Constructing Confidence Intervals for the Difference in Means and Proportions, Study notes of Calculus

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Parameter Estimation: Part III

Cyr Emile M’LAN, Ph.D.

[email protected]

Parameter Estimation: Part III

Introduction

Text Reference

Introduction to Probability and

Statistics for Engineers and Scientists, Chapter 7.

Reading Assignment

Sections 7.4-7.5, December

1st

Parameter Estimation: Part III

Constructing Confidence Interval for

μ

1

−

μ

2

and derive the following confidence intervals as we didbefore in other cases.Thus, a

α

confidence interval for

μ

1

μ

2

is

X

Y

z

α/

2

σ

2 1

n

σ

2 2

m

X

Y

z

α/

2

σ

2 1

n

σ

2 2

m

or equivalently

X

Y

z

α/

2

σ

2 1

n

σ

2 2

m

Parameter Estimation: Part III

Constructing Confidence Interval for

μ

1

−

μ

2

A

α

one-sided upper confidence interval for

μ

1

μ

2

is

X

Y

z

α/

2

σ

2 1

n

σ

2 2

m

A

α

one-sided lower confidence interval for

μ

1

μ

2

is

X

Y

z

α/

2

σ

2 1

n

σ

2 2

m

Parameter Estimation: Part III

Constructing Confidence Interval for

μ

1

−

μ

2

Solution

We have:

Var

(

X

−

Y

) =

σ

21

n

σ

22

m

=

σ

2

(

1 n

1

m

)

We know that the statistics,

S

2 1

and

S

2 2

(sample variances for the two

populations) are two unbiased estimates of

σ

2

. i.e., E

(

S

2 1

) =

σ

2

and

E

(

S

2 2

) =

σ

2

. In addition, under the normality assumption, we have

(

n

−

S

2 1

σ

2

∼

χ

2 n

−

1

and

(

m

−

S

2

2

σ

2

∼

χ

2 m

−

1

As a consequence, Var

(

S

2 1

) =

2

σ

4

n

−

1

and Var

(

S

2 2

) =

2

σ

4

m

−

1

The statistic

S

2 p

=

(

n

−

S

2 1

• (

m

−

S

2

2

n

m

−

2

is an unbiased estimator of

σ

2

and also have smaller variance than

both sample variance

S

2 1

and

S

2 2

, therefore better.

S

2 p

is called a

pooled sample variance

.

Parameter Estimation: Part III

Constructing Confidence Interval for

μ

1

−

μ

2

Indeed, we have:

E

(

S

2 p

)

=

(

n

−

E

(

S

2 1

)

• (

m

−

E

(

S

2 2

)

n

m

−

2

=

σ

2

Var

(

S

2 p

)

=

(

n

−

2

Var

(

S

2

1

)

• (

m

−

2

Var

(

S

2 2

)

(

n

m

−

2

)

2

=

2

σ

4

n

m

−

2

It can be shown that

(

n

m

−

2

)

S

2 p

σ

2

∼

χ

2 n

m

−

2

.

Hence,

The sampling distribution of the statistic,

T

is as follows:

T

=

(

X

−

Y

)

−

(

μ

1

−

μ

2

)

S

p

√

1 n

1

m

∼

T

n

m

−

2

.

Parameter Estimation: Part III

Small-Sample Confidence Intervals for

μ

−

μ

Example 7.

:

A farm-equipment manufacturer wants to compare the average dailydowntime for two sheet-metal stamping machines located in twodifferent factories. Investigation of company records for 10 randomlyselected days on each of the two machines gave the followingresults.

n

1

= 10

¯ x

1

= 12

min

s

2 1

= 6

n

2

= 10

¯ x

2

= 9

min

s

2 2

= 4

Assume that we have the common variance assumption holds.

Estimate the difference between the average daily downtime for thetwo sheet-metal stamping machines with confidence coefficient 0.95.What additional assumptions are necessary for the method used tobe valid?

(

12

−

±

2

.

101 (

.

√

1

10

1

10

= 3

±

2

.

101

.

Parameter Estimation: Part III

• p. 10/

A guide to Political Philosophy

You have two cows.

Socialism:

You give one to your neighbour.

Communism:

The government takes both

and give you the milk. Fascism:

The government takes both and

sells you the milk. Nazism:

The government takes both and

shoots you. Capitalism:

You sell one and buy a bull.

Trade Union:

They take both from you, shoot

one, milk the other one, and throw the milkaway.

Parameter Estimation: Part III

Constructing Confidence Interval for

p

Example 7.

:

(Example 7.2 revisited)

Find an approximate a

two-sided confidence

intervals for

p

Solution

A

two-sided confidence intervals for

p

is given by

̂ p

±

z

α/

2

√

̂

p

(

−

̂

p

)

n

.

Here

p

and

n

p

p

The desired 99% two-sided confidence interval is then

0

.

19

±

2

.

576

√

0

.

19(

−

0

.

200

= 0

.

19

±

0

.

0715

or

[

2615]

Parameter Estimation: Part III

Constructing Confidence Interval for

p

A better approximate

α

two-sided

confidence intervals for

p

is

̂ p

z

2 α/

2

2

n

±

z

α/

2

√

̂

p

(

−

̂ p

)

n

z

2 α/

2

4

n

2

1 +

z

2 α/

2

n

.

Example 7.23 revisited

:

The desired 99% two-sided confidence intervals isthen

[

2500]

Note that this interval is smaller than the interval [

2615]

derived previously.

Parameter Estimation: Part III

Large-Sample Confidence Intervals, cont.

Example 7.

:

A large firm made up of several companies hasinstituted a new quality-control inspection policy. Among60 artisans sampled in Company A, only 15 objected tothe new policy. Among 64 artisans sampled inCompany B, 20 objected to the policy. Estimate the truedifference between the proportions voicing no objectionto the new policy for the two companies with confidencelevel 0.98. Solution

:

Let

p

1

denote the proportion corresponding to company

A and

p

2

that of company B. A 98% CI for

p

1

p

2

is

(̂

p

1

−

̂

p

2

)

±

z

α/

2

√

̂ p

1

(

−

̂

p

1

)

n

1

̂ p

2

(

−

̂ p

2

)

n

2

.

Parameter Estimation: Part III

Large-Sample Confidence Intervals, cont.

Here

p

1

and

p

2

. We also

have

n

1

p

1

p

1

and

n

2

p

2

p

2

Hence, the desired 98% CI is then

(

.

25

−

0

.

±

2

.

33

√

0

.

25(

−

0

.

60

0

.

3125(

−

0

.

64

−

0

.

0625

±

0

.

1876

or

[

1251]

This interval contains zero. Thus, a zero value for thedifference between the proportions voicing no objectionto the new policy for the two companies ,

p

1

p

2

, is

“believable”

at the 98% level on the basis of the

observed data.

Parameter Estimation: Part III

Choice of the Sample Size

The length of the confidence interval,

Y

±

z

α/

2

σ

√

n

2

z

α/

2

σ

√

n

gives an idea of the precision or accuracy in the point

estimate

Y

Setting

z

α/

2

σ

n

to be

l

lead to the sample size

n

=

4

z

2 α/

2

σ

2

l

2

Hence,

the sample size required for the CI to have a

length

l

is

n

z

2 α/

2

σ

2

l

2

Parameter Estimation: Part III

Choice of the Sample Size, cont.

For a proportion

p

the sample size required for the CI to

have a length

l

is

n

=

4

z

2 α/

2

p

(

−

p

)

l

2

When

p

is unknown, we use the worst case-scenario

sample size

n

=

z

2 α/

2

l

2

which corresponds to setting

p

Parameter Estimation: Part III