Inferencing Between Two Samples: Z Tests & Confidence Intervals for Difference of Means - , Study notes of Data Analysis & Statistical Methods

Download Inferencing Between Two Samples: Z Tests & Confidence Intervals for Difference of Means - and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity!

Purdue University

Lecture 18: Inferences Based on Two Samples

Devore: Section 9.1-9.

Aug, 2006

Purdue University

z

Tests and Confidence Intervals for a Difference Between Two

Population Means

•

An example of such hypothesis would be

μ

1

−

μ

2

= 0

or

σ

1

σ

2

. It may also be appropriate to estimate

μ

1

−

μ

2

and

compute its

100(

−

α

)%

confidence interval

•

1. Assumptions

X

1 ,... , X

m

is a random sample from a population with

mean

μ

1

and variance

σ

(^12)

Y

1 ,... , Y

n

is a random sample from a population with mean

μ

2

and variance

σ

(^22)

3. The

X

and

Y

samples are independent of one another

Aug, 2006

Purdue University

•

The natural estimator of

μ

1

−

μ

2

is

¯

X

−

Y¯

. To standardize this

estimator, we need to find

E

(

¯

X

−

¯

Y

)

and

V

(

¯

X

−

Y¯

)

.

•

E

(

¯

X

−

Y¯

) =

μ

1

−

μ

2 , so

¯

X

−

¯

Y

is an unbiased estimator

of

μ 1 − μ 2.

•

E The proof is elementary:

(

¯

X

−

Y¯

) =

E

(

¯

X ) − E ( ¯

Y

) =

μ

1

−

μ

2

Aug, 2006

Purdue University

•

The standard deviation of

¯

X

−

¯

Y

is

σ

¯

X

−

Y¯

=

√

σ 12

m

+

σ (^22)

n

•

The proof is also elementary:

V

(

¯

X

−

¯

Y

) =

V

(

¯

X

) +

V

(

¯

Y

) =

σ

(^12)

m

+

σ

(^22)

n

The standard deviation is the root of the above expression

Aug, 2006

Purdue University

The Case of Normal Populations with Known Variances

•

As before, this assumption is a simplification.

•

Under this assumption,

Z

=

¯

X

−

Y¯

− ( μ 1 − μ 2 )

√

σ (^12)

m

+

σ (^22)

n

(1)

has a standard normal distribution

•

The null hypothesis

μ

1

−

μ

2

= 0

is a special case of the more

general

μ

1

−

μ

2

= ∆

0

. Replacing

μ

1

−

μ

2

in (1) with

∆

0

gives us a test statistic.

Aug, 2006

Purdue University

•

1. alternatives:The following summary considers all possible types of

H a : μ 1 − μ 2 > ∆ 0

has the rejection region

z

≥

z

α

H a : μ 1 − μ 2 < ∆ 0

has the rejection region

z

≤ −

z

α

H a : μ 1 − μ 2 6

= ∆

0

has the rejection region

z

≥

z

α/

2

or

z

≤ −

z

α/

2 .

Aug, 2006

Purdue University

Example

•

Consider Ex. 9.1 in Devore. Sample sizes are

m

= 20

and

n

= 25

. Note that

m

=

n

...it is not important now but will be

later...

•

exploratory data analysisNote that the normality suggestion is based on some

•

The hypotheses are

H 0 : μ 1 − μ 2

= 0

and

H a : μ 1 − μ 2 6

= 0

•

The test statistic is

z

=

¯x

−

¯y

√

σ (^12)

m

+

σ (^22)

n

Aug, 2006

Purdue University

•

For a level of significance

α

= 0

.

,

z

α/

2

=

z

. 005

= 2

.

and

the rejection regions is

z

≤ −

.

or

z

≥

.

.

•

The computed value of

z

-statistic is

−

.

which is well within

the rejection region. The

P

(^) -value for this rejection region is

2(

−

Φ(

.

66))

≈

0

which mean rejection at

any reasonable

level.

Aug, 2006

Purdue University

Fall 2006

Type II Error and the Choice of the Sample Size

•

H Consider the case of an upper-tailed alternative hypothesis

a : μ 1 − μ 2 > ∆ 0.

•

The rejection region is

¯x

−

¯y

≥ ∆ 0 + z α σ ¯

X

−

(^) Y¯

. Therefore,

P

(

Type II Error

) =

P

(

¯

X

−

¯

Y <

∆ 0 + z α σ ¯

X

−

Y¯

when

μ

1 −

μ

2

= ∆

′ )

•

Since

¯

X

−

¯

Y

is normally distributed under the alternative

μ

1

−

μ

2

= ∆

′

with mean

∆

′

and standard deviation

σ

¯

X

−

Y¯

=

√

σ (^12)

m

+

σ 22

n

, we have

β

(∆

′ ) = Φ

( z α − ∆ ′ − ∆ 0

σ

)

Aug, 2006

Purdue University

•

alternatives. In particular, ifSimilar results can be easily obtained for the other two possible

H a : μ 1 − μ 2 < ∆ 0

, we have

β

(∆

′ ) = 1

− Φ ( − z α − ∆ ′ − ∆ 0

σ

)

•

If

μ 1 − μ 2 6

= ∆

0 , the probability of Type II Error is

Φ

(

z

α/

2 − ∆ ′ − ∆ 0

σ ) − Φ ( − z

α/

2 − ∆ ′ − ∆ 0

σ

)

Aug, 2006

Purdue University

Example

•

probability of detecting a differenceConsider Example 9.3 from Devore. Suppose that the

.

between the two means

should be

.

. Can the

.

level test with

m

= 20

and

n

= 25

support this?

•

β For a two-sample test we have

(5) = Φ

(

.

−

−

0

.

) − Φ ( − 2.

−

−

0

.

)

=

.

•

β Because the rejection region is symmetric, we have

(

−

5) =

β

(5)

, and, therefore, the probability of detecting a

difference of

.

is

−

β

(5) =

.

.

•

We can conclude that slightly larger sample sizes are needed.

Aug, 2006

Purdue University

•

P To determine a sample size that satisfies

(

Type II Error when

μ

1

−

μ

2

= ∆

′ ) =

β

we need to solve

σ

(^12)

m

+

σ

(^22)

n

=

(∆

′ − ∆ 0 ) 2

( z α + z β

)

2

•

For two equal sample sizes this yields

m = n = ( σ

(^12)

+

σ

2 2 (^) )(

z

α

+

z

β (^) )

2

(∆

′ − ∆ 0 ) 2

Aug, 2006

Purdue University

Large-Sample Tests

•

unnecessary and variancesIn this case, the assumption of normality for the data is

σ

1 2 (^) ,

σ

(^22)

need not be known

•

This is because for large

n

the variable

Z

=

¯

X

−

Y¯

− ( μ 1 − μ 2 )

√

S

(^12)

m

+

S

(^22)

n

is approximately standard normal

Aug, 2006

Purdue University

•

Then, if the null hypothesis is

μ

1

−

μ

2

= ∆

0 , the test statistic

Z

=

¯

X

−

Y¯

−

∆

0

√

S

(^12)

m

+

S

(^22)

n

is approximately standard normal under the null hypothesis

•

This test is usually appropriate if both

m >

and

n >

Aug, 2006

Purdue University

Example

•

main competitor. If a study showed that a sample ofA company claims that its light bulbs are superior to those of its

n

1

= 40

of

its bulbs has a mean lifetime of

hours of continuous use

with a standard deviation of

hours , while a sample of

n

2

= 40

bulbs made by its main competitor had a mean lifetime

of

hours of continuous use with a standard deviation of

hours, does this substantiate the claim at the

0

.

level of

significance?

Aug, 2006

Purdue University

• H 0 : μ 1 − μ 2

= 0

and

H a : μ 1 − μ 2 > 0

•

Reject

H

0

if

Z >

.

•

Calculations:

z

=

−

√

27

2

40

+

31

2

40

= 1

.

•

Decision:

H

0

cannot be rejected at

α

= 0

.

; the

p

-value is

0

.

Aug, 2006

Purdue University

Confidence intervals for

μ

1

−

μ

2

•

Since the test statistic

Z

that we just described is exactly

normal when

σ

1 2

and

σ

(^22)

are known,

P

 

−

z

α/

2

< Z

=

¯

X

−

¯

Y − ( μ 1 − μ 2 )

√

σ 12

m

+

σ 22

n

< z

α/

2  

= 1

−

(^) α

•

The

100(

−

α

)%

CI is easy to derive from this probability

statement; it is

¯x

−

¯y

±

z

α/

2 σ

¯

X

−

Y¯

where

σ

¯

X

−

Y¯

is a square root expression.

Aug, 2006

Purdue University

•

If both

m

and

n

are large, CLT implies that the normality

assumption is not necessary and substitution of

s

i 2

for

σ

i 2

,

i

= 1

,

will produce an

approximately

100(

−

α

)%

CI

•

More precisely, such an interval is

¯x

−

¯y

±

z

α/

2 √

s

1 2

m

+

s

2 2

n

•

Again, this result should be used only if both

m

and

n

exceed

•

Note that this CI has a standard form of

θˆ

±

z

α/

2 σ

θˆ

Aug, 2006

Purdue University

Example

•

An experiment was conducted in which two types of engines,

A

and

B

, were compared. Gas mileage, in miles per gallon, was

measured.

experiments were conducted using engine type

A

and

were done for engine type

B

. The gasoline used and

engineother conditions were held constant. The average mileage for

A

was

mpg and the average for machine

B

was

mpg. Find an approximate

96%

CI on

μ

B

−

μ

A

, where

μ

A

and

μ

B

are population mean gas mileage for machines

A

and

B

,

respectively. Sample standard deviation are

and

for

machines

A

and

B

, respectively.

Aug, 2006

Purdue University

•

The point estimate of

μ

B

−

μ

A

is

¯x

B

−

¯x

A

= 42

−

36 = 6

.

For

α

= 0

.

, we find the critical value

z

. 02

= 2

.

.

•

Thus, the confidence interval is

±

.

√

+

= (

.

,

.

87)

Aug, 2006

Purdue University

The Two-Sample t-test

•

Both populations are normal, so thatAssumptions:

X

1 ,... , X

m

is a random

sample from a normal distribution and so is

Y

1 ,... , Y

n

.

constructing a normal probability plot of theThe plausibility of these assumptions can be judged by

x

i s and another of

the

y

i s.

Aug, 2006

Purdue University

•

standardized variableWhen the population distributions are both normal, the

T

=

¯

X

−

¯

Y − ( μ 1 − μ 2 )

√

S (^12)

m

+

S

22

n

has approximately

t

distribution with

ν

df

•

ν

can be estimated from data as

ν

=

(

s 12

m

+

s 22

n

)

2

( s 12 /m

) 2

m − 1 + ( s

22 /n

) 2

n −

1

•

ν

has to be rounded down to the nearest integer...why not up?

Aug, 2006

Purdue University

•

The

two-sample confidence interval for

μ

1

−

μ

2

with

confidence level

100(1

−

α

)%

is

¯x

−

¯y

±

t α/

2 ,ν

√

s

12

m

+

s

22

n

described earlier. A one-sided confidence bound can also be calculated as

•

The two-sample

t -test for testing

H 0 : μ 1 − μ 2

= ∆

0

is

conducted using the test statistic

t

=

¯x

−

¯y

−

∆

0

√

s 12

m

+

s 22

n

Aug, 2006

Purdue University

Fall 2006

Alternative hypothesis

Rejection region for approximate level

α

test

H a : μ 1 − μ 2 > ∆ 0 t ≥ t

α,ν

H a : μ 1 − μ 2 < ∆ 0 t

≤ −

t α,ν

H a : μ 1 − μ 2 6

= ∆

0

either

t

≥

t α/

2 ,ν

or

t

≤ −

t α/

2 ,ν

•

one-sample testA P-value can be computed exactly as we did before for the

Aug, 2006

Purdue University

Example

•

illustrate it:Consider example 9.6 in Devore. The following table helps to

Fabric Type

Sample Size

Sample Mean

Sample Standard Deviation

Cotton

10

51.71

.79

Triacetate

10

126.14

3.59

Aug, 2006