Role of estimation in statistics

2. Estimation:-
The entire process of using an estimator to produce an estimate
of the parameter is known as Estimation. Or estimation is a
process to estimate unknown value of population parameter with
the help of sample.
Estimator:-
Any statistic used to estimate a parameter is an estimator.
Estimate:-
Any specific value of a statistic is an estimate.
An estimate of a population parameter may be expressed in two
ways:
Point Estimation:-
A single number used to estimate a population parameter with
the help of a sample is called a point estimate. The process of
estimating a single number is known as point estimation.
Interval Estimation:-
A range of values used to estimate a population parameter with
the help of a sample is called aninterval estimate, and the
process of estimating with a range of values is known as interval
estimation.
Confidence Interval:-
The interval estimates based on specific confidence levels are
known as confidence intervals, and the upper and lower limits of
the intervals are known as confidence limits.

3. The Level of Confidence:-
The degree of probability associated with an interval estimate is
known as the confidence level or confidence coefficient.
CONFIDENCE INTERVAL FOR :
A (1-α)100% confidence interval for population mean when σ is
known and size of sample either small (n < 30) or large (n ≥ 30)
given as
𝑿̅ ± 𝒁∝
𝟐⁄ (
𝝈
√ 𝒏
)
Example-1:
We wish to estimate the mean serum indirect bilirubin level of
4-day-old infants.The mean for a sample of 16 infants was found
to be 5.98 mg/dl.Assuming bilirubin levels in 4-day-old infants
are approximately normally distributed with a standard deviation
of 3.5 mg/dl find the 95% confidence interval for mean.
Solution:
n = 16
𝑥̅ = 5.98
𝜎 = 3.5
 = 1 – 0.95 = 0.05
𝑍∝
2⁄ = 𝑍0.05
2
= 𝑍0.025 = 1.96
(From z- table)

4. 𝑥̅± 𝑍∝
2⁄ (
𝜎
√ 𝑛
)
5.98 ±1.96 (
3.5
√16
)
5.98 ± 1.715
4.265    7.695
CONFIDENCE INTERVAL FOR :
A (1-α)100% confidence interval for population mean when σ is
unknown and size of sample large (n ≥ 30) given as
𝑿̅ ± 𝒁∝
𝟐⁄ (
𝒔
√ 𝒏
)
Example-2:
In an automotive safety test conducted by the North
Carolina Highway Safety Research Center, the average tire
pressure in a sample of 62 tires was found to be 24 pounds per
square inch, and the standard deviation was 2.1 pounds per
square inch. Construct a 95 percent confidence interval for the
population mean. Also interpret the result.
Solution:
n = 62
𝑥̅ = 24
S = 2.1

5.  = 1 – 0.95 = 0.05
𝑍∝
2⁄ = 𝑍0.05
2
= 𝑍0.025 = 1.96
(From z- table)
𝑥̅± 𝑍∝
2⁄ (
𝑆
√ 𝑛
)
24 ±1.96 (
2.1
√62
)
24 ± 0.523
23.48 24.52 per square inch
The population tires will have pressure more than 23.48 but less
than 24.52 per square inch.
CONFIDENCE INTERVAL FOR :
A (1-α)100% confidence interval for population mean when σ is
unknown and size of sample small (n < 30) given as
𝑿̅ ± 𝒕∝
𝟐,(𝒏−𝟏)⁄ (
𝒔
√ 𝒏
)
Where 𝒔 = √
𝟏
𝒏−𝟏
{∑ 𝒙 𝟐 −
(∑ 𝒙) 𝟐
𝒏
} or 𝒔 = √
𝒏 ∑ 𝒙 𝟐−(∑ 𝒙) 𝟐
𝒏(𝒏−𝟏)

6. Example-3:
Seven homemaker were randomly sampled, and it was
determined that the distance they walked in their house work
had anaverage of 39.2 miles per week and a sample standard
deviation of 3.2 miles per week. Construct a 95% confidence
interval for the population mean.
Solution:
n = 7
𝑥̅ = 39.2
S = 3.2
 = 1 – 0.95 = 0.05
𝑡∝
2,(𝑛−1)⁄ = 𝑡0.05
2
(7−1)
= 𝑡0.025,6 = 2.447(From t- table)
𝑥̅± {𝑡∝
2,(𝑛−1)⁄ } (
𝑆
√ 𝑛
)
39.2±2.447(
3.2
√7
)
39.2 ± 2.9596
36.240    42.160miles

7. Example-4:
The contents of 7 similar containers of sulfuric acid are
9.8, 10.2, 10.4, 9.8, 10.0, 10.2 and 9.6 liters. Find a 95%
Confidence interval for the mean content of all such containers,
assuming an approximate normal distribution for container
contents.
Solution:
X X2
9.8 96.04
10.2 104.04
10.4 108.16
9.8 96.04
10.0 100
10.2 104.04
9.6 92.16
x
=70
x2
=
700.48
Where n = 7
x =70
x2
= 700.48
𝑥̅=
Σ𝑥
𝑛
=
70
7
= 10
𝑆 = √
𝑛Σ𝑥2−(Σ𝑥)2
𝑛( 𝑛−1)

8. = √
7(700.48) − (70)2
7 (7 − 1)
S = 0.283
 = 1 – 0.95 = 0.05
𝑡∝
2,(𝑛−1)⁄ = 𝑡0.05
2
(7−1)
= 𝑡0.025,6 = 2.447(From t- table)
𝑥̅± {𝑡∝
2⁄ ( 𝑛 − 1)} (
𝑆
√ 𝑛
)
10 ± 2.447 (
0.283
√7
)
10 ± 0.2617
9.74 10.26

9. Example-5:
The mean and S.D for the quality grade – point averages
are 2.6 and 0.3. How large a sample is required if we want to be
95% confidence that our estimate of  is within 0.05?
Solution:
 = 1 – 0.95 = 0.05
𝑍∝
2⁄ = 𝑍0.05
2
= 𝑍0.025 = 1.96
 = 0.3 and e = 0.05
𝑛 = [
𝑍∝
2⁄ 𝛿
𝑒
]
2
𝑛 = [
(1.96)(0.3)
0.05
]
2
n = 138.3
n  139
Do yourself the following problem:
*A hospital administrator wishes to estimate the mean
average of babies born in his hospital. How large a
sample of birth records should be take if he wants 95%
confidence level. He also wants the estimate to be
within to be ± 0.05 pound of the true means weight.
(n = 15.37 16)