Statistics for Business and
Economics, 14th Edition by James
T. McClave
Complete Chapter Solutions Manual
are included (Ch 1 to 15)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1.Statistics, Data, and Statistical Thinking
2.Methods for Describing Sets of Data
3.Probability
4.Random Variables and Probability Distributions
5.Sampling Distributions
6.Inferences Based on a Single Sample: Estimation with Confidence
Intervals
7.Inferences Based on a Single Sample: Tests of Hypotheses
8.Inferences Based on Two Samples: Confidence Intervals and Tests of
Hypotheses
9.Design of Experiments and Analysis of Variance
10.Categorical Data Analysis
11.Simple Linear Regression
12.Multiple Regression and Model Building
13.Methods for Quality Improvement: Statistical Process Control
14.Time Series: Descriptive Analyses, Models, and Forecasting
15.Nonparametric Statistics
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all chapters are included
in this document. (Complete Chapters included Ch15-1)
Chapter 15
Nonparametric Statistics
15.1 The sign test is preferred to the t-test when the population from which the sample is selected is not normal.
15.2 a. Since the normal distribution is symmetric, the probability that a randomly selected observation
exceeds the mean of a normal distribution is .5.
b. By the definition of "median," the probability that a randomly selected observation exceeds the
median of a normal distribution is .5.
c. If the distribution is not normal, the probability that a randomly selected observation exceeds the
mean depends on the distribution. With the information given, the probability cannot be determined.
d. By definition of "median," the probability that a randomly selected observation exceeds the median
of a non-normal distribution is .5.
15.3 a. P ( x ≥ 7 ) = 1 − P ( x ≤ 6 ) = 1 − .965 = .035
b. P ( x ≥ 5) = 1 − P ( x ≤ 4 ) = 1 − .637 = .363
c. P ( x ≥ 8 ) = 1 − P ( x ≤ 7 ) = 1 − .996 = .004
d. P ( x ≥ 10 ) = 1 − P ( x ≤ 9 ) = 1 − .849 = .151
μ = np = 15 (.5) = 7.5 and σ = npq = 15 (.5)(.5) = 1.9365
P ( x ≥ 10) ≈ P z ≥
(10 − .5) − 7.5 = P z ≥ 1.03 = .5 − .3485 = .1515
( ) (Using Table II, Appendix D)
1.9365
e. P ( x ≥ 15) = 1 − P ( x ≤ 14 ) = 1 − .788 = .212
μ = np = 25 (.5) = 12.5 and σ = npq = 25(.5)(.5) = 2.5
P ( x ≥ 15) ≈ P z ≥
(15 − .5) − 12.5 = P z ≥ .80 = .5 − .2881 = .2119
( ) (Using Table II, Appendix D)
2.5
H0 : η = 9
15.4 a.
Ha : η > 9
The test statistic is S = {Number of observations greater than 9} = 7.
The p-value = P ( x ≥ 7 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From Table
I,
883
, 884 Chapter 15
p − value = P ( x ≥ 7 ) = 1 − P ( x ≤ 6 ) = 1 − .828 = .172
Since the p-value is not less than α ( p = .172 </ .05) , H0 is not rejected. There is insufficient evidence
to indicate the median is greater than 9 at 𝛼 = .05.
H0 : η = 9
b.
Ha : η ≠ 9
S1 = {Number of observations less than 9} = 3 and
S2 = {Number of observations greater than 9} = 7
The test statistic S is the larger of of S1 and S2. Thus, S = 7.
The p-value = 2 P ( x ≥ 7 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From
Table I,
p − value = 2P ( x ≥ 7) = 2 (1 − P ( x ≤ 6) ) = 2 (1 − .828) = .344
Since the p-value is not less than α ( p = .344 </ .05) , H0 is not rejected. There is insufficient evidence
to indicate the median is different than 9 at 𝛼 = .05.
H 0 : η = 20
c.
H a : η < 20
The test statistic is S = {Number of observations less than 20} = 9.
The p-value = P ( x ≥ 9 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From Table
I,
p − value = P ( x ≥ 9 ) = 1 − P ( x ≤ 8) = 1 − .989 = .011
Since the p-value is less than α ( p = .011 < .05) , H0 is rejected. There is sufficient evidence to
indicate the median is less than 20 at 𝛼 = .05.
H 0 : η = 20
d.
H a : η ≠ 20
S1 = {Number of observations less than 20} = 9 and
S2 = {Number of observations greater than 20} = 1
The test statistic S is the larger of S1 and S2. Thus, S = 9.
The p-value = 2 P ( x ≥ 9 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From Table
I,
p − value = 2P ( x ≥ 9) = 2 (1 − P ( x ≤ 8) ) = 2 (1 − .989) = .022
Since the p-value is less than α ( p = .022 < .05) , H0 is rejected. There is sufficient evidence to
indicate the median is different than 20 at 𝛼 = .05.
Economics, 14th Edition by James
T. McClave
Complete Chapter Solutions Manual
are included (Ch 1 to 15)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1.Statistics, Data, and Statistical Thinking
2.Methods for Describing Sets of Data
3.Probability
4.Random Variables and Probability Distributions
5.Sampling Distributions
6.Inferences Based on a Single Sample: Estimation with Confidence
Intervals
7.Inferences Based on a Single Sample: Tests of Hypotheses
8.Inferences Based on Two Samples: Confidence Intervals and Tests of
Hypotheses
9.Design of Experiments and Analysis of Variance
10.Categorical Data Analysis
11.Simple Linear Regression
12.Multiple Regression and Model Building
13.Methods for Quality Improvement: Statistical Process Control
14.Time Series: Descriptive Analyses, Models, and Forecasting
15.Nonparametric Statistics
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all chapters are included
in this document. (Complete Chapters included Ch15-1)
Chapter 15
Nonparametric Statistics
15.1 The sign test is preferred to the t-test when the population from which the sample is selected is not normal.
15.2 a. Since the normal distribution is symmetric, the probability that a randomly selected observation
exceeds the mean of a normal distribution is .5.
b. By the definition of "median," the probability that a randomly selected observation exceeds the
median of a normal distribution is .5.
c. If the distribution is not normal, the probability that a randomly selected observation exceeds the
mean depends on the distribution. With the information given, the probability cannot be determined.
d. By definition of "median," the probability that a randomly selected observation exceeds the median
of a non-normal distribution is .5.
15.3 a. P ( x ≥ 7 ) = 1 − P ( x ≤ 6 ) = 1 − .965 = .035
b. P ( x ≥ 5) = 1 − P ( x ≤ 4 ) = 1 − .637 = .363
c. P ( x ≥ 8 ) = 1 − P ( x ≤ 7 ) = 1 − .996 = .004
d. P ( x ≥ 10 ) = 1 − P ( x ≤ 9 ) = 1 − .849 = .151
μ = np = 15 (.5) = 7.5 and σ = npq = 15 (.5)(.5) = 1.9365
P ( x ≥ 10) ≈ P z ≥
(10 − .5) − 7.5 = P z ≥ 1.03 = .5 − .3485 = .1515
( ) (Using Table II, Appendix D)
1.9365
e. P ( x ≥ 15) = 1 − P ( x ≤ 14 ) = 1 − .788 = .212
μ = np = 25 (.5) = 12.5 and σ = npq = 25(.5)(.5) = 2.5
P ( x ≥ 15) ≈ P z ≥
(15 − .5) − 12.5 = P z ≥ .80 = .5 − .2881 = .2119
( ) (Using Table II, Appendix D)
2.5
H0 : η = 9
15.4 a.
Ha : η > 9
The test statistic is S = {Number of observations greater than 9} = 7.
The p-value = P ( x ≥ 7 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From Table
I,
883
, 884 Chapter 15
p − value = P ( x ≥ 7 ) = 1 − P ( x ≤ 6 ) = 1 − .828 = .172
Since the p-value is not less than α ( p = .172 </ .05) , H0 is not rejected. There is insufficient evidence
to indicate the median is greater than 9 at 𝛼 = .05.
H0 : η = 9
b.
Ha : η ≠ 9
S1 = {Number of observations less than 9} = 3 and
S2 = {Number of observations greater than 9} = 7
The test statistic S is the larger of of S1 and S2. Thus, S = 7.
The p-value = 2 P ( x ≥ 7 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From
Table I,
p − value = 2P ( x ≥ 7) = 2 (1 − P ( x ≤ 6) ) = 2 (1 − .828) = .344
Since the p-value is not less than α ( p = .344 </ .05) , H0 is not rejected. There is insufficient evidence
to indicate the median is different than 9 at 𝛼 = .05.
H 0 : η = 20
c.
H a : η < 20
The test statistic is S = {Number of observations less than 20} = 9.
The p-value = P ( x ≥ 9 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From Table
I,
p − value = P ( x ≥ 9 ) = 1 − P ( x ≤ 8) = 1 − .989 = .011
Since the p-value is less than α ( p = .011 < .05) , H0 is rejected. There is sufficient evidence to
indicate the median is less than 20 at 𝛼 = .05.
H 0 : η = 20
d.
H a : η ≠ 20
S1 = {Number of observations less than 20} = 9 and
S2 = {Number of observations greater than 20} = 1
The test statistic S is the larger of S1 and S2. Thus, S = 9.
The p-value = 2 P ( x ≥ 9 ) where x is a binomial random variable with 𝑛 = 10 and 𝑝 = .5. From Table
I,
p − value = 2P ( x ≥ 9) = 2 (1 − P ( x ≤ 8) ) = 2 (1 − .989) = .022
Since the p-value is less than α ( p = .022 < .05) , H0 is rejected. There is sufficient evidence to
indicate the median is different than 20 at 𝛼 = .05.
