Business Statistics Communicating
with Numbers, 5th Edition
by Sanjiv Jaggia
Complete Chapter Solutions Manual
are included (Ch 1 to 20)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1. Data and Data Preparation
2. Tabular and Graphical Methods
3. Numerical Descriptive Measures
7. Sampling and Sampling Distributions
8. Interval Estimation
9. Hypothesis Testing
10. Statistical Inference Concerning Two Populations
11. Statistical Inference Concerning Variance
12. Chi-Square Tests
13. Analysis of Variance
14. Regression Analysis
15. Inference with Regression Models
16. Regression Models for Nonlinear Relationships
17. Regression Models with Dummy Variables
18. Forecasting with Time Series Data
19. Returns, Index Numbers, and Inflation
20. Nonparametric Tests
,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 Ch20-1)
Chapter 20. Nonparametric Tests
Solutions
1.
a. 𝐻0 : 𝑚 = 20; 𝐻𝐴 : 𝑚 ≠ 20
b.
Ranks of Ranks of
𝑥 𝑑 = 𝑥 − 20 |d| Rank Negative Positive
Differences Differences
25 5 5 3 3
18 -2 2 2 2
21 1 1 1 1
27 7 7 4 4
30 10 10 5 5
𝑇− = 2 +
𝑇 = 13
The value of the test statistic 𝑇 = 𝑇 + =13
c. Since the p-value= 0.188 > 0.05 = 𝛼, we do not reject 𝐻0 . At the 5% significance
level, we cannot conclude that the median differs from 20.
2.
a. 𝐻0 : 𝑚 ≤ 140; 𝐻𝐴 : 𝑚 > 140
b.
Ranks of Ranks of
𝑥 𝑑 = 𝑥 − 140 |d| Rank Negative Positive
Differences Differences
150 10 10 4 4
145 5 5 3 3
138 -2 2 2 2
155 15 15 6 6
141 1 1 1 1
152 12 12 5 5
𝑇 − =2 𝑇 + = 19
The value of the test statistic 𝑇 = 𝑇 + =19
20-1
, Chapter 20 - Nonparametric Tests
c. Since the p-value= 0.047 < 0.05 = 𝛼, we reject 𝐻0 . At the 5% significance level,
we conclude that the median is greater than 140.
3.
a. 𝐻0 : 𝑚 ≥ 10; 𝐻𝐴 : 𝑚 < 10
b.
Ranks of Ranks of
𝑥 𝑑 = 𝑥 − 10 |d| Rank Negative Positive
Differences Differences
8 -2 2 2 2
5 -5 5 5.5 5.5
11 1 1 1 1
7 -3 3 3 3
6 -4 4 4 4
5 -5 5 5.5 5.5
𝑇 − = 20 𝑇+ = 1
The value of the test statistic 𝑇 = 𝑇 + =1
c. Since the p-value= 0.029 < 0.05 = 𝛼, we reject 𝐻0 . At the 5% significance level,
we conclude that the median is less than 10.
4.
a. Given 𝑇 = 𝑇 + = 265 and T is assumed normally distributed, we calculate
𝑛(𝑛+1) 30(30+1) 𝑛(𝑛+1)(2𝑛+1) 30×(30+1)(2×30+1)
𝜇𝑇 = = = 232.50 and 𝜎𝑇 = √ =√ =
4 4 24 24
𝑇−𝜇𝑇 265−232.50
48.6184. Therefore, 𝑧 = = = 0.6685.
𝜎𝑇 48.6184
b. The p-value =𝑃(𝑍 ≥ 0.6685) = 0.2519
In Excel: =1-NORM.DIST(0.6685, 0, 1, TRUE)
In R: 1-pnorm(0.6685, 0, 1, lower.tail=TRUE)
c. Because the p-value= 0.2519 > 𝛼 = 0.05, we do not reject 𝐻0 . Thus, at the 5%
significance level, we cannot conclude that the median is greater than 150.
5.
a. 𝐻0 : 𝑚 = 100; 𝐻𝐴 : 𝑚 ≠ 100
20-2
with Numbers, 5th Edition
by Sanjiv Jaggia
Complete Chapter Solutions Manual
are included (Ch 1 to 20)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1. Data and Data Preparation
2. Tabular and Graphical Methods
3. Numerical Descriptive Measures
7. Sampling and Sampling Distributions
8. Interval Estimation
9. Hypothesis Testing
10. Statistical Inference Concerning Two Populations
11. Statistical Inference Concerning Variance
12. Chi-Square Tests
13. Analysis of Variance
14. Regression Analysis
15. Inference with Regression Models
16. Regression Models for Nonlinear Relationships
17. Regression Models with Dummy Variables
18. Forecasting with Time Series Data
19. Returns, Index Numbers, and Inflation
20. Nonparametric Tests
,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 Ch20-1)
Chapter 20. Nonparametric Tests
Solutions
1.
a. 𝐻0 : 𝑚 = 20; 𝐻𝐴 : 𝑚 ≠ 20
b.
Ranks of Ranks of
𝑥 𝑑 = 𝑥 − 20 |d| Rank Negative Positive
Differences Differences
25 5 5 3 3
18 -2 2 2 2
21 1 1 1 1
27 7 7 4 4
30 10 10 5 5
𝑇− = 2 +
𝑇 = 13
The value of the test statistic 𝑇 = 𝑇 + =13
c. Since the p-value= 0.188 > 0.05 = 𝛼, we do not reject 𝐻0 . At the 5% significance
level, we cannot conclude that the median differs from 20.
2.
a. 𝐻0 : 𝑚 ≤ 140; 𝐻𝐴 : 𝑚 > 140
b.
Ranks of Ranks of
𝑥 𝑑 = 𝑥 − 140 |d| Rank Negative Positive
Differences Differences
150 10 10 4 4
145 5 5 3 3
138 -2 2 2 2
155 15 15 6 6
141 1 1 1 1
152 12 12 5 5
𝑇 − =2 𝑇 + = 19
The value of the test statistic 𝑇 = 𝑇 + =19
20-1
, Chapter 20 - Nonparametric Tests
c. Since the p-value= 0.047 < 0.05 = 𝛼, we reject 𝐻0 . At the 5% significance level,
we conclude that the median is greater than 140.
3.
a. 𝐻0 : 𝑚 ≥ 10; 𝐻𝐴 : 𝑚 < 10
b.
Ranks of Ranks of
𝑥 𝑑 = 𝑥 − 10 |d| Rank Negative Positive
Differences Differences
8 -2 2 2 2
5 -5 5 5.5 5.5
11 1 1 1 1
7 -3 3 3 3
6 -4 4 4 4
5 -5 5 5.5 5.5
𝑇 − = 20 𝑇+ = 1
The value of the test statistic 𝑇 = 𝑇 + =1
c. Since the p-value= 0.029 < 0.05 = 𝛼, we reject 𝐻0 . At the 5% significance level,
we conclude that the median is less than 10.
4.
a. Given 𝑇 = 𝑇 + = 265 and T is assumed normally distributed, we calculate
𝑛(𝑛+1) 30(30+1) 𝑛(𝑛+1)(2𝑛+1) 30×(30+1)(2×30+1)
𝜇𝑇 = = = 232.50 and 𝜎𝑇 = √ =√ =
4 4 24 24
𝑇−𝜇𝑇 265−232.50
48.6184. Therefore, 𝑧 = = = 0.6685.
𝜎𝑇 48.6184
b. The p-value =𝑃(𝑍 ≥ 0.6685) = 0.2519
In Excel: =1-NORM.DIST(0.6685, 0, 1, TRUE)
In R: 1-pnorm(0.6685, 0, 1, lower.tail=TRUE)
c. Because the p-value= 0.2519 > 𝛼 = 0.05, we do not reject 𝐻0 . Thus, at the 5%
significance level, we cannot conclude that the median is greater than 150.
5.
a. 𝐻0 : 𝑚 = 100; 𝐻𝐴 : 𝑚 ≠ 100
20-2
