 and 7.3?

2) What percentage of weights fall above 7.7?

Solutions:

1) (

x

 2 + 2

s

=  (0. ), .

4 57 7 3 Approximately 95% of the weights fall between 5.7 and 7.3

2) (

x

 3 + 3

s

=  + Approximately 99.7% of the weights fall between 5.3 and 7.7

Approximately 0.3% of the weights fall outside (5.3, 7.7) Approximately (0.3/2)=0.15% of the weights fall above 7.7

74

A Note about the Empirical Rule

Note:

The empirical rule may be used to determine whether or not a set of data is approximately normally distributed 1. Find the mean and standard deviation for the data 2. Compute the actual proportion of data within 1, 2, and 3 standard deviations from the mean 3. Compare these actual proportions with those given by the empirical rule 4. If the proportions found are reasonably close to those of the empirical rule, then the data is approximately normally distributed

75

Chebyshev’s Theorem

Chebyshev’s Theorem:

The proportion of any distribution that lies within

k

standard deviations of the mean is at least 1  (1/

k

2 ), where

k

is any positive number larger than 1. This theorem applies to all distributions of data.

Illustration:

at least 1 1 2

x

76

Important Reminders!

 Chebyshev’s theorem is very conservative and holds for any distribution of data  Chebyshev’s theorem also applies to any population  The two most common values used to describe a distribution of data are

k

= 2, 3  The table below lists some values for

k

and 1 - (1/

k

2 ):

k k

2 ) 1.7

0.65

2 0.75

2.5

0.84

3 0.89

77

Example



Example:

At the close of trading, a random sample of 35 technology stocks was selected. The mean selling price was 67.75 and the standard deviation was 12.3. Use Chebyshev’s theorem (with

k

= 2, 3) to describe the distribution.

Solutions:

Using

k

=2: At least 75% of the observations lie within 2 standard deviations of the mean: (

x

 2

s

,

x

+ 2

s

) = ( 67 .

75  2 ( 12 .

3 ), 67 .

75 + 2 ( 12 .

3 ) = ( 43 .

15 , 92 .

35 ) Using

k

=3: At least 89% of the observations lie within 3 standard deviations of the mean: (

x

 3

s

,

x

+ 3

s

) = ( 67 .

75  3 ( 12 .

3 ), 67 .

75 + 3 ( 12 .

3 ) = ( 30 .

85 , 104.65

)

78

2.8 ~ The Art of Statistical Deception Good Arithmetic, Bad Statistics Misleading Graphs Insufficient Information

79

Good Arithmetic, Bad Statistics

• The mean can be greatly influenced by outliers –

Example

: The mean salary for all NBA players is $15.5 million Misleading graphs: 1. The frequency scale should start at zero to present a complete picture. Graphs that do not start at zero are used to save space.

2. Graphs that start at zero emphasize the size of the numbers involved 3. Graphs that are chopped off emphasize variation

80

35 30 25

Number of Cancellations

20 15 10 5 0

Flight Cancellations

1996 1998

Year

2000 2002

81

Number of Cancellations

30 29 28 27 35 34 33 32 31

Flight Cancellations

1996 1998

Year

2000 2002

82

Insufficient Information

•

Example

: An admissions officer from a state school explains that the average tuition at a nearby private university is $13,000 and only $4500 at his school. This makes the state school look more attractive.

– If most students pay the full tuition, then the state school appears to be a better choice – However, if most students at the private university receive substantial financial aid, then the actual tuition cost could be much lower!

83