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Estimation of Parameters
Umar khayam
Introduction
The process of drawing inferences about a
population on the basis information contained
in a sample taken from the population is called
statistical inference. Statistical inference is
divided into two major areas: estimation of
parameters and testing of hypothesis.
Major Areas of Inferential Statistics
Inferential
Statistics
Estimation
of
Parameters
Testing
of
Hypothesis
Estimation
Estimation is a procedure by which we obtain an
estimate of the true but unknown value of a
population parameter by using the sample
observations from the population. For example
we may estimate the mean and the variance of
population by computing the mean and the
variance of a sample drawn from the population.
Testing of hypothesis
Testing of hypothesis is a procedure which enable
us to decide on the basis of information obtained by
sampling whether to accept or reject any specified
statement or hypothesis regarding the value of the
parameter in a statistical problem.
We shall discuss estimation in this chapter and we shall deal with
testing of hypothesis in the next chapter
Estimate and Estimators
An estimate is a numerical value of the
unknown parameter obtained by applying a
rule or a formula , called an estimator, to a
sample of size n, taken from the population.
Categories of Estimation
Estimation
Point
Estimation
Interval
Estimation
Point estimate and Interval Estimate
When an estimate for the unknown population
parameter is expressed by a single value, it is
called a point estimate. An estimate expressed by
a range of values within which the true value of
the parameter is believed to lie, is referred to as
an interval estimate.
Example
A random sample of n = 3 has the elements
1, 3, and 5. Compute a point
estimates of
(i) The population mean
(ii) Population standard deviation
Solution
(i) The sample mean is
∑X
1+3+5
X= n
= 3
=3
Thus the point estimate of the population mean µ is 3.
(ii)
2
2
∑
X
∑X
2
S
=
X
X
n
n
1
1
3
9
5
25
9
35
=
35
3
-
9
3
2
= 11.67 – 9
S = 1.63
A random sample of n = 6 has the elements
1, 3, 6, 9, 12 and 5. Compute a point
estimates of
(i) The population mean
(ii) Population standard deviation
Estimation by confidence interval
A confidence interval estimate of the unknown
parameter Ө is an interval computed from a random
sample of n values with a statement of how confident
(90%, 95% or 99%) we are that the interval contains
the unknown parameter Ө. A confidence interval
estimate is in the form (L< Ө< U) , where L is the
lower confidence limit for Ө and U is the upper
confidence limit for Ө.
List of Formulae
(1) 95 % Confidence interval for µ with σ known.
X  1.96

n
(1) 90 % Confidence interval for µ with σ known.
X  1.645

n
(1) 99 % Confidence interval for µ with σ known.

X  2.58
n
Note: When σ is unknown, replace σ by s.
Exercise No 4 Page 263.
A soft drink machine is regulated so that the amount of
drink dispensed is approximately normally distributed
with a standard deviation equal to 1.5 deciliters. Find a
90% confidence interval for the mean of all drinks
dispensed by the machine if random sample of 36 drinks
had an average content of 22.5 deciliters.
Solution of Exercise No 4 Page 263.
Here
n =36 x =22.5 σ = 1.5
90% confidence interval for population mean µ

X  1.645
n
1.5
22.5  1.645
36
22.5  (2.4675 / 6)
or
or 22.5 – 0.41 , 22.5 + 0.41
or (22.09, 22.91)
Exercise No 5 Page 263
The heights of a random sample of 50
college students showed a mean of 174.5
cm and a standard deviation of 6.9 cm.
Construct a 95% confidence interval for the
mean height of all college students.
Solution of Exercise No 5 Page 263.
Here
n = 50
X = 174.5
S = 6.9
The 95% confidence interval for µ
X  1.96
or
or
or
or
174.5  1.96
174.5  1.91
S
n
6.9
50
174.5 – 1.91 , 174.5 +1.91
(172.59 , 176.41)
Exercise 6 page 263
A random sample of 100 automobile owners shows that
an automobile is driven on the average 23500 miles
per year , in the state of Virginia, with a standard
deviation of 3900 miles. Construct a 99% confidence
interval for the average number of miles an automobile is
driven annually in Virginia.
Solution of Exercise No 6 Page 263.
Here
n = 100 , X = 23500, S = 3900
Therefore 99% C.I. for µ is
S
X  2.58
n
or
3900
100
3900
23500  2.58
10
23500  1006.2
23500  2.58
or
or
or
23500-1006.2 , 23500+1006.2
or (22493.8 , 24506.2)
The Kryptonite Corporation personnel director
wishes to estimate the mean scores for a
proposed aptitude test that may be use in
screening applicants for clerical positions. The
population standard deviation is assumed to be
б=15 For a sample of 100 applicants the
sample mean scores is 75.6. Construct a 95%
confidence interval estimate of the true mean
Confidence Interval for the difference between the means
of two Populations (i.e. 1 – 2):
(1) 95 % Confidence interval for 1 – 2.
2
(2)
2
x1  x2   1.96 ó1  ó2
n1
n2
90 % Confidence interval for 1 – 2.
2
2
 x1  x2   1.645 ·ó1  ó2
n1
n2
(3) 99 % Confidence interval for 1 – 2.
2
2
x1  x2   2.58 ó1  ó2
n1
n2
Example 4 page 254
A standardized chemistry test was given to 75 boys and 50
girls. The girls made an average grade of 76 with a
standard deviation of 6, while the boys made an average
grade of 82 with a standard deviation of 8. find a 95%
confidence interval for the difference1  ,2 where  1 is
the mean score of all boys and  2 is the mean score of
all girls who might take this test.
Solution of Example 4 Page 254
Here
Boys
Girls
n1= 75
n2= 50
X2 = 76
X1 = 82
σ1 = 8
2
σ1 = 64
σ2 = 6
2
σ2 = 36
Solution of Example 4 Page 254 cont,
Therefore 95% confidence interval for
or
2
2
 x1  x2   1.96
ó1
ó
 2
n1
n2
82  76  1.96
64
36

75
50
or
6  1.96 0.853  0.72
or
6  1.96 1.573
or
6  2.458
or 6-2.458 , 6+2.458
or ( 3.542 , 8.458)
1  2
Exercise 14 page 264
A random sample of size 25 taken from a
population with standard deviation 5 has a
mean 80. A second sample of size 36 taken
from a different population with a standard
deviation 3 has a mean 75. Find a 90%
confidence interval for 1  2
Solution of Exercise 14 Page 264
Here
Population I
Population II
n1= 25
n2= 36
X1 = 80
X2 = 75
σ1 = 5
2
σ1 = 25
σ2 = 3
2
σ2 = 9
Solution of Exercise 14 Page 264
Therefore 90% confidence interval for 1  2
2
2
ó1
ó2
x1  x2   1.645

n1
n2
80  75  1.645
25
9

25
36
or
or
5  1.645 1  0.25
or
5  1.645 1.25
or
5  1.839
or (5-1.839, 5+1.839)
or (3.161, 6.839)
Exercise 15 page 264
Two kinds of thread are being compared for strength.
Fifty pieces of each type of thread are tested under similar
conditions. Brand A had an average tensile strength of
87.2 kilograms with a standard deviation of 6.3
kilograms, while brand B had an average tensile strength
of 78.3 kilograms with a standard deviation of 5.6
kilograms. Construct a 99% confidence interval for the
difference of the population means.
Solution of Exercise 15 Page 264
Here
Brand B
Brand A
n1= 50
n2= 50
X1 = 87.2
X2 = 78.3
S1 = 6.3
2
S1 = 39.69
S2 = 5.6
2
S2 = 31.36
Solution of Exercise 15 Page 264 cont;
Therefore 99% confidence interval for 1  2
2
2
 x1  x2   2.58 S1  S2
n1
n2
or 87.2  78.3  2.58 39.69  31.36
50
50
or 8.9  2.58 0.7938  0.6272
or
8.9  2.58 1.421
or
8.9  3.08
or
(8.9 – 3.08 , 8.9 + 3.08)
or ( 5.82 , 11.98)
· A study was made to estimate the difference in
salaries of collage professor in the private and
state colleges of Virginia. A random sample of
100 professor in private collages showed an
average 9month salary of 25000 with a
standard deviation of 1200. A random sample
of 200 professor in state collages showed an
average salary of 26000 with a standard
deviation of 1400 find a 98% for the difference
between the average salaries
What is population proportion ‘’p’’
· In many situation one may be interested in
estimating certain characteristic of the
population. Most common case is that of a
binary characteristic. For example in opinion
polls the answer is in the form of Yes or No the
result of an examination may be pass or fail.
Such a binary characteristic of the population is
generally referred to as population proportion
and is denoted by “p”
List of Formulae
(1) 95 % Confidence interval for population proportion P.
ˆ  1.96
p
ˆ 1  p
ˆ
p
n
(1) 90 % Confidence interval for population proportion P.
pˆ  1.645
pˆ 1  pˆ 
n
(1) 99 % Confidence interval for population proportion P.
pˆ  2.58
pˆ 1  pˆ 
n
Example 8 page 268
In a random sample of 500 people eating lunch at a
hospital cafeteria on various Fridays, it was found that
x=160 preferred seafood. Find a 95% confidence interval
for the actual proportion of people who eat seafood on
Friday at this cafeteria.
Solution:
p̂ = X / n = 160 /500 = 0.32
Here
Solution of Example 8 page 268 cont;
Therefore 95% confidence interval for P
pˆ  1.96
or
pˆ 1  pˆ 
n
0.32  1.96
0.321  0.32 
500
0.320.68
0.32  1.96
500
0.32  0.041
or
or
or ( 0.32 – 0.041 , 0.32+0.041)
or ( 0.28 , 0.36)
Exercise 2 page 273
A random sample of 400 cigarette smokers is selected and
86 are found to have preference for Brand A. Find the
90% confidence interval for the fraction of the population
of cigarette smokers who prefer brand A.
Solution: Here
X = 86 and n=400
p̂ = X / n = 86 / 400 = 0.215
Solution of Exercise 2 page 273 cont;
Therefore the 90% confidence interval for P
pˆ  1.645
or
pˆ 1  pˆ 
n
0.215  1.645
0.2151  0.215
400
0.215  1.645
0.2150.785
400
or
or 0.215  0.034
or (0.215- 0.034 , 0.215 + 0.034)
or (0.181 , 0.249)
Exercise 4 page 273
A sample of 75 college students is selected and 16 are
found to have cars on campus. Use a 99% confidence
interval to estimate the fraction of students who have cars
on campus.
Solution:
Here n=75 and X= 16
p̂= X/n = 16/75 = 0.21
Solution of Exercise 4 page 273
Therefore 99% confidence interval for P
ˆ 1  p
ˆ
p
ˆ
p  2.58
n
or
0.21  2.58
0.211  0.21
78
0.21  0.12
or
or
(0.21 – 0.12 , 0.21+0.12)
or (0.09 , 0.33)
Confidence Interval for the difference between two
Population proportions P1 – P2:
(1) 90% confidence Interval for P1 – P2
 pˆ1  pˆ 2   1.645
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
(2) 95% confidence Interval for P1 – P2
 pˆ1  pˆ 2   1.96
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
(3) 99% confidence Interval for P1 – P2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 
ˆ ˆ 
p1  p2  2.58
n1

n2
Example 11 page 272
A poll is taken among the residents of a city and the
surrounding county to determine the feasibility of a
proposal to construct a civic centre. If 2400 of 5000 city
residents favor the proposal and 1200 of 2000 county
residents favor it, find a 90% confidence interval for the
true difference in the fractions favoring the proposal to
construct the civic centre.
Solution of Example 11 page 272
p̂ 1= 244/500= 0.48 and p̂ 2 = 1200/200 = 0.60
Here
Therefore 90% confidence interval for P1 – P2
ˆ
ˆ
ˆ
ˆ
or
 pˆ1  pˆ 2   1.645 p1 1  p1   p2 1  p2 
n1
or
0.48  0.60  1.645
n2
0.481  0.48 0.601  0.60

5000
2000
or 0.48  0.60  1.645 0.480.52  0.600.40
5000
2000
or  0.12  1.645 0.00005  0.00012
or  0.12  0.0214
or ( -0.12 – 0.0214 , -.0.12+ 0.0214)
or (-0.1414, -0.0986)
Exercise 10 page 274
In a study to estimate the proportion of residents in a
certain city and its suburbs who favor the construction of
a nuclear power plant, it is found that 52 of 100 urban
residents favor the construction while only 34 of 125
suburban residents are in favor. Find a 95% confidence
interval for the difference between the proportion of urban
and suburban residents who favor construction of the
nuclear plant.
Solution of Exercise 10 page 274
p̂ 2= 34/125 =0.272
p̂ 1 =52/100 = 0.52
Here
Therefore 95% confidence interval for P1 – P2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
0.52  0.272  1.96 0.521  0.52  0.2721  0.272
 pˆ1  pˆ 2   1.96
or
or
or
100
125
0.52(0.48) 0.272(0.728)
0.248  1.96

100
125
0.248  1.96 0.0025  0.0016
0.248  0.1225
or (0.248- 0.1225 , 0.248+0.1225)
or (0.1255 , 0. 3705)
Exercise 12 Page 274
A geneticist is interested in the proportion of males and
females in the population that have a certain minor blood
disorders. In a random sample of 100 males, 24 are found
to be afflicted, where as 13 of 100 females tested appear
to have the disorders. Compute a 99% confidence interval
for the difference between the proportion of males and
females that have this blood disorder.
Solution of Exercise 10 page 274
Here p̂ 1 = 24/100 = 0.24 p̂ 2 = 13/100 = 0.13
Therefore 99% confidence interval for P1 – P2
 pˆ1  pˆ 2   2.58
or
or
or
0.24  0.13  2.58
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
0.241  0.13 0.241  0.13

100
100
0.11  2.58
0.240.76 0.130.87

100
100
0.13  2.58 0.0018  0.0011
0.13  0.1389
or
or ( 0.13- 0.1389 , 0.13 + 0.1389)
or (-0.0089 , 0.2689)