Bsic Statistics Presentation
Document Sample


INTRODUCTION TO
STATISTICS
Md. Mortuza Ahmmed
Applications of Statistics
Agriculture
Business and economics
Marketing Research
Education
Medicine
Variable
Qualitative
Variable
Independe Dependent
nt variable variable
Discrete Continuous
variable variable
Quantitative
Variable
Scales of Measurement
Nominal Ordinal
scale Scale
Ratio Interval
scale scale
FREQUENCY TABLE
Rating of Relative
Tally marks Frequency
Drink 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
160
150
140
120
100
100
80
60 56
40
25
20
0
Muslim Hindu Christians Others
COMPONENT BAR DIAGRAM
300
250
200
Section D
Section C
150
Section B
100 Section A
50
0
Male Female
MULTIPLE BAR DIAGRAM
100
90
80
70
60 Section A
50 Section B
40 Section C
Section D
30
20
10
0
Male Female
PIE CHART
Religion of students
Muslim Hindu Christians Others
8%
15%
46%
31%
LINE GRAPH
Share price of BEXIMCO
7000
6400
6000
5600
5000
5000 4500
4000
3000
3000
2000
1000
0
July August September October November
HISTOGRAM
20
18
16
14
12
10
8
6
4
2
0
BAR DIAGRAM VS. HISTOGRAM
Histogram Bar diagram
Area gives frequency Height gives frequency
Bars are adjacent to Bars are not adjacent
each others to each others
Constructed for Constructed for
quantitative data qualitative data
STEM AND LEAF PLOT
Stem Leaf
1 1479
2 13479
3 1379
4 1347
5 1349
6 1347
SCATTER DIAGRAM
300
250
200
Supply
150
100
50
0
0 5 10 15 20 25 30
Price
COMPARISON AMONG THE GRAPHS
Graph Advantages Disadvantages
Shows percent of total Use only discrete data
Pie chart
for each category
Can compare to normal Use only continuous data
Histogram
curve
Compare 2 or 3 data sets Use only discrete data
Bar diagram
easily
Compare 2 or 3 data sets Use only continuous data
Line graph
easily
Shows a trend in the data Use only continuous data
Scatter plot
relationship
Stem and Leaf Handle extremely large Not visually appealing
Plot data sets
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.
Arithmeticmean (AM)
Geometric mean (GM)
Harmonic mean (HM)
Median
Mode
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 15 18 16 14 15 15 12 17 90 95
CALCULATION OF AM FOR GROUPED DATA
x f f.x
0 05 00
1 10 10
2 05 10
3 10 30
4 05 20
10 02 20
Total N = 37 90
AM = 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
Best measure of central
Type of Variable
tendency
Nominal Mode
Ordinal Median
Interval/Ratio (not
Mean
skewed)
Interval/Ratio (skewed) Median
WHEN WE ADD OR MULTIPLY EACH
VALUE BY SAME AMOUNT
Data Mean Mode Median
Original 6, 7, 8, 10, 12, 14, 12.2 14 13
data Set 14, 15, 16, 20
Add 3 to 9, 10, 11, 13, 15, 15.2 17 16
each 17, 17, 18, 19, 23
value
Multiply 2 12, 14, 16, 20, 24, 24.4 28 26
to each 28, 28, 30, 32, 40
value
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 AM X HM
a and b,
= (a + b) / 2 . 2ab /
AM = (a + b) / 2 (a + b)
GM = (ab) ^ ½ = ab
= (GM) 2
HM = 2 / (1 / a + 1 / b)
= 2ab / (a + b)
EXAMPLE
For any two numbers, AM = 10 and
GM = 8. Find out the numbers.
(ab)^ ½ = 08 (a - b)2 = (a + b)2 – 4ab
ab = 64 = (20)2 – 4 .64
= 144
(a + b) / 2 = 10
a + b = 20 . . . . .(1) => 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 √ab = 4√3 (a - b)2
= (GM)2/ HM =>ab = 48 = (a + b)2 – 4ab
= (4√3) 2 / 6 = (16)2 – 4 . 48
=8 (a + b) / 2 = 8
= 64
=> a + b = 16 …(1) a - 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 values
Capable of further algebraic
treatment
AM ≥ GM ≥ HM
For any two numbers a & b (√a - √b) 2 ≥ 0
AM = (a + b) / 2 a + b – 2(ab)^1/2 ≥ 0
GM = (ab)^1/2 a + b ≥ 2(ab)^1/2
HM = 2 / (1 / a + 1 / b) (a + b) / 2 ≥ (ab)^1/2
= 2ab / (a + b) => AM ≥ GM
Multiplying both sides by 2(ab)^1/2 / (a + b)
(ab)^1/2 ≥ 2ab / (a + b)
GM ≥ HM
So, AM ≥ GM ≥ HM
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