# Student Solutions Manual for Fundamental Of Statistics 4th Edition By Michael Sullivan – Test Bank

Chapter 11

Inferences on Two Samples

Section 11.1

1. Independent

2. Dependent

3. Since the members of the two samples are

married to each other, the sampling is

dependent. The data are quantitative, since a

numeric response was given by each subject.

4. Because the 100 subjects are randomly

allocated to one of two groups, the sampling is

independent, and it is from a population of

quantitative data. The data are quantitative,

since a numeric response was given by each

subject.

5. Because the two samples were drawn at

separate times with two different groups of

people, the sampling is independent. The data

are qualtitative, since each respondent gave a

“Yes” or “No” response.

6. Because the 1,030 subjects gave their opinion

on each of the parties, the sampling is

dependent. The data are qualtitative, since

each respondent gave a response of

“Favorable” or “Unfavorable” for each party.

7. Two independent populations are being

studied: students who receive the new

curriculum and students who receive the

traditional curriculum. The people in the two

groups are completely independent, so the

sampling is independent. The data are

quantitaitve, since it is the test score for each

respondent.

8. Because the subjects are measured more than

once, the sampling is dependent. The reaction

time data are quantitative.

9. (a) The hypotheses are 01 2 Hp p : = versus

11 2 Hp p : > .

(b) The two sample estimates are

1

1

1

368 ˆ 0.6802

541

x

p n == ≈ and

2

2

2

351 ˆ 0.5919

593

x

p n == ≈ .

The pooled estimate is

1 2

1 2

368 351 ˆ 0.6340

541 593

x x

p n n

+ + == ≈ + + .

The test statistic is

1 2

0

1 2

ˆ ˆ

1 1 ˆ ˆ (1 )

0.6802 0.5919

1 1 0.6340(1 0.6340) 541 593

3.08

p p

z

p p n n

− =

− +

− =

− +

≈

(c) This is a right-tailed test, so the critical

value is 0.05 z z 1.645 α = = .

(d) 0 -value ( 3.08)

1 0.9990 0.0010

P Pz = ≥

=− =

Since 0 0.05 z z =>= 3.08 1.645 and

P-value 0.0010 0.05 = <= α , we reject

H0 . There is sufficient evidence to

conclude that 1 2 p p > .

10. (a) The hypotheses are 01 2 Hp p : = versus

11 2 Hp p : < .

(b) The two sample estimates are

1

1

1

109 ˆ 0.2295

475

x p n == ≈ and

2

2

2

78 ˆ 0.24

325

x p n == = .

The pooled estimate is

1 2

1 2

109 78 ˆ 0.2338

475 325

Chapter 11: Inferences on Two Samples

Copyright © 2014 Pearson Education, Inc.

382

The test statistic is

1 2

0

1 2

ˆ ˆ

1 1 ˆ ˆ (1 )

0.2295 0.24

1 1 0.2338(1 0.2338) 475 325

0.34 [Tech: 0.35]

p p

z

p p n n

− =

− +

− =

− +

≈− −

(c) This is a left-tailed test, so the critical

value is 0.05 z z 1.645 − =− =− α .

(d) 0 P Pz -value ( 0.35) 0.3632 = ≤− =

[Tech: 0.3649]

Since 0 0.05 z z =− >− =− 0.34 1.645 and

P-value 0.3632 0.05 = >= α , we do not

reject H0 . There is not sufficient evidence

to conclude that 1 2 p p < .

11. (a) The hypotheses are 01 2 Hp p : = versus

11 2 Hp p : ≠ .

(b) The two sample estimates are

1

1

1

28 ˆ 0.1102

254

x p n == ≈ and

2

2

2

36 ˆ 0.1196

301

x p n == ≈ .

The pooled estimate is

1 2

1 2

28 36 ˆ 0.1153

254 301

x x p n n

+ + == ≈ + + .

The test statistic is

1 2

0

1 2

ˆ ˆ

1 1 ˆ ˆ (1 )

0.1102 0.1196

1 1 0.1153(1 0.1153) 254 301

0.35 [Tech: 0.34]

p p

z

p p n n

− =

− +

− =

− +

≈− −

(c) This is a two-tailed test, so the critical

values are / 2 0.025 z z 1.96 ± =± =± α .

(d) 0 -value 2 ( 0.35)

2 0.3632

0.7264 [Tech: 0.7307]

P Pz = ⋅ ≤−

= ⋅

=

Since 0z = −0.35 falls between

0.025 − =− z 1.96 and 0.025 z = 1.96 and

P-value 0.7264 0.05 = >= α , we do not

reject H0 . There is not sufficient

evidence to conclude that 1 2 p p ≠ .

12. (a) The hypotheses are 01 2 Hp p : = versus

11 2 Hp p : ≠ .

(b) The two sample estimates are

1

1

1

804 ˆ 0.9199

874

x p n == ≈ and

2

2

2

902 ˆ 0.9455

954

x p n == ≈ .

The pooled estimate is

1 2

1 2

804 902 ˆ 0.9333

874 954

x x p n n

+ + == ≈ + + .

The test statistic is

1 2

0

1 2

ˆ ˆ

1 1 ˆ ˆ (1 )

0.9199 0.9455

1 1 0.9333(1 0.9333) 874 954

2.19

p p

z

p p n n

− =

− +

− =

− +

≈ −

(c) This is a two-tailed test, so the critical

values are / 2 0.025 z z 1.96 ± =± =± α .

(d) 0 -value 2 ( 2.19)

2 0.0143

0.0286

P Pz = ⋅ ≤−

= ⋅

=

Since 0 0.025 z z =− <− =− 2.19 1.96 and

P-value 0.0286 0.05 = <= α , we reject

H0 . There is sufficient evidence to

conclude that 1 2 p p ≠ .

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