Estimation of Parameters Umar khayam. Introduction The process of drawing inferences about a...

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

Interval Estimation

Point 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 = = = 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

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.

(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

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

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

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.

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)


n

SX 96.1

50

9.696.15.174

91.15.174

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.

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)


n

SX 58.2

100

390058.223500

10

390058.223500

2.100623500

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

22

1

21

21

ó·ó645.1

nnxx

2

22

1

21

21

óó96.1

nnxx

2

22

1

21

21

óó58.2

nnxx

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

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

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

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

Solution of Exercise 14 4

Here

Population I Population II

n1= 25 n2= 36

X1 = 80 X2 = 75

σ1 = 5 σ2 = 3

σ1 = 25 σ2 = 92 2


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


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 S2 = 5.6

S1 = 39.69 S2 = 31.362 2

Solution of Exercise 15 Page 264 cont;


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

· 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.



n

ppp

ˆ1ˆ96.1ˆ

n

ppp

ˆ1ˆ645.1ˆ

n

ppp

ˆ1ˆ58.2ˆ

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

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

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

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

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

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

Confidence Interval for the difference between two Population proportions P1 – P2:

(1) 90% 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

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

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


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.


Here =52/100 = 0.52 = 34/125 =0.272


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

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.


Here = 24/100 = 0.24 = 13/100 = 0.13


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

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