Ch. 8 EstimationCh. 9 Hypothesis Testing

Confidence Intervals & Hypothesis Tests:

differences between means

of normal populations

1. Matched pairs/ dependent samples

2. Independent samples

Matched pairs/ dependent samples

Sample members are chosen in pairs, one member from each population.

Members of a pair should be:

very similar to one anotherOR

the same member “before” and “after” a specific event.

Paired t-test

We have n matched pairs of observations,

(x1 ,y1 ), (x2 ,y2 ), … , (xn ,yn )

from populations with mean µx and µy .

Consider the n differences di = xi – yi

d = sample mean of differences

sd = sample standard deviation of differences

µd = µx - µy

_

A drug company wants to compare the effect of a new drug for treating

male-pattern baldness with Rogaine.

The new drug is applied to a 4 square inch area on the left side of the scalp of six bald men, and Rogaine is applied to

an equal area on the right.

Find the 95% confidence interval.

Paired t-test

L,U = d tα/2sd

Confidence Intervalfor matched pairs:

di = Xi new - yi Rogainefor i = 1, . . . , n.

+_

We need normality for small sample problems.

sd =sd

n

sd =di − d ( )2

i=1

n∑

n −1

df = n – 1

95% Confidence Interval

202

314

84

578

874

602

185

251

25

412

414

444

17

63

59

166

460

158

Joe

Fred

Harry

Chester

Max

Duncan

New Rogaine dname95% Confidence Interval

sd = 161.165

sd = 65.80

95% Confidence Interval

n=6df=n-1df=5

d = 153.833_

L,U = d tα/2sd

L,U = 153.833 2.571 65.80( )

95% Confidence Intervaldf = n−1 = 5

+_

+_

= (-15.27 , 322.94)

Difference between pop. means:Independent samples

Use the differences between sample means

as the base point for inference.

E(X-Y) = E(X) – E(Y) = µX - µy

_ _ _ _

All linear combinationsof jointly normally distributedrandom variables are always

normally distributed.

If X and Y are normally distributed,

so is (X – Y)

_ _

_ _

µ1µ 2 x 1x 2

x 1 − x 2µ1 − µ 2

x 1 − x 2µ1 − µ2( )0acceptance region rejection

regionrejection region

In general,

However,random sampling independence

so covariance = 0

Var(X-Y) = Var(X) + Var(Y) – 2 Cov(X,Y)

Var(X-Y) = Var(X) + Var(Y)_ _ _ _

_ _ _ _ _ _

For independent X and Y:

Var(X-Y) = Var(X) + Var(Y)

= σ2 + σ2

_ _

___ ___

_ _

x y

nx ny

σx-y = σ2 + σ2__ __nx ny

________

√

C.V.L

,C.V.U =

µ1 − µ2( )

0zα/2σx

1−x 2

+_

Difference between pop. means:Independent samples, σ known

U.S. companies claim that the Japanese are “dumping” steel in the U.S. by selling it at a lower average price than the average price in Japan. The price of Japanese steel is normally distributed and has a known variance of 180 yen in Japan and 160 yen in U.S. at current rate of exchange. A sample of 10 sales in the U.S. had an avg. price of 200 yen while 20 sales in Japan averaged 210 yen.

At the 1% level of significance, test the null hypothesis that Japan is “dumping” steel in the U.S.

Difference between pop. means:Independent samples, σ known

H0: µJapan ≤ µU.S.

H1: µ Japan > µU.S.

H0: µJapan − µU.S. ≤ 0

H1: µJapan − µU.S. > 0

raw score test statistic:

X Japan − X

US

average pricedifference

Under the “effective” null hypothesis:

E X

Japan− X

US( )= µJapan

− µUS

= 0

X Jap an −X US

2

σ =X Japan

2

σ +X US

2

σ

mean:

variance:

σ

X Japan − X US

2 = Japan

2σn

Japan

+ US

2σn

US

X Japan

− X US

A sample of 10 sales in the U.S. showan average price of 200 yen while a sampleof 20 sales in Japan average 210 yen.

α = 0.01

0 C.V.

raw score space:

0 Zα

α = 0.01

z

standardized space:

α = 0.01

z

First, carry out the test in standardized space:

For α = 0.01, the normal table yields Zα = 2.33

z =x

Japan− x

US( )− µJapan

− µUS( )

σJapan

2

nJapan

+ σUS

2

nUS

z =210 − 200( )− 0( )

180

20+160

10

= 2

Z = 2falls to the left of

Zα = 2.33 .

Do not rejectthe null hypothesis

X Japan

− X US

α = 0.01

0 C.V.

Next, carry out the test in raw score space:

C.V .= (µ

Japan− µ

US) + Zα

σJapan

2

nJapan

+σ

US

2

nUS

C.V .= (0) + 2.33

180

20+

160

10= 11.65

X Japan

− X US

α = 0.01

Testing in raw score space (continued):

X Japan

− X US

=210 −200 = 10

10 < 11.65raw score test statistic value < critical value (C.V.)

Do not reject the null hypothesis.

= 2.33

α = 0.01

= 2.33Z = 2

p-value = 0.0228

Since p-value > α , then do not reject null hypothesis.

Now, carry out the test using the p-value:Normal table shows P (0 < z < 2) = 0.4772 so for a one-tail test the p-value = 0.0228 .

Z = 2 is thestandardizedtest statisticvalue.

The p-value is the probabilityof observing what youobserve or somethingmore extreme.

Difference between pop. means:Independent samples, σ unknown (use s)

Domino Pizza is choosing between the Ford Escort and Honda Civic. If the difference for time-in-repair after 2

years is more than 15 days in favor of Honda, Honda will be cheaper.

At α=0.1, test the null hypothesis that the time-in-repair difference is less than or equal

to 15 days.

Ford: nf = 100, sf = 8.6, xf = 26

Honda: nh = 100, sh = 7.2, xh = 10

_

_

H0 :µFord – µHonda ≤ 15

H1:µFord – µHonda > 15

s x 1 − x 2=

8.62

100+

7.22

100

sx 1 − x 2= s1

2

n1+ s2

2

n2

= 1.122

H0 :µFord – µHonda ≤ 15

H1:µFord – µHonda > 15

C.V. = 15 + 2.33 1.122( )C.V. = 17.614

C.V. = µF −µH( )0 + tαsx 1−x 2

acceptance region rejection region

15

17.614

x F − x H

16