Basic Statistics for Class 11, B.COm, BSW, B.A, BBA, MBA

1. STATISTICS
FOR SOCIAL
WORK
By Gaurav Rana, Assistant Professor
SHRI RAM COLLEGE OF COMMERCE

2. Introduction • Statistics deals with the collection, analysis,
interpretation and presentation of numerical
data.
• Using its various methods, effort is made to find
the causes behind it with the help of qualitative
and quantitative facts of an economic problem.
• Our wants are unlimited but the resources used
in the production of goods that satisfy our wants
are limited and scarce. Scarcity is the root of all
economic problems.
• Statistics finds economic relationships using
data and verifies them.
• Statistical methods help analyse economic
problems and formulate policies to solve them.

3. Collection of
Data
Survey is a method of gathering
information from individuals.
The questionnaire should not be too
long
The questionnaire should be easy to
understand and avoid ambiguous or
difficult words
The questions should be arranged in
an order such that the person
answering should feel comfortable
The series of questions should move
from general to specific.
Instrument used in surveys is
questionnaire/ interview schedule.
It is advisable to conduct a try-out
with a small group which is known as
Pilot Survey or Pre-testing of the
questionnaire.

4. Mode of
Data
Collection
Personal Interviews
Mailing Questionnaire
Telephone Interviews

5. Mode of Data
Collection
CENSUSANDSAMPLESURVEYS
Population and Sample
Sampling Error
Difference between the sample
estimate and the corresponding
population parameter
Non Sampling Error
Sampling Bias
Non-Response Errors
Errors in Data Acquisition
Random Sampling
Non-Random Sampling

6. Organisation
of Data
CLASSIFICATIONOFDATA
Qualitative
Classification
Population
(Gender)
Male (Marital
Status)
Married (Income
and Age)
Unmarried (Height
and Marks)
Female (Marital
Status)
Married (Income
and Age)
Unmarried (Height
and Marks)
Quantitative
Classification
Chronological
Classification
Time Period
Spatial
Classification
Geographical
Country
States
Regions

7. Variables:
Continuous
and
Discrete
• A continuous variable can take
any numerical value. It may take
integral values (1, 2, 3, 4, ...),
fractional values (1/2, 2/3, 3/4,
...), and values that are not exact
fractions ( 2 =1.414, 3 =1.732).
• A discrete variable can take only
certain values. Its value changes
only by finite “jumps”.

8. WHAT IS A
FREQUENCY
DISTRIBUTION?
• A frequency distribution is a comprehensive
way to classify raw data of a quantitative
variable.
Lower
Class
Limit
Class
Mid
Point
Upper
Class
Limit
Class
interval
CLASS FREQUENCY LOWER
LIMIT
UPPER
LIMIT
CLASS
MARK
0-10 1 0 10 5
10-20 5 10 20 15
20-30 8 20 30 25
30-40 4 30 40 35
40-50 2 40 50 45

9. How to
prepare a
Frequency
Distribution?
Should we have equal or unequal sized class
intervals?
How many classes should we have?
What should be the size of each class?
How should we determine the class limits?
• Inclusive class intervals
• Exclusive class intervals
How should we get the frequency for each
class?
Exclusive Class
Intervals
Inclusive class
intervals **
0-10 0-10
10-20 11-20
20-30 21-30
**Adjustment in Class Interval

10. Question?? • Prepare a frequency distribution by exclusive method taking
class interval of 20 from the following data.
528 574 568 508 555
551 567 520 544 552
532 549 597 562 502
545 501 537 573 582

11. Question??
Solution:-
• Prepare a frequency distribution by exclusive method taking
class interval of 20 from the following data.
528 574 568 508 555
551 567 520 544 552
532 549 597 562 502
545 501 537 573 582
Class Interval Frequency
501-520 4
521-540 3
541-560 6
561-580 5
581-600 2

12. Question?? • Prepare a frequency distribution by inclusive method taking
class interval of 20 from the following data.
528 574 568 508 555
551 567 520 544 552
532 549 597 562 502
545 501 537 573 582

13. Question??
Solution:-
• Prepare a frequency distribution by inclusive method taking
class interval of 20 from the following data.
528 574 568 508 555
551 567 520 544 552
532 549 597 562 502
545 501 537 573 582
Class Interval Frequency
500-520 3
520-540 4
540-560 6
560-580 5
580-600 2

14. Measures of
Central
Tendency
ARITHMETIC MEAN
MEDIAN
MODE
Mean:TheAverage of the numbers
Median: Denoting to a value lying at the midpoint of a frequency
distribution
Mode:The number which appears most often in a set of numbers

15. Arithmetic
Mean
Ungrouped
Data
Direct
Method
Assumed
Mean
Method
Step
Deviation
Method
Grouped
Data
Discrete
Series
Direct
Method
Assumed
Mean
Method
Step
Deviation
Method
Continuous
Series
Direct
Method
Step
Deviation
Method

16. Arithmetic Mean:- Individual Series
(Ungrouped Data)
Direct
Method
Assumed
Mean
Step
Deviation

17. Arithmetic Mean:- Individual Series
(Ungrouped Data)
1.Calculate Arithmetic Mean from the data showing marks of students in a
class in an economics test: 40, 50, 55, 78, 58.
2. The following data shows the weekly income of 10 families.
Family: A,B,C,D,E,F,G,H,I,J and Income Respectively
850,700,100,750,5000,80,420,2500,400,360

18. Solution 2. The following data shows the weekly income of 10 families.
Family: A,B,C,D,E,F,G,H,I,J and Income Respectively
850,700,100,750,5000,80,420,2500,400,360
Families Income d = X - 850 d’ = (X– 850)/ 10
A 850 0 0
B 700 - 150 -15
C 100 -750 -75
D 750 -100 10
E 5000 4150 415
F 80 -770 -77
G 420 -430 -43
H 2500 1650 165
I 400 -450 -45
J 360 -490 -49
N = 10 11160 2660 266
1. Mean = 11160/10
= 1116
2. Assumed Mean
= 850 + (2660)/10 =
1116
3. Step Deviation
Method = 850 +
(266/10) * 10 =
1116

19. Arithmetic Mean:- Discrete Series
(Grouped Data)
Direct
Method
Assumed
Mean
Step
Deviation

20. Arithmetic Mean:- Discrete Series
(Grouped Data)
1. Students in a housing colony collected donation in only three sizes: 100
Rupees 200 Rupees and 300 Rupees and the number of donation collected are
respectively 200 50 and 10.

21. Solution:
1. Students in a housing colony collected donation in
only three sizes: 100 Rupees 200 Rupees and 300
Rupees and the number of donation collected are
respectively 200 50 and 10.
Donatio
n Size
(X)
Donatio
n
Collecte
d (f)
fX d = (X-
200)
d’ = (X–
200)/10
0
fd fd’
100 200 20000 -100 -1 -20000 -200
200 50 10000 0 0 0 0
300 10 3000 100 1 1000 10
260 33000 0 -19000 -190
Direct Method =
33000/260 = 126.92
Assumed Mean = 200
+ (-19000/260) =
126.92
Step Deviation = 200
+ (-190/260) * 100

22. Arithmetic Mean:- Continuous Series
(Grouped Data)
Direct
Method
Step
Deviation

23. Arithmetic Mean:- Continuous Series
(Grouped Data)
1. Calculate average marks of the following students using (a) Direct method
(b) Step deviation method.
Marks No. of
Students
0-10 5
10-20 12
20-30 15
30-40 25
40-50 8
50-60 3
60-70 2

24. Solution:-1. Calculate average marks of the following students using
(a) Direct method (b) Step deviation method.
Marks No. of
Students
Mid value
(m)
fm d’ fd’
0-10 5 5 25 -3 -15
10-20 12 15 180 -2 -24
20-30 15 25 375 -1 -15
30-40 25 35 875 0 0
40-50 8 45 360 1 8
50-60 3 55 165 2 6
60-70 2 65 130 3 6
Sum 70 2110 0 -34
Direct Method =
2110/70 = 30.14
Step Deviation
Method = 35 + (-
34/70) * 10 = 30.14

25. How to
Calculate
Weighted
Arithmetic
Mean
• (W1P1 +W2P2 + WnPn)/
W1 + W2 +Wn…

26. Median
• Computation of median
• Example: Suppose we have the following observation in a data set: 5, 7, 6, 1, 8,
10, 12, 4, and 3.
• Arranging the data, in ascending order you have:
• Example: The following data provides marks of 10 students. You are required
to calculate the median marks. 25, 28, 29, 30, 32, 33, 35, 42, 45, 33
• Median = (32+33)/2 = 32.5 Marks
1 3 4 5 6 7 8 10 12
Median
25 28 29 30 32 33 33 35 42 45
Median? Median?

27. Median
• Discrete Series
• Example: The frequency distribution of the number of persons and their
respective incomes (in Rs) are given below. Calculate the median income.
• Solution
Income 10 20 30 40
No. of Persons 2 4 10 4
Income 10 20 30 40
No. of Persons 2 4 10 4
Cumulative
frequency (cf)
2 6 16 20
Median = (N+1)th/2 = (20+1)/2 = 10.5th Item. Therefore 16 is median as it
include 10.5th item in it. So the median income is Rs 30.

28. Median
• Continuous Series
• Example: Following data relates to daily wages of persons working in a
factory. Compute the median daily wage.
• Solution
Daily Wages 55-60 50-55 45-50 40-45 35-40 30-35 25-30 20-25
No. of Persons 7 13 15 20 30 33 28 14
Daily Wages 20-25 25-30 30-35 35 (L) -40 40-45 45-50 50-55 55-60
No. of Persons 14 28 33 30 (f) 20 15 13 7
cf. 14 42 75 (cf) 105 125 140 153 160
Step 1: N/2th Item = 160/2 = 80, and 80 falls under 35-40
Step 2: 35 + (80 -75)/30 * (40-35) = 35.83
L = 35
F = 3-
Cf = 75
H = Class Interval

29. Median
• Quartiles
• Q1
• In Case of discrete Series (N+1)/4
• Q3
• In Case of discrete Series 3(N+1)/4
• Percentiles
• In Case of discrete Series (N+1)/100 and Multiply with Percentile
• Example 90 Percentile = 90(N+1)/100 in discrete series.

30. Mode
• Discrete Series
• Example: Look at the following discrete series:
• Continuous Series
Variable 10 20 30 40 50
Frequency 2 8 20 10 5
Mode Mode

31. Mode
• Continuous Series
• Example: Calculate the value of modal worker family’s monthly income from
the following data:
Income Per Month cf Income
group
Frequency
Less than 50 97 45 - 50 97-95 = 2
Less than 45 95 40 - 45 95-90 = 5
Less than 40 90 35 - 40 90 – 80 = 10
Less than 35 80 30 - 35 80-60 = 20
Less than 30 60 25 - 30 60 – 30 = 30 Value of mode lies
between
Less than 25 30 20 -25 30-12 = 18
Less than 20 12 15 - 20 12 – 4 = 8
Less than 15 4 10 -15 4
Mode =
25 + 12/ (12 + 10) * 5
= 27.273
L = Lowest Value in
Class interval = 25,
D1 = 30 – 18 = 12(Mode
with next lower)
D2 = 30 – 20 = 20 (Mode
with next higher)
h = Class interval at
mode.

32. Measures of
Dispersion
• Calculation of family
income of three friends
• SITA, GEETA, MITA

33. Measures of
Dispersion
Dispersion is the extent
to which values in a
distribution differ from
the average of the
distribution.
• Calculation of family
income of three friends
• SITA, GEETA, MITA
Family Member Sita Geeta Mita
1 12000 7000 0
2 14000 10000 7000
3 16000 14000 8000
4 18000 17000 10000
5 N/A 20000 50000
6 N/A 22000 N/A
Total Income 60000 90000 75000
Avg. Income 15000 15000 15000

34. Measures of
Dispersion
• Range
• Quartile Deviation
• Mean Deviation
• Standard Deviation

35. Measures of
Dispersion
• Range: It is the difference
between the largest (L) and the
smallest value (S) in a
distribution.
• R = L – S
• Calculate range. of the
following observations: 20, 25,
29, 30, 35, 39, 41, 48, 51, 60
and 70
• R = 70 – 20 = 50

36. Measures of
Dispersion
• Quartile Deviation: Semi
Inter Quartile Range.
• Q.D = (Q3– Q1)/ 2
• Calculate Q.D. of the following
observations: 20, 25, 29, 30, 35, 39,
41, 48, 51, 60 and 70
• Q1 = (N + 1)/4 = (11+1)/4 = 3rd Item
• Q3 = 3(N+1)/4 = 3(11+1)/4 = 9th Item.
• Q.D. = (9th – 3rd)/2 = 51 -29)/2 = 11

37. Measures of
Dispersion
• Quartile Deviation: for
frequency distribution.
• Step 1. Q1 = n/4, 40/4 = 10th
Value falls in 10-20.
• Step 2. Q1 = 10+{(10-5)/8} * 10
= 16.25
• Step 3. Q3 = 3n/4 = 120/4 = 30th
Value falls in 40-60
• Step 4. Q3 = 40 + {(30 – 29)/7}*
20 = 42.87
• Step 5. (Q3-Q1)/2,
• = (42.87 – 16.25)/2 = 13.31
C-I Frequency cf
0-10 5 5
10-20 8 13
20-40 16 29
40-60 7 36
60-90 4 40
n = 40

38. Measures of
Dispersion
• Mean Deviation
• M.D = Sum |d|/ n
• Calculate the mean deviation of the
following values; 2, 4, 7, 8 and 9 through
mean and median.
• Step 1. Sum of x/ n = 30/5 = 6
• Step 2. Sum of |d|/n = 12/5 = 2.4
• Step 1. Calculate median = 7
• Step 2. Sum of |d|/n = 11/5 = 2.2
X |d| d =|X - Median|
2 4 5
4 2 3
7 1 0
8 2 1
9 3 2
12 11

39. Measures of
Dispersion
• Mean Deviation for
Continuous Series
• M.D = Sum f|d|/ Sumf
C.I f m.
p
fm |d| = m.p
- Mean
f|d|
10-20 5 15 75 25.5 127.5
20-30 8 25 200 15.5 124
30-50 16 40 640 .5 8
50-70 8 60 480 19.5 156
70-80 3 75 225 34.5 103.5
40 162
0
519
M.D. = 519/40 = 12.975

40. Measures of
Dispersion
• Mean Deviation from
Median
• M.D(median) = Sum f|d|/
Sumf
C.I f m.p |d| = m.p
- Median
f|d|
20-30 5 25 25 125
30-40 10 35 15 150
40-60 20 50 0 0
60-80 9 70 20 180
80-90 6 85 35 210
50 665
M.D. = 665/50 = 13.3

41. Measures of
Dispersion
• Standard Deviation for
ungrouped data
I. Actual Mean Method
II. Assumed Mean Method
III. Direct Method
IV. Step-Deviation Method

42. Measures of
Dispersion
• Standard Deviation from
Actual Mean Method
• S.D = (Sumd^2/n)^.5
• n = Total no. of observation
• S.D = (1270/5)^.5 =
15.937
X D (x – Mean) d2
5 -19 361
10 -14 196
25 1 1
30 6 36
50 26 676
0 1270

43. Measures of
Dispersion
• Standard Deviation from
Assumed Mean Method
• S.D =
• {(Sumd^2/n) – (Sumd/n)^2}^.5
• n = Total no. of observation
• S.D = {(1275/5) – (-
5/5)^2)^.5 = 15.937
X d(x-Mean) d2
5 -20 400
10 -15 225
25 0 0
30 5 25
50 25 625
-5 1275

44. Measures of
Dispersion
• Standard Deviation from
Direct Method
• S.D =
• {(SumX^2/n) – (Mean)^2}^.5
• n = Total no. of observation
• S.D = {4150/5 – (24)^2}^.5
= 15.937
X X2
5 25
10 100
25 625
30 900
50 2500
120 4150

45. Measures of
Dispersion
• Standard Deviation from
Step Deviation Method
• Step 1. : c here is a common factor
• Step 2. Calculate Mean = Sum x’/n
• n = Total no. of observation
• Step 3. Calculate S.D
• S.D = {(50.80/5)* 5}^.5
• = 15.937
x x’ d’ = (x’ -
X)
d’2
5 1 -3.8 14.44
10 2 -2.8 7.84
25 5 .2 .04
30 6 1.2 1.44
50 10 5.2 27.04
24 50.80

46. Measures of
Dispersion
• Standard Deviation from
Step Deviation Method
(alternative Method)
• S.D = (51/5 – (-1/5)^2)^.5 * 5
x d =
(x-
25)
d’ = (d /5) d’2
5 -20 -4 16
10 -15 -3 9
25 0 0 0
30 5 1 1
50 25 5 25
-1 51

47. Measures of
Dispersion
• Standard Deviation for
Grouped data
I. Actual Mean Method
II. Assumed Mean Method
III. Step-Deviation Method

48. Measures of
Dispersion
• Standard Deviation for
Grouped data
I. Actual Mean Method
C.I f m fm d fd fd2
10-20 5 15 75 -25.5 -127.5 3251.25
20-30 8 25 200 -15.5 -124 1922
30-50 16 40 640 -.5 -8 4
50-70 8 60 480 19.5 156 3042
70-80 3 75 225 34.5 103.5 3570
40 1620 0 11790
Mean = Sum fm/
sum f
S.D =
(11790/40)^.5 =
17.168

49. Measures of
Dispersion
• Standard Deviation for
Grouped data
I. Assumed Mean Method
C.I f m d fd fd2
10-20 5 15 -25 -125 3125
20-30 8 25 -15 -120 1800
30-50 16 40 0 0 0
50-70 8 60 20 160 3200
70-80 3 75 35 105 3675
40 20 11800
S.D = {(11800/40) –
(20/40)^2}^.5 = 17.168

50. Measures of
Dispersion
• Standard Deviation for
Grouped data
I. Step Deviation Method
C.I f m d d’ fd fd2
10-20 5 15 -25 -5 -25 125
20-30 8 25 -15 -3 -24 72
30-50 16 40 0 0 0 0
50-70 8 60 20 4 32 128
70-80 3 75 35 7 21 147
40 4 472
S.D = {(472/40) –
(4/40)^2}^.5 = 17.168

51. Absolute
Relative
Measures
• Coefficient of Range
• = L –S/ L+S
• Coefficient of Quartile
Deviation
• = Q3-Q1/Q3+Q1
• Coefficient of Mean
Deviation.

52. Absolute
Relative
Measures
• Coefficient ofVariation
• = {Standard Deviation/
Arithmetic Mean} * 100

53. Questions??

54. Thank You!!