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 ≠ .