## Mean of the Differences between Matched Pairs

Section 9 – 5A: Testing a Claim about the Mean of the Differences between Matched Pairs Test H 0: µd = 0 (there is no difference in Population Means ...
Author: Polly McKenzie
Section 9 – 5A:

Testing a Claim about the Mean of the Differences between Matched Pairs Test H 0: µd = 0 (there is no difference in Population Means of the matched pairs) against ud > 0

or

H1: ud < 0

or

H1: µd ≠ 0

at a significance level of α with DF = n − 1 Requirements 1. The sample data consists consists of matched pairs. 2. The samples are simple random samples. 3. The number of matched pairs is greater than 30 or the population of the differences is normal. Notation for the Samples of Two Population Means of Matched Pairs d = the individual differences between each matched pair. n = the number of matched pairs in the sample

µd = the average (mean) of all the differences d in the population of matched pairs. d = the mean( average) of all the differences in the sample of matched pairs. sd = the standard deviation of all the differences in the sample of matched pairs.

Testing a Claim about the Mean of the Differences between Matched Pairs Dependent Samples with H 0: µd = 0 Test Statistic:

t=

(d )

⎛ sd ⎞ ⎜ ⎟ ⎝ n⎠

with DF = n − 1

Section 9 – 5A Lecture

Page 1 of 1

Testing a Claim about the Mean of the Differences between Matched Pairs Example 1(Left Tail Test) Eight Folsom Lake college students selected at random were tested for resting heart rates. They were then shown an exciting 30 minute video lecture of Mr. Eitelʼs statistics class. A second reading of their heart rate was taken after the video The table below shows the heart rates for the before and after readings. Use a α = .05 significance level to test the claim that watching the exciting statistics video can lower heart rates. Assume the population of differences is normal. Student

A

B

C

D

E

F

G

H

Before

40

45

67

43

56

72

81

92

After

38

42

65

44

52

72

80

88

Difference = d After – Before

–2

–3

2

–1

–4

0

–1

–4

A general review of the differences would lead you to claim that the heart rate may be lower after the statistics video. H 0: ud = 0 The results from putting the list of differences into the calculator. H1: ud < 0

d = −1.625

α = .05

n=8

Left Tail Test of H 0: ud = 0

Test Statistic: (d ) t= ⎛ sd ⎞ ⎜ ⎟ ⎝ n⎠

DF = 7

Reject H0

α = .05

sd = 2.066

Do Not Reject H0

t= t

t = –1.895

(−1.625) ⎛ 2.066 ⎞ ⎜ ⎟ ⎝ 8 ⎠

t = −2.22 Conclusion based on H 0 :

Reject H0

Conclusion based on the problem: There is sufficient evidence at the α = .05 level to support the claim that watching the exciting statistics video can lower heart rates.

Section 9 – 5A Lecture

Page 2 of 2

t Distribution: Critical t Values Degrees of Freedom 7

Area In One Tail (Right Tail) 0.100

0.050

0.025

0.010

0.005

1.415

1.895

2.365

2.998

3.499

Section 9 – 5A Lecture

Page 3 of 3

Testing a Claim about the Mean of the Differences between Matched Pairs Example 2 (Right Tail Test) A PE Teacher felt that hearing a recorded pep talk from Hulk Hogan would increase the number of sit ups a group of 12 years olds could do in 5 minutes. A random group of 12 year olds were selected and the number of sit ups they could do was recorded. The next day they listened to a recorded pep talk from Hulk Hogan and a second test was given. Use a α = .01 significance level to test the claim that hearing a recorded pep talk from Hulk Hogan can help increase the number of sits ups a 12 year old can do in 5 minutes. Assume the population of differences is normal. Student

A

B

C

D

E

F

G

H

I

J

Before

23

12

25

32

15

16

21

17

11

14

After

24

13

20

34

16

16

22

12

15

10

1

1

5

2

1

0

1

–5

4

–3

Difference = d After – Before

A general review of the differences would lead you to guess that the number of sit ups may increase after listening to the recording of Hulk Hogan. H 0: ud = 0 The results from putting the list of differences into the calculator. H1: ud > 0

d = .7

α = .01

sd = 2.94

n = 10

Right Tail Test of H 0: ud = 0

Test Statistic: d t= sd n

DF = 9 Do Not Reject H0

Reject H0

α = .01

t= t

.7 2.94 10

t = 2.821 t = .75 Conclusion based on H 0 :

Do not Reject H0

Conclusion based on the problem: There is not sufficient evidence at the α = .01 level to reject the hypothesis that hearing a recorded pep talk from Hulk Hogan will not increase the number of sits ups a 12 year old can do in 5 minutes. or There is not sufficient evidence at the α = .01 level to support the claim that hearing a recorded pep talk from Hulk Hogan can help increase the number of sits ups a 12 year old can do in 5 minutes. Section 9 – 5A Lecture

Page 4 of 4

t Distribution: Critical t Values Degrees of Freedom 9

Area In One Tail (Right Tail) 0.100

0.050

0.025

0.010

0.005

1.383

1.833

2.262

2.821

3.250

Section 9 – 5A Lecture

Page 5 of 5

Testing a Claim about the Mean of the Differences between Matched Pairs Example 3 (Two Tail Test) Folsom Lake College students must take an assessment test before they can attend their first math class. A random group of 9 students were given the Math 100 placement test and their scores were recorded. They were then allowed to retake the exact same test 4 times and the final score was recorded. Use a α = .05 significance level to test the claim that taking the test 4 times will not change the placement score. Assume the population of differences is normal. Student

A

B

C

D

E

F

G

H

I

Before

32

45

33

28

26

27

32

20

21

After

38

49

35

44

32

36

35

25

28

6

4

2

16

6

9

3

5

7

Difference = d After – Before

A general review of the differences would lead you to claim that there is a change in the placement score after taking the test 4 times. H 0: ud = 0 The results from putting the list of differences into the calculator. H1: ud ≠ 0

d = 6.44

α = .05 so α = .025

n=9

Two Tail Test of H 0: ud = 0 DF = 8

Reject H0

α 2 = .025

Test Statistic: (d ) t= ⎛ sd ⎞ ⎜ ⎟ ⎝ n⎠

Reject H0

Do Not

α 2 = .025

Reject H0

t t = –2.306

sd = 4.15

t = 2.306

t=

(6.44 )

⎛ 4.15 ⎞ ⎜ ⎟ ⎝ 9 ⎠

t = 4.66 Conclusion based on H 0 :

Reject H0

Conclusion based on the problem: There is sufficient evidence at the α = .01 level to support the claim that taking the test 4 times will change the placement score.

Section 9 – 5A Lecture

Page 6 of 6

t Distribution: Critical t Values Degrees of Freedom 8

Area In One Tail (Right Tail) 0.100

0.050

0.025

0.010

0.005

1.397

1.860

2.306

2.896

3.355

Section 9 – 5A Lecture

Page 7 of 7