CHAPTER 1 EQT 271 (part 1)BASIC STATISTICS

Basic Statistics

1.1 Statistics in Engineering1.2 Collecting Engineering Data1.3 Data Presentation and Summary1.4 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.

the methods of statistics allow scientists and engineers to design valid experiments and to draw reliable conclusions from the data they produce

•Specifically, statistical techniques can be a powerful aid in designing new products and systems, improving existing designs, and improving production process.

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. Tabular2. 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; and3. line chart.

Types of Graph Qualitative Data

Bar chart

Pie chart

Line chart

Frequency table

Bar Chart: used to display the frequency distribution in the graphical form.

Observation FrequencyMalay 33Chinese9Indian 6Others 2

Pie Chart: used to display the frequency distribution. It displays the ratio of the observations

Line chart: used to display the trend of observations. It is a very popular display for the data which represent time. Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

10 7 5 10 39 7 260 316 142 11 4 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; and3. ogive.

Frequency Distribution Weight (Rounded decimal point) Frequency

60-62 563-65 1866-68 4269-71 2772-74 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 DistributionWhen 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.

•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.

WeightFrequenc

y60-62 563-65 1866-68 4269-71 2772-74 8Total 100

Table 1: Weight of 100 male students in XYZ university

• 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.

Weight Frequency60-62 563-65 1866-68 4269-71 2772-74 8Total 100

Variable Frequencycolumn

Third class (Interval Class)

Lower Limit of the fourth class

Frequencyof the third class.

Upper limit of the fifth class

Table 1.: Weight of 100 male students in XYZ university

• 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 MarkClass midpoint or mark = (Lower Limit + Upper Limit)/2- Finding The Number of ClassesNumber of classes = - Finding Class Width For Interval Classclass 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.

1 3.3log n

Example 1:From Table 1: Class Boundary

Weight (Class Interval)

Class Boundary Frequency

60-62 59.5-62.5 563-65 62.5-65.5 1866-68 65.5-68.5 4269-71 68.5-71.5 2772-74 71.5-74.5 8Total 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

Cumulative Frequency

60-62 5 59.5-62.55

63-65 18 62.5-65.55 + 18 = 23

66-68 42 65.5-68.523 + 42 = 65

69-71 27 68.5-71.565 + 27 =92

72-74 8 71.5-74.592 + 8 = 100

100

Exercise 1:

Given a raw data as below:27 27 27 28 27 24 25 2826 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 value2. 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 n3. Determine class width, i = (max-min)/num. of class4. Determine class limit.5. Find class boundaries and class mid points6. Count frequency for each class7. 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 2829 23 35 25 29 35 29 2723 32 31 32 24 35 21 3535 22 33 24 39 43

1.Construct a frequency distribution and cumulative frequency distribution table.2.Draw a histogram

Data Summary•Summary statistics are used to summarize a set of observations.a)Measures of Central TendencyMeanMedianModeb)Measures of DispersionRangeVarianceStandard deviationc)Measures of PositionZ scoresPercentilesQuartilesOutliers

a)Measures of Central TendencyMean• Mean of a sample is the sum of the sample data

divided by the total number sample. • Mean for ungrouped data is given by:

• Mean for group data is given by:

x

nx

xornnforn

xxxx n

_

21_

,...,2,1,.......

ffx

orf

xfx n

ii

n

iii

1

1

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 5.110

x

Example 3 (Grouped Data):

Use the frequency distribution of weights 100 male students in XYZ university, to find the mean.

Weight Frequency60-6263-6566-6869-7172-74

51842278

Solution :Weight (Class

IntervalFrequency, f

60-6263-6566-6869-7172-74

51842278

?fx

xf

100

6745 45.67

Class Mark, x

6164677073

fx

305115228141890584

6745100

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:

1

1

2

whereL = the lower class boundary of the median classc = the size of median class intervalF the sum of frequencies of all classes lower than the median class

the fre

j

j

j

j

fF

x L cf

f

quency of the median class

Example 4 (Ungrouped data):n is oddFind 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 = 4n is evenFind the median for data 4,6,3,2,5,7 (n = 6)Rearrange the data : 2,3,4,5,6,7Median = (4+5)/2 = 4.5

Example 5 (Grouped Data):The sample median for frequency distribution as in example 3Solution:

Weight (Class Interval

Frequency, f

60-6263-6566-6869-7172-74

51842278

12 ?j

j

fF

x L cf

Median class

]42

232

100

[35.65

73.67

Cumulative Frequency,

F

Class Boundary

5236592100

59.5-62.562.5-65.565.5-68.568.5-71.571.5-74.5

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

1

1 2

When data has been grouped in classes and a frequency curveis drawnto fit the data, the mode is the value of x corresponding to the maximum point on the curve, that is

ˆ

the lower c

x L c

L

1

2

lass boundary of the modal classc = the size of the modal class interval

the difference between the modal class frequency and the class before itthe difference between the modal class frequency a

nd the class after it

*the class which has the highest frequency is called the modal class

• Mode for grouped data

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 belowSolution:

Weight (Class

Interval

Frequency, f

60-6263-6566-6869-7172-74

51842278

Total 100

Mode class

1

1 2

ˆ ?x L c

Class Boundary

59.5-62.562.5-65.565.5-68.568.5-71.571.5-74.5

])2742()1842(

)1842([35.65

35.67

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

For population1

)( 22

nxx

S

nx

2

2 )(

The variance for the grouped data:•For sample

or

•For population

or

22

2

1fx n x

Sn

22

2

( )

1

fxfx

nSn

nxnfx

22

2n

nfx

fx

22

2

)(

• 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.

The positive square root of the variance is the standard deviation

22 2( )

1 1x x fx n x

Sn n

Example 8 (Ungrouped data)Find the variance and standard deviation of the sample data : 3, 5, 2, 6, 5, 9, 5, 2, 8, 6

43.59

9.489

)1.56()1.58()1.52()1.55(

)1.59()1.55()1.56()1.52()1.55()1.53(2222

222222

2

s

1)( 2

2

nxx

S

33.243.52

ss

Example 9 (Grouped data)Find the variance and standard deviation of the sample data below:

Weight (Class

Interval

Frequency, f

60-6263-6566-6869-7172-74

51842278

Total 100

22

2

( )

1

fxfx

nSn

Class Mark,

x

fx

6164677073

305115228141890584

6745

37214096448949005329

1860573728

18853813230042632

455803

2x 2fx

61.899

75.85299

1006745455803

2

93.261.8 s

Exercise 3The defects from machine A for a sample of products were organized into the following:

What is the mean, median, mode, variance and standard deviation? Or Calculate each possible measure of central tendency and dispersion.

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

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)

a)Construct a frequency distribution table.b)Find the mean, variance and standard deviation, mode and median. c)Construct a histogram.

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

Rules of Data DispersionBy using the mean 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, ii. 95% of the data will lie within two standard deviation of the mean,iii. 99.7% of the data will lie within three standard deviation of the mean,

x

)( sx

)2( sx

)3( sx

Example 10The 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:

]52,28[1240 sx]64,16[12.2402 sx

]76,4[12.3403 sx

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

QuartilesDivide data sets into fourths or four equal parts.

Smallest data value Q1 Q2 Q3

Largest data value

25% of data

25% of data

25% of data

25% of data

thnQ )1(411

thnmedianQ )1(212

thnQ )1(433

The positions are integersExample: 5, 8, 4, 4, 6, 3, 8 (n=7)

1.Put them in order: 3, 4, 4, 5, 6, 8, 82.Calculate the quartiles

,2)17(411 placendthQ ,4)17(

212 placeththQ

placeththQ 6)17(433

3, 4, 4, 5, 6, 8, 8,41Q,52 Q 83 Q

The positions are not integersExample: 5, 12, 10, 4, 6, 3, 8, 14 (n=8)

1.Put them in order: 3, 4, 5, 6, 8, 10, 12, 142.Calculate the quartiles

25.4)45(25.0425.2)18(411 ththQ

,7)68(5.065.4)18(212 ththQ

5.11)1012(75.01075.6)18(433 ththQ

3, 4, 5, 6, 8, 10, 12, 14

Example 11The 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 5.221.14 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

495.1)14.185.1(5.014.15.2)110(411 ththQ

,675.3)41.394.3(5.041.35.5)110(212 ththQ

0425.4)02.411.4(25.002.425.8)110(433 ththQ

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 QuartilesStep 1: Determine the first and third quartiles of data.Step 2: Compute the interquartile range (IQR).Step 3: Determine the fences. Fences serve as cut off points for determining outliers.

Step 4: If data value is less than the lower fence or greater than the upper fence, considered outlier.

3 1IQR Q Q

1

3

Lower Fence 1.5( )Upper Fence 1.5( )

Q IQRQ IQR

Example 12

2.47 3.97 3.94 4.11 5.221.14 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

32625.2)5475.2(5.1495.1

)(5.11

IQRQfenceLower

5475.2495.10425.413 QQIQR

,495.11Q 0425.43 Q

86375.7)5475.2(5.10425.4

)(5.13

IQRQfenceUpper

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

1 3MINIMUM Q Q MAXIMUMM

Example 13

2.47 3.97 3.94 4.11 5.221.14 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

,97.0Minimum

,495.11Q

,675.32 QM

0425.43 Q

,22.5Maximum

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.

Lower fence

Upper fence

Outlier Outlier

Interpreting Boxplot

- symmetric

- Skewed left because the tail is to the left

- Skewed right because the tail is to the right

Characteristics Of Skewed Distributions

                                                                  Mean/Median Versus Skewness

TO CONSTRUCT BOXPLOTStep 1: Determine the lower and upper fences:

Step 2: Draw vertical lines at .Step 3: Label the lower and upper fences.Step 4: Draw a line from to the smallest data value that is larger than the lower fence. Draw a line from 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 (*).

1

3

Lower Fence 1.5( )Upper Fence 1.5( )

Q IQRQ IQR

1 3, and Q M Q

3Q1Q

Example 14

2.47 3.97 3.94 4.11 5.221.14 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

,495.11Q ,675.32 QM 0425.43 Q

5475.2495.10425.413 QQIQR

32625.2)5475.2(5.1495.1

)(5.11

IQRQfenceLower

86375.7)5475.2(5.10425.4

)(5.13

IQRQfenceUpper

1Q M 3QfenceLower

fenceUpper

- The distribution is skewed left