Section 8.3 Testing the Difference Between Means (Dependent Samples) © 2012 Pearson Education, Inc. All rights reserved. 1 of 70

Section 8.3

Testing the Difference Between Means (Dependent Samples)

© 2012 Pearson Education, Inc. All rights reserved. 1 of 70

Section 8.3 Objectives

• Perform a t-test to test the mean of the differences for a population of paired data


• The test statistic is the mean of these differences.

t-Test for the Difference Between Means

• To perform a two-sample hypothesis test with dependent samples, the difference between each data pair is first found:– d = x1 – x2 Difference between entries for a data pair

ddd

n Mean of the differences between paired

data entries in the dependent samples



Three conditions are required to conduct the test.1. The samples must be randomly selected.2. The samples must be dependent (paired).3. Both populations must be normally distributed.If these requirements are met, then the sampling

distribution for is approximated by a t-distribution with n – 1 degrees of freedom, where n is the number of data pairs.

d

d-t0 t0μd


Symbols used for the t-Test for μd

The number of pairs of data

The difference between entries for a data pair, d = x1 – x2

d The hypothesized mean of the differences of paired data in the population

n

d

Symbol Description


Symbols used for the t-Test for μd

Symbol Descriptiond The mean of the differences between the paired

data entries in the dependent samples

The standard deviation of the differences between the paired data entries in the dependent samples

ddn

222

( )( )

1 1d

ddd d nsn n

sd



• The test statistic is

• The standardized test statistic is

• The degrees of freedom are d.f. = n – 1.

.d

d

dt

s n

.ddn


t-Test for the Difference Between Means (Dependent Samples)

1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.

2. Specify the level of significance.

3. Determine the degrees of freedom.

4. Determine the critical value(s).

State H0 and Ha.

Identify α.

Use Table 5 in Appendix B if n > 29 use the last row (∞) .

d.f. = n – 1

In Words In Symbols



5. Determine the rejection region(s).

6. Calculate and

7. Find the standardized test statistic.

d .ds dn

d2

22( )

( )1 1d

ddd d nsn n

d

d

dt

s n

In Words In Symbols



8. Make a decision to reject or fail to reject the null hypothesis.

9. Interpret the decision in the context of the original claim.

If t is in the rejection region, reject H0.

Otherwise, fail to reject H0.

In Words In Symbols


Example: t-Test for the Difference Between Means

Athletes 1 2 3 4 5 6 7 8

Height (old) 24 22 25 28 35 32 30 27

Height (new) 26 25 25 29 33 34 35 30

A shoe manufacturer claims that athletes can increase their vertical jump heights using the manufacturer’s new Strength Shoes®. The vertical jump heights of eight randomly selected athletes are measured. After the athletes have used the Strength Shoes® for 8 months, their vertical jump heights are measured again. The vertical jump heights (in inches) for each athlete are shown in the table. Assuming the vertical jump heights are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10? (Adapted from Coaches Sports Publishing)


Solution: Two-Sample t-Test for the Difference Between Means

• H0:

• Ha:

• α = • d.f. = • Rejection Region:

0.108 – 1 = 7

μd ≥ 0 μd < 0 (claim)

d = (jump height before shoes) – (jump height after shoes)



Before After d d2

24 26 –2 4

22 25 –3 9

25 25 0 0

28 29 –1 1

35 33 2 4

32 34 –2 4

30 35 –5 25

27 30 –3 9

Σ = –14 Σ = 56

14 1.758

dn

d

2

2

2

( 14)56

(

88 1

2.12

)

3

1

1

d

dd nsn




• H0:

• Ha:

• α = • d.f. = • Rejection Region:

• Test Statistic:

0.108 – 1 = 7

µd ≥ 0 μd < 0 (claim)

• Decision:


At the 10% level of significance, there is enough evidence to support the shoe manufacturer’s claim that athletes can increase their vertical jump heights using the new Strength Shoes®.

Reject H0.


1.75 0 2.3332.1213 8

d

d

dt

s n

t ≈ –2.333

To use the calculator for testing the difference between the means of dependent samples, enter the data in two columns and form a third Column ( in which you calculate the difference for each pair ( You can then performa one-sample t-test on the difference column ().

Section 8.4

Testing the Difference Between Proportions


Section 8.4 Objectives

• Perform a z-test for the difference between two population proportions p1 and p2


Two-Sample z-Test for Proportions

• Used to test the difference between two population proportions, p1 and p2.

• Three conditions are required to conduct the test.1. The samples must be randomly selected.2. The samples must be independent.3. The samples must be large enough to use a

normal sampling distribution. That is,n1p1 ≥ 5, n1q1 ≥ 5, n2p2 ≥ 5, and n2q2 ≥ 5.


Two-Sample z-Test for the Difference Between Proportions

• If these conditions are met, then the sampling distribution for is a normal distribution.

• Mean:• A weighted estimate of p1 and p2 can be found

by using

• Standard error:

1 2ˆ ˆp p

1 21 2ˆ ˆp p p p

1 2ˆ ˆ

1 2

1 1 , where 1 .p p pq q pn n

1 21 1 1 2 2 2

1 2, where and ˆ ˆ

x xp x n p x n p

n n



• The test statistic is .• The standardized test statistic is

where

1 2 1 2

1 2

( ) ( )ˆ ˆ

1 1

p p p pz

pqn n

1 2ˆ ˆp p

1 2

1 2 and .1

x xp q p

n n

1 1 2 2Note: , , , and must be at least 5.n p n q n p n q



1. State the claim. Identify the null and alternative hypotheses.

2. Specify the level of significance.

3. Determine the critical value(s).

4. Determine the rejection region(s).

5. Find the weighted estimate of p1 and p2. Verify that

State H0 and Ha.

Identify α.

Use Table 4 in Appendix B.

1 2

1 2

x xp

n n

In Words In Symbols


n1p, n1q, n2 p,

and n2q are at least 5.


6. Find the standardized test statistic and sketch the sampling distribution.

7. Make a decision to reject or fail to reject the null hypothesis.

8. Interpret the decision in the context of the original claim.

1 2 1 2

1 2

( ) ( )ˆ ˆ

1 1

p p p pz

pqn n

If z is in the rejection region, reject H0.

Otherwise, fail to reject H0.

In Words In Symbols


Example: Two-Sample z-Test for the Difference Between Proportions

In a study of 150 randomly selected occupants in passenger cars and 200 randomly selected occupants in pickup trucks, 86% of occupants in passenger cars and 74% of occupants in pickup trucks wear seat belts. At α = 0.10 can you reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks? (Adapted from National Highway Traffic Safety Administration)

Solution: 1 = Passenger cars 2 = Pickup trucks


Solution: Two-Sample z-Test for the Difference Between Means

• H0:

• Ha:

• α =

• n1= , n2 =

• Rejection Region:

0.10150 200

p1 = p2 (claim)p1 ≠ p2



1 2

1 2

129 148 0.7914

150 200

x xp

n n

1 1 1 129ˆx n p 2 2 2 148ˆx n p

1 0.7914 01 .2086q p


Note:


1 2 1 2

1 2

0.86 0.74 0

1 10.7914 0.2086

150 200

2.

ˆ ˆ

1

73

1

p p p pz

pqn n



• H0:

• Ha:

• α =

• n1= , n2 =


• Test Statistic:

0.10150 200

2.73z p1 = p2 (claim) p1 ≠ p2

• Decision:At the 10% level of significance, there is enough evidence to reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks.

Reject H0


Example: Two-Sample z-Test for the Difference Between Proportions

A medical research team conducted a study to test the effect of a cholesterol-reducing medication. At the end of the study, the researchers found that of the 4700 randomly selected subjects who took the medication, 301 died of heart disease. Of the 4300 randomly selected subjects who took a placebo, 357 died of heart disease. At α = 0.01 can you conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo? (Adapted from New England Journal of Medicine)Solution: 1 = Medication 2 = Placebo


z0–2.33

0.01


• H0:

• Ha:

• α =

• n1= , n2 =


0.014700 4300

p1 ≥ p2 p1 < p2 (claim)



1 2

1 2

301 357 0.0731

4700 4300

x xp

n n

11

1

3010.064

470ˆ

0

xp

n

1 0.0731 01 .9269q p

22

2

3570.083

430ˆ

0

xp

n

1 1

2 2

4700(0.0731) 5 4700(0.9269) 5

4300(0.0731

Note:

) 5 4300(0.926

9) 5

n p n q

n p n q



1 2 1 2

1 2

0.064 0.083 0

1 10.0731 0.92

ˆ

694700 4300

3.

ˆ

1

6

1

4

p p p pz

pqn n


z0–2.33

0.01


• H0:

• Ha:

• α =

• n1= , n2 =


• Test Statistic:

0.014700 4300

3.46z p1 ≥ p2 p1 < p2 (claim)

–2.33–3.46

• Decision:At the 1% level of significance, there is enough evidence to conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo.

Reject H0


Section 8.4 Summary

• Performed a z-test for the difference between two population proportions p1 and p2


Documents

Section 8.3 Testing the Difference Between Means (Dependent Samples) © 2012 Pearson Education, Inc. All rights reserved. 1 of 70