Transcript chapter 1 - UniMAP Portal

CHAPTER 1 EQT
271 (part 1)
BASIC
STATISTICS
Basic Statistics
1.1
1.2
1.3
1.4
Statistics in Engineering
Collecting Engineering Data
Data Presentation and Summary
Probability Distributions
- Discrete Probability Distribution
- Continuous Probability Distribution
1.5 Sampling Distributions of the Mean and
Proportion
• Statistics is the area of science that deals with
collection, organization, analysis, and interpretation of
data.
• A collection of numerical information is called
statistics.
• Because many aspects of engineering practice involve
working with data, obviously some knowledge of
statistics is important to an engineer.
•Specifically, statistical techniques can be a powerful aid in
designing new products and systems, improving existing
designs, and improving production process.

the methods of statistics
allow
scientists
and
engineers to design valid
experiments and to draw
reliable conclusions from
the data they produce
Basic Terms in Statistics
Population
- Entire collection of individuals which are characteristic being
studied.
 Sample
- A portion, or part of the population interest.
 Variable
- Characteristics which make different values.
 Observation
- Value of variable for an element.
 Data Set
- A collection of observation on one or more variables.

• Direct observation
- The simplest method of obtaining data.
- Advantage: relatively inexpensive
- Disadvantage: difficult to produce useful information
since it does not consider all aspects regarding the issues.
• Experiments
- More expensive methods but better way to produce data
- Data produced are called experimental
•
Surveys
- Most familiar methods of data collection
- Depends on the response rate
•
Personal Interview
- Has the advantage of having higher
expected response rate
- Fewer incorrect respondents.
Grouped Data Vs Ungrouped Data
Grouped data - Data that has been
organized into groups (into a frequency
distribution).
Ungrouped data - Data that has not been
organized into groups. Also called as raw data.
Graphical Data Presentation
• Data can be summarized or presented in two ways:
1. Tabular
2. Charts/graphs.
• The presentations usually depends on the type (nature)
of data whether the data is in qualitative (such as
gender and ethnic group) or quantitative (such as
income and CGPA).
Data Presentation of Qualitative Data
• Tabular presentation for qualitative data is usually in the form
of frequency table that is a table represents the number of
times the observation occurs in the data.
*Qualitative :- characteristic being studied is
nonnumeric.
Examples:- gender, religious affiliation or eye color.
• The most popular charts for qualitative data are:
1. bar chart/column chart;
2. pie chart; and
3. line chart.
Types of Graph
Qualitative Data
Line chart
Bar chart
Pie chart
Frequency table
Observation Frequency
Malay
33
Chinese
9
Indian
6
Others
2
Bar Chart: used to display the frequency distribution in the
graphical form.
Pie Chart: used to display the frequency distribution. It displays
the ratio of the observations
Malay
Chinese
Indian
Others
Line chart: used to display the trend of observations. It is a very
popular display for the data which represent time.
Jan
10
Feb
7
Mar
5
Apr
10
May Jun Jul
39
7 260
Aug
316
Sep
142
Oct
11
Nov
4
Dec
9
Data Presentation Of Quantitative Data
• Tabular presentation for quantitative data is usually in the form of
frequency distribution that is a table represent the frequency of
the observation that fall inside some specific classes (intervals).
*Quantitative : variable studied are numerically.
Examples:balanced in accounts, ages of students, the life of an
automobiles batteries such as 42 months).
• Frequency distribution: A grouping of data into mutually
exclusive classes showing the number of observations in
each class.
• There are few graphs available for the graphical presentation of
the quantitative data.
• The most popular graphs are:
1. histogram;
2. frequency polygon; and
3. ogive.
Frequency Distribution
Weight (Rounded decimal point)
Frequency
60-62
5
63-65
18
66-68
42
69-71
27
72-74
8
Histogram: Looks like the bar chart except that the horizontal
axis represent the data which is quantitative in nature. There is no
gap between the bars.
Frequency Polygon: looks like the line chart except that the
horizontal axis represent the class mark of the data which is
quantitative in nature.
Ogive: line graph with the horizontal axis represent the upper
limit of the class interval while the vertical axis represent the
cummulative frequencies.
Constructing Frequency Distribution
When summarizing large quantities of raw data, it is often useful to distribute
the data into classes. Table 1 shows that the number of classes for Students`
weight.
Weight
60-62
63-65
66-68
69-71
72-74
Total
Frequency
5
18
42
27
8
100
Table 1: Weight of 100 male students in
XYZ university
• A frequency distribution for quantitative data lists all the classes and the
number of values that belong to each class.
• Data presented in the form of a frequency distribution are called grouped
data.
• For quantitative data, an interval that includes all the values that fall
within two numbers; the lower and upper class which is called class.
 Class is in the first column for frequency distribution table.
 Classes always represent a variable, non-overlapping; each value is
belong to one and only one class.
• The numbers listed in second column are called frequencies, which
gives the number of values that belong to different classes.
Frequencies denoted by f.
Table 1.: Weight of 100 male students in XYZ university
Variable
Third class
(Interval Class)
Lower Limit
of the fourth class
Weight
60-62
63-65
66-68
69-71
72-74
Total
Frequency
5
18
42
27
8
100
Upper limit of the fifth class
Frequency
column
Frequency
of the third
class.
• The class boundary is given by the midpoint of the upper limit of
one class and the lower limit of the next class.
• The difference between the two boundaries of a class gives the
class width; also called class size.
Formula:
- Class Midpoint or Mark
Class midpoint or mark = (Lower Limit + Upper Limit)/2
- Finding The Number of Classes
Number of classes =
1  3.3log n
- Finding Class Width For Interval Class
class width , i = (Largest value – Smallest value)/Number of classes
* Any convenient number that is equal to or less than the smallest values in the
data set can be used as the lower limit of the first class.
Example 1:
From Table 1: Class Boundary
Weight (Class
Interval)
Class
Boundary
Frequency
60-62
59.5-62.5
5
63-65
62.5-65.5
18
66-68
65.5-68.5
42
69-71
68.5-71.5
27
72-74
71.5-74.5
8
Total
100
Cumulative Frequency Distributions
• A cumulative frequency distribution gives the total number of values that fall
below the upper boundary of each class.
• In cumulative frequency distribution table, each class has the same lower
limit but a different upper limit.
Table 2: Class Limit, Class Boundaries, Class Width , Cumulative Frequency
Weight
(Class
Interval)
Number of
Students, f
Class
Boundaries
60-62
5
59.5-62.5
63-65
18
62.5-65.5
66-68
42
65.5-68.5
69-71
27
68.5-71.5
72-74
8
100
71.5-74.5
Cumulative
Frequency
5
5 + 18 = 23
23 + 42 = 65
65 + 27 =92
92 + 8 = 100
Exercise 1:
Given a raw data as below:
27
27
27
28
27
24
25
28
26
28
26
28
31
30
26
26
a) How many classes that you recommend?
b) What is the class interval?
c) Build a frequency distribution table.
d) What is the lower boundary for the first class?
HOW TO CONSTRUCT HISTOGRAM?
Prepare the frequency distribution table by:
1. Find the minimum and maximum value
2. Decide the number of classes to be included in your frequency
distribution table.
-
Usually 5-20 classes. Too small- may not able to see any pattern
OR
-
Sturge’s rule, Number of classes= 1+3.3log n
3. Determine class width, i = (max-min)/num. of class
4. Determine class limit.
5. Find class boundaries and class mid points
6. Count frequency for each class
7. Draw histogram
Exercise 2:
The data below represent the waiting time (in minutes)
taken by 30 customers at one local bank.
25
31
20
30
22
32
37
28
29
23
35
25
29
35
29
27
23
32
31
32
24
35
21
35
35
22
33
24
39
43
1. Construct a frequency distribution and cumulative frequency
distribution table.
2. Draw a histogram
Data Summary
•
a)



b)



c)




Summary statistics are used to summarize a set of observations.
Measures of Central Tendency
Mean
Median
Mode
Measures of Dispersion
Range
Variance
Standard deviation
Measures of Position
Z scores
Percentiles
Quartiles
Outliers
a) Measures of Central Tendency
Mean
• Mean of a sample is the sum of the sample data divided by the
total number sample.
•
Mean for ungrouped data is given by:
_
x
x1  x2  .......  xn x

x
, for n  1,2,..., n or x 
n
n
_
• Mean for group data is given by:
n

x
fx
fx

or
f

f

i 1
n
i 1
i i
i
Example 2 (Ungrouped data):
Mean for the sets of data 3,5,2,6,5,9,5,2,8,6
Solution :
35 2 6595 28 6
x
 5.1
10
Example 3 (Grouped Data):
Use the frequency distribution of weights 100 male students in
XYZ university, to find the mean.
Weight
Frequency
60-62
63-65
66-68
69-71
72-74
5
18
42
27
8
Solution :
Weight (Class
Interval
Frequency, f
Class Mark,
x
fx
60-62
63-65
66-68
69-71
72-74
5
18
42
27
8
61
64
67
70
73
305
1152
2814
1890
584
100
fx
6745

x
? 

67
.
45
f
100
6745
Median of ungrouped data:
• The median depends on the number of observations in the data,
n . If n is odd, then the median is the (n+1)/2 th observation of the
ordered observations.
• But if n is even, then the median is the arithmetic mean of the
n/2 th observation and the (n+1)/2 th observation.
Median of grouped data:
 f


F

j 1 
2
x  Lc

f


j


where
L = the lower class boundary of the median class
c = the size of median class interval
Fj 1  the sum of frequencies of all classes lower than the median class
f j  the frequency of the median class
Example 4 (Ungrouped data):
n is odd
Find the median for data 4,6,3,1,2,5,7 ( n = 7)
Rearrange the data : 1,2,3,4,5,6,7
(median = (7+1)/2=4th place)
Median = 4
n is even
Find the median for data 4,6,3,2,5,7 (n = 6)
Rearrange the data : 2,3,4,5,6,7
Median = (4+5)/2 = 4.5
Example 5 (Grouped Data):
The sample median for frequency distribution as in example 3
Solution:
Median
class
Weight
(Class
Interval
Frequency, f
Cumulative
Frequency,
F
Class
Boundary
60-62
63-65
66-68
69-71
72-74
5
18
42
27
8
5
23
65
92
100
59.5-62.5
62.5-65.5
65.5-68.5
68.5-71.5
71.5-74.5
 f


F

j 1 
2
x  Lc
?
fj




100
 23
 65.5  3[ 2
]  67.73
42
Mode
• Mode of ungrouped data:
 The value with the highest frequency in a data set.
 It is important to note that there can be more than one
mode and if no number occurs more than once in the set,
then there is no mode for that set of numbers
• Mode for grouped data
When data has been grouped in classes and a frequency curveis drawn
to fit the data, the mode is the value of x corresponding to the maximum
point on the curve, that is
 1 
xˆ  L  c 




2
 1
L  the lower class boundary of the modal class
c = the size of the modal class interval
1  the difference between the modal class frequency and the class before it
 2  the difference between the modal class frequency and the class after it
*the class which has the highest frequency is called the modal class
Example 6 (Ungrouped data)
Find the mode for the sets of data 3, 5, 2, 6, 5, 9, 5, 2, 8, 6
Mode = number occurring most frequently = 5
Example 7 Find the mode of the sample data below
Solution:
Weight
(Class
Interval
Mode class
Frequency
Class
,f
Boundary
60-62
63-65
66-68
69-71
72-74
5
18
42
27
8
59.5-62.5
62.5-65.5
65.5-68.5
68.5-71.5
71.5-74.5
Total
100
 1 
(42  18)
xˆ  L  c 
  ?  65.5  3[ (42  18)  (42  27) ]
 1   2 
 67.35
b) Measures of Dispersion
• Range = Largest value – smallest value
• Variance: measures the variability (differences) existing in a
set of data.
The variance for the ungrouped data:
For sample

S
2
( x  x)


2
n 1
For population
2 
2
(
x


)

n
The variance for the grouped data:
• For sample
S
2
fx


2
2
nx
or
S 
2
n 1

2
(
fx
)

fx 2 
n
n 1
• For population

2


fx 2  nx 2
n
or
2 

2
(
fx
)

2
fx 
n
n
 The positive square root of the variance is the standard
deviation

S
 ( x  x)
n 1
2

 fx
2
2
nx
n 1
• A large variance means that the individual scores (data) of
the sample deviate a lot from the mean.
• A small variance indicates the scores (data) deviate little
from the mean.
Example 8 (Ungrouped data)
Find the variance and standard deviation of the
sample data : 3, 5, 2, 6, 5, 9, 5, 2, 8, 6

2
(
x

x
)
S2  
n 1
(3  5.1) 2  (5  5.1) 2  (2  5.1) 2  (6  5.1) 2  (5  5.1) 2  (9  5.1) 2
2
2
2
2

(
5

5
.
1
)

(
2

5
.
1
)

(
8

5
.
1
)

(
6

5
.
1
)
s2 
9
48.9

 5.43
9
s  s 2  5.43
 2.33
Example 9 (Grouped data)
Find the variance and standard deviation of the sample data below:
Weight
(Class
Interval
S 
2
Frequency Class
,f
Mark,
x
60-62
63-65
66-68
69-71
72-74
5
18
42
27
8
Total
100

( fx)
fx 
n
n 1
2
2
61
64
67
70
73
fx
305
1152
2814
1890
584
6745
x2
3721
4096
4489
4900
5329
fx 2
18605
73728
188538
132300
42632
455803
67452
455803 
100  852.75  8.61

99
99
s  8.61  2.93
Exercise 3
The defects from machine A for a sample of products were
organized into the following:
Defects
(Class Interval)
Number of products get
defect, f (frequency)
2-6
1
7-11
4
12-16
10
17-21
3
22-26
2
What is the mean, median, mode, variance and standard
deviation? Or
Calculate each possible measure of central tendency and
dispersion.
Exercise 4 (submit on Friday)
The following data give the sample number of iPads sold by a
mail order company on each of 30 days. (Hint : 5 number of
classes)
8 25
11
15
29
22
10
5
17
21
22 13
26
16
18
12
9
26
20
16
23 14
19
23
20
16
27
9
21
14
a) Construct a frequency distribution table.
b) Find the mean, variance and standard deviation, mode and
median.
c) Construct a histogram.
Rules of Data Dispersion
By using the mean x and standard deviation, we can find the
percentage of total observations that fall within the given interval
about the mean.
Empirical Rule
Applicable for a symmetric bell shaped distribution / normal
distribution.
There are 3 rules:
i. 68% of the data will lie within one standard deviation of the
mean,( x  s )
ii. 95% of the data will lie within two standard deviation of the
mean, ( x  2 s )
iii. 99.7% of the data will lie within three standard deviation of
the mean, ( x  3s )
Example 10
The age distribution of a sample of 5000 persons is bell shaped with
a mean of 40 yrs and a standard deviation of 12 yrs. Determine the
approximate percentage of people who are 16 to 64 yrs old.
Solution:
x  s  40  12  [ 28,52]
x  2 s  40  2.12  [16,64]
x  3s  40  3.12  [4,76]
Approximately 68% of the measurements will fall between 28 and
52, approximately 95% of the measurements will fall between 16
and 64 and approximately 99.7% to fall into the interval 4 and 76.
c) Measures of Position
• To describe the relative position of a certain data
value within the entire set of data.
 z scores
 Percentiles
 Quartiles
 Outliers
Quartiles
Divide data sets into fourths or four equal parts.
Smallest
data value
25%
of data
Q1
Q2
25%
of data
Q3
25%
of data
1
Q1   (n  1)th
4
1
Q 2  median   (n  1)th
2
3
Q3   (n  1)th
4
Largest
data value
25%
of data
The positions are integers
Example: 5, 8, 4, 4, 6, 3, 8 (n=7)
1. Put them in order: 3, 4, 4, 5, 6, 8, 8
2. Calculate the quartiles
1
Q1   (7  1)th  2nd place,
4
3
Q3   (7  1)th  6th place
4
1
Q 2   (7  1)th  4th place,
2
3, 4, 4, 5, 6, 8, 8
Q1  4,
Q 2  5, Q3  8
The positions are not integers
Example: 5, 12, 10, 4, 6, 3, 8, 14 (n=8)
1. Put them in order: 3, 4, 5, 6, 8, 10, 12, 14
2. Calculate the quartiles
3, 4, 5, 6, 8, 10, 12, 14
1
Q1   (8  1)th  2.25th  4  0.25(5  4)  4.25
4
1
Q 2   (8  1)th  4.5th  6  0.5(8  6)  7,
2
3
Q3   (8  1)th  6.75th  10  0.75(12  10)  11.5
4
Example 11
The following data represent the number of inches of rain in
Chicago during the month of April for 10 randomly years.
2.47
3.97
3.94
4.11
5.22
1.14
4.02
3.41
1.85
0.97
Determine the quartiles.
Solution:
0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22
1
Q1   (10  1)th  2.5th  1.14  0.5(1.85  1.14)  1.495
4
1
Q 2   (10  1)th  5.5th  3.41  0.5(3.94  3.41)  3.675,
2
3
Q3   (10  1)th  8.25th  4.02  0.25(4.11  4.02)  4.0425
4
Outliers
• Extreme observations
• Can occur because of the error in measurement of a variable,
during data entry or errors in sampling.
Checking for outliers by using Quartiles
Step 1:
Determine the first and third quartiles of data.
Step 2:
Compute the interquartile range (IQR).
IQR  Q3  Q1
Step 3:
Determine the fences. Fences serve as cut off points for
determining outliers. Lower Fence  Q  1.5( IQR)
1
Step 4:
Upper Fence  Q3  1.5( IQR)
If data value is less than the lower fence or greater than the
upper fence, considered outlier.
Example 12
2.47 3.97 3.94 4.11 5.22
1.14 4.02 3.41 1.85 0.97
Determine whether there are outliers in the
data set.
Solution:
0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22
Q1  1.495, Q3  4.0425
IQR  Q3  Q1  4.0425  1.495  2.5475
Lower fence  Q1  1.5( IQR )
Upper fence  Q3  1.5( IQR )
 1.495  1.5(2.5475)
 2.32625
 4.0425  1.5(2.5475)
 7.86375
Since all the data are not less than -2.32625 and not
greater than 7.86375, then there are no outliers in the
data
The Five Number Summary
• Compute the five-number summary
MINIMUM Q1 M Q3 MAXIMUM
Example 13
2.47 3.97 3.94 4.11 5.22
1.14 4.02 3.41 1.85 0.97
Compute all the five-number summary.
Solution:
0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22
Minimum  0.97,
Q1  1.495,
M  Q 2  3.675,
Q3  4.0425
Maximum  5.22,
BOXPLOT
• The five-number summary can be used to create a
simple graph called a boxplot.
• Form the boxplot, you can quickly detect any
skewness in the shape of the distribution and see
whether there are any outliers in the data set.
Outlier
Outlier
Lower
fence
Upper
fence
Interpreting Boxplot
- symmetric
- Skewed left
because the tail is
to the left
- Skewed right
because the tail
is to the right
Mean/Median Versus Skewness
TO CONSTRUCT BOXPLOT
Step 1: Determine the lower and upper fences:
Lower Fence  Q1  1.5( IQR)
Upper Fence  Q3  1.5( IQR)
Step 2: Draw vertical lines at Q1 , M and
. Q3
Step 3: Label the lower and upper fences.
Step 4: Draw a line from Q1 to the smallest data value that is larger
than the lower fence. Draw a line from Q3 to the largest data value
that is smaller than the upper fence.
Step 5: Any data value less than the lower fence or greater than
the upper fence are outliers and mark (*).
Example 14
2.47 3.97 3.94 4.11 5.22
1.14 4.02 3.41 1.85 0.97
-Sketch the boxplot and interpret the shape
of the boxplot.
Solution:
0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22
Q1  1.495, M  Q 2  3.675, Q3  4.0425
IQR  Q3  Q1  4.0425  1.495  2.5475
Lower fence  Q1  1.5( IQR )
Upper fence  Q3  1.5( IQR )
 1.495  1.5(2.5475)
 2.32625
 4.0425  1.5(2.5475)
 7.86375
Lower
fence
Q1
M Q3
- The distribution is skewed left
Upper
fence