estimation of parameters umar khayam. introduction the process of drawing inferences about a...
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Estimation of Parameters
Umar khayam
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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.
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Major Areas of Inferential Statistics
Inferential Statistics
Estimation of
Parameters
Testing of
Hypothesis
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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.
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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
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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.
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Categories of Estimation
Estimation
Interval Estimation
Point Estimation
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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.
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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
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Solution
(i) The sample mean is
X = = = 3
Thus the point estimate of the population mean µ is 3.
(ii)
S = -
= - = 11.67 – 9
S = 1.63
∑ X
n1+3+5
3
X X
1 1
3 9
5 25
9 35
2 ∑ X ∑ X
n n
2 2
35 9
3 32
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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
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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.
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List of Formulae (1) 95 % Confidence interval for µ with σ known.
(1) 90 % Confidence interval for µ with σ known.
(1) 99 % Confidence interval for µ with σ known.
Note: When σ is unknown, replace σ by s.
nX
96.1
nX
645.1
nX
58.2
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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.
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Solution of Exercise No 4 Page 263.
Here
n =36 x =22.5 σ = 1.5
90% confidence interval for population mean µ
or
or 22.5 – 0.41 , 22.5 + 0.41
or (22.09, 22.91)
nX
645.1
36
5.1645.15.22
)6/4675.2(5.22
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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.
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Here
n = 50 X = 174.5 S = 6.9
The 95% confidence interval for µ
or
or
or 174.5 – 1.91 , 174.5 +1.91
or (172.59 , 176.41)
Solution of Exercise No 5 Page 263.
n
SX 96.1
50
9.696.15.174
91.15.174
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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.
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Here n = 100 , X = 23500, S = 3900
Therefore 99% C.I. for µ is
or
or
or
or 23500-1006.2 , 23500+1006.2
or (22493.8 , 24506.2)
Solution of Exercise No 6 Page 263.
n
SX 58.2
100
390058.223500
10
390058.223500
2.100623500
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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
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Confidence Interval for the difference between the means of two Populations (i.e. 1 – 2):
(1) 95 % Confidence interval for 1 – 2.
(2) 90 % Confidence interval for 1 – 2.
(3) 99 % Confidence interval for 1 – 2.
2
22
1
21
21
ó·ó645.1
nnxx
2
22
1
21
21
óó96.1
nnxx
2
22
1
21
21
óó58.2
nnxx
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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 , where is
the mean score of all boys and is the mean score of
all girls who might take this test.
21 12
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Here
Boys Girls
n1= 75 n2= 50
X1 = 82 X2 = 76
σ1 = 8 σ2 = 6
σ1 = 64 σ2 = 36
Solution of Example 4 Page 254
2 2
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Solution of Example 4 Page 254 cont,
Therefore 95% confidence interval for
or
or
or
or
or 6-2.458 , 6+2.458
or ( 3.542 , 8.458)
21
50
36
75
6496.17682
2
22
1
21
21
óó96.1
nnxx
72.0853.096.16 573.196.16
458.26
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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 21
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Solution of Exercise 14 Page 264
Here
Population I Population II
n1= 25 n2= 36
X1 = 80 X2 = 75
σ1 = 5 σ2 = 3
σ1 = 25 σ2 = 92 2
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Therefore 90% confidence interval for
or
or
or
or
or (5-1.839, 5+1.839)
or (3.161, 6.839)
Solution of Exercise 14 Page 264
36
9
25
25645.17580
2
22
1
21
21
óó645.1
nnxx
25.01645.15 25.1645.15
839.15
21
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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.
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Solution of Exercise 15 Page 264
Here
Brand B Brand A
n1= 50 n2= 50
X1 = 87.2 X2 = 78.3
S1 = 6.3 S2 = 5.6
S1 = 39.69 S2 = 31.362 2
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Solution of Exercise 15 Page 264 cont;
Therefore 99% confidence interval for
or
or
or
or
or (8.9 – 3.08 , 8.9 + 3.08)
or ( 5.82 , 11.98)
50
36.31
50
69.3958.23.782.87
2
22
1
21
21
SS58.2
nnxx
6272.07938.058.29.8
421.158.29.8 08.39.8
21
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· 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
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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”
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List of Formulae
(1) 95 % Confidence interval for population proportion P.
(1) 90 % Confidence interval for population proportion P.
(1) 99 % Confidence interval for population proportion P.
n
ppp
ˆ1ˆ96.1ˆ
n
ppp
ˆ1ˆ645.1ˆ
n
ppp
ˆ1ˆ58.2ˆ
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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:
Here = X / n = 160 /500 = 0.32 p̂
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Solution of Example 8 page 268 cont;
Therefore 95% confidence interval for P
or
or
or
or ( 0.32 – 0.041 , 0.32+0.041)
or ( 0.28 , 0.36)
n
ppp
ˆ1ˆ96.1ˆ
500
32.0132.096.132.0
500
68.032.096.132.0
041.032.0
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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
= X / n = 86 / 400 = 0.215 p̂
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Solution of Exercise 2 page 273 cont;
Therefore the 90% confidence interval for P
or
or
or
or (0.215- 0.034 , 0.215 + 0.034)
or (0.181 , 0.249)
n
ppp
ˆ1ˆ645.1ˆ
400
215.01215.0645.1215.0
400
785.0215.0645.1215.0
034.0215.0
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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
= X/n = 16/75 = 0.21p̂
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Solution of Exercise 4 page 273
Therefore 99% confidence interval for P
or
or
or (0.21 – 0.12 , 0.21+0.12)
or (0.09 , 0.33)
n
ppp
ˆ1ˆ58.2ˆ
78
21.0121.058.221.0
12.021.0
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Confidence Interval for the difference between two Population proportions P1 – P2:
(1) 90% confidence Interval for P1 – P2
(2) 95% confidence Interval for P1 – P2
(3) 99% confidence Interval for P1 – P2
2
22
1
1121
ˆ1ˆˆ1ˆ645.1ˆˆ
n
pp
n
pppp
2
22
1
1121
ˆ1ˆˆ1ˆ96.1ˆˆ
n
pp
n
pppp
2
22
1
1121
ˆ1ˆˆ1ˆ58.2ˆˆ
n
pp
n
pppp
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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.
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Solution of Example 11 page 272
Here = 244/500= 0.48 and = 1200/200 = 0.60
Therefore 90% confidence interval for P1 – P2
or
or
or
or
or
or ( -0.12 – 0.0214 , -.0.12+ 0.0214)
or (-0.1414, -0.0986)
2p̂1p̂
2
22
1
1121
ˆ1ˆˆ1ˆ645.1ˆˆ
n
pp
n
pppp
2000
60.0160.0
5000
48.0148.0645.160.048.0
0214.012.0
2000
40.060.0
5000
52.048.0645.160.048.0
00012.000005.0645.112.0
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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.
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Solution of Exercise 10 page 274
Here =52/100 = 0.52 = 34/125 =0.272
Therefore 95% confidence interval for P1 – P2
or
or
or
or (0.248- 0.1225 , 0.248+0.1225)
or (0.1255 , 0. 3705)
1p̂ 2p̂
2
22
1
1121
ˆ1ˆˆ1ˆ96.1ˆˆ
n
pp
n
pppp
125
272.01272.0
100
52.0152.096.1272.052.0
125
)728.0(272.0
100
)48.0(52.096.1248.0
0016.00025.096.1248.0
1225.0248.0
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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.
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Solution of Exercise 10 page 274
Here = 24/100 = 0.24 = 13/100 = 0.13
Therefore 99% confidence interval for P1 – P2
or
or
or
or
or ( 0.13- 0.1389 , 0.13 + 0.1389)
or (-0.0089 , 0.2689)
1p̂ 2p̂
2
22
1
1121
ˆ1ˆˆ1ˆ58.2ˆˆ
n
pp
n
pppp
100
13.0124.0
100
13.0124.058.213.024.0
100
87.013.0
100
76.024.058.211.0
0011.00018.058.213.0
1389.013.0