basic statistics presentation
DESCRIPTION
TRANSCRIPT
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