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# Ch8: Hypothesis Testing (2 Samples)

## 1. Ch8: Hypothesis Testing (2 Samples)

8.1 Testing the Difference Between Means

(Independent Samples, 1 and 2 Known)

8.2 Testing the Difference Between Means

(Independent Samples, 1 and 2 Unknown)

8.3 Testing the Difference Between Means

(Dependent Samples)

8.4 Testing the Difference Between Proportions

Larson/Farber

1

## 2. 8.1 Two Sample Hypothesis Test

Compares two parameters from two populations.

Two types of sampling methods:

Independent (unrelated) Samples

•Sample 1: Test scores for 35 statistics students.

•Sample 2: Test scores for 42 biology students who do not study statistics.

Dependent Samples (paired or matched samples)

Each member of one sample corresponds to a member of the other sample.

•Sample 1: Resting heart rates of 35 individuals before drinking coffee.

•Sample 2: Resting heart rates of the same individuals after drinking two cups of coffee.

Larson/Farber

2

## 3. Stating a Hypotheses in 2-Sample Hypothesis Test

Null hypothesis• A statistical hypothesis H0

• Statement of equality ( , =, or ).

• No difference between the

parameters of two populations.

H0: μ1 = μ2

Ha: μ1 ≠ μ2

OR

Alternative Hypothesis (Ha)

(Complementary to Null Hypothesis)

• A statement of inequality (>, , or <).

• True when H0 is false.

H0: μ1 ≤ μ2

Ha: μ1 > μ2

OR

H0: μ1 ≥ μ2

Ha: μ1 < μ2

Regardless of which hypotheses you use, you always assume there is no

difference between the population means, or μ1 = μ2.

Larson/Farber

3

## 4. Two Sample z-Test for the Difference Between Means (μ1 and μ2.)

Three conditions are necessary1. The samples must be randomly selected.

2.

The samples must be independent.

3.

Each population must have a normal distribution with a known

population standard deviation OR each sample size must be at least 30.

Sampling distribution for x1 x2 (difference of sample means) is a normal with:

Mean: x x x x 1 2

1

2

1

Standard error:

2

x x

1

Test Statistic:

x1 x2

Sampling

distribution

1

x2

1

1

2

1 2

σ x x

1

2

2

12 22

n1

n2

Standardized Test Statistic:

x1 x2 1 2

12 22

z

where x x

x x

n1 n2

1

σ x x

Larson/Farber

2

x2

2

2

•For large samples: use s1 and s2 in place of 1 and 2.

x x •For small samples: use a two-sample z-test if populations

are normally distributed & pop. std deviations are known.

1

2

4

## 5. Using a Two-Sample z-Test for the Difference Between Means (Independent Samples 1 and 2 known or n1 and n2 30 )

Using a Two-Sample z-Test for theDifference Between Means (Independent

Samples 1 and 2 known or n1 and n2 30 )

In Words

In Symbols

1.

State the claim mathematically. Identify the

null and alternative hypotheses.

State H0 and Ha.

2.

Specify the level of significance.

Identify .

3.

Sketch the sampling distribution.

4.

Determine the critical value(s).

Use Table 4 in Appendix B.

5.

6.

Determine the rejection region(s).

Find the standardized test statistic.

z

7.

Make a decision to reject or fail to

reject the null hypothesis.

8.

Interpret the decision in the context of

the original claim.

Larson/Farber

x1 x2 1 2

x x

1

2

If z is in the rejection region,

reject H0.

5

## 6. Example1: Two-Sample z-Test for the Difference Between Means

A consumer education organization claims that there is a difference in the meancredit card debt of males and females in the United States. The results of a random

survey of 200 individuals from each group are shown below. The two samples are

independent. Do the results support the organization’s claim? Use α = 0.05.

• H0 : μ 1 = μ 2

Females (1)

Males (2)

Ti83/84

Stat-Tests

• Ha: μ1 ≠ μ2

3:2-SampZTest

x1 $2290

x2 $2370

• .05

s1 = $750

s2 = $800

• n1= 200 , n2 = 200

• Rejection Region:

n1 = 200

n2 = 200

z

0.025

-1.96

Larson/Farber

1.96

2

750

200

0.025

0

(2290 2370) 0

Z

2

1.03

800

200

Decision: Fail to reject H0

At the 5% level of significance, there is not

enough evidence to support the

organization’s claim that there is a difference

in the mean credit card debt of males and 6

## 7. Example2: Using Technology to Perform a Two-Sample z-Test

The American Automobile Association claims that the average daily cost for mealsand lodging for vacationing in Texas is less than the same average costs for

vacationing in Virginia. The table shows the results of a random survey of

vacationers in each state. The two samples are independent. At α = 0.01, is there

enough evidence to support the claim?

Texas (1)

Virginia (2)

• H0: μ1 ≥ μ2

x2 $252

x1 $248

• Ha : μ 1 < μ 2

TI-83/84set up:

Calculate

s1 = $15

n1 = 50

s2 = $22

n2 = 35

0.01

0

Larson/Farber

z

Decision: Fail to reject H0

At 1% level, Not enough

evidence to support AAA’s claim.

7

## 8. 8.2 Two Sample t-Test for the Difference Between Means (1 or 2 unknown)

8.2 Two Sample t-Test for the DifferenceBetween Means ( 1 or 2 unknown)

• If ( 1 or 2 is unknown and samples are taken from normally-distributed) OR

If ( 1 or 2 is unknown and both sample sizes are greater than or equal to 30)

THEN a t-test may be used to test the difference between the population

means μ1 and μ2.

• Three conditions are necessary to use a t-test for small independent samples.

1.

The samples must be randomly selected.

2.

The samples must be independent.

3.

Each population must have a normal distribution.

Test Statistic:

t

x1 x2 1 2

x x

1

Larson/Farber

2

8

## 9. Two Sample t-Test for the Difference Between Means

The standard error and the degrees of freedom of the sampling distributiondepend on whether the population variances and are equal.

• Equal Variances

• UnEqual Variances

• Information from the two samples is

• Standard Error is:

combined to calculate a pooled

estimate of the standard deviation ˆ

s12 s22

2

2

.

x

x

n

1

s

n

1

s

1 1 2 2

n1 n2

ˆ

n1 n2 2

1

The standard error for the

sampling distribution of

x1 x2

is

x x ˆ 1 1

n1 n2

1

2

d.f = smaller of n1 – 1 or n2 – 1

2

d.f.= n1 + n2 – 2

Larson/Farber

9

## 10. Normal or t-Distribution?

.## 11. Two-Sample t-Test for the Difference Between Means - Independent Samples (1 or 2 unknown)

Two-Sample t-Test for the DifferenceBetween Means - Independent Samples

( 1 or 2 unknown)

In Words

In Symbols

State H0 and Ha.

1.

State the claim mathematically. Identify the

null and alternative hypotheses.

2.

Specify the level of significance.

3.

Identify the degrees of freedom and sketch

the sampling distribution.

d.f. = n1+ n2 – 2 or

d.f. = smaller of n1 – 1 or n2 – 1.

4.

5.

Determine the critical value(s).

Determine the rejection region(s).

Use Table 5 in Appendix B.

6.

Find the standardized test statistic.

Identify .

t

x1 x2 1 2

x x

1

7.

Make a decision to reject or fail to reject

the null hypothesis and interpret the

decision in the context of the original claim

Larson/Farber

2

If t is in the rejection region,

reject H0. Otherwise, fail to

reject H0.

11

## 12. Example: Two-Sample t-Test for the Difference Between Means

The braking distances of 8 Volkswagen GTIs and 10 Ford Focuses were testedwhen traveling at 60 miles per hour on dry pavement. The results are shown

below. Can you conclude that there is a difference in the mean braking distances

of the two types of cars? Use α = 0.01. Assume the populations are normally

distributed and the population variances are not equal. (Consumer Reports)

• H0: μ1 = μ2

GTI (1)

Focus (2)

• Ha: μ1 ≠ μ2

x2 143ft

x1 134ft

• 0.01

0.005

0.005

s1 = 6.9 ft

s2 = 2.6 ft

t

• d.f. = 8 – 1 = 7

n1 = 8

n2 = 10

t

Stat-Test

4: 2-SampTTest

Larson/Farber

Pooled:

No 4th ed

(134 143) 0

6.92 2.62

8

10

-3.499 0

3.499

Fail to Reject H0

3.496 • Decision:

At the 1% level of significance, there is not

enough evidence to conclude that the mean

braking distances of the cars are different.

12

## 13. Example: Two-Sample t-Test for the Difference Between Means

A manufacturer claims that the calling range (in feet) of its 2.4-GHz cordless telephoneis greater than that of its leading competitor. You perform a study using 14 randomly

selected phones from the manufacturer and 16 randomly selected similar phones from

its competitor. The results are shown below. At α = 0.05, can you support the

manufacturer’s claim? Assume the populations are normally distributed and the

population variances are equal.

Manufacturer (1)

Competition (2)

H 0 = μ 1 ≤ μ2

Ha = μ1 > μ2 (Claim)

= .05

d.f. = 14 + 16 – 2 = 28

1.701

x2 1250ft

s2 = 30 ft

n1 = 14

n2 = 16

n1

x x

0.05

0

x1 1275ft

s1 = 45 ft

1

n1 n2 2

2

t

1 s12 n2 1 s2 2

14 1 45

2

1

1

n1

n2

16 1 30

2

1

1

13.8018

14 16

14 16 2

Decision: Reject H0

At 5% level of significance there

Stat-Test

x

x

1275

1250

0

1

2

1

2

is enough evidence to support

t

1.811 4: 2-SampTTest

4th ed

x x

13.8018

theLarson/Farber

manufacturer’s

claim.

Pooled: Yes 13

1

2

## 14. 8.3 t-Test for the Difference Between Means (Paired Data/Dependent Samples)

• To perform a two-sample hypothesis test with dependent samples, thedifference between each data pair is first found:

d = x1 – x2 Difference between entries for a data pair

The test statistic is the mean d of these differences.

d d Mean of the differences between paired data

n

entries in the dependent samples

Three conditions are required to conduct the test.

1. The samples must be randomly selected.

-t0

μd

t0

2. The samples must be dependent (paired).

3. Both populations must be normally distributed.

If these conditions 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.

Larson/Farber

d

14

## 15. Symbols used for the t-Test for μd

SymbolDescription

n

The number of pairs of data

d

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

d

d

sd

Degrees of Freedom (d.f.) = n - 1

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

Mean of the differences between paired data entries in dependent samples

(Test Statistic) d d

n

The standard deviation of the differences between the paired data

entries in the dependent samples

2

2 ( d )

d

(d d )

n

sd

n 1

n 1

2

(Standardized Test Statistic)

Larson/Farber

d d

t

sd n

15

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

In WordsIn Symbols

1.

State the claim mathematically.

Identify the null and alternative hypotheses.

State H0 and Ha.

2.

Specify the level of significance.

Identify .

3.

Identify the degrees of freedom and sketch the

sampling distribution.

d.f. = n – 1

4.

Determine critical value(s) & rejection region

5.

Calculate d and sd . Use a table.

Use Table 5 in Appendix B

If n > 29 use the last row (∞) .

d d

n

6.

Find the standardized test statistic.

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

null hypothesis and interpret the decision in

the context of the original claim.

Larson/Farber

t

( d )

d

(d d )

n

sd

n 1

n 1

2

2

2

d d

sd n

If t is in the rejection region, reject

H0. Otherwise, fail to reject H0.

16

## 17. Example: t-Test for the Difference Between Means

A golf club manufacturer claims that golfers can lower their scores by using themanufacturer’s newly designed golf clubs. Eight golfers are randomly selected,

and each is asked to give his or her most recent score. After using the new clubs

for one month, the golfers are again asked to give their most recent score. The

scores for each golfer are shown in the table. Assuming the golf scores are

normally distributed, is there enough evidence to support the manufacturer’s

claim at α = 0.10?

Golfer

1

2

3

4

5

6

7

8

Score (old)

89

84

96

82

74

92

85

91

Score (new)

83

83

92

84

76

91

80

91

d = (old score) – (new score)

μd ≤ 0

• H 0:

μd > 0

• H a:

• 0.10

0

• d.f. = 8 – 1 = 7

4th ed

& Sd calculations

on next slide

dLarson/Farber

Rejection

Region

0.10

1.415

t

d d

1.625 0

1.498

sd n 3.0677 8

Decision: Reject H0

t At the 10% level of significance, the results of

this test indicate that after the golfers used the

17

new clubs, their scores were significantly lower.

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

d = (old score) – (new score)Old

89

New

83

d

6

d2

36

84

96

82

83

92

84

1

4

–2

1

16

4

74

92

85

76

91

80

–2

1

5

4

1

25

91

91

Larson/Farber

0

0

Σ = 13 Σ = 87

d 13

d

1.625

n

8

2 ( d )

d

n

sd

n 1

2

(13) 2

87

8

8 1

3.0677

18

## 19. 8.4 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.

• If these conditions are met, then the sampling distribution for pˆ1 pˆ 2 is normal

Standardized Test Statistic

• Mean: pˆ pˆ p1 p2 (Test Statistic)

• Find weighted estimate of p1 and p2 using z ( pˆ1 pˆ 2) ( p1 p2)

1

p

2

x1 x2

, where x1 n1 pˆ1 and x2 n2 pˆ 2

n1 n2

1 1

• Standard error: pˆ pˆ pq

n1 n2

1

Larson/Farber

2

1 1

pq

n1 n2

where p x1 x2 and q 1 p

n1 n2

Note: n1 p, n1q, n2 p, and n2q must be at least 5.

19

## 20. Two-Sample z-Test for the Difference Between Proportions

In WordsIn Symbols

1.

State the claim. Identify the null

alternative hypotheses.

2.

Specify the level of significance.

3.

Determine the critical value(s).

4.

Determine the rejection region(s).

and

State H0 and Ha.

Identify .

Use Table 4 in Appendix B.

x1 x2

n1 n2

5.

Find the weighted estimate of

6.

Find the standardized test statistic.

z

7.

Make a decision to reject or fail to reject the

null hypothesis and interpret decision in the

context of the original claim.

If z is in the rejection region,

reject H0. Otherwise, fail to

reject H0.

Larson/Farber

p1 and p2.

p

( pˆ1 pˆ 2) ( p1 p2)

1 1

pq

n1 n2

20

## 21. Example1: Two-Sample z-Test for the Difference Between Proportions

In a study of 200 randomly selected adult female(1) and 250 randomly selected adultmale(2) Internet users, 30% of the females and 38% of the males said that they plan to

shop online at least once during the next month. At α = 0.10 test the claim that there is a

difference between the proportion of female and the proportion of male Internet users who

plan to shop online.

x1 n1 pˆ1 60

= .10

n1 = 200 n2 = 250

x2 n2 pˆ 2 95

H0 : p1 = p2

0.05

0.05

Ha : p1 ≠ p2

x1 x2

60 95

p

0.3444

Z

-1.645 0

1.645

n1 n2 200 250

pˆ1 pˆ 2 p1 p2

0.30 0.38 0

q 1 p 1 0.3444 0.6556

z

1 1

pq

n1 n2

1.77

0.3444 0.6556

1

1

200 250

Note:

n1 p 200(0.3444) 5

n1q 200(0.6556) 5

n2 p 250(0.3444) 5

n2q 250(0.6556) 5

Decision: Reject H0

At the 10% level of significance, there is enough evidence to conclude that

there is a difference between the proportion of female and the proportion of

male Internet

Larson/Farber

4th edusers who plan to shop online.

Stat-Tests

5: 2-PropZTest

21

## 22. Example2: Two-Sample z-Test for the Difference Between Proportions

A medical research team conducted a study to test the effect of a cholesterol reducingmedication(1). 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(2), 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? (New England Journal of Medicine)

x

301

= .01

n1 = 4700 n2 = 4300

ˆ1 1

p

0.064

n1

4700

H0 : p1 ≥ p2

x

357

0.01

Ha : p1 < p2

ˆ 2

p

0.083

-2.33

z

pˆ1 pˆ 2 p1 p2

1

1

pq

n1 n2

3.46

z

0

0.064 0.083 0

0.0731 0.9269

1

1

4700 4300

2

p

n2

4300

x1 x2

301 357

0.0731

n1 n2

4700 4300

q 1 p 1 0.0731 0.9269

Note:

n1 p 4700(0.0731) 5

n1q 4700(0.9269) 5

n2 p 4300(0.0731) 5

n2q 4300(0.9269) 5

Decision: Reject H0

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

Stat-Tests

Larson/Farber

4th ed who took the placebo.

22

than for those

5: 2-PropZTest