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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 11th Chemistry Guide Pdf Chapter 14 Haloalkanes and Haloarenes Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 14 Haloalkanes and Haloarenes

11th Chemistry Guide Haloalkanes and Haloarenes Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The IUPAC name of is
a) 2 – Bromo pent – 3 – ene
b) 4 – Bromo pent – 2 – ene
c) 2 – Bromo pent – 4 – ene
d) 4 – Bromo pent – 1 – ene
Answer:
b) 4 – Bromo pent – 2 – ene

Question 2.
Of the following compounds, which has the highest boiling point?
a) n – Butyl chloride
b) Isobutyl chloride
c) t – Butyl chloride
d) n – Propyl chloride
Answer:
a) n – Butyl chloride

Question 3.
Arrange the following compounds in increasing order of their density
A) CCl₄
B) CHCl₃
C) CH₂Cl₂
D) CH₃Cl
a) D < C < B < A
b) C > B > A > D
c) A < B < C < D
d) C > A > B > D
Answer:
a) D < C < B < A

Question 4.
With respect to the position of – Cl in the compound CH3 – CH = CH – CH₂ – Cl, it is classified as
a) Vinyl
b) Allyl
c) Secondary
d) Aralkyl
Answer:
b) Allyl

Question 5.
What should be the correct IUPAC name of diethyl chloromethane?
a) 3 – Chloro pentane
b) 1 – Chloropentane
c) 1 – Chloro – 1, 1 – diethyl methane
d) 1- Chloro-1-ethyl propane
Answer:
a) 3 – Chloro pentane

Question 6.
C – X bond is strongest in
a) Chloromethane
b) Iodomethane
c) Bromomethane
d) Fluoromethane
Answer:
d) Fluoromethane

Question 7.
In the reaction X is ______.

Answer:
b)

Question 8.
Which of the following compounds will give racemic mixture on nucleophilic substitution by OH^– ion?
i)

ii)

iii)
a) (i)
b) (ii) and (iii)
c) (iii)
d) (i) and (ii)
Answer:
c) (iii)

Question 9.
The treatment of ethyl formate with excess of RMgX gives
a)

b)
c) R – CHO
d) R – O – R
Answer:
c) R – CHO

Question 10.
Benzene reacts with Cl₂ in the presence of FeCl₃ and in absence of sunlight to form
a) Chlorobenzene
b) Benzyl chloride
c) Benzal chloride
d) Benzene hexachloride
Answer:
a) Chlorobenzene

Question 11.
The name of C₂F₄C₁₂ is
a) Freon – 112
b) Freon – 113
c) Freon – 114
d) Freon – 115
Answer:
c) Freon – 114

Question 12.
Which of the following reagent is helpful to differentiate ethylene dichloride and ehtylidene chloride?
a) Zn / methanol
b) KQH / ethanol
c) aqueous KOH
d) ZnCl₂ / Con HCl
Answer:
c) aqueous KOH

Question 13.
Match the compounds given in Column I with suitable items given in Column II:

Code
a) A → 2 B → 4 C → 1 D → 3
b) A → 3 B → 2 C → 4 D → 1
c) A → 1 B → 2 C → 3 D → 4
d) A → 3 B → 1 C → 4 D → 2
Answer:
d) A → 3 B → 1 C → 4 D → 2

Question 14.
Assertion:
Inmonohaloarenes, electrophilic substitution occurs at ortho and para positions.
Reason:
Halogen atom is a ring deactivator.
Assertion and Reason type questions.
Directions:
In the following questions, a statement of assertion (A) is followed by a statement of reason (R) mark the correct choice as
(i) If both assertion and reason are true and reason is the correct explanation of assertion.
(ii) If both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) If assertion is true but reason is false.
(iv) If both assertion and reason are false.
a) (i)
b) (ii)
c) (iii)
d) (iv)
Answer:
b) (ii)

Question 15.
Consider the reaction,
CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr This reaction will be the fastest in
a) ethanol
b) methanol
c) DMF (N, N’ – dimethyl formamide)
d) water
Answer:
c) DMF (N, N’ – dimethyl formamide)

Question 16.
Freon – 12 manufactured from tetrachloro methane by
a) Wurtz reaction
b) Swarts reaction
c) Haloform reaction
d) Gattermann reaction
Answer:
b) Swarts reaction

Question 17.
The most easily hydrolysed molecules under S_N¹ condition is
a) allyl chloride
b) ethyl chloride
c) isopropyl chloride
d) benzyl chloride
Answer:
a) allyl chloride

Question 18.
The carbon cation formed in S_N¹ reaction of alkyl halide in the slow step is
a) sp³ hybridized
b) sp² hybridized
c) sp hybridized
d) none of these
Answer:
b) sp² hybridized

Question 19.
The major products obtained when chlorobenzene is nitrated with HNO₃ and con H₂SO₄
a) 1 – chloro – 4 – nitrobenzene
b) 1 – chloro – 2 – nitrobenzene
c) 1 – chloro – 3 – nitrobenzene
d) 1 – chloro – 1 – nitrobenzene
Answer:
a) 1 – chloro – 4 – nitrobenzene

Question 20.
Which one of the following is most reactive towards nucleophilic substitution reaction?
a)

b)

c)

d)
Answer:
d)

Question 21.
Ethylidene chloride on treatment with aqueous KOH gives
a) acetaldehyde
b) ethylene glycol
c) formaldehyde
d) glycoxal
Answer:
a) acetaldehyde

Question 22.
The raw material for Raschig process
a) chloro benzene
b) phenol
c) benzene
d) anisole
Answer:
c) benzene

Question 23.
Chloroform reacts with nitric acid to produce
a) nitro toluene
b) nitro glycerine
c) chloropicrin
d) chloropicric acid
Answer:
c) chloropicrin

Question 24.
Acetone X, X is
a) 2 – propanol
b) 2 – methyl – 2 – propanol
c) 1 – propanol
d) acetonol
Answer:
b) 2 – methyl – 2 – propanol

Question 25.
Silverpropionate when refluxed with Bromine in carbon tetrachloride gives
a) propionic acid
b) chloroethane
c) Bromo ethane
d) chloro propane
Answer:
c) bromo ethane

II. Write brief answer to the following questions:

Question 26.
Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides.
i) CH₃ – CH = CH – Cl

ii) C₆H₅CH₂I

iii)

iv) CH₂ = CH – Cl
Answer:
i) CH₃ – CH = CH – Cl = Allylic halide

ii) C₆H₅CH₂I = Benzylic halide

iii) = Alkyl halide

iv) CH₂ = CH – Cl = Vinyl halide

Question 27.
Why chlorination of methane is not possible in dark?
Answer:
The reaction of chlorine and methane is a free radical reaction under the influence of light energy. Chlorine molecule first split into two Cl atoms or radicals. These are both very reactive species in contact with methane they form methyl radical and HCl.

Methyl radical further reacts with Cl to give CH₃Cl and another Cl atom thus of a chain reaction. So this reaction takes place only under the influence of light. Hence the reaction does not take place in dark condition.

Question 28.
How will you prepare n propyl iodide from n – propyl bromide?
Answer:
Finkelstein reaction,
nCH₃ – CH₂ – CH₂ – Br + NaI n – CH₃ – CH₂ – CH₂ – I + NaBr
n – propyl iodide                                              n- propyl bromide

Question 29.
Which alkyl halide from the following pair is
i) chiral
ii) undergoes faster S_N² reaction?

Answer:

It contains one chiral carbon atom.
2 – bromo butane undergoes S_N² mechanism faster than 1- Chloro butane.

Question 30.
How does chlorobenzene react with sodium in the presence of ether? What is the name of the reaction?
Answer:
Haloarenes react with sodium metal in dry ether, two aryl groups combine to give biaryl products.
This reaction is called fittig reaction.
C₆H₅Cl + 2Na + Cl – C₆H₅ C₆H₅ – C₆H₅ + 2NaCl
Chlorobenzene                                      Biphenyl

Question 31.
Give reasons for polarity of C – X bond in halo alkane.
Answer:
Carbon halogen bond is a polar bond as halogens are more electro negative than carbon. The carbon atom exhibits a partial positive charge (δ⁺) and halogen atom a partial negative charge (δ^–)

The C -X bond is formed by overlap of sp³ orbital of carbon atom with half-filled p- orbital of the halogen atom. The atomic size of halogen increases from fluorine to iodine, which increases the C – X bond length. Larger the size, greater is the bond length, and weaker is the bond formed. The bond strength of C – X decreases from C – F to C – I in CH₃X.

Question 32.
Why is it necessary to avoid even traces of moisture during the use of Grignard reagent?
Answer:
Grignard reagents are mostly reactive and react with the source of product to give hydrocarbons. Even alcohols, amines, H₂O are sufficiently acidic to convert them to corresponding hydrocarbons.
R Mg X + H₂O → RH +

Due to its high reactivity, it is necessary to avoid even traces of moisture from Grignard reagent.

Question 33.
What happens when acetyl chloride is treated with an excess of CH₃MgI?
Answer:

Question 34.
Arrange the following alkyl halide in increasing order of bond enthalpy of RX.
CH₃Br, CH₃F, CH₃Cl, CH₃I
Answer:
The order is:
CH₃I < CH₃Br < CH₃Cl < CH₃F.

Question 35.
What happens when chloroform reacts with oxygen in the presence of sunlight?
Answer:
2 CHCl₃ + O₂ → 2 COCl₂ + 2 HCl

Question 36.
Write down the possible isomers of C₅H₁₁Br and give their IUPAC and common names.
Answer:
C₅H₁₁Br – Possible isomers
1. CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – Br → 1 – bromo pentane

2. → 2 – bromo pentane

3. → 3 – bromo pentane

4.  → 1 – bromo 2, 2 – dimethyl propane

5. → 1 – bromo 3 – methyl butane

6. → 2 – bromo 3 – methyl butane

7. → 2 – bromo 2 – methyl butane

8. → 1 – bromo 2- methyl butane

9. → (2S) – 1 – bromo 2 – methyl butane

10. → (2R) – 1 – bromo 2 – methyl butane

Question 37.
Mention any three methods of preparation of haloalkanes from alcohols.
Answer:
Haloalkanes are prepared by the following methods.
1) From alcohols:
Alcohols can be converted into halo alkenes by reacting it with any one of the following reagent.
1. Hydrogen halide
2. Phosphorous halides
3. Thionyl chloride.

a) Reaction with hydrogen halide:

Mixture of con. HCl and anhydrous ZnCl₂ is called Lucas Reagent.

The order of reactivity of halo acids with alcohol is in the order HI > HBr > HCl.
The order of reactivity of alcohols with halo acid is tertiary > secondary > primary.

b) Reaction with phosphorous halides:
Alcohols react with PX₅ or PX₃ to form haloalkanes.
Example:
CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl
Ethane                      Chloro ethane

3CH₃CH₂OH + PCl₃ → 3 CH₃CH₂Cl + H₃PO₃
Ethanol                       Chloro ethane

c) Reaction with Thionyl chloride(Sulphonyl Chloride)
CH₃CH₃OH + SOCl₂ CH₃CH₂Cl + SO₂↑ + HCl↑
Ethanol                                      Chloro ethane

Question 38.
Compare S_N¹ and S_N² reaction mechanisms.
Answer:

Question 39.
Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction.

Answer:

Question 40.
Discuss the aromatic nucleophilic substitutions reaction of chlorobenzene.
Answer:
The halogen of haloarenes can be substituted by OH^–, NH₂^– or CN^– with appropriate nucleophilic reagents at high temperature and pressure.
Example:
(i) Chlorobenzene reacts with ammonium at 250 and at 50 atm to give aniline.
C₆H₅Cl + 2NH₃ C₆H₅NH₂ + NH₄Cl
Chlorobenzene                  Aniline

(ii) Chlorobenzcne reacts with CuCN in presence of pyridine at 250 to give phenyl cyanide.
C₆H₅Cl + CuCN C₆H₅CN + CuCl
Chlorobenzene                Phenyl cyanide

(iii) Dows process:
C₆H₅Cl + NaOH C₆H₅OH + NaCl
Chlorobenzene                   Phenol
This reaction is known as Dow’s process.

Question 41.
Account for the following:
(i) t – butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n – butyl chloride reacts with S_N² mechanism.
(ii) p – dichloro benzene has higher melting point than those of o – and m – dichloro benzene.
Answer:
(i) t – butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n – butyl chloride reacts with S_N² mechanism.
It general, S_N¹ reaction proceeds through the formation, of carbocation, The tert- butyl chloride readily loses Cl ion to form stable 3° carbocation. Therefore, it reacts with aqueous KOH by S_N¹ mechanism as:

On the other hand n-Butyl chloride does not undergo ionization to form n-Butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by S_N² mechanism, which occurs is one step through a transition state involving nucleophilic attack of OH^– ion from the backside with simultaneous expulsion of Cl^– ion from the front side.

S_N¹ mechanism follows the reactivity order as 3° > 2°> 1° while S_N² mechanism follows the reactivity order as 1° > 2° > 3°. Therefore, tert-butyl chloride (3°) reacts by S_N¹ mechanism while n-butyl chloride (1°) reacts by S_N² mechanism. (ii) p – dichloro benzene has higher melting point than those of o – and m – dicholoro benzene. The higher melting point of p – isomer is due to its symmetry which leads to more close packing of its molecules in the crystal lattice and consequently strong intermolecular attractive force which requires more energy for melting. p – Dihalo benzene > o – Dichloro benzene> m – Dichioro benzene
Melting point: 323 K 256 K 249 K

Question 42.
In an experiment methyl iodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed.
a) Name of the product and write the equation for the reaction.
b) Why all the reagents used in the reaction should be dry? Explain.
c) How is acetone prepared from the product obtained in the experiment?
Answer:
a) Name of the product and write the equation for the reaction.
CH₃I + Mg CH₃MgI

b) Why all the reagents used in the reaction should be dry? Explain.
All the reagents used in the reaction should be dry because reagent reacts with H20 to produce alkane. This is the reason that everything has to be very dry during the preparation of Grignard reagents.
CH₃ – MgI + H₂O → CH₄ +
Methane

c) How is acetone prepared from the product obtained in the experiment?

Question 43.
Write a chemical reaction useful to prepare the following.
i) Freon – 12 from Carbon tetrachloride
ii) Carbon tetrachloride from carbon disulphide.
Answer:
i) Freon – 12 from Carbon tetrachloride:
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride.
CCl₄ + 2HF 2HCl + CCl₂F₂
Carbon tetrachloride            Freon – 12

ii) Carbon tetrachloride from carbon disulphide.
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl3 as catalyst giving carbon tetra chloride.
CS₂ + 3 Cl₂ CCl₄ + S₂Cl₂
Carbon disulfide                Carbon tetrachloride

Question 44.
What are Freons? Discuss their uses and environmental effects.
Answer:
The chloro fluoro derivatives of methane and ethane are called freons.
Nomenclature:
Freon is represented as Freon – cba
Where a = number of carbon atoms – 1;
b = number of hydrogen atoms + 1
a = total number of fluorine atoms

Uses:
i) Freons are used as a refrigerants in refrigerators and air conditioners.
ii) It is used as a propellant for aerosols and foams
iii) It is used as propellant for foams to spray out deodorants, shaving creams, and insecticides.

Question 45.
Predict the products when bromo ethane is treated with the following.
i) KNO₂ ii) AgNO₂
Answer:
i) KNO₂:
Bromo ethane reacts with alcoholic solution of NaNO2 or KNO2 to form ethyl nitrite.
CH₃CH₂Br + KNO₂ → CH₃CH₂ – O – N = O + KBr
Bromoethane Ethyl nitrite

ii)AgNO₂:
Bromo ethane reacts with alcoholic solution of AgNO2 to form nitro ethane.
CH₃CH₂Br + AgNO₂ → CH₃CH₂ NO₂ + AgBr
Bromoethane                Nitro ethane

Question 46.
Explain the mechanism of S_N¹ reaction by highlighting the stereochemistry behind it.
Answer:

In S_N¹ reactions, if the alkyl halide is optically active, the product obtained in a racemic mixture. The intermolecular carbocation formed in slowest step being sp2 hybridised is planar species. Therefore the attack of the nucleophile OH on it, can occur from both the faces with equal case forming a mixture of two enantiomers. Thus S_N¹ reaction of optically active alkyl halides are accompained by racemisation.

Question 47.
Write short notes on the following.
i) Raschig process
ii) Dows process
iii) Darzen’s process
Answer:
i) Raschig process:
Chloro benzene is commercially prepared by passing a mixture of benzene vapour, air and HCl overheated cupric chloride, this reaction is called the Raschig process,

ii) Dows Process:
C₆H₅Cl + NaOH C₆H₅OH + NaCl
This reaction is known as Dows process.

iii) Darzen’s process:
CH₃CH₂OH + SOCl CH₃CH₂Cl + SO₂↑ + HCl↑
Ethanol Chloro ethane
This reaction is known as Darzen’s process.

Question 48.
Starting from CH₃MgI, How will you prepare the following?
i) Acetic acid
ii) Acetone
iii) Ethyl acetate
iv) Iso propyl alcohol
v) Methyl cyanide
Answer:
i) Acetic acid:
Solid carbon dioxide reacts with methyl magnesium iodide to form addition product which on hydrolysis yields aceti acid.

ii) Acetone:
Acetyl chloride reacts with methyl magnesium iodide and followed by acid hydrolysis to give acetone.

iii) Ethyl Acetate:
Ethyl chloroformate reacts with methyl magnesium iodide to form ethyl acetate.

iv) Isopropyl alcohol:
Aldehydes (Acetaldehyde) other than formaldehyde, react with methyl magnesium iodide to give addition product which on hydrolysis yields isopropyl alcohol.

v) Methyl cyanide:
Methyl magnesium iodide reacts with cyanogen chloride to give methyl cyanide.

Question 49.
Complete the following reactions.
i) CH₃ – CH = CH₂ + HBr
ii) CH₃ – CH₂ – Br + NaSH
iii) C₆H₅Cl + Mg
iv) CHCl₃ + HNO₃
v) CCl₄ + H₂O
Answer :
i) CH₃ – CH = CH₂ + HBr CH₃ – CH₂ – CH₂ – Br
Propene                                                   n – propyl bromide

ii) CH₃ – CH₂ – Br + NaSH CH₃ – CH₂ – SH + NaBr
Propyl bromide                                     Ethanethiol

iii) C₆H₅Cl (Chloro benzene) + Mg C₆H₅MgCl (Phenyl magnesium chloride)

iv) CHCl₃ + HNO₃ CCl₃NO₂ + H
Chloroform                               Chloropicrin

v) CCl₄ (Carbon tetrachloride) + H₂O COCl₂ (Carbonyl chloride) + 2HCl

Question 50.
Explain the preparation of the following compounds.
i) DDT
ii) Chloroform
iii) Biphrnyl
iv) Chloropicrin
v) Freon – 12
Answer:
i) DDT:
DDT can be prepared by heating a mixture of chlorobenzene with chloral (Trichloro acetaldehyde) in the presence of con.H₂SO₄.

ii) Chloroform:
Preparation:
Chloroform is prepared in the laboratory by the reaction between ethyl alcohol with bleaching powder followed by the distillation of the product chloroform. Bleaching powder act as a source of chlorine and calcium hydroxide. This reaction is called haloform reaction. The reaction proceeds in three steps as shown below.

Step – 1: Oxidation
CH₃CH₂OH + Cl₂ → CH₃CHO + 2HCl
Ethyl alcohol           Ethanal (Acetaldehyde)

Step – 2: Chlorination
CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl
Acetaldehyde        Trichloro acetaldehyde

Step – 3: Hydrolysis
2CCl₃CHO + Ca(OH)₂ → 2CHCl₃ + (HCOO)₂ Ca
Chloral                         chloroform

iii) Biphenyl:
Chloro benzene react with sodium metal in dry ether, to give biphenyl. This reaction is called fitting reaction.
C₆H₅Cl + 2 Na + Cl – C₆H₅ C₆H₅ – C₆H₅ + 2NaCl
Chloro benzene                                          Biphenyl

iv) Chloropicrin:
Chloroform reacts with nitric acid to form chloropicrin. (Trichloro nitro methane)
CHCl₃ + HNO₃ CCl₃NO₂ + H₂O
Chloroform                   Chloropicrin

v) Freon – 12
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride
CCl₄ + 2 HF 2 HCl + CCl₂F₂
Carbon tetrachloride                     Freon – 12

Question 51.
An organic compound (A) with molecular formula C₂H₅Cl reacts with KOH gives compounds (B) and with alcoholic KOH gives compound (C). Identify (A), (B), (C).
Answer:

Question 52.
Simplest alkene (A) reacts with HCl to form compound (B). Compound (B) reacts with ammonia to form compound (C) of molecular formula C₂H₇N. Compound (C) undergoes carbylamine test. Identify (A),
(B) and (C).
Answer:
CH₂ = CH₂ + HCl → C₂H₅Cl
(A) Ethylene              (B) Ethyl chloride

C₂H₅Cl + NH₃    →    C₂H₅NH₂ + HCl
(C) Ethyl chloride      (B) Ethyl amine

Question 53.
A hydrocarbon C₃H₆(A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C₃H₆O. What are the (A), (B) and (C). Explain the reactions.
Answer:

Question 54.
Two isomers (A) and (B) have the same molecular formula C₂H₄Cl₂. Compound (A) reacts with aqueous KOH gives compound (C) of molecular formula C₂H₄O. Compound (B) reacts with aqueous KOH gives compound (D) of molecular formula C₂H₆O₂. Identify (A), (B), (C) and (D).
Answer:

11th Chemistry Guide Haloalkanes and Haloarenes Additional Questions and Answers

I. Choose the best answer:

Question 1.
Which of the following is 1° alkyl halide
a) R – CH₂ – X
b) R₂CHX
C) R₃C – X
d) R – H
Answer:
a) R – CH₂ – X

Question 2.
2° halide among the following
a) isopropyl chloride
b) iso – butyl chloride
c) n – propyl chloride
d) n – butyl chloride
Answer:
a) isopropyl chloride

Question 3.
Ethylidene dibromide is
a) CH₃CH₂Br
b) Br – CH₂ – CH₂ – Br
c) CH₃ – CHBr₂
d) CH₂ = CH Br
Answer:
c) CH₃ – CHBr₂

Question 4.
Which of the following is gemdihalide?
a) CH₃CHBrCH₂Br
b) CH₃CHBr₂
c) CH₃CHBrCH₂CH₂Br
d) BrCH₂CH₂Br
Answer:
b) CH₃CHBr₂

Question 5.
Vicinal dihalide is
a) CH₃CH₂Br
b) Br – CH₂ – CH₂ – Br
c) CH₃ – CHBr₂
d) CH₂ = CHBr
Answer:
b) Br – CH₂ – CH₂ – Br

Question 6.
The reagent used to get alkyl halide from alcohol
a) PCl₅
b) SOCl₂
c) Both a and b
d) Cl₂
Answer:
c) Both a and b

Question 7.
For the preparation of alkyl halides from alcohols which among the following cannot be used
a) PCl₅
b) SOCl₂
c) PCl₃
d) NaCl
Answer:
d) NaCl

Question 8.
In the preparation of alkyl halide from alkane and halogen which of the following reaction involved
a) Electrophilic addition
b) Nucleophilic addition
c) Electrophilic substitution
d) Nucleophilic substitution
Answer:
a) Electrophilic addition

Question 9.
In the preparation of alkyl halide from alkane and halogen which of the following reaction involved
a) Free radical substitution
b) Nucleophilic addition
c) Electrophilic substitution
d) Nucleophilic substitution
Answer:
a) Free radical substitution

Question 10.
Grignard reagent is formed when alkyl halide reacts with which one of the following
a) Mg in alcohol
b) Mg in acid
c) Mg in dry ether
d) MgO
Answer:
c) Mg in dry ether

Question 11.
When alkyl halide reacts with moist Ag₂O gives
a) alcohol
b) ether
c) alkane
d) Alkene
Answer:
a) alcohol

Question 12.
Alkyl halide on reduction with Zn + HCl gives
a) alcohol
b) alkene
c) alkane
d) ether
Answer:
c) alkane

Question 13.
Cyanide is formed as the major product of the reaction when alkyl halide is treated with one of the following :
a) AgNO₂
b) KNO₂
c) AgCN
d) KCN
Answer:
c) AgCN

Question 14.
Which of the reactions are most common in alkyl halides
a) Nucleophilic addition
b) Electrophilic addition
c) Nucleophilic substitution
d) Electrophilic substitution
Answer:
c) Nucleophilic substitution

Question 15.
Treatment of ammonia with excess of ethyl chloride will yield
a) Diethyl amine
b) Ethane
c) Tetra ethyl ammonium chloride
d) methyl amine
Answer:
c) Tetra ethyl ammonium chloride

Question 16.
In chloro ethane the carbon bearing halogen is bonded to ________.
a) three, primary
b) two, secondary
c) one, tertiary
d) two, primary
Answer:
d) two, primary

Question 17.
Which of the fallowing is used as refrigerant?
a) CH₃COCH₃
b) CCl₄
c) C₂H₅Cl
d) CF₄
Answer:
c) C₂H₅Cl

Question 18.
SN¹reaction occurs through the intermediate formation of
a) carbocation
b) carbanion
c) free radicals
d) transition
Answer:
a) carbocation

Question 19.
The rate of SN² reaction is maximum when the solvent is
a) Methyl alcohol
b) Water
c) Dimethyl sulphoxide
d) Benzene
Answer:
c) Dimethyl sulphoxide

Question 20.
The most reactive nucleophile among the following is
a) CH₃O^–
b) C₆H₅O^–
c) (CH₃)₂CHO^–
d) (CH₃)₃CO^–
Answer:
a) CH₃O^–

Question 21.
The correct order of reactivity towards nucleophilic substitution reaction is
a) CH₃F > CH₃Cl > CH₃Br > CH₃I
b) CH₃I > CH₃Br > CH₃Cl > CH₃F
C) CH₃I > CH₃Cl > CH₃Br > CH₃F
d) CH₃I > CH₃Br > CH₃F > CH₃Cl
Answer:
b) CH₃I > CH₃Br > CH₃Cl > CH₃F

Question 22.
In SN₂ reactions the order of reactivity of the halides.
CH₃X, C₂H₅X , n – C₃H₇X, n- C₄H₉X is
a) CH₃X > C₂H₅X > n – C₃H₇X > n – C₄H₉X
b) C₂H₅X > n – C₃H₇X > n – C₄H₉X > CH₃X
c) C₂H₅X > n – C₃H₇X > n – C₄H₉X < CH₃X
d) n – C₄H₉X > n – C₃H₇X > C₂H₅X > CH₃X
Answer:
a) CH₃X > C₂H₅X > n – C₃H₇X > n – C₄H₉X

Question 23.
SN² mechanism proceeds through the formation of a
a) carbocation
b) transition state
c) free radical
d) carbanion
Answer:
b) transition state

Question 24.
In Dow’s process the starting raw material is
a) Phenol
b) Chloro benzene
c) Aniline
d) Diazobenzene
Answer:
b) Chloro benzene

Question 25.
Chloro benzene is prepared commercially by
a) Dow’s process
b) Decon’s process
c) Raschig process
d) Etard’s process
Answer:
c) Raschig process

Question 26.
Chloro benzene is ________ reactive than benzene towards electrophilic substitution and directs incoming electrophile to the ______ position.
a) more, ortho & para
b) less, ortho & para
c) more, meta
d) less, meta
Answer:
b) less, ortho & para

Question 27.
The raw material for raschig; process is
a) chloro benzene
b) phenol
c) benzene
d) anisol
Answer:
c) benzene

Question 28.
Chloro benzene on treatment with sodium in dry ether gives diphenyl. The name of the reaction is
a) Fitting reaction
b) Wurtz fittig reaction
c) Wurtz reaction
d) Sandmeyer reaction
Answer:
a) Fitting reaction

Question 29.
An organic compound which produces a bluish green coloured flame on heating in the presence of copper is,
a) chloro benzene
b) benzaldehyde
c) aniline
d) benzoic acid
Answer:
a) chloro benzene

Question 30.
The raw materials for the commercial manufacture of DDT are
a) chloro benzene and chloroform
b) chloro benzene and chloro methane
c) chloro benzene and chloral
d) chloro benzene and iodoform
Answer:
c) chloro benzene and chloral

Question 31.
Iodoform is used as
a) anaesthetic
b) antiseptic
c) analgesic
d) anti febrin
Answer:
b) antiseptic

Question 32.
The following is used in paint removing
a) CHCl₃
b) CH₂Cl₂
c) CCl₄
d) CH₃CI
Answer:
b) CH₂Cl₂

Question 33.
In fire extinguishers, following is used
a) CHCl₃
b) CS₂
c) CCl₄
d) CH₂Cl₂
Answer:
c) CCl₄

Question 34.
The following is used for metal cleaning and finishing
a) CHCl₃
b) CHI₃
c) CH₂Cl₂
d) C₆H₆
Answer:
c) CH₂Cl₂

Question 35.
First chlorinated insecticide
a) DDT
b) Gammaxene
c) Iodoform
d) Freon
Answer:
a) DDT

Question 36.
The following is used as anaesthetic
a) C₂H₄
b) CHCl₃
c) CH₂Cl₂
d) DDT
Answer:
b) CHCl₃

Question 37.
Freon – 12 is
a) CF₃Cl
b) CHCl₂F
c) CF₂Cl₂
d) DDT
Answer:
c) CF₂Cl₂

Question 38.
The name of DDT
a) p, p’ – dichloro diphenyl trichloro ethane
b) p, p’ – dichloro diphenyl trichloro ethene
C) p, p’ – dichloro diphenyl tnchloro benzene
d) p, p’ – tetra chloro ethane
Answer:
a) p, p’ – dichloro diphenyl trichloro ethane

Question 39.
Freon R – 22 is
a) CHClF₂
b) CCl₂F₂
c) CH₃Cl
d) CH₂Cl₂
Answer:
a) CHClF₂

Question 40.
Molecular formula of DDT has
a) 5 Cl atoms
b) 4 Cl atoms
c) 3 Cl atoms
d) 2 Cl atoms
Answer:
a) 5 Cl atoms

Question 41.
What is DDT among the following
a) Green house gas
b) A fertilizer
c) Bio degradable pollutant
d) Non – Bio degradable pollutant
Answer:
d) Non – Bio degradable pollutant

Question 42.
The IUPAC name of (CH₃)₃CHCH₂Br is
a) 1 – bromo – 2 – methyl propane
b) 2 – bromo – 2 -methyl propane
c) 1 – bromo – 1 – methyl propane
d) 2 – bromo – 1 -methylpropane
Answer:
a) 1 – bromo – 2 – methyl propane

Question 43.
IUPAC name of allyl chloride is
a) 1 – chloro ethane
b) 3 – chloro- 1 – propyne
c) 3 – chloro – 1 – propene
d) 1 – chloro propane
Answer:
c) 3 – chloro – 1 – propene

Question 44.
The number of structural isomers possible with the formula C₄H₉Cl are
a) 5
b) 4
c) 3
d) 2
Answer:
b) 4

Question 45.
Density is highest for
a) CH₃Cl
b) CH₂Cl₂
c) CHCl₃
d) CCl₄
Answer:
d) CCl₄

Question 46.
C₂H₅OH X. In this reaction ‘X’ is
a) Ethanol
b) Ethylene chloride
c) ethylidene chloride
d) ethyl chloride
Answer:
d) ethyl chloride

Question 47.
Thionyl chloride is preferred in the preparation of chloro compound from alcohol since
a) Both the byproducts are gases and they escape out leaving product in pure state
b) It is a chlorinating agent
c) It is an oxidising agent
d) All other reagents are unstable
Answer:
a) Both the byproducts are gases and they escape out leaving product in pure state

Question 48.
The only alkene which gives primary alkyl halides on hydro halogenation
a) C₂H₄
b) C₃H₆
c) C₄H₈
d) C₅H₁₀
Answer:
a) C₂H₄

Question 49.
– OH cannot be replaced by – Cl if we use
a) PCl₅
b) PCl₃
c) S₂Cl₂
d) SOCl₂
Answer:
c) S₂Cl₂

Question 50.
In the hydrohalogenation of ethylene for adding HCl, the catalyst used is
a) Anhydrous AlCl₃
b) Conc. Sulphuric acid
c) Dilute Sulphuric acid
d) Anhydrous ZnCl₂
Answer:
a) Anhydrous AlCl₃

Question 51.
Which one of the following has the lowest boiling point?
a) CH₃Cl
b) C₂H₅Cl
c) C₂H₅Br
d) C₂H₅I
Answer:
a) CH₃Cl

Question 52.
Chloroethane is reacted with alcoholic potassium hydroxide. The product formed is
a) C₂H₆O
b) C₂H₆
c) C₂H₄
d) C₂H₄O
Answer:
c) C₂H₄

Question 53.
What is X in the following reaction? C₂H₅Cl + X → C₂H₅OH + KCl
a) KHCO3
b) alc. KOH
c) aq. KOH
d) K₂CO₃
Answer:
c) aq. KOH

Question 54.
Which of the following acids will give maximum yield of alkyl chloride in Hunsdiecker reaction
a) CH₃CH₂CH₂COOH
b) (CH₃)₂CHCOOH
c) (CH₃)₃CCOOH
d) C₆H₅CH (CH₃)COOH
Answer:
a) CH₃CH₂CH₂COOH

Question 55.
In the reaction sequence
C₂H₅Cl + KCN X. What is the molecular formula of X is
a) C₂H₅CN
b) C₂H₅NC
c) C₂H₅OH
d) C₂H₄O
Answer:
a) C₂H₅CN

Question 56.
Ethyl chloride on heating with silver cyanide forms a compound X. The functional isomer of X is
a) C₂H₅NC
b) C₂H₅NCN
c) H₃C – NH – CH₃
d) C₂H₅NH₂
Answer:
b) C₂H₅NCN

Question 57.
With Zn – Cu couple and C₂H₅OH, ethyl Iodide reacts to give
a) ethers
b) diethyl ether
c) Iodoform
d) Ethane
Answer:
d) Ethane

Question 58.
Ethyl bromide on boiling with alcoholic solutions of sodium hydroxide forms
a) Ethane
b) ethylene
c) ethyl alcohol
d) all of these
Answer:
b) ethylene

Question 59.
Following major compound is formed when ethyl chloride react with silver nitrite
a) Nitro ethane
b) Ethyl nitrite
c) Ethylene
d) Acetaldehyde
Answer:
b) Ethyl nitrite

Question 60.
Which of the following represents Williamson’s synthesis?
a) CH₃COOH + PCl₃ →
b) CH₃ – CH₂ – Cl + CH₃COOH →
c) CH₃ – CH₂ – ONa + CH₃ – CH₂ – Cl →
d) CH₃ – CH₂ – OH + Na →
Answer:
c) CH₃ – CH₂ – ONa + CH₃ – CH₂ – Cl →

Question 61.
The reaction of alkyl halide with benzene in presence of anhydrous A1Cl₃ gives alkyl benzene the reaction is known as
a) Friedel – craft’s reaction
b) Carbylamine reaction
c) Gattermann reaction
d) Wurtz reaction
Answer:
a) Friedel – craft’s reaction

Question 62.
A Grignard’s reagent reacts with water to give
a) ether
b) alkanes
c) amine
d) Alcohol
Answer:
b) alkanes

Question 63.
C₂H₅Cl + Mg → C₂H₅ MgCl in this reaction the solvent is
a) C₂H₅OH
b) Water
c) Dry ether
d) Acetone
Answer:
c) Dry ether

Question 64.
For a nucleophilic substitution reaction the rate was found in the order RI > RBr > RCl > RF then the reaction could be
a) S_N¹ only
b) S_N² only
c) Either S_N¹ or S_N²
d) Neither S_N¹ or S_N²
Answer:
c) Either S_N¹ or S_N²

Question 65.
S_N² reaction leads to
a) inversion of configuration
b) retention of configuration
c) partial racemisation
d) no racemisation
Answer:
a) inversion of configuration

Question 66.
Which of the following alkyl halide is hydrolysed by S_N¹ mechanism
a) CH₃Cl
b) CH₃ – CH₂ – Cl
c) CH₃ – CH₂ – CH₂ – Cl
d) (CH₃)₃CCl
Answer:
d) (CH₃)₃CCl

Question 67.
S_N¹ reaction is favoured by
a) non – polar solvents
b) Bulky group on the carbon atom attached to the halogen atom
c) Small groups on the carbon atom attached to halogen atom
d) All of the above
Answer:
b) Bulky group on the carbon atom attached to the halogen atom

Question 68.
Which of the following is not stereospecific
a) S_N¹
b) S_N²
c) E₂
d) Addition of Br₂ to ethylene in CCl₄
Answer:
a) S_N¹

Question 69.
Which of the following factors does not favour S_N¹ mechanism
a) Strong base
b) Polar solvent
c) Low. conc. of nucleophile
d) 3° halide
Answer:
c) Low. conc. of nucleophile

Question 70.
The order of reactivity of various alkyl halides toward SN1 reaction is
a) 3° > 2° > 1°
b) 1° > 2° > 3°
c) 3° = 2° = 1°
d) 1° > 3° > 2°
Answer:
a) 3° > 2° > 1°

Question 71.
In aryl halides carbon atom holding halogen is
a) sp² hybridised
b) sp hybridised
c) sp³ hybndised
d) sp³d hybridised
Answer:
a) sp² hybridised

Question 72.
Chloro benzene can be prepared by reacting benzene diazonlum chloride with
a) HCl
b) Cu₂Cl₂ / HCl
c) Cl₂ / AlCl₃
d) HNO₂
Answer:
b) Cu₂Cl₂ / HCl

Question 73.
+ Cl₂ X, X is
a) Chloro benzene
b) m – dichloro benzene
c) benzene hexachioride
d) p – dichlorobenzene
Answer:
a) Chloro benzene

Question 74.
The following is an example of Sandmeyer reaction
a) C₆H₅N₂⁺ Cl^– C₆H₅Cl
b) C₆H₅N₂⁺ Cl^– C₆H₅OH
c) C₆H₅N₂⁺ Cl^– C₆H₅F
d) C₆H₅N₂⁺ Cl^– C₆H₅Cl
Answer:
a) C₆H₅N₂⁺ Cl^– C₆H₅Cl

Question 75.
Chlorobenzene on reaction with CH₃Cl in presence of AlCl₃ gives
a) toulene
b) m – chloro toulene
c) only o – chloro toulene
d) mixture of o – and p – chlorotoulene
Answer:
d) mixture of o – and p – chlorotoulene

Question 76.
2C₆H₅Cl + 2Na → X, X is
a) toulene
b) biphenyl
C) phenyl ethane
d) 1 – chloro – 2 – phenyl ethane
Answer:
b) biphenyl

Question 77.
Chlorobenzene on fusing with solid NaOH gives
a) benzene
b) benzoic acid
c) phenol
d) benzene chloride
Answer:
c) phenol

Question 78.
Chlorobenzene on nitration gives major product of
a) 1 – chloro – 4 – nitro benzene
b) 1 – chloro – 3 – nitro benzene
c) 1, 4 – dinitro benzene
d) 2, 4, 6 – tri nitro benzene
Answer:
a) 1 – chloro – 4 – nitro benzene

Question 79.
The reaction C₆H₅I + 2 Na + CH₃I → C₆H₅CH₃ + 2 NaI is
a) Wurtz reaction
b) Fittig reaction
c) Wurtz-Fittig reaction
d) Sandmeyer reaction
Answer:
c) Wurtz-Fittig reaction

Question 80.
R – Cl + Nal R – I + NaCl. This reaction is
a) Wurtz reaction
b) Fittig reaction
c) Finkelstein reaction
d) Frankland reaction
Answer:
c) Finkelstein reaction

Question 81.
C₂H₅OH + SOCl₂ x + y + z. In this reaction x, y and z respectively are
a) C₂H₄Cl₂, SO₂, HCl
b) C₂H₅Cl, SO₂, HCl
c) C₂H₅Cl, SOCl, HCl
d) C₂H₄, SO₂, Cl₂
Answer:
b) C₂H₅Cl, SO₂, HCl’

Question 82.
C₂H₅Cl + AgOH → A + AgCl.
A + CH₃COCl → C + HCl. “C” is
a) Ethyl acetate
b) Methyl acetate
c) butanone – 2
d) propanone
Answer:
a) Ethyl acetate

Question 83.
The compound (B) in the below reaction is:
C₂H₅Cl A B
a) ethylene chloride
b) acetic acid
c) propionic acid
d) ethyl cyanide
Answer:
c) propionic acid

Question 84.
Chloro ethane reacts with X to form diethyl ether. What is X?
a) NaOH
b) H₂SO₄
c) C₂H₅ONa
d) Na₂S₂O₃
Answer:
c) C₂H₅ONa

Question 85.
1 – chlorobutane on reaction with alcoholic potash gives
a) 1 – butene
b) 1 – butanol
c) 1 – butyne
d) 2 – butanol
Answer:
a) 1 – butene

Question 86.
Propane nitrile may be prepared by heating
a) Propyl alcohol with KCN
b) ethyl chloride with KCN
c) Propyl chloride with KCN
d) ethyl chloride with KCN
Answer:
d) ethyl chloride with KCN

Question 87.
CH₃CH = CH₂ A B, B is
a) propanol – 2
b) propanal – 1
c) propanol – 1
d) propanal – 2
Answer:
c) propanol – 1

Question 88.
Y is

Answer:
c)

Question 89.
The correct order of increasing boiling points is
a) 1 – chloropropane < isopropylchloride < 1 – chlorobutane
b) isopropylchloride < 1 – chloropropane < 1 – chlorobutane
c) 1 – chlorobutane < isopropylchloride < 1 – chloropropane
d) 1 – chlorobutane < 1 – chloropropane < isopropylchloride
Answer:
b) isopropylchloride < 1 – chloropropane < 1 – chlorobutane

Question 90.
The correct order of decreasing S_N² reactivity
a) RCH₂X > R₂CHX > R₃CX
b) RCH₂X > R₃CX > R₂CHX
c) R₂CHX > R₃CX > RCH₂X
d) R₃CX > R₂CHX > RCH₂X
Answer:
a) RCH₂X > R₂CHX > R₃CX

II. Very short question and answer (2 Marks):

Question 1.
What are haloalkanes? Give example.
Answer:
Mono halogen derivatives of alkanes are called haloalkanes. Haloalkanes are represented by general formula R – X, Where, R is an alkyl group (C_nH_{2n + 1}) – and X is a halogen atom (X = F, Cl, Br or I). Haloalkanes are further classified into primary, secondary, tertiary haloalkane on the basis of type of carbon atom to which the halogen is attached.
Example:

Question 2.
How will you convert methane into tetra chloro methane?
Answer:
Chlorination of methane gives different products which have differences in the boiling points. Hence, these can be separated by fractional distillation.

Question 3.
What is Finkelstein reaction?
Answer:
Chloro or bromoalkane on heating with a concentrated solution of sodium iodide in dry acetone gives iodo alkanes. This reaction is called as Finkelstein reaction.
CH₃CH₂Br + NaI CH₃CH₂I + NaI
Bromo ethane Iodoethane

Question 4.
What is Swartz reaction?
Answer:
Chloro or bromo alkanes on heating with metallic fluorides like AgF or SbF₃ gives fluoro alkanes. This reaction is called Swarts reaction.
CH₃CH₂Br + AgF CH₃CH₂F + AgBr
Bromo ethane Fluoro ethane

Question 5.
What is Hunsdiecker reaction?
Answer:
Silver salts of fatty acids when refluxed with bromine in CCl₄ gives bromo alkane.
CH₃CH₂COOAg + Br₂ CH₃CH₂Br + CO₂ + AgBr
Silver propionate Bromo ethane

Question 6.
How is ehtyl magnesium bromide prepared from ethyl bromide?
Answer:
When a solution of ethyl bromide in ether is treated with magnesium, we get ethyl magnesium bromide.
CH₃ – CH₂ – Br + Mg CH₃CH₂MgBr
Ethyl bromide Ethyl magnesium bromide

Question 7.
How will you convert ethyl bromide into ethyl lithium?
Answer:
Ethyl bromide reacts with active metals like sodium, lead etc in the presence of dry ether to form ethyl lithium.
CH₃ CH₂ Br + 2Li CH₃ CH₂Li + LiBr
Ethyl bromide Ethyl Lithium

Question 8.
How is TEL prepared from ethyl bromide?
Answer:
When ethyl bromide reacts with Na / Pb alloy to give TEL.
4CH₃ CH₂ Br + 4Na/Pb → (CH₃CH₂)₄Pb + 4NaBr + 3Pb
Ethyl bromide Sodium-lead alloy Tetra ethyl lead (TEL)

Question 9.
What are organic metallic compounds? Give example.
Answer:
Organo metallic compounds are organic compounds in which there is a direct carbon-metal bond
Example:
CH₃ Mg I – Methyl magnesium iodide
CH₃CH₂Mg Br – Ethyl magnesium bromide

Question 10.
How is methane prepared from Grignard reagent?
Answer:
Compounds like water, alcohols and amines which contain active hydrogen atom react with Grignard reagents to form alkanes.
CH₃MgI + HO – H → CH₄ + Mg I (OH)

CH₃MgI + C₂H₅OH          CH₄ + MgI (OC₂H₅)
Ethyl alcohol                Methane

Question 11.
What are Haloarenes? Give a suitable example.
Answer:
Haloarenes are the compounds in which the halogen is directly attached to the benzene ring.
Example:

Question 12.
How is chloro benzene prepared from benzene by direct halogenation?
Answer:
Chloro benzene is prepared by the direct chlorination of benzene in the presence of lewis acid catalyst like FeCl₃.

Question 13.
Write a note on Sand Meyer reaction.
Answer:
When aqueous solution of benzene diazonium chloride is warmed with Cu₂Cl in HCl gives chioro benzene.

Question 14.
How is Iodo benzene prepared from benzene diazonium chloride?
Answer:
Iodo benzene is prepared by warming benzene diazonium chloride with an aqueous KI solution.
C₆H₅N₂Cl + Kl          C₆H₅I + N₂ + KCl
Benzene dizonium chloride        Iodo benzene

Question 15.
What happens when ethylidene dichloride is treated with Zinc dust in methanol?
Answer:
Gem dihalides and vic – dihalides on treatment with zinc dust in methanol give alkenes.
CH₃ – CHCl₂ + Zn CH₂ = CH₂ + ZnCl₂
Ethylidene dichloride                     Ethylene

Question 16.
How will you convert chloroform into methylene chloride?
Answer:
a) Reduction of chloroform in the presence of Zn + HCl gives methylene chloride.
CHCl₃ (chloroform) CH₂Cl (Methylene chloride) + HCl
b) Reduction of chloroform using H2/Ni
CHCl₃ (chloroform) CH₂Cl₂ (Methylene chloride) + HCl

Question 17.
Write the chlorination reaction of methane.
Answer:
Chlorination of methane gives methylene chloride
CH₄ CH₃Cl CH₂Cl₂
Methane                                               Methylene chloride

Question 18.
How is chloroform prepared from carbon tetrachloride?
Answer:
Carbon tetrachloride is reduced by iron powder in dilute HCl medium to form chloroform
CCl₄ (carbon tetrachloride) + 2[H] CHCl₃ (chloroform) + HCl

III. Short question and answers (3 Marks):

Question 1.
write IUPAC name of the following.
i)

ii)

iii)
Answer:
i)

ii)

iii)

Question 2.
Write the structure of the following compounds.
i) 1 – Bromo – 4 – ethyl cyclohexane
ii) 1, 4 – Dichlorobut – 2 – ene
iii) 2 – chloro – 3 – methyl pentane
Answer:
i) 1 -Bromo – 4 – ethyl cyclohexane:

ii) 1, 4 – Dichlorobut – 2 – ene
– Cl – CH₂ – CH = CH – CH₂ – Cl

iii) 2 – chloro – 3 – methyl pentane

Question 3.
Write any three methods of preparation of chloro ethane from ethanol.
Answer:
a) Reaction with hydrogen halide:
Ethanol is heated with HCl in presence of anhydrous ZnCl₂ to give ethyl chloride.
CH₃CH₂OH + HCl CH₃CH₂Cl + H₂O
Ethanol Chloroethane

b) Reaction with phosphorous halides:
Ethanol reacts with PCl₅ or PCl₃ it gives ethyl chloride.
CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl
Ethanol                      Chloro ethane

3 CH₃CH₂OH + PCl₃ → 3CH₃CH₂Cl + H ₃PO₃
Ethanol                        Chloro ethane

C) Reaction with thionyl chloride (Sulphonyl chloride):
When ethanol reacts with SOCl₂ in presence of pyridine, it gives chloro ethane.
CH₃CH₂OH + SOCl₂ CH₃CH₂Cl + SO₂ + HCl
Ethanol                                            Chloro ethane

Question 4.
What happens when haloalkane reacts with aqueous alkali or most silver oxide?
Answer:
Ethyl bromide reacts with aqueous solution of KOH or moist silver oxide (Ag₂O/H₂O) to form ethanol.
CH₃ – CH₂ – Br + KOH (aq) CH₃CH₂ – OH + KBr
Bromoethane                                   Ethyl alcohol

CH₃ – CH₂ – Br   +   AgOH(aq)    CH₃CH₂ – OH + AgBr
Bromoethane         Moist silver oxide            Ethanol

Question 5.
Explain the ammonolysis reaction of bromo ethane.
Answer:
Ethyl bromide reacts with alcoholic ammonia solution to form ethylamine.
CH₃ – CH₂ – Br + H – NH₂ → CH₃ CH₂ – NH₂ + HBr
Bromo ethane                        Ethyl amine

However, with excess of ethyl bromide, secondary and tertiary amines along with quaternary ammonium salts are obtained.

Question 6.
What happens when bromoethane reacts with alcoholic KCN and alcoholic AgCN?
Answer:
Reaction with alcoholic KCN:
Ethyl bromide reacts with alcoholic KCN solution to form ethyl cyanides.
CH₃ – CH₂ – Br + KCN → CH₃ – CH₂ – CN + KBr
Bromoethane                    Ethyl cyanide

Reaction with alcoholic AgCN:
Haloalkanes react with alcoholic AgCN solution to form alkyl isocyanide.
CH₃CH₂Br + AgCN → CH₃CH₂NC + AgBr
Bromoethane           Ethyl isocyanide

Question 7.
How are the following compounds prepared from bromo ethane?
i) Ethane thiol
(ii) Diethyl ether
Answer:
i) Ethane thiol:
Ethyl bromide reacts with sodium or potassium hydrogen sulphide to form thio alcohols.
CH₃CH₂Br CH₃CH₂SH + NaBr
Bromo ethane                           Ethane thiol

ii) Diethyl ether:
Ethyl bromide, when boiled with sodium alkoxide gives diethyl ether. This method can be used to prepare mixed (unsymmetrical) ethers also.
CH₃ – CH₂Br +      NaOCH₂CH₃          CH₃CH₂ – O – CH₂CH₃ + NaBr
Bromo ethane        Sodium ethoxide                 diethyl ether

Question 8.
How is ethane prepared from the following compounds?
i) Bromo ethane ii) Iodo ethane
Answer:
i) From bromo ethane:
Ethyl bromide is reduced to ethane by treating with H₂ in the presence of metal catalyst like nickel, palladium etc or with hydroiodic acid in the presence of red phosphrous.
Ni(or)Pd
CH₃CH₂Br + H₂ CH₃ – CH₃ + HBr
Bromo ethane                              Ethane

ii) From iodo ethane:
Iodo ethane is reduced with H₂ in presence of Red P it gives ethane.
CH₃CH₂I + HI CH₃ – CH₃ + I₂
Iodo ethane                         Ethane

Question 9.
Write the uses of carbon tetra chloride.
Answer:

1. Carbon tetrachloride is used as dry cleaning agent.
2. It is used as a solvent for oils, fats and waxes
3. As the vapour of CCl₄ is non-combustible, it is used under the name pyrene for extinguishing the fire in oil or petrol.

Question 10.
Write the uses of Grignard reagents.
Answer:

1. Grignard reagents are synthetically very useful compounds. These reagents are converted to various organic compounds like alcohols, carboxylic acids, aldehydes and ketones.
2. The alkyl group being electron rich acts as a carbanion or a nucleophile.
3. They would attack polarized molecules at a point of low electron density. The following reactions illustrate the synthetic uses of Grignard reagent.

Question 11.
What is Balz – Schiemann reaction?
Answer:
Fluoro benzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produces fluorobenzene. This reaction is called Balz – schiemann reaction.
C₆H₅N₂Cl + HBF₄ C₆H₅N₂ + BF₄^– C₆H₅ F + BF₃ + N₂
Benzene diazonium chloride                                          Fluorobenzene

Question 12.
Write the uses of chloro benzene.
Answer:

• Chloro benzene is used in the manufacture of pesticides like DDT
• It is used as high boiling solvent in organic synthesis.
• It is used as fibre – swelling agent in textile processing.

Question 13.
How is ethylidene dichloride prepared from
(i) Acetaldehyde
(ii) Acetylene
Answer:
i) Treating acetaldehyde with PCl₅:
CH₃CHO (Acetaldehyde) + PCl₅ → CH₃CHCl₂ (Ethylidene dichloride) + POCl

ii) Adding hydrogen chloride to acetylene

Question 14.
How is ethylene dichloride prepared from (i) Chloride (ii) PCl₅?
Answer:
i) Addition of chlorine to ethylene

ii) Action of PCl₅ (or HCl) on ethylene glycol

Question 15.
What happens when the following compounds are treated with alcoholic KOH?
i) Ethylidene dichloride
ii) Ethylene dichloride
Answer:
i) Ethylidene dichloride:

ii) Ethylene dichloride:

Question 16.
Write the uses of methylene chloride.
Answer:
Methylene chloride is used as

1. aerosol spray propellant
2. solvent in paint remover
3. process solvent in the manufacture of drugs
4. a metal cleaning solvent

Question 17.
How are the following compounds prepared from chlorobenzene?
i) Benzene
ii) Phenyl magnesium chloride
Answer:
i) C₆H₆ (Benzene):
Chlorbenzene undergoes reduction with Ni – Al alloy in the presence of NaOH gives benzene.
C₆H₅Cl + 2(H) C₆H₆ + HCl
Chloro benzene              Benzeneide

ii) Phenyl Magnesium chloride:
Chloro benzene reacts with magnesium to form phenyl magnesium chloride in tetra hydrofuran (THF).
C₆H₅Cl (Chloro benzene) + Mg C₆H₅MgCl (Phenyl magnesium chloride)

IV. Long question and answers (5 Marks):

Question 1.
Explain the S_N² mechanism of haloalkanes with suitable example.
Answer:
S2 stands for bimolecular nucleophilic substitution
“S” stands for substitution
“N” stands for nucleophilic
“2” stands for bimolecular (two molecules are involved in the rate determining step)
The rate of S_N² reaction depends upon the concentration of both alkyl halide and the nucleophile.
Rate of reaction = k₂ [alkyihalide] [nucleophile]
This S_N² reaction follows second order kinetics and occurs in one step.

This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the back side and the halide ion leaves from the front side. The carbon at which substitution ocurs ha inverted configuration during the course of reaction just as an umbrella has tendency to invert in a wind storm. This inversion of configuration is called Walden inversion; after paul walden who first discovered the inversion of configuration of a compound in S_N² reaction.

We understand S_N² reaction mechanism by taking an example of reaction between chloromethane and aqueous KOH. S_N² reaction of an optically active haloalkane is always accompanied by inversion of configuration at the asymmetric centre.

Question 2.
Explain the S_N¹ mechanism of haloalkanes with suitable examples.
Answer:
S_N¹ stands for unimolecular nucleophilic substitution
‘S’ stands for substitution
‘N’ stands for nucleophilic
‘1’ stands for unimolecular (one molecule is involved in the rate determining step)
The rate of the following S_N¹ reaction depends upon the concentration of aikyl halide (RX) and is independent of the concentration of the nucleophile (OH-).
Rate of the reaction = k[alkyl halide]
R – Cl + OH^– → R – OH + Cl^–
This S_N¹ reaction follows first order kinetics and occurs in two steps.
We understand S_N¹ reaction mechanism by taking a reaction between tertiary hutyl bromide with aqueous KOH.

This reaction takes place in two steps aas shown below.

Step – 1:
Formation of carbocation:

t – butyl bromide

The polar C – Br bond breaks forming a carbocation and bromide ion. This step is slow and hence it is the rate determining step.

Step 2:
Nucleophilic attack on carbocation
The carbocation immediately reacts with the nucleophile. This step is fast and hence does not affect the rate of the reactions.

As shown above, the nucleophilic reagent OH~ can attack carbocation from both the sides, they will be mirror image of each other.

In the above example the substrate tert-butyl bromide is not optically active, hence the obtained product is optically inactive. If halo alkane substrate is optically active then, the product obtained will be optically inactive racemic mixture. As nucleophilic reagent OH- can attack carbocation from both the sides, to form equal proportion of dextro and levorotatory optically active isomers which form optically inactive racemic mixture.

Example :
Hydrolysis of optically active 2- bromo butane gives racemic mixture of ±butan-2-ol.

Question 3.
Explain the E₂ reaction mechanism with suitable example.
Answer:
E₂ stands for bimolecular elimination reaction
‘E’ stands for elimination
‘2’ stands for bimolecular
The rate of E₂ reaction depends on the concentration of alkyl halide and base

Rate = k₂ [alkyl halide] [base]
It is therefore, a second-order reaction generally primary alkyl halide undergoes this reaction in the presence of alcoholic KOH. E₂ is a one-step process in which the abstraction of the proton from the p carbon and expulsion of halide from the carbon occurs simultaneously. The mechanism is shown below.

Question 4.
Explain the E₁ reaction mechanism with suitable example.
Answer:
E₁ stands for unimolecular elimination reaction
‘E’ stands for elimination ‘1’ stands for unimolecular
The rate of E₁ reaction depends on the concentration of alkyl halide only and hence
rate = k [alkyl halide]
Generally, tertiary alkyl halide which undergoes elimination reaction by this mechanism in the presence of alcoholic KOH following first order kinetics is a two step process. The mechanism is shown by taking following example.

step – 1:
Heterolytic fission to yield a carbocation:

Step – 2:
Elimination of a proton from the β – carbon to produce an alkene:

Question 5.
How are the following compounds prepared from Methyl magnesium iodide?
(i) Ethanol
(ii) Tert-butyl alcohol
(iii) Acetaldehyde
Answer:
(i) Ethanol:
Formaldehyde reacts with methyl magnesium iodide to give addition products which on hydrolysis yield ethanol.

(ii) Tert-butyl alcohol:
Acetone reacts with methyl magnesium iodide to give an additional product which on hydrolysis yields tert butyl alcohol.

(iii) Acetaldehyde:
Ethyl formate reacts with methyl magnesium iodide and followed by acid hydrolysis, it forms Acetaldehyde.

Question 6.
How is ethylene dichloride converted into
i) Acetaldehyde
ii) Ethylene glycol
ii) Ethylene
Answer:
i) Acetaldehyde:
(Hydrolysis with aqueous NaOH or KOH)
Ethylidene chloride reacts with aqueous KOH to give acetaldehyde.

ii) Ethylene glycol :
Ethylene chloride reacts with aqueous KOH, it gives glycol.

iii) Ethylene:
Ethylene dichloride is heated with Zinc in presence of methanol gives ehtylene.

Question 7.
How are the following compounds prepared from chloroform?
(i) Phosgene
(ii) Methylene chloride
(iii) Methyl isocyanide
Answer:
(i) Phosgene:
Chloroform undergoes oxidation in the presence of light and air to form phosgene (carbonyl chloride)

Since phosgene is very poisonous, its presence makes chloroform unfit for use as an anaesthetic.

(ii) Methylene chloride:
Chloroform undergoes reduction with zinc and HCl in the presence of ethyl alcohol to form methylene chloride.
CHCl₃ (Chloroform) + 2(H) CH₂Cl₂ (Methylene chloride)+ Cl₂

(iii) Methyl isocyanide:
Chloroform reacts with aliphatic or aromatic primary amine and alcoholic caustic potash, to give foul-smelling alkyl isocyanide (carbylamines).
CH₃NH2     +     CHCl₂ + 3KOH → CH₃NC + 3KCl + 3H₂O
Methylamine Chloroform          Methyl isocyanide

Question 8.
Discuss the aromatic electrophilic substitutions reaction of chlorobenzene.
Answer:

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