Estimation part I

1. NADEEM
UDDIN
ASSOCIATE
PROFESSOR
OF STATISTICS
ESTIMATION
PART-I

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:

3. 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 an interval estimate, and the
process of estimating with a range of values
is known as interval estimation.

4. 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.
The Level of Confidence:-
The degree of probability associated with an
interval estimate is known as the confidence
level or confidence coefficient.

5. 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
𝑿 ± 𝒁∝
𝟐
𝝈
𝒏

6. 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.

7. Solution:
n = 16
𝑥 = 5.98
𝜎 = 3.5
 = 1 – 0.95 = 0.05
𝑍∝
2
= 𝑍0.05
2
= 𝑍0.025 = 1.96 (From z- table)
𝑥 ± 𝑍∝
2
𝜎
𝑛
5.98 ±1.96
3.5
16
5.98 ± 1.715
4.265    7.695

8. 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
𝑿 ± 𝒁∝
𝟐
𝒔
𝒏

9. 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.

10. Solution:
n = 62
𝑥 = 24
S = 2.1
 = 1 – 0.95 = 0.05
𝑍∝
2
= 𝑍0.05
2
= 𝑍0.025 = 1.96(From z-table)
𝑥 ± 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.

11. 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
𝑿 ± 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)

12. Example-3:
Seven homemaker were randomly
sampled, and it was determined that the
distance they walked in their house work
had an average 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.

13. Solution:
n = 7
𝑥 = 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)
𝑥 ± 𝑡∝
2,(𝑛−1)
𝑆
𝑛
39.2±2.447
3.2
7
39.2 ± 2.9596
36.240    42.160miles

14. 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:

15. 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
𝑥=
Σ𝑥
𝑛
=
70
7
= 10
𝑆 =
𝑛Σ𝑥2− Σ𝑥 2
𝑛 𝑛−1
S =
7 700.48 − 70 2
7 7−1
S = 0.283
 = 1 – 0.95 = 0.05

16. 𝑡∝
2,(𝑛−1)
= 𝑡 0.05
2
(7−1)
= 𝑡0.025,6 = 2.447
(From t- table)
𝑥 ± 𝑡∝
2
𝑛 − 1
𝑆
𝑛
10 ± 2.447
0.283
7
10 ± 0.2617
9.74 10.26

17. 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?

18. 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