CAIIB ABM Module A Unit 2 : Sampling Methods

CAIIB ABM Module A Unit 2: Sampling Methods (New Syllabus) 

IIBF has released the New Syllabus Exam Pattern for CAIIB Exam 2023. Following the format of the current exam, CAIIB 2023 will have now four papers. The CAIIB Paper 1 (Advanced Bank Management) includes an important topic called “Sampling Methods”. Every candidate who are appearing for the CAIIB Certification Examination 2023 must understand each unit included in the syllabus.

In this article, we are going to cover all the necessary details of CAIIB Paper 1 (ABM) Module A (Statistics ) Unit 2 : Sampling Methods, Aspirants must go through this article to better understand the topic, Sampling Methods and practice using our Online Mock Test Series to strengthen their knowledge of Banker Customer Relationship. Unit 2: Sampling Methods

Sampling

Sampling is a process used in statistical analysis in which a predetermined number of observations are taken from a larger population. The methodology used to sample from a larger population depends on the type of analysis being performed.

Types of sampling

There are two methods of selecting from populations

• Non- random or judgement sampling
• Random or probability sampling

Random Sampling

A probability sampling method is any method of sampling that utilizes some form of random selection. In order to have a random selection method, you must set up some process or procedure that assures that the different units in your population have equal probabilities of being chosen.

Type of Random Sampling

There are four main type of random sampling

i)Simple Random Sampling (SRS)

ii)Stratified Sampling

iii)Cluster Sampling

iv)Systematic Sampling

• Simple Random Sampling (SRS):Simple Random Sampling selects samples by methods that allow each possible sample to have an equal probability of being picked and each item in the entire population to have an equal chance or being included in the sample.
• Systematic Sampling:In systematic sampling, elements are selected from the population at a uniform level that is measured in time, order, or space. If we wanted to interview every twentieth student on a college campus, we would choose a random starting point in the first twenty names in the student directory and then pick every twentieth name thereafter.
• Stratified Sampling: To use stratified sampling, we divide the population into relatively homogenous groups, called strata. Then we use one of two approaches. Either we select at random from each stratum a specified number of elements corresponding to the proportion of that stratum in the population as a whole or we draw an equal number of elements from each stratum and give weight to the results according to the stratum’s proportion of total population.
• Cluster Sampling: In cluster sampling, we divide the population into groups or clusters and then select a random sample of these clusters. We assume that these individual clusters are representative of the population as a whole. If a market Research team is attempting to determine by sampling the average number of television sets per household in a large city, they could use a city map and divide the territory into blocks and then choose a certain number of blocks (clusters) for interviewing. Every household in each of these blocks would be interviewed. A well designed cluster sampling procedure can produce a more precise sample at considerably less cost than that of simple random sampling.

 

Sampling distribution

Sampling Distribution is the distribution of all possible values of a statistic from all possible samples of a particular size drawn from the population.

Each sample we draw from a population would have its own means or measure of central tendency and standard deviation. Thus, the statistics we compute for each sample, would vary & be different for each random sample taken.

Sample Distribution Table

SAMPLES, their DATA & Mean

CONCEPT of STANDARD ERROR

• Standard deviation of the distribution of the sample means is called the standard error of the mean.
• Similarly standard error of the proportion is the standard deviation of the distribution of the sample proportions.
• g. We take the average height of college girls in India across various samples. We would calculate mean height of each sample. Obviously there is some variability in observed mean. This variability in sampling statistics results from the sampling error due to chance.
• Thus the standard deviation of the sampling distribution of means measures the extent to which the means vary because of a chance error in the sampling process. Thus the standard deviation of distribution of a sample statistic is known as the Standard error of the statistic.
• Thus, a standard error indicates not only the size of the chance error but also the accuracy we are likely to get if we use the sample statistic to estimate a population statistic.

 

Sampling From Normal Population

Finite Populations:

 μ= 162.40

x̄ = 162.40

This is not coincidence. The mean of the sample means is the same as the population mean, whenever we use simple random sampling.

Example -Bank calculate that its individual saving account have a mean of Rs.2000 and SD of 600. bank takes a sample of 100 account. Calculate the Standard error?

What is the probability that the sample lie between 1900 & 2050.

σX¯=σ/ √n

= 600 /10

=60

Probability associated with a standard normal variable

Standard Error Of The Mean For Infinite Populations:

Example: Bank calculate that its individual saving account have a mean of Rs.5000 and SD of 600. bank takes a sample of 100 account. Calculate the Standard error?

What is the probability that the sample lie between 1900 & 2050.

STEP 1 : Standard deviation of error

σX¯=σ/ √n

= 600 /10

=60

Central Limit Theorem

• The mean of the sampling distribution of the mean will equal the population mean regardless of the sample size, even if the population is not normal.
• As the sample size increases, the sampling distribution of the mean will approach normality, regardless of the shape of the population distribution.
• This relationship between the shape of the population distribution & the shape of sampling distribution of the mean is called the Central Limit Theorem.

 

• Actually a sample doesnot have to be very large for the sampling distribution of the mean to approach normal
• Statistician use the normal distribution as an approximation to the sampling distribution whenever the sample size is atleast 30, but the sampling distribution whenever the sample size is atleast 30.
• The significance of the CLT is that it permits us to use sample statistics to make interference about population parameters without knowing anything about the shape of the frequency distribution of that population.

Example:

Bank distribution has a mean of Rs.19000 & standard deviation of Rs.2000. If we draw a random sample of 30 tellers, What is the probability that their earning will average more than Rs.19750 annually?

STEP 1 : Calculate Standard error 

= 2000/ √30

= 2000/ 5.477

= 365.16

STEP 2 : Z value & Standard Normal Probability Distribution

X = 19750

= 19750-19000/365.16

= 750/ 365.16

=2.05

Finite Population Multiplier

Standard Error Of The Mean For Finite Populations

N = size of population

N = size of the sample

 Example:

We are interested in a population of 20 textile companies of the same size, all of which are experiencing excessive labour turnover. Standard deviation of the distribution of annual turnover is 75 employees. If we sample 5 of these textile companies, without replacement then compute the standard error of mean?

= 75/ √5 [√(20-5/20-1)]

= 33.54 * 0.888

= 29.8

Numerical on Sampling

Q1. A sack contains 3 pink balls and 7 green balls. What is probability to draw one pink ball and two green balls in one draw?

(a)

(b)

(c)

(d)

(e)

Ans(b)

Out of (3+ 7) = 10 balls, three (one pink & two green) balls are expected to be drawn

So, required probability =

=

=

=

 

Q2.A sack contains 4 black balls 5 red balls. What is probability to draw 1 black ball and 2 red balls in one draw?

(a) 11/19

(b) 10/21

(c)12/22

(d) 19/11

Ans: B

Solution :

Out of 9, 3 (1 black & 2 red) are expected to be drawn)

Hence sample space

n(S) = 9c3

= 9!/(6!×3!)

= 362880/4320

= 84

Now out of 4 black ball 1 is expected to be drawn hence

n(B) = 4c1

= 4

Same way out of 5 red balls 2 are expected be drawn hence

n(R) = 5c2

= 5!/(3!×2!)

= 120/12

= 10

Then P(B U R) = n(B)×n(R)/n(S)

i.e 4×10/84 = 10/21

PDF-CAIIB PAPER 1 Module A Unit 2-Sampling Methods (Ambitious_Baba)

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