Point and Interval Estimation:
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, but it may include simple random sampling or
systematic sampling.
Population: The entire data is called as population
Sample: a small part or quantity intended to show what
the whole is like or a part from the population is called as
sample.
Term Used in Sampling
Population (N) Sample (n)
Population Mean ( μ ¿ Sample Mean ( x )
Population Variance (σ 2
¿ Sample Variance ( s )
2
Population Standard Sample Standard Deviation (s)
Deviation (σ )
Population Proportion (P) Sample Proportion (p)
Standard Error (SE)
1. Standard Error of Mean
2. Standard Error of Proportion
1. Standard Error of Mean
With Replacement Without Replacement
√
σ σ N −n
SE ( x ¿ = √ n SE ( x ¿ = √n N −1
√ N −n
N −1
= Finite Population Parameter Factor
Note ***: If σ is not available then it can be obtained by
using the formula
σ= s √ n
n−1
Q. If the population SD is known to be 5 for the
population containing 80 units. What is the standard
error of sample mean for a sample of size 25 without
replacement?
Given:
Population Standard Deviation (σ ) = 5
Population (N) = 80
Sample (n) = 25
To find: Standard Error of Sample Mean without
replacement
Solution:
Standard Error of Sample Mean without replacement
SE ( x ¿ = σ
√n √N −n
N −1
SE ( x ¿ = 5
√
80−25
√ 25 80−1
SE ( x ¿ = 5 55
5 79√
SE ( x ¿ = 0.83
Q. A random sample of size 5 is taken from the
population containing 100 units. If the sample
observations are 10, 12, 13, 7, 18. Find the estimate of
the standard error of sample mean.
Solution:
2
x ( x−x ) (x−x )
10 -2 4
12 0 0
13 1 1
7 -5 25
18 6 36
∑ (x−x )2 =66
Mean = (10 + 12 + 13 + 7 + 18) / 5 = 12
∑
√ √
2
(x−x ) 66
S.D = n
= 5
= 3.63
Sample Standard Deviation (s) = 3.63
Sample (n) = 5
Population (N) = 100
σ is not available then it can be obtained by using the
formula
σ= s √ n
n−1
σ= 3.63 √ 5
5−1
σ= 3.63 √ 5
4
σ = 4.06
Standard Error of Sample Mean without replacement
SE ( x ¿ = σ
√n √ N −n
N −1
SE ( x ¿ = 4.06 100−5
√ 5 100−1 √
SE ( x ¿ = 4.06 95
√ 5 99 √
SE ( x ¿ = 1.78
2. Standard Error of Proportion
With Replacement Without Replacement
SE ( p ¿ = √ pq
n
SE ( p ¿ = √ √
pq
n
N−n
N −1
Q. A random sample of 200 articles taken from a large
batch of articles containing 15 defective articles. What
is the estimate of the standard error of the sample
proportion of defective articles?
Solution:
Sample (n) = 200
15
p = 200 = 0.075
q=1–p
q = 1 – 0.075
q = 0.925
Standard Error of Proportion with replacement:
SE ( p ¿ = √ pq
n
SE ( p ¿ =√ 0.075∗0.925
200
SE ( p ¿ = 0.0186
Q. A factory produces 60000 pairs of shoes on a daily
basis. From a sample of 600 pairs, 3% was found to be
inferior quality. Estimate the standard error of
proportion.
Solution:
Population (N) = 60000
Sample (n) = 600
3
p = 100 = 0.3
q=1–p
q = 1 – 0.3
q = 0.7
Standard Error of Proportion without replacement:
SE ( p ¿ = √ √ pq
n
N−n
N −1
SE ( p ¿ =√ √
0.3∗0.7 60000−600
600 60000−1
SE ( p ¿ =√ √ 0.21 59400
600 59999
SE ( p ¿ = 0.0069
Estimate:
1. Point Estimate
2. Interval Estimate
Point Estimate: It provides single value that is used to
approximate an unknown population parameter.
It ignores the Standard Error.
Sample Mean ( x ) = Population Mean ( μ ¿
Sample Proportion (p) = Population Proportion (P)
Sample Standard Deviation (s) = Population Standard
Deviation (σ )
σ= s √ n
n−1
Sample Variance ( s ) = Population Variance (σ )
2 2
n
σ
2
=s. 2
n−1
Interval Estimate:
Steps:
Step1: Compute Sample Parameter (Mean and
Proportion)
Step 2. Calculate Standard Error
Step 3. Find Interval as ¿ - Z SE, x + Z SE]
α α
Note**** Value of Z
Confidence Interval Value of Z α
90% 1.64
95% 1.96
98% 2.33
99% 2.58
99.73% 3.00
Confidence Interval:
How much confidence you are that error will not happen
90% means there will be no error
10% means there will be error
Chances of error is also called as level of significance or
degree of freedom (kitne error ke liye hum tayar hai)
Q. The marks obtained by the group of people of 15
students in an examination have a mean 55 and
variance 49. What are 99% confidence limit for the
mean of population of marks assuming it to be
normal? Given that the upper 0.5 % value of the
distribution with n – 1 = 14 degree of freedom is 2.58
Given:
Sample (n) = 15
Sample Mean ( x ) = 55p
Sample Variance ( s ) = 49
2
Sample Standard Deviation (s) = √ Sample Variance
Sample Standard Deviation (s) = √ 49
Sample Standard Deviation (s) = 7
σ is not available then it can be obtained by using the
formula
σ= s√ n
n−1
σ =7√ 15
15−1
σ =7√ 15
14
σ √ 15 7
SE = √ n = 7. √14∗ √15 = √14 1.8708
=¿
Interval is given as ¿ - Z SE, x + Z SE]
α α
= ¿ - 2.58∗1.8708 , 55 + 2.58∗1.8708]
= ¿ , 59.8246]
Q. A pharmaceutical company wants to estimate the
mean life of a particular drug under typical weather
condition. A random samples of 81 bottles yield the
following information:
Sample Mean = 23 months
Population Variance = 6.25 months
Find the interval estimate with a confidence level of
98%
Given Z value for the level of significance of 10% for
two tailed test = 1.645 and the level of significance for
2% for two tailed test = 2.33.
Solution:
Sample (n) = 81
Sample Mean ( x ) = 23
Population Variance (σ ) = 6.25 2
Population Standard Deviation (σ ) = √ 6.25
Population Standard Deviation (σ ) = 2.5
σ
SE = √ n
2.5
SE = √ 81 = 0.278
Interval is given as ¿ - Z SE, x + Z SE]
α α
= ¿ - 2.33∗0.278 , 23 + 2.33¿ 0.278]
= ¿ , 23.64774 ]
Estimation of Sample Size:
For Estimation of Mean For Estimation of Proportion
n = ¿, Where E is Estimated z2 pq
n= E2
, Where E is Estimated
Error Error
Q. In measuring Reaction time, physiological estimate
that the standard deviation is 1.08 second. What
would be the size of sample in order to be 99%
confidence that the error of the estimated mean would
not exceed 0.18 seconds? Given Z value for the
confidence 99% for two tailed test is 2.58.
Given:
σ = 1.08
Z = 2.58
E = 0.18
Size of Sample (n) = ¿
n=¿
n = 239.6
n = 240
Q. The incidence of the particular disease in an area is
such that the 20% of the people in that area suffer
from it. What size of sample is to be taken to ensure
that the error of estimation of the proportion should
not be more than 5% with 95% confidence?
p = 20/100 = 0.2
q = 1 – p = 1 – 0.2 = 0.8
E = 0.05
Z = 1.96
2
z pq
Size of Sample (n) = E2
2
1.96 ∗0.2∗0.8
n= 0.05 2
0.614656
n= 0.0025
n = 245.86
n = 246
