11- 1

Chapter

Eleven

McGraw-Hill/Irwin

© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

11- 2Chapter Eleven

Two-Sample Tests of Two-Sample Tests of HypothesisHypothesisGOALS

When you have completed this chapter, you will be able to:

TWOConduct a test of hypothesis regarding the difference in two population proportions.

THREE

Conduct a test of hypothesis about the mean difference between paired or dependent observations.

ONE

Conduct a test of hypothesis about the difference between two independent population means.

11- 3Chapter Eleven continued

Two Sample Tests of Two Sample Tests of HypothesisHypothesis

GOALSWhen you have completed this chapter, you will be able to:

FOUR

Understand the difference between dependent and independent samples.

11- 4

Comparing two populations

Does the

distribution of the differences in sample

means have a

mean of 0?

Comparing two populations

If both samples contain at least 30 observations we use the z distribution as the test statistic.

No assumptions about the shape of the populations are required.

The samples are from independent populations.

The formula for computing the value of z is:

2

22

1

21

21

n

s

n

s

XXz

11- 5

EXAMPLE 1

with a standard deviation of $7,000 for a sample of 35 households. At the .01 significance level can we conclude the mean income in Bradford is more?

Two cities, Bradford and Kane are separated only by the Conewango River. There is competition between the two cities. The local

paper recently reported that the mean household income in Bradford is $38,000 with a standard deviation of $6,000 for a sample of 40 households. The same article reported the mean income in Kane is $35,000

11- 6

Example 1 continued

Step 2

State the level of significance. The .01 significance level is

stated in the problem.

Step 3

Find the appropriate test statistic. Because both

samples are more than 30, we can use z as the test statistic.

Step 1 State the null and

alternate hypotheses.H0: µB < µK

H1: µB > µK

Step 4 State the decision rule.The null hypothesis is rejected if z is greater than 2.33 or p < .01.

11- 7

Example 1 continued

98.1

35

)000,7($

40

)000,6($

000,35$000,38$22

z

Step 5: Compute the value of z and make a decision.

The p(z > 1.98) is .0239 for a one-tailed test of significance.

Because the computed Z of 1.98

< critical Z of 2.33, the p-value of .0239 > of .01, the decision is to not reject the null hypothesis. We cannot conclude that the mean household income in Bradford is larger.

11- 8

21

21

nn

XXpc

Two Sample Tests of ProportionsTwo Sample Tests of Proportions investigate whether two samples came from populations with an equal proportion of successes.

The two samples are pooled using the following formula.

where X1 and X2 refer to the number of successes in the respective samples of n1 and n2.

The value of the test statistic is computed from the following formula.

21

21

)1()1(

n

pp

n

pp

ppz

cccc

where X1 and X2 refer to the number of successes in the respective samples of n1 and n2.

11- 9

Example 2

Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than 5 days last year, while a sample of 300 unmarried workers

showed 35 missed more than five days. Use a .05 significance level.

11- 10

Example 2 continued

The null and the alternate hypothesesH0: U < M H1: U > M

The null hypothesis is rejected if the computed value of z is greater than 1.65 or the p-value < .05.

The pooled proportion

250300

2235

cp

= .1036

11- 11

Example 2 continued

10.1

250

)1036.1(1036.

300

)1036.1(1036.250

22

300

35

z

The p(z > 1.10) = .136 for a one-tailed test of significance.

Because a calculated z of 1.10 < a critical z of 1.96, p of .136 > of .05, the null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers.

11- 12

Small Sample Tests of Means

The required assumptions1. Both populations must follow

the normal distribution.2. The populations must have

equal standard deviations.3. The samples are from

independent populations.

Small Sample Tests of MeansSmall Sample Tests of MeansThe t distribution is used as the test statistic if one or more of the samples have less than 30 observations.

11- 13

Small sample test of means continued

2

)1()1(

21

222

2112

nn

snsns p

21

2

21

11

nns

XXt

p

Step Two: Determine the value of t from the following formula.

Finding the value of the test statistic requires two steps.

Step One: Pool the sample standard deviations.

11- 14

Example 3

A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg.

A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the .05 significance level can the EPA conclude that the mpg is higher on the imported cars?

11- 15

Example 3 continued

Step 1 State the null and

alternate hypotheses. H0: µD > µI

H1: µD < µI

Step 2

State the level of significance. The .05 significance level is

stated in the problem.

Step 3

Find the appropriate test statistic. Both samples are less than 30, so we use the t distribution.

11- 16

Example 3 continued

918.921215

)9.3)(112()4.2)(115(

2

))(1())(1(

22

21

222

2112

nn

snsns p

Step 4

The decision rule is to reject H0 if t<-1.708 or if p-value < .05. There are n-1 or 25

degrees of freedom.

Step 5 We compute the pooled variance.

11- 17

Example 3 continued

640.1

12

1

15

1312.8

7.357.33

11

21

2

21

nns

XXt

p

We compute the value of t as follows.

11- 18

Since a computed z of –1.64 > critical z of –1.71, the p-value of .0567 > of .05, H0 is not rejected. There is insufficient sample evidence to claim a higher mpg on the imported cars.

P(t < -1.64) = .0567 for a one-tailed t-test.

Example 3 continued

11- 19

Hypothesis Testing Involving Paired Observations

Dependent samples are samples that are paired or related in some fashion.

Independent samples are samples that are not related in any way.

If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices.

If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program.

11- 20

Hypothesis Testing Involving Paired Observations

td

s nd

/

dsd

Use the following test when the samples are dependent:

where is the mean of the differences is the standard deviation of the differencesn is the number of pairs (differences)

11- 21

EXAMPLE 4

An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random sample of eight cities revealed the following information. At the .05 significance level can the testing agency conclude that there is a difference in the rental charged?

City Hertz ($)

Avis ($)

Atlanta 42 40

Chicago 56 52

Cleveland 45 43

Denver 48 48

Honolulu 37 32

Kansas City 45 48

Miami 41 39

Seattle 46 50

11- 22

Example 4 continued

Step 4

H0 is rejected if

t < -2.365 or t > 2.365;

or if p-value < .05.

We use the t distribution with n-1 or 7 degrees of freedom.

Step 2 The stated

significance level is .05.

Step 3 The appropriate

test statistic is the paired t-test.

Step 1Ho: d = 0H1: d = 0

Step 5Perform the

calculations and make a decision.

11- 23

Example 4 continued

City Hertz Avis d d2

Atlanta 42 40 2 4

Chicago 56 52 4 16

Cleveland 45 43 2 4

Denver 48 48 0 0

Honolulu 37 32 5 25

Kansas City 45 48 -3 9

Miami 41 39 2 4

Seattle 46 50 -4 16

11- 24

Example 4 continued

00.18

0.8

n

dd

1623.3

188

878

1

222

n

n

dd

sd

894.081623.3

00.1

ns

dt

d

11- 25

Example 4 continued

P(t>.894) = .20 for a one-tailed t-test at 7 degrees of freedom.

Because 0.894 is less than the critical value, the p-value of .20 > a of .05, do not reject the null hypothesis. There is no difference in the mean amount charged by Hertz and Avis.

11- 26

Advantage of dependent samples:Reduction in variation in the sampling distribution

Disadvantage of dependent samples:Degrees of freedom

are halved

Comparing dependent and independent samples

The same subjects

measured at two different points

in time.

Two types of dependent samplesTwo types of dependent samples

Matched or paired

observations