Estimation

Statistical Methods Usha A. Kumar, IIT Bombay

Interval Estimation

In many practical problems, the exact knowledge of the parameter may not be necessary. It is quite adequate if an interval along with a probability statement is specified such that the probability that the random interval will cover the unknown parameter is a specified number.

Confidence intervals

Let The population distribution be normal. The value of the population standard

deviation be known.



µ X

1−α

Z0 α

2Z− α

2Z

α2

α2


Random interval

nX

nX ZZ σµσ

αα22

+≤≤−

nZX σ

α2

±

Random Interval

For , the random interval is

The probability is 0.95 that the random interval includes or covers the true value of .


0.05α =

1.96 , 1.96X Xn nσ σ − +

µ

Confidence Interval for If after observing we

compute the observed sample mean and then substitute in the random interval, the resulting fixed interval is called the confidence interval for .


µ

1 1 2 2, , , ,n nX x X x X x= = =

100(1 )%α−

xx

µ

Confidence Interval for The confidence interval can be

expressed as


µ

2 2

2 2

, is a 100(1- )% CI for

or

with 100(1- )% confidence.

x xn n

x xn n

Z Z

Z Z

α α

α α

σ σ α µ

σ σµ α

− +

− ≤ ≤ +


ExampleSuppose it is known that the life of electric tubes manufactured by a company is normally distributed with a standard deviation of 50 hours. A random sample of 49 pieces observed for life estimation showed an average life of 1180 hrs. Compute a 95 percent confidence interval for the population mean of tubes manufactured by this company.


ExampleA new drug discovered to decrease blood pressure was tested on 100 patients which shows a mean decrease of 18 units and a sample standard deviation of 6 units. Find the 99% confidence interval for the mean decrease in blood pressure due to the new drug.

Large sample Confidence Intervals

If n is sufficiently large, the standardized variable

has approximately a standard normal distribution. This implies that

is a large sample confidence interval for with confidence level approximately . This formula is valid regardless of the shape of the population distribution.


/XZ

nµ

σ−

=

/ 2x znασ

±

µ100(1 )%α−


Large sample Confidence Intervals - unknown

Consider the standardized variable

Both vary in value from one sample to another. However, for large n the substitution of S for adds little extra variability and hence this variable also has approximately a standard normal distribution.

σ

/XZs n

µ−=

and X µ

σ


ExampleA market survey is conducted to ascertain the proportion of smokers smoking a particular brand A. Out of 100 smokers surveyed, 64 were found to be smoking that brand. Find a 95% confidence interval for the proportion of smokers smoking brand A.

Confidence interval for Population proportion

Let p be the proportion of “successes” in a population.

If


10 and 10,np nq≥ ≥

/ 2 / 2 1(1 ) /p pP z z

p p nα α α −− < < ≈ − −

Confidence interval for Population proportion

A confidence interval for a population proportion p with confidence level approximately is

The approximate confidence limits are


100(1 )%α−2 2 2 2

/ 2 / 2 / 2 / 2/ 2 / 22 2

2 2/ 2 / 2

ˆ ˆ ˆ ˆˆ ˆ2 4 2 4,

1 ( ) / 1 ( ) /

z z z zpq pqp z p zn n n n n n

z n z n

α α α αα α

α α

+ − + + + +

+ +

/ 2ˆ ˆˆ pqp znα±


ExampleThe length of time required for persons taking the civil service test is assumed to be normally distributed. A random sample of 16 persons taking the test is conducted and their test times are recorded, yielding an average test time of 60 minutes with a standard deviation of 12 minutes. Find a 95 percent confidence interval for the population mean test time.


The t distribution

The population has a normal distribution.

The value of the population standard deviation is unknown.

Result

When is the mean of a random sample of n from a normal distribution with mean , the rv

has a probability distribution called a t distribution with n-1 degrees of freedom.


X

µ

/XTS n

µ−=


The t Distribution

Developed by British statistician, William Gosset

A family of distributions -- a unique distribution for each value of its parameter, degrees of freedom (d.f.)

Symmetric, Unimodal, Mean = 0, Flatter than Z


The density of T

The density of T is

where v is the degrees of freedom.

( 1) / 22[( 1) / 2]( ) 1 ,( / 2)

vv tf t t

vv vπ

− + Γ +

= + −∞ < < ∞ Γ


-3 -2 -1 0 1 2 3

Standard Normalt (d.f. = 25)

t (d.f. = 1)t (d.f. = 5)

Confidence Interval for

Let be the sample mean and sample standard deviation computed from the results of a random sample from a normal population with mean . Then a confidence interval for is


µ

and x s

µ100(1 )%α− µ

/ 2, 1 / 2, 1,n ns sx t x tn nα α− −

− +


Example

A random sample of 20 teletype operators indicates that their salaries fluctuate quite a bit. The sample standard deviation of their daily salaries is Rs. 90. Construct a 90 percent confidence interval on the population standard deviation of the daily salaries.

Result Let be a random

sample from a normal distribution with parameters . Then the rv

has a chi squared probability distribution with n-1 df.


1 2, , , nX X X

2 and µ σ

22

2 2

( )( 1) iX Xn Sσ σ

∑ −−=


Selected χ2 Distributions

df = 3

df = 5

df = 10

0


Chi square density functionThe continuous random variable X has a chi-squared distribution, with degrees of freedom, if its density function is given by

where is a positive integer.

/ 2 1 / 2/ 2

1( ) , 02 ( / 2)

v xvf x x e x

v− −= >

Γ

v

v

Confidence Interval for


22 21 / 2, 1 / 2, 12

2 22

2 2/ 2, 1 1 / 2, 1

2 2

2 2/ 2, 1 1 / 2, 1

2

( 1) 1

( 1) ( 1)

Thus

( 1) ( 1), is a 100(1- )% confidence

interval for .

n n

n n

n n

n SP

orn S n S

n s n s

α α

α α

α α

χ χ ασ

σχ χ

αχ χ

σ

− − −

− − −

− − −

−< < = −

− −< <

− −

2.σ

Estimation

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