Basic Statistics Presentation

INTRODUCTION TO STATISTICS

Md. Mortuza Ahmmed

Applications of Statistics

Agriculture

Business and economics

Marketing Research

Education

Medicine

Qualitative Variable

Dependent variable

Continuous variable

Quantitative Variable

Discrete variable

Independent variable

Scales of Measurement

Ordinal Scale

Interval scale

Ratio scale

Nominal scale

FREQUENCY TABLE

Rating of

DrinkTally marks Frequency

Relative

Frequency

P IIII 05 05 / 25 = 0.20

G IIII IIII II 12 12 / 25 = 0.48

E IIII III 08 08 / 25 = 0.32

Total 25 1.00

SIMPLE BAR DIAGRAM

Muslim Hindu Christians Others0

20

40

60

80

100

120

140

160150

100

56

25

COMPONENT BAR DIAGRAM

Male Female0

50

100

150

200

250

300

Section DSection CSection BSection A

MULTIPLE BAR DIAGRAM

Male Female0

10

20

30

40

50

60

70

80

90

100

Section ASection BSection CSection D

PIE CHART

46%

31%

15%

8%

Religion of studentsMuslim Hindu Christians Others

LINE GRAPH

July August September October November0

1000

2000

3000

4000

5000

6000

7000

5000

5600

6400

3000

4500

Share price of BEXIMCO

HISTOGRAM

0

2

4

6

8

10

12

14

16

18

20

BAR DIAGRAM VS. HISTOGRAM

Histogram Bar diagram

Area gives

frequency

Height gives

frequency

Bars are adjacent

to each others

Bars are not

adjacent to each

others

Constructed for

quantitative data

Constructed for

qualitative data

STEM AND LEAF PLOT

Stem

Leaf

1

1 4 7 9

2

1 3 4 7 9

3

1 3 7 9

4

1 3 4 7

5

1 3 4 9

6

1 3 4 7

SCATTER DIAGRAM

0 5 10 15 20 25 300

50

100

150

200

250

300

Price

Su

pp

ly

COMPARISON AMONG THE GRAPHS

Graph Advantages Disadvantages

Pie chartShows percent of total

for each category

Use only discrete data

HistogramCan compare to normal

curve

Use only continuous

data

Bar diagramCompare 2 or 3 data

sets easily

Use only discrete data

Line graphCompare 2 or 3 data

sets easily

Use only continuous

data

Scatter plotShows a trend in the

data relationship

Use only continuous

data

Stem and Leaf

Plot 

Handle extremely large

data sets

Not visually appealing

MEASURES OF CENTRAL TENDENCY

A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data.

Arithmetic mean (AM)Geometric mean (GM)Harmonic mean (HM)

MedianMode

ARITHMETIC MEAN

It is equal to the sum of all the values in the data set divided by the number of values in the data set.

PROBLEMS Find the average of the values 5, 9, 12,

4, 5, 14, 19, 16, 3, 5, 7.

The mean weight of three dogs is 38 pounds.  One of the dogs weighs 46 pounds.  The other two dogs, Eddie and Tommy, have the same weight.  Find Tommy’s weight.

On her first 5 math tests, Zany received scores 72, 86, 92, 63, and 77.  What test score she must earn on her sixth test so that her average for all 6 tests will be 80? 

AFFECT OF EXTREME VALUES ON AM

Staff 1 2 3 4 5 6 7 8 9 10

Salary

1518

16

14 15 15 12 17 90 95

CALCULATION OF AM FOR GROUPED DATA

x f f.x0 05 001 10 102 05 103 10 304 05 2010 02 20

Total N = 37 90AM = 90 / 37

= 2.43

MEDIAN

1 3 2 MEDIAN = 2

1 2 3

1 4 3 2 MEDIAN = (2 +

3) / 2 = 2.5

1 2 3 4

MODE

WHEN TO USE THE MEAN, MEDIAN AND MODE

Type of VariableBest measure of central tendency

Nominal Mode

Ordinal Median

Interval/Ratio (not skewed)

Mean

Interval/Ratio (skewed) Median

WHEN WE ADD OR MULTIPLY EACH VALUE BY SAME AMOUNT

Data Mea

n

Mod

e

Media

n

Original

data Set

6, 7, 8, 10, 12,

14, 14, 15, 16,

20

12.2 14 13

Add 3 to

each

value

9, 10, 11, 13,

15, 17, 17, 18,

19, 23

15.2 17 16

Multiply

2 to

each

value

12, 14, 16, 20,

24, 28, 28, 30,

32, 40

24.4 28 26

MEAN, MEDIAN AND MODE FOR SERIES DATA

For a series 1, 2, 3 ….n, mean = median = mode = (n + 1) / 2

So, for a series 1, 2, 3 ….100,

mean = median = mode = (100 + 1) / 2 = 50.5

GEOMETRIC MEAN

HARMONIC MEAN

AM X HM = (GM) 2

For any 2 numbers a and b,

AM = (a + b) / 2

GM = (ab) ^ ½

HM = 2 / (1 / a + 1

/ b) = 2ab / (a + b)

AM X HM

= (a + b) / 2 . 2ab / (a + b)

= ab

= (GM) 2

EXAMPLE

For any two numbers, AM = 10 and GM = 8. Find out the

numbers. (ab)^ ½ = 08

Þab = 64

(a + b) / 2 = 10

Þa + b = 20 . . . . .(1)

(a - b)2 = (a + b)2 – 4ab

= (20)2 – 4 .64

= 144

=> a - b = 12 . . . .(2)

Solving (1) and (2) (a, b) = (16, 4)

EXAMPLE

For any two numbers, GM = 4√3 and HM = 6. Find out AM and the numbers.

AM

= (GM)2/ HM

= (4√3) 2 / 6

= 8

√ab = 4√3

=>ab = 48

(a + b) / 2 = 8

=> a + b = 16 …(1)

(a - b)2

= (a + b)2 – 4ab

= (16)2 – 4 . 48

= 64

Þa - b = 8 ...(2)

Solving (1) & (2) (a, b) = (12, 4)

CRITERIA FOR GOOD MEASURES OF CENTRAL TENDENCY

Clearly defined

Readily comprehensible

Based on all observations

Easily calculated

Less affected by extreme valuesCapable of further algebraic treatment

AM ≥ GM ≥ HMFor any two numbers a & b

AM = (a + b) / 2

GM = (ab)^1/2

HM = 2 / (1 / a + 1 / b)

= 2ab / (a + b)

(√a - √b) 2 ≥ 0Þa + b – 2(ab)^1/2 ≥ 0Þa + b ≥ 2(ab)^1/2 Þ(a + b) / 2 ≥ (ab)^1/2

=> AM ≥ GM

Multiplying both sides by 2(ab)^1/2 / (a + b)

(ab)^1/2 ≥ 2ab / (a + b)

ÞGM ≥ HM

So, AM ≥ GM ≥ HM