Chapter 7 L2

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Chapter 7

Hypothesis Testing Two

PopulationsWeek 10

L3: - Hypothesis Testing Between two means

(2 populations, variance unknown but not equal)- Hypothesis Testing Between two means(2 populations, variance unknown large samples)

- Hypothesis Testing- proportion

( 2 populations) 1September 2013

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Learning Objectives

At the end of the lesson student should beable to

Carry out hypothesis testing fordifference in means for two normalpopulations ( variances unknown)

Carry out hypothesis testing for

difference in means for difference in twoproportion

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Null Hypothesis is the same as before:

TWO V RI NCES RE UNKNOWN ND UNEQU L

Test about the difference

Test statistic:

)/()/(

)()(

2

2

21

2

1

2121

nSnS

xx

T

Case 2:2

2

2

1

But the degree of freedom is given by

1

)/(

1

)/(

)(

2

2

2

2

2

1

2

1

2

1

2

2

2

2

1

2

1

n

ns

n

ns

n

s

n

s

If is not an integer, round down to the nearest integer.

021

3

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Test about the difference

Case 2:2

2

2

1

021

Critical region:

Alternative Hypothesis Rejection Criteria ( Reject H0)

TEST BOUT TWO NORM L ME NS

WHEN TWO V RI NCES RE UNKNOWN

vv tTortT ,2/,2/ 211: H

vtT ,211: H

vtT ,211: H

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A two sided confidence interval for

when variance is unknown :

)%1(100 21

2

2

2

1

2

1,2/21

21

2

22

1

21

,2/21

n

s

n

stxx

n

s

n

stxx

Confidence interval:

Case 2: 22

2

1

Value of is from the formula earlier5

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6

Example 1:

Two companies manufacture a rubber material intended for use in

an automotive application. 25 samples of material from each

company are tested, and the amount of wear after 1000 cycles are

observed. For company 1, the sample mean and standard deviation

of wear are

and for company 2, we obtain

cycles1000/9.1andcycles1000/12.20 11 mgsmgx

Do the sample data support the claim that the two companiesproduce material with different mean wear? Assume each population

is normally distributed but unequal variances?

cycles1000/9.7andcycles1000/64.11 22 mgsmgx

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7

2. State the null hypothesis H0 and appropriate alternative

hypothesis, H1

0: 210 H 0: 211 H

3. Determine the appropriate test statistic

)/()/(

)()(

2

2

21

2

1

2121

nSnS

xxT

1. Identify the parameter of interest

Parameter of interest; the difference between the true average

Variances unknown and unequal21

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8

5. State the rejection region for the statistic

479.2ifHjectRe 0 T

4. Critical value given 0.01 479.226,01.0, tt v

2 22 2 2 2

1 2

1 2

2 2 2 22 22 2

1 2

1 2

1 2

(1.9) (7.9)

25 2526.77

(1.9) (7.9)

25 25

24 241 1

26

s s

n n

s s

n n

n n

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9

7. Make a decision

Decision: since we reject H0

Enough evidence to support that

479.222.5 T

21

6. Compute the value of the test statistic

22.5

25

)9.7(

25

)9.1(

0)64.1112.20(

22

T

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10

Exercise:

The following data represent the running times of films produced

by 2 motion-picture companies. Test the hypothesis that theaverage running time of films produced by company 2 exceeds the

average running time of films produced by company 1 by 10

minutes against the one-sided alternative that the difference is less

than 10 minutes? Use 0.01 and assume the distributions of

times to be approximately normal with unequal variances.

Time

Company

X1 102 86 98 109 92

X2 81 165 97 134 92 87 114

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11


hypothesis, H1

10: 120 H 10: 121 H


)/()/(

)()(

2

2

21

2

1

1212

nSnS

xxT

1. Identify the parameter of interest

Parameter of interest; the difference between the true averageVariances unknown and unequal

12

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12

5. State the rejection region for the statistic

998.2ifHjectRe 0 T

4. Critical value given 0.01 998.27,01.0, tt v

38.76/)7/333.913(4/)5/8.78(

)7/333.9135/8.78(

1

)/(

1

)/(

)(

22

2

2

2

2

2

2

1

2

1

2

1

2

2

2

2

1

2

1

n

ns

n

ns

n

s

n

s

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13

7. Make a decision

Decision: since , we fail to reject H0

Not enough evidence that

998.222.0 T

1012

22.07/333.9135/8.78

10)4.97110(

)/()/(

)()(

2

2

21

2

1

1212

nSnS

xxT

6. Compute the value of the test statistic

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Test statistic:

Test about the difference between two means (Large Samples)

0

2

2

2

1

2

1

2121 Hunder)1,0(~)()(

N

n

S

n

S

xxZ

Null Hypothesis:

vs210: H 211211211 :or:or,: HHH


211: H

211: H

211: H

2/2/ zZorzZ

zZ

zZ

Case 3:

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A confidence interval for is)%1(100 21

2

2

2

1

2

1

2/2121

2

2

2

1

2

1

2/21

n

S

n

Szxx

n

S

n

Szxx

The procedures used in hypothesis testing for 2 populationsare the same as in the hypothesis testing for one population.

Conclusion of the test could also be based on the confidence

Interval and the p-value.

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Example 3:

A study wants to investigate if the mean price of a new Proton car has

changed in 2012 compared to 2011. A sample of cars were taken duringthese two years and summary information as given below:

i. Perform the test for the above investigation. What is your conclusion?

ii. Find the two sided 95% CI for the difference between the two means in

2011 and 2012. What is your conclusion comparing with part (i).

Year 2011 2012

Mean car price RM 45,000 RM 48,000

Standard deviation of car

prices

RM 1,500 RM 2,000

Number of cars 150 200

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Solution

1. From the problem context, identify the parameter of interest

21


hypothesis, H1

210 : H 211: H


)/()/(

)()(

2

2

21

2

1

2121

nSnS

xxZ

To test the difference between the true average large sample

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6. Compute any necessary sample quantities, substitute these into

the equation for the test statistic, and compute the value

,200,150,2000,1500,000,48,45000212121

nnSSXX

Compute the value of the test statistic:

04.16

)200/2000()150/1500(000,48000,45

)/()/()()(

22

2

2

21

2

1

2121

nSnS

xxZ

Find the p - value

P-value = 0)11(2))04.16(1(2 Since p-value < 0.05, we reject the null hypothesis. We conclude that the mean price

of car in 2012 has change significantly compared to 2011.

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From formula:

2

2

2

1

2

1

2/2121

2

2

2

1

2

1

2/21)()()(

n

S

n

Szxx

n

S

n

Szxx

Find a 95% CI on the difference in means.

200

2000150

150096.14800045000200

2000150

150096.1480004500022

21

22

32.263367.336621

Since 1 2 0is not in the interval then we reject H0.

Both propellants are not the same mean burning rate.

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Exercise:

In a test to compare the effectiveness of two drugs designed

to lower cholesterol levels, 75 randomly selected patientswere given drug A and 100 randomly selected patients were

given drug B. Those given drug A reduced their cholesterol

levels by an average of 40mg with standard deviation 12mg,

and those given drug B reduced their cholesterol levels by anaverage 42mg with standard deviation of 15mg.

i. Perform the test for the above investigation. Can you

conclude that the mean reduction using drug B is greater

than that of drug A?ii. Find the two sided 95% CI for the difference between the

two means of reductions using drug A and drug B. What is

your conclusion comparing with part (i).

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TEST FOR TWO PROPORTIONS


Null Hypothesis:210: ppH ):,:(or 210210 ppHppH

2/2/ zZorzZ 211: ppH

zZ211: ppH

211: ppH zZ

Test Problems about two proportions:

Test statistic:

21

21

21

2

22

1

11

n1

n1

21

ppwhen)1,0(NZ

nn

XXp,

n

Xp,

n

Xp;

))(p1(p

ppZ

21

21

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A two sided confidence interval for is:)%1(100 21 pp

ULppLL

n

pp

n

ppzppUL

n

pp

n

ppzppLL

21

2

22

1

112/21

2

22

1

112/21

)1()1()(

)1()1()(


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23

Solution

Step 1 Problem: Test about two proportions large samples

Step 2 H0: pA= pB H1: pA> pB

Example 4:In a study on the effects of sodium restricted diets on hypertension, 24

out of 55 hypertensive patients were on sodium restricted diets, and 36

out of 149 non-hypertensive patients were on sodium restricted diets.

i. Test the hypothesis that the proportion of patients on sodium

restricted diets is higher for hypertensive patients at a=0.05.

ii. What is the P-value for this test?

iii. Construct a two sided 95% CI and comment.

Step 3 Determine the appropriate test statistic

BA

BA

B

B

B

A

A

A

BA

BA

BA

nn

XXp,

n

Xp,

n

Xpwhere

ppifN(0,1)))(p(1p

ppZ

n1

n1

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24

Step 4 Critical value given 0.05 z= z 0.05 = 1.645

645.1jectRe 0 ZifH

Step 5 Rejection region for the statistic

Step 6 Compute the value of the test statistic:

29.014955

3624

where,9455.4

149

1

55

1)30.01(30.0

24.044.0

24.0,44.0,36,24,149,55

0

pz

ppxxnn BABABA

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25

Step 7.Make a decision

Since z0> 1.65, then we reject H0. It means that enough

evidence to claim that the proportion of patients on

hypertension is higher than non hypertension patients

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Example 5:

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26

Solution

Step 1 Problem: Test about two proportions large samples

Step 2 H0: p1= p2 H1: p1> p2

Example 5:

A vote is to be taken among residents of a town and the

surrounding county to determine whether a proposed chemical

plant should be constructed. If 120 of 200 town voters favour the

proposal and 240 of 500 county residents favour it, would youagree that the proportion of town voters favouring the proposal is

higher than the proportion of county voters? Use 0.05


21

21

2

22

1

11

21

n1

n1

21

nn

XXp,

n

Xp,

n

Xpwhere

ppifN(0,1)))(p(1p

ppZ

21

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27

Step 4 Critical value given 0.05 z = z 0.05 = 1.645

645.1jectRe 0 ZifH

Step 5 Rejection region for the statistic


500,200,240,120 2121 nnXX

21

21

2

22

1

11

21

n1

n1

21

nn

XXp,

n

Xp,

n

Xpwhere

ppifN(0,1)))(p(1p

ppZ

21

2.9

500200

240120,

500

240,

200

120 21

Z

ppp

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28


Decision: since , we reject H0 and agree

that the proportion of town voters favouring the

proposal is higher than the proportion of county voters.

645.19.2 Z

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TEST FOR TWO V RI NCES


):

or,:or:vs:

2

2

2

11

2

2

2

11

2

2

2

11

2

2

2

10

H

HHH

2

2

2

11: H

2

2

2

11: H

2

2

2

11: H

Test Problems about two variances:

Test statistic:

29

2

2

2

10

s

s

F

1,1,2/101,1,2/0 2121or

nnnn fFfF

1,1,0 21 nnfF

1,1,0 21 nnfF

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A two sided confidence interval for is:)%1(100 2

2

2

1


30

1,1,2/22

2

1

22

2

1

1,1,2/122

2

1

1212

nnnn fs

s

fs

s

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Example 6:

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31

Solution

Step 1 Problem: Test about two population variances

Step 2

Example 6:A random sample of 12 air pollution index at UTP station produced a

variance 0.0340 while a random sample of another 13 air pollution index

at Tronoh station produced a variance 0.0525.

i. Are the population variances equal?. Use a = 0.05.

ii. Find the 95% two-sided confidence interval on the ratio of two

variances.


2

2

2

11

2

2

2

10 :vs: HH

2

2

21

0s

sF

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Step 4 Critical value a=0.05, fa/2, 12,13 = 3.15

Step 5 Rejection region: Reject IF


32.015.3

11or15.3

13,12,025.0

13,12,975.0013,12,025.00 f

fFfF

0H

648.00525.0

0340.02

2

2

10

s

sF


Since 0.32

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A 95% two sided confidence interval for is:222

1


33

041.2206.0

)15.3(0525.0

0340.0

15.3

1

0525.0

0340.0

1

2

2

2

1

2

2

2

1

1,1,2/2

2

2

12

2

2

1

1,1,2/

2

2

2

1

1,1,2/2

2

2

1

2

2

2

11,1,2/12

2

2

1

12

12

1212

nn

nn

nnnn

fss

fss

fs

sf

s

s

September 2013

Documents

Chapter 7 L2