Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.1

Question 1.
What is transportation problem?
Solution:
The objective of transportation problem is to determine the amount to be transported from each origin to each destinations such that the total transportation cost is minimized.

Question 2.
Write mathematical form of transportation problem.
Solution:
Let there be m origins and n destinations. Let the amount of supply at th i th origin is a_i. Let the demand at j th destination is b_j.
The cost of transporting one unit of an item from origin i to destination j is C_ij and is known for all combination (i,j). Quantity transported from origin i to destination j be x_ij.

The objective is to determine the quantity x_ij to be transported over all routes (i,j) so as to minimize the total transportation cost. The supply limits at the origins and the demand requirements at the destinations must be satisfied.
The above transportation problem can be written in the following tabular form:

Now the linear programming model representing the transportation problem is given by
The objective function is Minimize Z = \(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\), \(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) c_ij x_ij
Subject to the constraints
\(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) = x_ij = a_i, i = 1, 2 …….. m (supply constraints)
\(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\) = x_ij = b_j, i = 1, 2 …….. n (demand constraints)
x_ij ≥ 0 for all i, j (non- negative restrictions)

Question 3.
What is feasible solution and non degenerate solution in transportation problem?
Solution:
Feasible Solution:
A feasible solution to a transportation problem is a set of non-negative values x_ij (i = 1, 2, … m, j = 1, 2, … n) that satisfies the constraints.

Non degenerate basic feasible solution:
If a basic feasible solution to a transportation problem contains exactly m + n – 1 allocations in independent positions, it is called a Non degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

Question 4.
What do you mean by balanced transportation problem?
Solution:
In a transportation problem if the total supply equals the total demand (Σa_i = Σb_j) then it is said to be balanced transportation problem.

Question 5.
Find an intial basic feasible solution of the following problem using north west corner rule.

Solution:
Here total Supply 19 + 37 + 34 = 90
Total demand = 16 + 18 + 31 + 25 = 90
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem
∴ we can final an initial basic feasible solution to due given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (Q₁, D₁). Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column’s exhausted.
(i.e) x₁₁ = min (19, 16) = 16

Now the cell in the north west corner is (O₁, D₂) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
(i.e) x₁₂ = min (3, 18) = 3

Reduced transportation table is

Now the cell in the north west corner is (O₂, D₂)
x₂₂ = min (37, 15) = 15

Now the cell in the north west corner is (O₂, D₃)
x₂₃ = min (22, 31) = 9

Now the cell in the north west corner is (O₃, D₃)
x₃₃ = min (34, 9) = 9

Thus we have the following allocations

Transportation schedule:
O₁ → D₁, O₁, → D₂, O₂ → D₂; O₂ → D₃ O₃ → D₃; O₃ → D₄
= (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)
= 80 + 9 + 105 + 198 + 63 + 125
= 580

Question 6.
Determine an intial basic feasible solution of the following transportation problem by north west corner method.

Solution:
Here total capacity (a_i) = 30 + 40 + 50 = 120
Total demand (b_j) = 35 + 28 + 32 + 25 = 120
(i.e) Total capacity = Total demand
∴ The given problem is balanced transportation.
∴ We can find an initial basic feasible solution to the given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (chennai, Bangalore)
x₁₁ = min (30, 35) = 30
Reduced transportation table is

Now the cell in the North west corner is (Madurai, Bangalore)
x₂₁ = min(40, 5) = 5

From the above table we can choose the cell in the North west corner. Here the cell is (Nasik, Madurai)
x₂₂ = min (35, 28) = 28

From the above table we can choose the cell in the North west corner. Here the cell is (Bhopal, Madurai)
x₂₂ = min (7, 32) = 7

From the above table we can choose the cell in the North west corner. Here the cell is (Trichy, Bhopal)
x₃₃ = min (50, 25) = 25
Reduced transportation table is

x₃₄ = min (25, 25) = 25
Thus we have the following allocations

Transportation Schedule:
Chennai → Bangalore; Madurai → Bangalore;
Madurai → Naisk; Madurai → Bhopal
Trichy → Bhopal; Trichy → Delhi
The total transportation cost =
(30 × 6) + (5 × 5) + (28 × 11) + (7 × 9) + (25 × 7) + (25 × 13)
= 180 + 25 + 308 + 63 + 175 + 325
= 1076

Question 7.
Obtain an initial basic feasible solution to the following transportation problem by using least-cost method.

Solution:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total demand = 30 + 25 + 45 = 100
(i.e) Total supply = Total demand
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cos is 4 corresponds to the cell (O₂, D₃)
Allocate min (35, 45) = 35 units to this cell

The least cost corresponds to the cell (O₁, D₃)
Allocate min (25, 10) = 10 units to this cell

The least cost is 6 corresponds to the cell (O₃, D₂)
Allocate min (40, 25) = 25 units to this cell

The least cost is 7 corresponds to the cell (O₃, D₁)
Allocate min (15, 30) = 15

Thus we have the following allocations

Transportation Schedule:
O₁ → D₁; O₁, → D₃; O₂ → D₃; O₃ → DI; O₃ → D₂
Total Transportation cost
= (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 135 + 50 + 140 + 105 + 150
= 580

Question 8.
Explain vogel’s approximation method by obtaining initial feasible solution of the following transportation problem.

Solution:
Here Σ ai = 6 + 1 + 10 = 17
Σ bj = 7 + 5 + 3 + 2 = 17
Σ ai = Σ bj
(i.e) Total supply = Total Demand
∴ The given problem is balanced transportation problem
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns.

Choose the largest difference. Here the largest difference is 6 which corresponds to column D₄
In this column choose the least cost. Here the least cost corresponds to (O₂, D₄)
Allocate min (1, 2) = 1 unit to this cell the reduced transportation table is

choose the largest difference 5 which corresponds to column D₂. Here the least cost corresponds to (O₁, D₂).
Allocate min (6, 5) = 5 units in this cell

Choose the largest difference 5 which corresponds to row O₁. Here the least cost corresponds to (O₁, D₁)
Allocate min (1, 7) = 1 unit in this cell

Choose the largest difference 4 which corresponds to row O₃. Here least cost corresponds to (10, 6) = 6 units in this cell.

Choose the largest difference 6 which corresponds to row O₃. Here the least cost corresponds to (O₃, D₄).
Allocate min (4, 1) = 1

Thus we have the following allocations

Transportation schedule
O₁ → D₁; O₁ → D₂; O₂ → D₄;
O₃ → D₁; O₃ → D₃; O₃ → D₄
Total transportation cost:
= (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 9.
Consider the following transportation problem.

Determine initital basic feasible solution by VAM
Solution:
Given Transportation problem is

Here Σ ai = Σ bj = 100
∴ The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns

Choose the largest difference. Here the difference is 3 which corresponds to D₂
In this column choose the least cost. Here the least cos corresponds to (O₃, D₂)
Allocate min (20, 40) = 20 units to this cell

Choose the largest difference is 4 which corresponds to column D₃. In this column choose the least cost. Here the least cost corresponds to (O₁, D₃).
Allocate min (30, 20) = 20 units to this cell

Choose the largest different is 3 which corresponds to column D₂. In this column choose the least cost. Here the least cost corresponds to (O₂, D₂)
Allocate min (50, 20) = 20 units to this cell

Choose the largest difference is 2 which corresponds to column D₄. In this column choose the least cost. Here the least cost corresponds to (O₂, D₄).
Allocate min (30, 10) = 10 units to this cell

Allocate min (20, 30) = 20 units to this cell

Here
x₁₁ = 10
x₁₃ = 20
x₂₁ = 20
x₂₂ = 20
x₂₄ = 10
x₃₂ = 20
Transportation schedule
O₁ → D₁; O₁ → D₃; O₂ → D₁;
O₂ → D₂; O₂ → D₄; O₃ → D₂
The transportation cost
= (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)
= 50 + 60 + 80 + 100 + 40 + 40
= 370

Question 10.
Determine basic feasible solution to the following transportation problem using North west Corner rule.

Solution:
Here total supply = 4 + 8 + 9 = 21
Total demand = 3 + 3 + 4 + 5+ 6 = 21
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (P, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.

Form the above table we can choose the cell in North west corner. Here the cell is (P,B)
x = min (1, 3) = 1

From the above table we can choose the cell in north west corner. Here the cell is (Q, B)
x = min (2, 8) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (Q, C)

From the above table, we can choose the cell in North west corner. Here the cell is (Q, D)
x = min (2, 5) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (R, D)
x = min (9, 3) = 3

Thus we have the following table

Transportation Schedule:
P → A; P → B; Q → B; Q → C; Q → D R → D; R → E
Total transportation cost:
= (3 × 2) + (1 × 11) + (2 × 4) + (4 × 7) + (2 × 2) + (3 × 8) + (6 × 2)
= 6 + 11 + 8 + 28 + 4 + 24 + 72
= 153

Question 11.
Find the initial basic feasible solution of the following transportation problem:

Using
(i) North West corner tule
(ii) Least Cost method
(iii) Vogel’s approximation method
Solution:
(i) North west corner rule:
Here the total supply = 10 + 10 + 10 = 30
Total demand = 7 + 12 + 11 = 30
(i.e) Total supply = Total demand
The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (A, I)
x₁₁ = min (7, 10) = 7

From the above table we can choose the cell in the north west corner. Here the cell is (B, I)
x = min (3, 12) = 3

From the above table we can choose the cell in the North west corner. Here the cell is (B, II)
x = min (9, 10) = 9

Here the cell in the North west corner is (C, II)
x = min (11, 1) = 1

Thus we have the following allocations

Transportation schedule:
A → I; B → I; B → II; C → II; C → III
Total transportation cost:
= (7 × 1) + (3 × 0) + (9 × 4) + (1 × 1) + (10 × 5)
= 7 + 0 + 36 + 1 + 50
= 94

(ii) Least cost method:
The given transportation table is

Here Total supply = Total demand = 30
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cost is 0 corresponds to the cell (B, I)
Allocate min (12, 10) = 10 units to this cell

The least cost 1 corresponds to the cell (C, II)
Allocate min (11, 10) = 10 units to this cell

Here the least cost 2 corresponds to the cell (B, III)
Allocate min (2, 10) = 2 units to this cell.

Here the least cost is 5 corresponds to the cell (C, III)

Transportation schedule:
A → III; B → I; B → III; C → II; C → III
Total transportation cost:
= (7 × 6) + (10 × 10) + (2 × 2) + (10 × 1) + (1 × 5)
= 42 + 0 + 4 + 10 + 5
= 61

(iii) Vogel’s approximation method:
Here Σ ai = Σ bj = 30
(i.e) Total supply = Total demand
∴ This given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.


A → I; B → I; B → III; C → I; C → II
Total transportation cost:
= (7 × 1) + (2 × 0) + (10 × 2) + (1 × 3) + (10 × 1)
= 7 + 0 + 20 + 3 + 10
= 40

Question 12.
Obtain an initial basic feasible solution to the following transportation problem by north west corner method.

Solution:
Here the total available = 250 + 300 + 400 = 950
Total Required = 200 + 225 + 275 + 250 = 950
(i.e) Total Available = total required
∴ The given problem is balanced transportation problem.
we can find an initial basic feasible solution to the given problem.
From the above table, we can choose the cell in the North west corner. Here the cell is (A, D).
x₁₁ = min (250, 200) = 200

From the above table we can choose the cell in North west corner. Here the cell is (A, E)
x = min (50, 225) = 50

From the above table, the north west corner cell is (B, E)
x = min (300, 175) = 175

From the above table, the north west corner cell is (B, F)
x = min (125, 275) = 125

Here the north west corner cell is (C, F)
x = min (400, 150) = 150

Transportation schedule:
A → D; A → E; B → E; B → F; C → G
Total Transportation cost:
= (200 × 11) + (50 × 13) + (175 × 18) + (125 × 14) + (150 × 13) + (250 × 10)
= 2200 + 650 + 3150 + 1750 + 1950 + 2500
= 12,200

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Miscellaneous Problems

Question 1.
Using three yearly moving averages, Determine the trend values from the following data.

Solution:

Question 2.
From the following data, calculate the trend values using fourly moving averages.

Solution:

Question 3.
Fit a straight line trend by the method of least squares to the following data.

Solution:

Therefore, the required equation of the straight line trend is given by
y = a + bx
y = 55.9875 + 0.830 x
⇒ y = 55.9875 + 0.83 (\(\frac { x-1983.5 }{0.5}\))
The trend values can be obtained by
When x = 1980
y = 55.9875 + 0.83 (\(\frac { 1980-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-7)
= 55.9875 – 5.81
= 50.1775
When x = 1981
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-5)
= 55.9875 – 4.15
= 51.8375
When x = 1982
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-3)
= 55.9875 – 2.49
= 53.4975
When x = 1983
y = 55.9875 + 0.83 (\(\frac { 1983-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-1)
= 55.9875 – 0.83
= 55.1575
When x = 1984
y = 55.9875 + 0.83 (\(\frac { 1984-1983.5 }{0.5}\))
55.9875 + 0.83 (1)
= 56.8175
when x = 1985
y = 55.9875 + 0.83 (\(\frac { 1985-1983.5 }{0.5}\))
= 55.9875 + 0.83 (3)
= 55.9875 + 2.49
= 58.4775
when x = 1986
y = 55.9875 + 0.83 (\(\frac { 1986-1983.5 }{0.5}\))
= 55.9875 + 0.83 (5)
= 55.9875 + 4.15
= 60.1375
when x = 1987
y = 55.9875 + 0.83 (\(\frac { 1987-1983.5 }{0.5}\))
= 55.9875 + 0.83 (7)
= 55.9875 + 5.81
= 61.7975

Question 4.
Fit a straight line trend by the method of least squares to the following data.

Solution:

Lasperyre’s price Index number


Hence Fisher’s Ideal Index satisfies Time reversal test

Question 5.
Using the following data, construct Fisher’s Ideal Index Number and Show that it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Factor reversal test

Hence Fisher’s Ideal Index satisfies Factor reversal test.

Question 6.
Compute the consumer price index for 2015 on the basis of 2014 from the following data.

Solution:

Question 7.
An Enquiry was made into the budgets of the middle class families in a city gave the following information.

What changes in the cost of living have taken place in the middle class families of a city?
Solution:

Conclusion:
The cost of living has increased up to 26.10% in 2011 as compared to 2010.

Question 8.
From the following data, calculate the control limits for the mean and range chart.

Solution:

UCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 51 + 0.577(6.5)
= 51 + 3.7505
= 54.7505
= 54.75
CL = \(\bar { \bar x}\) = 51
UCL = \(\bar { \bar x}\) – A₂\(\bar { R}\)
= 51 – 0.577(6.5)
= 51 – 3.7505
= 47.2495
= 47.25
The control limits for Range chart is
UCL = D₄\(\bar { R}\)
= 2.114(6.5)
= 13.741
CL = \(\bar { R}\) = 6.5
LCL = D₃\(\bar { R}\) = 0(6.5) = 0

Question 9.
The following data gives the average life(in hours) and range of 12 samples of 5 lamps each. The data are

Construct control charts for mean and range Comment on the control limits.
Solution:

UCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.6675
= 1614.1675
= 1614.17
CL = \(\bar { \bar x}\) = 1367.5
LCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 1367.5 – 0.577(427.5)
= 1367.5 – 246.6675
= 1120.8325
= 1120.83
The control limits for Range chart is
UCL = D₄\(\bar { R}\)
= 2.115(427.5)
= 904.1625
= 904.16
CL = \(\bar { R}\) = 427.5
LCL = D₃\(\bar { R}\)
= 0(427.5)
= 0

Question 10.
The following are the sample means and I ranges for 10 samples, each of size 5. Calculate ; the control limits for the mean chart and range chart and state whether the process is in control or not.

Solution:

UCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 4.982 + 0.577(0.36)
= 4.982 + 0.20772
= 5.18972
= 5.19
CL = \(\bar { \bar x}\) = 4.982
LCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 4.982 – 0.577(0.36)
= 4.982 – 0.20772
= 4.77428
= 4.774
The control limits for range chart is
UCL = D₂\(\bar { R}\) = 2.115(3.6)
= 7.614
CL = \(\bar { R}\) = 3.6
LCL = D₃\(\bar { R}\)
= 0(0.36) = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.4

Question 1.
A time series is a set of data recorded
(a) Periodically
(b) Weekly
(c) successive points of time
(d) all the above
Solution:
(d) all the above

Question 2.
A time series consists of
(a) Five components
(b) Four components
(c) Three components
(d) Two components
Solution:
(b) Four components

Question 3.
The components of a time series which is attached to short term fluctuation is
(a) Secular trend
(b) Seasonal variations
(c) Cyclic variation
(d) Irregular variation
Solution:
(d) Irregular variation

Question 4.
Factors responsible for seasonal variations are
(a) Weather
(b) Festivals
(c) Social customs
(d) All the above
Solution:
(d) All the above

Question 5.
The additive model of the time series with the components T, S, C and I is
(a) y = T + S + C × I
(b) y = T + S × C × I
(c) y = T + S + C + I
(d) y = T + S × C + I
Solution:
(c) y = T + S + C + I

Question 6.
Least square method of fitting a trend is
(a) Most exact
(b) Least exact
(c) Full of subjectivity
(d) Mathematically unsolved
Solution:
(a) Most exact

Question 7.
The value of ‘b’ in the trend line y = a + bx is
(a) Always positive
(b) Always negative
(c) Either positive or negative
(d) Zero
Solution:
(c) Either positive or negative

Question 8.
The component of a time series attached to long term variation is trended as
(a) Cyclic variation
(b) Secular variations
(c) Irregular variation
(d) Seasonal variations
Solution:
(b) Secular variations

Question 9.
The seasonal variation means the variations occurring with in
(a) A number of years
(b) within a year
(c) within a month
(d) within a week
Solution:
(b) within a year

Question 10.
Another name of consumer’s price index number is:
(a) Whole-sale price index number
(b) Cost of living index
(c) Sensitive
(d) Composite
Solution:
(b) Cost of living index

Question 11.
Cost of living at two different cities can be compared with the help of
(a) Consumer price index
(b) Value index
(c) Volume index
(d) Un-weighted index
Solution:
(a) Consumer price index

Question 12.
Laspeyre’s index = 110, Paasche’s index = 108, then Fisher’s Ideal index is equal to:
(a) 110
(b)108
(c) 100
(d) 109
Solution:
(d) 109
Hint:
p²₀₁ = 110; p^p₀₁ = 108
Fisher’s Ideal Index = \(\sqrt { 110×108 }\) = \(\sqrt { 11880 }\) = 108.99 = 109

Question 13.
Most commonly used index number is:
(a) Volume index number
(b) Value index number
(c) Price index number
(d) Simple index number
Solution:
(c) Price index number

Question 14.
Consumer price index are obtained by:
(a) Paasche’s formula
(b) Fisher’s ideal formula
(c) Marshall Edgeworth formula
(d) Family budget method formula
Solution:
(d) Family budget method formula

Question 15.
Which of the following Index number satisfy the time reversal test?
(a) Laspeyre’s Index number
(b) Paasche’s Index number
(c) Fisher Index number
(d) All of them.
Solution:
(c) Fisher Index number

Question 16.
While computing a weighted index, the current period quantities are used in the:
(a) Laspeyre’s method
(b) Paasche’s method
(c) Marshall Edgeworth method
(d) Fisher’s ideal method
Solution:
(b) Paasche’s method

Question 17.
The quantities that can be numerically measured can be plotted on a
(a) p – chart
(b) c – chart
(c) x bar chart
(d) np – chart
Solution:
(c) x bar chart

Question 18.
How many causes of variation will affect the quality of a product?
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c) 2

Question 19.
Variations due to natural disorder is known as
(a) random cause
(b) non-random cause
(c) human cause
(d) all of them
Solution:
(a) random cause

Question 20.
The assignable causes can occur due to
(a) poor raw materials
(b) unskilled labour
(c) faulty machines
(d) all of them
Solution:
(d) all of them

Question 21.
A typical control charts consists of
(a) CL, UCL
(b) CL, LCL
(c) CL, LCL, UCL
(d) UCL, LCL
Solution:
(c) CL, LCL, UCL

Question 22.
\(\bar { x}\) chart is a
(a) attribute control chart
(b) variable control chart
(c) neither Attribute nor variable control chart
(d) both Attribute and variable control chart
Solution:
(b) variable control chart

Question 23.
R is calculated using
(a) x_max – x_min
(b) x_min – x_max
(c) \(\bar { x}\)_max – \(\bar { x}\)_min
(d) \(\bar {\bar x}\)_max – \(\bar {\bar x}\)_min
Solution:
(a) x_max – x_min

Question 24.
The upper control limit for x chart is given by
(a) \(\bar { x}\) + A₂\(\bar { R}\)
(b) \(\bar {\bar x}\) + A₂R
(c) \(\bar {\bar x}\) + A₂\(\bar { R}\)
(d) \(\bar { x}\) + A₂\(\bar {\bar R}\)
Solution:
(c) \(\bar {\bar x}\) + A₂\(\bar { R}\)

Question 25.
The LCL for R chart is given by
(a) D₂\(\bar { x}\)
(b) D₂\(\bar {\bar R}\)
(c) D₃\(\bar {\bar R}\)
(d) D₃\(\bar { x}\)
Solution:
(d) D₃\(\bar { x}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.3

Question 1.
Define Statistical Quality Control.
Solution:
The term Quality means a level or standard of a product which depends Material, Manpower, Machines, and Management (4M’s). Quality Control ensures the quality specifications all along them from the arrival of raw materials through each of their processing to the final delivery of goods. Quality Control is a powerful technique used to diagnose the lack of quality in any of the raw materials, processes, machines etc. It is essential that the end products should possess the qualities that the consumer expects from the manufacturer.

Question 2.
Mention the types of causes for variation in a production process.
Solution:
There are two causes of variation which affects the quality of a product, namely
1. Chance Causes (or) Random causes
2. Assignable Causes

Question 3.
Define Chance Cause.
Solution:
These are small variations which are natural and inherent in the manufacturing process. The variation occurring due to these causes is beyond the human control and cannot be prevented or eliminated under any circumstances, the minor causes which do not affect the quality of the products to an extent are called as Chance Causes (or) Random causes. For example Rain, floods, power cuts, etc.,

Question 4.
Define Assignable cause.
Solution:
The second type of variation which is present in any production process is due to non-random causes. The assignable causes may occur in at any stage of the process, right from the arrival of the raw materials to the final delivery of the product. Some of the important factors of assignable causes are defective raw materials, fault in machines, unskilled manpower, worn out tools, new operation, etc.

Question 5.
What do you mean by product control?
Solution:
Product Control means that controlling the quality of the product by critical examination through sampling inspection plans. Product Control aims at a certain quality level to be guaranteed to the customers. It attempts to ensure that the product sold does not contain a large number of defective items. Thus it is concerned with classification of raw materials, semi-finished goods or finished goods into acceptable on rejectable products.

Question 6.
What do you mean by process control?
Solution:
In Process Control the proportion of defective items in the production process is to be minimized and it is achieved through the technique of control charts.

Question 7.
Define a control chart.
Solution:
A control chart is essentially a graphic device for presenting data so as to directly reveal the frequency and extent of variations from established standards or goals. Control charts are simple to construct and easy to interpret and they tell the manager at a glance whether or not the process is in control, i.e within the tolerance limits.

Question 8.
Name the control charts for variables.
Solution:
(i) Charts for mean (\(\bar { X}\))
(ii) Charts for Range (R)

Question 9.
Define mean chart.
Solution:
The mean chart (\(\bar { X}\) chart) is used to show, the quality average of the samples taken from the given process. The \(\bar { X}\) charts are usually required for decision making to accept or reject the process.
Procedure for \(\bar { X}\)
i. Let X₁, X₂, X₃, etc. be the samples selected each containing “n” observations usually (n = 4.5 or 6)

ii. Calculate mean for each samples \(\bar { X}\)₁, \(\bar { X}\)₂, \(\bar { X}\)₃, ……… by using \(\bar { X}\)_i = \(\frac { ΣX_i }{n}\), i = 1, 2, 3, 4, ….
where Σx_i = total f “n” values included in the sample X₁.

iii. Find the mean (\(\bar { \bar X}\)) of the sample means
\(\bar { \bar X}\) = \(\frac { Σ \bar X }{number of samples}\)
where Σ\(\bar { X}\) = total of all the sample means.

Question 10.
Define R chart.
Solution:
The R chart is used to show the variability or dispersion of the samples taken from the given process. R charts are also required for decision making to accept of reject the process.
Procedure for R-Charts
Calculate R = Xmax – Xmin
Let R₁, R₂, R₃ ………….. be the ranges of the “n” samples. The average range is given by
\(\bar { X}\) = \(\frac { ΣR }{n}\)

Question 11.
What are the uses of statistical quality control?
Solution:
(i) The role of statistical quality control is to collect and analyse relevant data for the purpose of detecting whether the process is under control or not.
(ii) The value of quality control lies in the fact that assignable causes in a process can be quickly detected. Infact the variations are often discovered before the product becomes defective.
(iii) Statistical quality control is only diagnostic. It tells us whether the standard is being maintained or not.
(iv) This technique is used in almost all production industries such as automobile textile, electrical equipment, biscuits, both soaps, chemicals, Petroleum products, etc.
(v) The purpose for which SQC are used in two fold namely (a) process control (b) product control.
The main purpose of SQC is to device statistical techniques which would help in elimination of assignable causes and bring the production process under control.

Question 12.
Write the control limits for the mean chart.
Solution:
The calculation of control limits for x chart in two different cases is

Question 13.
Write the control limits for the R chart.
Solution:
The calculation of control limits for R chart in two different cases are

The values of A₂, D₂ and D₄ are given in the table.

Question 14.
A machine is set to deliver packets of a given j weight. Ten samples of size five each were recorded. Below are given relevant data:

Solution:
Calculate the control limits for mean chart and the range chart and then comment on the state of control, (conversion factors for n = 5, A₂ = 0.58, D₃ = 0 and D₄ = 2.115)

UCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 16.2 + (0.58)(7.4)
= 16.2 + 4.292
= 20.492
= 20.49
CL = \(\bar { \bar X}\) = 16.2
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 16.2 – (0.58) (7.4)
= 16.2 – 4.292
= 11.908
= 11.91
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= (2.115)(7.4)
= 15.651
= 15.65
CL = \(\bar { R}\) = 7.4
LCL = D₃ \(\bar { R}\) = (0)(7.4) = 0

Question 15.
Ten samples each of size five are difawn at regular intervals from a manufacturing process. The sample means (X) and their ranges (R) are given below:

Calculate the control limits in respect of \(\bar { X}\) chart. (Given A₂ = 0.58, D₃ = 0 and D₄ = 2.115 ) Comment on the state of control.
Solution:

UCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 46.2 + (0.58)(6.8)
= 46.2 + 3.944
= 50.144
= 50.14
CL = \(\bar { \bar X}\) = 46.2
LCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 46.2 – (0.58)(6.8)
= 46.2 – 3.944
= 42.256
= 42.26
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= (2.115)(6.8)
= 14.382
= 14.38
CL = \(\bar { R}\) = 6.8
LCL = D₃\(\bar { R}\) = (0)(6.8) = 0

Question 16.
Construct X and R charts for the following data:

Solution:


\(\bar { R}\) = \(\frac { 144 }{8}\) = 18
UCL = \(\bar { \bar x}\) + A₂ \(\bar { R}\)
= 37.71 + (0.58)(18)
= 37.71 + 10.44
= 48.15
CL = \(\bar { \bar x}\) = 37.71
LCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 37.71 – (0.58)(18)
= 37.71 – 10.44
= 27.27
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = 2.115 (18) = 38.07
CL = \(\bar { R}\) = 18
LCL = D₃ \(\bar { R}\) = 0(18) = 0

Question 17.
The following data show the values of sample mean (\(\bar {x}\)) and its range (R) for the samples of size five each. Calculate the values for control limits for mean, range chart and determine whether the process is in control.

Solution:

UCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 10.66 + (0.58)(6.3)
= 10.66 + 3.654 = 14.314
= 14.31
CL = \(\bar { \bar x}\) = 10.66
LCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 10.66 – (0.58)(6.3)
= 10.66 – 3.654
= 7.006
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = 2.115 (6.3)
= 13.3245
= 13.32
CL = \(\bar { R}\) = 6.3
LCL = D₃ \(\bar { R}\) = 0(6.3) = 0
Conclusion: Since all the points of sample range is within UCL of R chart, the process is in control.

Question 18.
A quality control inspector has taken ten ” samples of size four packets each from a potato chips company. The contents of the sample are given below, Calculate the control limits for mean and range chart.

(Given for n = 5, A₂ = 0.58, D₃ = 0 and D₄ = 2.115)
Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 12.5 + (0.58)(0.37)
= 12.5 + 0.2146 = 12.7146
= 12.71
CL = \(\bar { \bar X}\) = 12.5
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\) = 12.5 – (0.58) (0.37)
= 12.5 – 0.2146 = 12.2854
= 12.29
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = (2.115)(0.37) = 0.78255
= 0.78
CL = \(\bar { R}\) = 0.37
LCL = D₃\(\bar { R}\) = (0)(0.37) = 0

Question 19.
The following data show the values of sample means and the ranges for ten samples of size 4 each. Construct the control chart for mean and range chart and determine whether the process is in control.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 30.1 +(0.73)(20.1)
= 30.1 + 14.673 = 44.773
= 44.77
CL = \(\bar { \bar X}\) = 30.1
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 30.1 – (0.73) (20.1)
= 30.1 – 14.673 = 15.427
= 15.43
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= 2.28(20.1) = 45.828
= 45.83
CL = \(\bar { R}\) = 20.1
LCL = D₃ \(\bar { R}\) = 0(20.1) = 0

Question 20.
In a production process, eight samples of size 4 are collected and their means and ranges are given below. Construct mean chart and range chart with control limits.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 13.25 + (0.73) (3.12)
= 13.25 + 2.2776 = 15.5276
= 15.53
CL = \(\bar { \bar X}\) = 13.25
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 13.25 – (0.73)(3.12)
= 13.25 – 2.2776 = 10.972
= 10.97
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= 2.28(3.12) = 7.11984
= 7.12
CL = \(\bar { R}\) = 3.12
LCL = D₃\(\bar { R}\) = 0(3.12) = 0

Question 21.
In a certain bottling industry the quality control inspector recorded the weight of each of the 5 bottles selected at random during each hour of four hours in the morning.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 41 + (0.58)(4)
41 + 2.32 = 43.32
CL = \(\bar { \bar X}\) = 41
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 41 – (0.58)(4)
= 41 – 2.32
= 38.68
The control limits for range chart is
UCL = D₄ \(\bar { R}\) = 2.115(4)
= 8.46
CL = \(\bar { R}\) = 4
LCL = D₂ \(\bar { R}\) = 0(4) = 0
Conclusion: Since all the points of sample mean and Range are within the control limits, the process is in control.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.2

Question 1.
Define Index Number.
Solution:
Index Numbers are the indicators which reflect the changes over a specified period of time in price of different commodities, production, sales, cost of living etc.

“An Index Number is a device which shows by its variations the change in a magnitude which is not capable of accurate measurements in it-self or of direct valuation in practice”. – Whel-don

“An Index number is a statistical measure of fluctuations in a variable arranged in the form of a series and using a base. Period for maxing H comparisons” – Lawrence J Kalpan.

Question 2.
State the uses of Index Number.
Solution:
(1) Index number is an important tool for formulating decision and management p policies.
(2) It helps in studying the trends and tendencies.
(3) It determines the inflation and deflation in an Economy.

Question 3.
Mention the classification of Index Number.
Solution:
Index number can be classified as follows,
(i) Price Index Number:
It measures the general changes in the retail or wholesale price level of a particular or group of commodities.

(ii) Quantity Index Number:
These are indices of measure the changes in the quantity of goods manufactured in a factory.

(iii) Cost of living Index Number:
These are intended to study the effect of change in the price level on the cost of living of different classes of people.

Question 4.
Define Laspeyre’s Price index number.
Solution:
Laspeyre’s Price index number
P\(_{ 01 }^{L}\) ⥪ \(\frac { Σp_1q_0 }{Σp_0q_0}\) × 100
where P₁ = Current year price
p₀ = base year price
q₀ = base year quantity

Question 5.
Explain Paasche’s price index number.
Solution:
Paasches price index number
P\(_{ 01 }^{L}\) = \(\frac { Σp_1q_1 }{Σp_0q_1}\) × 100
where P₁ = current year price
q₁ = current year quantity
p₀ = base year price
q₀ = base year quantity

Question 6.
Write note on Fisher’s price index number.
Solution:
Fishers price index number
P^F = \(\sqrt { P^L×P^P}\)
P\(_{ 01 }^{F}\) = \(\sqrt { \frac{Σp_1p_0×Σp_1q_1}{Σp_0q_0×Σp_0q_1}}\) × 100
where P₁ = current year price
q₁ = current year quantity
p₀ = base year price
q₀ = base year quantity

Question 7.
State the test of adequacy of index number.
Solution:
There are two tests which are used to test the adequacy for an index number. The two tests are as follows
(i) Time Reversal Test
(ii) Factory Reversal Test

Question 8.
Define Time Reversal Test.
Solution:
It is an important test for testing the consistency of a good index number. This test maintains time consistency by working both forward and backward with respect to time (here time refers to base year and current year). Symbolically the following relationship should be satisfied. P₀₁ × p₁₀ = 1

Fisher’s index number formula satisfies the above relationship
P\(_{ 01 }^{F}\) = \(\sqrt { \frac{Σp_1p_0×Σp_1q_1}{Σp_0q_0×Σp_0q_1}}\)

when the base year and current year are interchanged, we get
P\(_{ 10 }^{F}\) = \(\sqrt { \frac{Σp_0q_1×Σp_0q_0}{Σp_1q_1×Σp_1q_0}}\)
P\(_{ 01 }^{F}\) × P\(_{ 10 }^{F}\) = 1

Question 9.
Explain factor reversal test.
Solution:
This is another test for testing the consistency of a good index number. The product of price index number and quantity index number from the base year to the current year should be equal to the true value ratio. That is, the ratio between the total value of current period and total
value of the base period is known as true value ratio. Factor Reversal Test is given by,

where P₀₁ is the relative change in price
Q₀₁ is the relative change in quantity.

Question 10.
Define true value ratio.
Solution:
\(\frac { Σp_1q_1 }{Σp_0q_0}\) is the ratio of the total value in the current period to the total value in the base period and this ratio is called the true value ratio.

Question 11.
Discuss about cost of Living Index Number.
Solution:
Cost of living index numbers are generally designed to represent the average change over time in the prices paid by the ultimate consumer for a specified quantity of goods and services cost of living index number is also known as consumer price index number.

It is well known that a given change in the level of prices (retail) affects the cost of living of different classes of people in different manners. The general index number fails to reveal this. Therefore it is essential to construct a cost of living index number which helps us in determining the effect of rise and fall in prices on different classes of consumers living in different areas.

Question 12.
Define family budget method.
Solution:
In this method, the weights are calculated by multiplying prices and quantity of the base year.
(i.e.) V = Σp₀q₀. The formula is given by,
Cost of Living Index Number = \(\frac { Σpv }{Σv}\)
where P = \(\frac { p_1 }{p_0}\) × 100 is the price relative.
v = Σp₀q₀ is the value relative.

Question 13.
State the uses of cost of Living Index Number.
Solution:
(i) It indicates whether the real wages of workers are rising or falling for a given time.
(ii) It is used by the administrators for regulating dearness allowance or grant of bonus to the workers.

Question 14.
Calculate by a suitable method, the index number of price from the following data:

Solution:

Question 15.
Calculate price index number for 2005 by (a) Laspeyre’s (b) Paasche’s method

Solution:

Question 16.
Calculate price index number for 2005 by (a) Laspeyre’s (b) Paasche’s method

Solution:

Question 17.
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Question 18.
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Question 19.
Calculate Fisher’s index number to the following data. Also show that it satisfies Time Reversal Test.

Solution:

Question 20.
Th following are the group index numbers and the group weights of an average working class family’s budget. Construct the cost of living index number:

Solution:

Question 21.
Construct the cost of living Index number for 2015 on the basis of 2012 from the following data using family budget method.

Solution:

Question 22.
Calculate the cost of living index by aggregate expenditure method:

Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.1

Question 1.
Define Time series.
Solution:
A Time Series consists of data arranged chronologically – Croxton & Cowden
When quantitative data are arranged in the order of their occurrence, the resulting series is called the Time Series – Wessel & Wallet.

A time series consists of a set of observations arranged in chronological order (either ascending or descending). Times Series has an important objective to identify the variations and try to eliminate the variations and also helps us to estimate or predict the future values.

Question 2.
What is the need for studying time series?
Solution:
Time series analysis is one of the statistical methods used to determine the patterns in data collected for a period of time. Generally, each of us should know about the past data to observe and understand the changes that have taken place in the past and current time. One can also identify the regular or irregular occurrence of any specific feature over a time period in a time series data.

Question 3.
State the uses of time series.
Solution:
It helps in the analysis of the past behavior.
It helps in forecasting and for future plans.
It helps in the evaluation of current achievements.
It helps in making comparative studies between one time period and others.
Therefore time series helps us to study and analyze the time related data which involves in business fields, economics, industries, etc.

Question 4.
Mention the components of the time series.
Solution:
There are four types of components in a time series. They are as follows
(i) Secular Trend
(ii) Seasonal variations
(iii) Cyclic variations
(iv) Irregular variations

Question 5.
Define secular trend.
Solution:
It is a general tendency of time series to increase or decrease or stagnates during a long period of time, an upward tendency is usually observed in population of a country, production, sales, prices in industries,income or individuals etc., A downward tendency is observed in deaths, epidemics, prices of electronic gadgets, water sources, mortality rate etc. It is not necessarily that the increase or decrease should be in the same direction throughout the given period of time.

Question 6.
Write a brief note on seasonal variations
Solution:
As the name suggests, tendency movements are due to nature which repeat themselves periodically in every seasons. These variations repeat themselves in less than one year time. It is measured in an interval of time. Seasonal variations may be influenced by natural force, social customs and traditions. These variations are the results of such factors which uniformly and regularly rise and fall in the magnitude. For example, selling of umbrellas’ and raincoat in the rainy season, sales of cool drinks in summer season, crackers in Deepawali season, purchase of dresses in a festival season, sugarcane in Pongal season.

Question 7.
Explain cyclic variations.
Solution:
These variations are not necessarily uniformly periodic in nature. That is, they may or may not follow exactly similar patterns after equal intervals of time. Generally one cyclic period ranges from 7 to 9 years and there is no hard and fast rule in the fixation of year for a cyclic period. For example, every business cycle has a Start-Boom – Depression. Recover, maintenance during booms and depressions, changes in government monetary policies, changes in interest rates.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
These variations do not have particular pattern and there is no regular period of time of their occurrences. These are accidently changes which are purely random or unpredictable. Normally they are short-term variations, but its occurrence sometimes has its effect so intense that they may give rise to new cyclic or other movements of variations. For example floods, wars, earthquakes, Tsunami, strikes, lockouts etc.

Question 9.
Define seasonal index.
Solution:
Seasonal Index for every season (monthly or quarterly) is calculated as follows
Seasonal Index (S.I) = \(\frac { Seasonal Average }{Grand Average}\) × 100
If the data is given monthwise
Seasonal Index = \(\frac { Monthly Average }{Grand Average}\) × 100
If quarterly data is given
Seasonal Index = \(\frac { Quarterly Average }{Grand Average}\) × 100

Question 10.
Explain the method of fitting a straight line.
Solution:
(i) The straight line trend is represented by the equation Y = a + bX …………. (1)
Where Y is the actual value, X is time, a, b are constants.
(ii) The constants ‘a and ‘b’ are estimated by solving the following two normal Equations
ΣY = n a + b ΣX ………… (2)
ΣXY = a ΣX + b ΣX² …………… (3)
Where ‘n’ = number of years given in the data.
(iii) By taking the mid-point of the time as the ori-gin, we get ΣX = 0
(iv) When ΣX = 0, the two normal equations reduces to
ΣY = na + b(0) ; a = \(\frac { ΣY }{n}\) = \(\bar { Y}\)
ΣXY = a(0) + bΣX² ; b = \(\frac { ΣXY }{ΣX^2}\)
The constant ‘a’ gives the mean of Y and ‘b gives the rate of change (slope),
(v) By substituting the values of ‘a and ‘b’ in the trend equation (1), we get the Line of Best Fit.

Question 11.
State the two normal equations used in fitting a straight line.
Solution:
The constants ‘a’ and ‘b’ are estimated by solving the following two normal Equations
ΣY = n a + b ΣX ………… (2)
ΣXY = a ΣX + b ΣX² …………. (3)
Where ‘n’ = number of years given in the data.

Question 12.
State the different methods of measuring trend.
Solution:
Following are the methods by which we can measure the trend.
(i) Freehand or Graphic Method
(ii) Method of Semi-Averages
(iii) Method of Moving Averages

Question 13.
Compute the average seasonal movement for the following series

Solution:
Computation of seasonal. Index by the method of simple averages.

Question 14.
The following figures relates to the profits of a commercial concern for 8 years

Find the trend of profits by the method of three yearly moving averages.
Solution:

Question 15.
Find the trend of production by the method of a five-yearly period of moving average for the following data:

Solution:
I Computation of five – yearly moving averages

Question 16.
The following table gives the number of small- scale units registered with the Directorate of Industries between 1985 and 1991. Show the growth on a trend line by the free hand method.

Solution:

Question 17.
The annual production of a commodity is given as follows:

Fit a straight line trend by the method of least squares
Solution:

Therefore the required equation of the straight line
Y = a + bx
Y = 169.428 + 3.285 X
⇒ Y = 169.428 + 3.285 (x – 1998)

Question 18.
Determine the equation of a straight line which best fits the following data

Compute the trend values for all years from 2000 to 2004
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 54 + 5.4X
Y = 54 + 5.4 (x – 2002)
The trend value can be obtained by when x = 2000
Yt = 54 + 5.4 (2000 – 2002)
Y = 54+ 5.4 (-2)
= 54 – 10.8
= 43.2
When x = 2001
Yt = 54 + 5.4 (2001 – 2002)
Y = 54 + 5.4 (-1)
= 54 – 5.4
= 48.6
When x = 2002
Yt = 54 + 5.4 (2002 – 2002)
V = 54 + 5.4 (0)
= 54
When x = 2003
Yt = 54 + 5.4 (2003 – 2002)
Y = 54 + 5.4 (1)
= 54 + 5.4
= 59.4
When x = 2004
Yt = 54 + 5.4 (2004 – 2002)
Y = 54 + 5.4(2)
= 54 + 10.8
= 64.8

Question 19.
The sales of a commodity in tones varied from January 2010 to December 2010 as follows: in year 2010 Sales (in tones)

Fit a trend line by the method of semi-average.
Solution:
Since the number of years is even (twelve). We can equally divide the given data it two equal parts and obtain averages of first six months and last six.

Question 20.
Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2002, 2003 and 2004.

Solution:

Question 21.
Use the method of monthly averages to find T the monthly indices for the following data of production of a commodity for the years 2002, 2003 and 2004.

Solution:
Computation of Seasonal Index by the method of simple averages

Question 22.
The following table shows the number of salesmen working for a certain concern:

Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 48.8 + 2X
(i.e) Y = 48.8 + 2 (x – 1994)
When x = 1992
Yt = 48.8 + 2 (1992 – 1994)
Y = 48.8 + 2 (-2)
= 48.8 – 4
= 44.8
When x = 1993
Yt = 48.8 + 2 (1993 – 1994)
Y = 48.8 + 2 (-1)
= 48.8 – 2
= 46.8
When x = 1994
Yt = 48.8 + 2 (1994 – 1994)
Y = 48.8 + 2(0)
= 48.8
When x = 1995
Yt = 48.8 + 2 (1995 – 1994)
Y = 48.8 + 2(1)
= 50.8
When x = 1996
Yt = 48.8 + 2 (1996 – 1994)
Y = 48.8 + 2 (2)
= 48.8 + 4
= 52.8
When x = 1997
Yt = 48.8 + 2 (1997 – 1994)
Y = 48.8 + 2 (3)
= 48.8 + 6
= 54.8

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

Question 1.
Explain the types of sampling.
Solution:
There are two types of sampling. They are
(1) Non-Random sampling or Non-probability sampling
(2) Random sampling or probability sampling Random sampling refers to selection of samples from the population in a random manner. A random sample is one where each and every item in the population has an equal chance of being selected.

“Every member of a parent population has had equal chances of being included “Dr. Yates.

“A random sample is a sample selected in such a way that every item in the population has an equal chance of being included”-Harver.

The different types of probability sampling are
(1) sampling random sampling
(2) stratified random sampling
(3) systematic sampling

Question 2.
Write short note on sampling distribution and standard error.
Solution:
sampling distribution:
Sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.

For instance if we draw a sample a size n from a given finite population of N, then the total number of possible samples is
Nc_n = \(\frac { N! }{n!(N-n)!}\) = k(say)

Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its standard Error abbreviated as S.E. The standard Error (S.E) of some of the well-known statistics, for large samples, are given below, where n is the samples size, σ² is the population variance.

Question 3.
Explain the procedures of testing of hypothesis
Solution:
The following are the steps involved in hypothesis testing problems:
1. Null hypothesis: Set up the null hypothesis H₀

2. Alternative hypothesis: Set up the alternative hypothesis. This will enable us to decide whether we have to use two tailed test or single tailed test (right or left tailed)

3. Level of significance: Choose the appropriate level of significant (a) depending on the reliability of the estimates and permissible risk. This is to be fixed before sample is drawn, i.e., a is fixed in advance.

4. Test statistic : Compute the test statistic
Z = \(\frac { t-E(t) }{\sqrt{var(t)}}\) = \(\frac { t-E(t) }{S.E(t)}\) N(0, 1) as n → ∞

5. Conclusion: We compare the computed value of Z in step 4 with the significant value or critical value or table value Zα at the given level of significance.
(i) If |Z | < Zα i.e., if the calculated value of is less than critical value we say it is not significant. This may due to fluctuations of sampling and sample data do not provide us sufficient evidence against the null hypothesis which may therefore be accepted.

(ii) If |Z |> Zα i.e., if the calculated value of Z is greater than critical value Zα then we say it is significant and the null hypothesis is rejected at level of significance α.

Question 4.
Explain in detail about the test of significance for single mean.
Solution:
Let xi , (i = 1, 2, 3, …, n) is a random sample of size from a normal population with mean µ and variance σ² then the sample mean is distributed normally with mean and variance
\(\frac { σ^2 }{n}\), i.e \(\bar { x }\) N(µ, \(\frac { σ^2 }{n}\))

Thus for large samples, the standard normal variate corresponding to \(\bar { x }\) is
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\) N (0, 1)

Under the null hypothesis that the sample has been drawn from a population with mean and variance σ², i.e., there is no significant difference between the sample mean (\(\bar { x }\)) and the population mean (α), the test statistic (for large samples) is:
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\)

Question 5.
Determine the standard error of proportion for a random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad?
Solution:
sample size n = 500
No. of bad pine apples = 65
sample proportion = P = \(\frac { 65 }{500}\) = 0.13
Q = 1 – p ⇒ Q = 1 – 0.13
∴ Q = 0.87
The S.E for sample proportion is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.13)(0.87) }{500}}\)
= \(\sqrt{\frac { 0.1131 }{500}}\) = \(\sqrt{0.0002262}\)
= 0.01504
∴ S.E = 0.015
Hence the standard error for sample proportion is S.E = 0.015

Question 6.
A sample of 100 students are drawn from a school The mean weight and variance of the sample are 67.45 kg and 9 kg respectively find (a) 95% and (b) 66% confidence intervals for estimating the mean weight of the students.
Solution:
sample size n = 100
The sample mean = \(\bar { x }\) = 67.45
The sample variance S² = 9
The sample standard deviation S = 3
S.E = \(\frac { S }{√n}\) = \(\frac { 3 }{\sqrt{100}}\) = \(\frac { 3 }{10}\) = 0.3

(a) The 95% confidence limits for µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E < µ < \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (1.96 × 0.3) ≤ µ ≤ 67.45 + (1.96 × 0.3) 67.45 – 0.588 ≤ µ ≤ 67.45 + 0.588
66.862 ≤ µ ≤ 68.038
The confidential limits is (66.86, 68.04)

(b) The 99% confidence limits for estimating µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E ≤ µ ≤ \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (2.58 × 0.3) ≤ µ ≤ 67.45 + (2.58 × 0.3)
67.45 – 0.774 ≤ µ ≤ 67.45 + 0.774
66.676 ≤ µ ≤ 68.224
∴ The 99% confidence limits is (66.68, 68.22)

Question 7.
The mean I.Q of a sample of 1600 children was 99. it is likely that this was a random sample from a population with mean I.Q 100 and standard deviation 15? (Test at 5% level of significance)
Solution:
sample size n = 1600
\(\bar { x }\) = 99
sample mean
Population mean µ = 100
population S.D σ = 15
under the Null hypothesis H₀ : µ = 100
Alternative hypothesis H₁ : µ = 100 (two tails)
Level of significance µ = 0.05

z = -2.666
z = -2.67
Calculated value |z| = 2.67
critical value at 5% level of significance is
z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is greater than table value i.e z ⇒ z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is rejected. Therefore we conclude that the sample mean differs, significantly from the population mean.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3

Choose the correct Answer:

Question 1.
A ……………… may be finite or infinite according as the number of observation or items in it is finite or infinite
(a) Population
(b) census
(c) parameter
(d) none of these
Solution:
(a) population

Question 2.
A …………….. of statistical individuals in a population is called a sample.
(a) Infinite set
(b) finite subset
(c) finite set
(d) entire set
Solution:
(b) finite subset

Question 3.
A finite subset of statistical individuals in a population is called ………………
a) a sample
(b) a population
(c) universe
(d) census
Solution:
(a) a sample

Question 4.
Any statistical measure computed from sample data is known as ……………..
(a) Parameter
(b) random sample
(c) Infinite measure
(d) uncountable
Solution:
(b) random sample

Question 5.
A ………………. is one where each item in the universe has an equal chance of known opportu¬nity of being selected
(a) Parameter
(b) random sample
(c) statistic
(d) entire data
Solution:
(b) random sample

Question 6.
A random sample is a sample selected in such a way that every item in the population has an equal chance of being included
(a) Harper
(b) fisher
(c) karl pearson
(d) Dr. yates
Solution:
(a) Harper

Question 7.
Which one of the following is probability sampling
(a) Purposive sampling
(b) judgement sampling
(c) sample random sampling
(d) Convenience sampling
Solution:
(c) sample random sampling

Question 8.
In simple random sampling of drawing any unit, the probability of drawing any unit at the draw is ?
(a) \(\frac { n }{N}\)
(b) \(\frac { 1 }{N}\)
(c) \(\frac { N }{n}\)
(d) n
Solution:
(b) \(\frac { 1 }{N}\)

Question 9.
In ……………. the heterogeneous groups divided into homogeneous groups
(a) Non-probability sample
(b) a sample random sample
(c) a stratified random sample
(d) Systematic sample
Solution:
(c) a stratified random sample

Question 10.
Errors in sampling are of
(a) Two types
(b) three types
(c) four types
(d) five types
Solution:
(a) Two types

Question 11.
The method of obtaining the most likely value of the population parameter using statistic is called
(a) estimate
(b) estimate
(c) biased estimate
(d) standard error
Solution:
(d) standard error

Question 12.
An estimator is a sample statistic used to estimate a
(a) population parameter
(b) biased estimate
(c) sample size
(d) census
Solution:
(a) population parameter

Question 13.
…………… is a relative property, which states that one estimate is efficient relative to another.
(a) efficiency
(b) sufficiency
(c) unbiased
(d) consistency.
Solution:
(a) efficiency

Question 14.
If probability p[|\(\bar { θ }\) – θ|< ∈|< ∈|] 1 → µ as n → α for any positive then \(\bar { θ }\) is said to estimator of θ
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(d) Consistent

Question 15.
An estimator is said to be ………….. if it contains all the information in the data about the parameter it estimates.
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(b) sufficient

Question 16.
An estimate of a population parameter given by two numbers between which the parameter would be expected to lie called an ………….. interval estimate of the parameter
(a) point estimate
(b) interval estimate
(c) standard error
(d) confidence
Solution:
(b) interval estimate

Question 17.
A ……………… is a statement or an assertion about the population parameter
(a) hypothesis
(b) statistic
(c) sample
(d) census
Solution:
(a) hypothesis

Question 18.
Type I error is
(a) Accept H₀ when it is true
(b) Accept H₀ when it is false
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(c) Reject H₀ when it is true

Question 19.
Type II error is?
(a) Accept H₀ when it is wrong
(b) Accept H₀ when it is when it is true
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(a) Accept H₀ when it is wrong

Question 20.
The standard error of sample mean is?
(a) \(\frac { σ }{\sqrt{2n}}\)
(b) \(\frac { σ }{n}\)
(c) \(\frac { σ }{√n}\)
(d) \(\frac { σ^2 }{√n}\)
Solution:
(c) \(\frac { σ }{√n}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2

Question 1.
Mention two branches of statistical inference?
Solution:
(i) Estimation (ii) Testing of Hypothesis

Question 2.
What is an estimator?
Solution:
Any sample statistic which is used to estimate an unknown population parameter is called an estimator ie., an estimator is a sample statistic used to estimate a population parameter.

Question 3.
What is an estimate?
Solution:
When we observe a specific numerical value of our estimator, we call that value is an estimate. In other words, an estimate is a specific value of a statistic.

Question 4.
What is point estimation?
Solution:
When a single value as an estimate, the estimate is called a point estimate of the population parameter. In other words, an estimate of a population parameter given by a single number is called as point estimation.

Question 5.
What is interval estimation?
Solution:
Generally, there are situation where point estimation is not desirable and we are interested in finding limits within which the parameter would be expected to lie is called an interval estimation.

Question 6.
What is confidence interval?
Solution:
Let us choose a small value of a which is known as level of significance (1% or 5%) and determine two constants says c₁ and c₂ such that p(c₁ < θ < c₂|t) = 1 – α

The quantities c₁ and c₂ so determined are known as the confidence Limits and the interval [c₁, c₂] with in which the unknown value of the population parameter is expected to lie is known as confidence interval.(1 – α)is called as confidence coefficient.

Question 7.
What is null hypothesis? Give an example.
Solution:
According to prof R. A. fisher, “Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true”, and it is denoted by H₀.

For example: If we want to find the population mean has a specified value µ₀, then the null hypothesis H₀ is set as follows H₀ : µ = µ₀

Question 8.
Define alternative hypothesis.
Solution:
Any hypothesis which is complementry to the null hypothesis is called as the alternative hypothesis and is usually denoted by H₁.

For example: If we want to test the null hypothesis that the population has specified mean µ i.e., H₀ : µ = µ₀ then the alternative hypothesis could be any one among the following:
(i) H₁ : µ ≠ µ₀ (µ > or µ < µ₀)
(ii) H₁ : µ > µ₀
(iii) H₁ : µ < µ₀

Question 9.
Define critical region.
Solution:
A region corresponding to a test statistic in the sample space which tends to rejection of H₀ is called critical region or region of rejection.

Question 10.
Define critical value.
Solution:
The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depend upon.
(i) The level of significance
(ii) The alternative hypothesis whether it is two-tailed or single tailed

Question 11.
Define level of significance.
Solution:
The probability of type 1 error is known as level. of significance and it is denoted by The level of significance is usually employed in testing of hypothesis are 5% and 1%. The level of significance is always fixed in advanced before collecting the sample information.

Question 12.
What is type I error
Solution:
There is every chance that a decision regarding a null hypothesis may be correct or may not be correct. The error of rejecting H₀ when it is true is called type I error.

Question 13.
What is single tailed test.
Solution:
When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as one-tailed test. Here H₁ : µ > µ₀ and H₁ : µ < µ₀ are known as one tailed alternative.

Question 14.
A sample of 100 items, draw from a universe with mean value 64 and S.D 3, has a mean value 63.5. Is the difference in the mean significant?
Solution:
sample size n = 100 ; sample mean \(\bar { x}\) = 63.5
sample SD S = 3;
population mean µ = 64 population SD σ = 3
Null Hypothesis H₀ : µ = 64 (the sample has been drawn from the population mean µ = 64 and SD σ = 3)
Alternative Hypothesis H₁ : µ ≠ 64 (two tail) i.e the sample has not been drawn from the population mean µ = 64 and SD σ = 3
The level of significance α = 5% = 0.05
Test statistic
z = \(\frac { 63.5-64}{\frac{3}{\sqrt{100}}}\) = \(\frac { -0.5 }{(\frac{3}{10})}\) = \(\frac { -0.5 }{0.3}\) = -1.667
|z| = 1.667
∴ calculated z = 1.667
critical value at 5% level of
significance is z_{\(\frac { α }{2}\)} = 1.96
Inference:
At 5% level of significance Z < Z_{\(\frac { α }{2}\)} since the calculated value is less than the table value the null hypothesis is accepted.

Question 15.
A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height of 67.39 inches and standard deviation 1.30 inches?
Solution:
sample size n = 400; sample mean \(\bar { x}\) = 67.47 inches
sample SD S = 1.30 inches population mean
µ = 67.39 inches
population SD σ = 1.30 inches
Null Hypothesis H₀ : µ = 67.39 inches (the sample has been drawn from the population mean µ = 67.39 inches; population SD σ = 1.30 inches)
Alternative Hypothesis H₁ = µ ≠ 67.39 inches(two tail)
i.e the sample has not been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches
The level of significance α = 5% = 0.05
Test static:

Thus the calculated and the significant value or Z_{\(\frac { α }{2}\)} = 1.96
table value comparing the calculated and table values
Z_{\(\frac { α }{2}\)} (i.e.,) 1.2308 < 1.96
Inference: since the calculated value is less than value i.e Z > Z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is accepted Hence we conclude that the data doesn’t provide us any evidence against the null hypothesis. Therefore, the sample has been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches.

Question 16.
The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Solution:
sample size n = 100
sample mean \(\bar { x}\) = 72
sample SD S = 8
population mean µ = 76
under the Null hypothesis H₀ : p = 76
Against the alternative hypothesis H₀ : µ ≠ 76 (two mail)
Level of significance µ = 0.05
Test statistic:

since alternative hypothesis is of two tailed test we can take |Z| = 5.
∴ critical value 5% level of significance is z > z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is less than table value i.e z > z_{\(\frac { α }{2}\)} at 5% level of significance the null hypothesis H0 is rejected Therefore, we conclude that there is significant difference between the sample mean and population mean µ = 76 and SD σ = 8.

Question 17.
The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance.
Solution:
Sample size n = 50
Sample mean \(\bar { x}\) = 1800
Sample SD S = 100
population mean µ = 1850
Null hypothesis H₀ : µ = 1850
Level of significance µ = 0.01

= -3.5355
∴ z = -3.536
calculated value |z| = 3.536
Critical value at 1% level of significance is z_{\(\frac { α }{2}\)} = 2.58
Inference:
Since the calculated value is greater than table (ie) z > z_{\(\frac { α }{2}\)} at 1% level of significance, the null hypothesis is rejected, therefore we conclude that we is rejected, Therefore we conclude that we can not support we conclude that we can support the claim of 0.01 of significance.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1

Question 1.
What is population?
Solution:
The group of individuals considered under study is called as population. The word population here refers not only to people but to all items that have been chosen for the study.

Question 2.
What is sample?
Solution:
A selection of a group of observation/individuals/population in such a way that is represents the population is called as sample.

Question 3.
What is statistic?
Solution:
Statistic: Any statistical measure computed from sample is known as statistic.

Question 4.
Define parameter.
Solution:
Parameter: The statistical constants of the population like mean (µ),variance (σ²) are referred as population parameters.

Question 5.
What is sampling distribution of a statistic?
Solution:
Sampling distribution of a statistic is the frequency distribution which is formed with various of a statistic computed from different samples of the same size drawn from the same population.

Question 6.
What is standard error?
Solution:
The standard deviation of the sampling distribution of a statistic is known as its Standard Error abbreviated as S.E.

Question 7.
Explain in detail about simple random sampling with a suitable example.
Solution:
In this technique the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. Simple random sampling may be done, with or without replacement of a samples selected. In a simple random sampling with replacement there is a possibility of selecting the same sample any number of times. So, simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units ,the probability of drawing any unit at the first draw is \(\frac { 1 }{N}\), the probability of drawing any unit in the second draw from among the available (N – 1) units is \(\frac { 1 }{(N-1)}\), and so on.

For example, if we want to select 10 students, out of 100 students, then we must write the names/roll number of all the 100 students on slips of the same size and mix them, then we make a blindfold selection of 10 students. This method is called unrestricted random sampling, because units are selected from the population without any restriction. This method is mostly used in lottery draws. If the population or universe is infinite, this method is inapplicable.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
In stratified random sampling, first divide the population into sub-populations, which are called strata. Then,the samples are selected from each of the strata through random techniques. The collection of all the samples from all strata gives the stratified random samples.

When the population is heterogeneous or different segments or groups with respect to the variable or characteristic under study, then stratified Random sampling methos is studied. First, the population is divided into homogeneous number of sun-groups of strata before the sample is drawn. A sample from each stratum at random. Following steps are involved for selecting random sample in a stratified random sampling method.

(a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.

(b) After the population is stratified,a sample of a specified size is drawn at random from each stratum using Lottery Method or table of random number method.

Question 9.
Explain in detail about systematic random sampling with example.
Solution:
In a systematic sampling, randomly select the first sample from the first k units. Then every k th members, starting with the first selected sample, is included in the sample.

Systematic sampling is a commonly used technique,if the complete and up-to-date list of the sampling units is available. We can arrange the items in numerical, alphabetical, selecting the first at random, the rest being automatically selected according to some pre-determined pattern. A systematic is formed by selecting every item from the population, where k refers to the sample interval. The sampling interval can be determined by divided the size of the population by the size of the sample to be chosen.

That is K = \(\frac { N }{n}\), where k is an integer.
k = sampling interval, size of the population, sample size
Procedure for selection of samples by systematic sampling method

(i) If we want to select a sample of 10 students from a class of 100 students,the sampling interval is Calculated as k = \(\frac { N }{n}\) = \(\frac { 100 }{10}\) = 10
Thus sampling interval = 10 denotes that for every 10 samples one sample

(ii) The first sample is selected from the first 10(sampling interval) samples through selection procedures.

(iii) If the selected first random sample is 5, then the rest of the samples are automatically selected by incriminating the value of the sampling interval 9k = 10. i.e, 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.

Question 10.
Explain in detail about sampling error.
Solution:
Sampling Error:
Error, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors, sampling errors are inherent in the method of sampling. They may arise accidentally without any bias or prejudice. Sampling Errors arise primarily due to the following reasons:
(a) Faulty selection of the sample instead of correct sample by defective sampling technique.
(b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
(c) In area surveys,while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as faulty demarcation of sampling units.

Question 11.
Explain in detail about non-sampling error.
Solution:
Non-sampling Errors:
The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments (tape, scale) are called Non¬sampling errors. It may arise in the following ways:
(a) Due to neglience and carelessness of the part of either investigator or respondents
(b) Due to lack of trained and qualified investigators.
(c) Due to framing of a wrong questionnaire.
(d) Due to apply wrong statistical measure.
(e) Due to incomplete investigation and sample survey.

Question 12.
State any two merits of simple random sampling.
Solution:
Merits:
1. Personal bias is completely eliminated.
2. This method is economical as it saves time, money and labour.
3. The method requires minimum knowledge about the population in advance.

Question 13.
State any three merits of stratified random sampling.
Solution:
Merits:
(a) A random stratified sample is superior to a sample random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
(b) A stratified random sample can be kept small in size without losing its accuracy
(c) it is easy to administer, if the population under study is sub divided
(d) It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had the geographical areas.

Question 14.
State any two demerits of systematic random sampling.
Solution:
Demerits:
1. Systematic samples are not random samples.
2. If N is not multiple of n-then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

Question 15.
State any two merits for systematic random sampling.
Solution:
Merits:
1. This is simple and convenient method.
2. This method distributes the sample more evenly over the entire listed population.
3. The time and work is reduced much.

Question 16.
Using the following Tippet’s random number table

Draw a sample of 10 three digit numbers which are even numbers.
Solution:
There are many ways to select 10 random samples from the given Tippets random number number table since the population size is three digit numbers, Here the door numbers must be even (ie) the unit digit must be even. Here we consider column wise selection of random numbers starting from first column.

So the first sample is 416 and other 9 samples are 056, 664, 952, 748, 524, 914, 154, 340 and 140.
Tippets random number Table

Question 17.
A wholesaler in apples claims that only 4% of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning of good apples.
Solution:
sample size = 600; Number of success = 600 – 36
= 564
sample proportion p = \(\frac { 564 }{600}\) = 0.94
600
population proportion (p) = probability of getting good apple
= 96%
= \(\frac { 96 }{100}\) {∵ 4% of the apples 100 are defective}
P = 0.96
Q = 1 – p = 1 – 0.96
Q = 0.04
The S.E for a sample proporation is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.96)(0.04) }{600}}\)
\(\sqrt{\frac { 0.0384 }{600}}\) = \(\sqrt{0.000064}\)
∴ S.E = 0.008
Hence the standard error foe sample proportion is S.E = 0.008

Question 18.
A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Tamil Nadu State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. Calculate standard error of mean.
Solution:
Given n = 1000; \(\bar{x}\) = 119 lbs (pounds)
s = 30 lbs is known in this problem.
since σ is unknown, so we consider \(\bar{σ}\) = s and µ = 120 lbs
S.E = \(\frac { \bar{σ} }{√n}\) = \(\frac { s }{√n}\) = \(\frac { 30 }{\sqrt{1000}}\)
= \(\frac { 30 }{31.623}\) = 0.9487
Therefore the standard error for the average weight of large group of students of 120 lbs is 0.9487

Question 19.
A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. Calculate the suitable standard error that this sample is taken from a population with standard deviation 3?
Solution:
Sample size n = 60
Sample S.D S = 2.5
population S.D a = 3
The standard error for sample S.D is given by
\(\sqrt{\frac { σ^2 }{2n}}\) = \(\sqrt{\frac { (3)^2 }{2(60)}}\) = \(\frac { 3 }{\sqrt{120}}\)
= \(\frac { 3 }{10.954}\) = 0.27387
= 0.2739
Thus standard error for sample S.D = 0.2739

Question 20.
In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village?
Solution:
sample size n = 400
case (i):
sample proporation of vegetarian p = \(\frac { 3 }{10.954}\) = \(\frac { 230 }{400}\)
p = 0.575
q = 1 – p
= 1 – 0.575
q = 0.425
Sample error S.E= \(\sqrt{\frac { pq }{n}}\)
= \(\sqrt{\frac { 0.575×0.425 }{400}}\) = \(\sqrt{\frac { 0.223125 }{400}}\)
\(\sqrt{0.0005578125}\)
S.E = 0.2361

Case(ii):
sample size n = 400
since both vegetarian and non- vegetarian foods are equally popular in that village
sample proparation of vegetarian p = \(\frac { 1 }{2}\) = 0.5
q = 1 -p ⇒ q = 1 – 0.5
q = 0.5