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Solutions for Statistics Informed Decisions Using Data, 7th Edition Sullivan (All Chapters included)

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Complete Solutions Manual for Statistics Informed Decisions Using Data, 7th Edition by Michael Sullivan III ; ISBN13: 9780138317409....(Full Chapters included Chapter 1 to 15)...1. Data Collection 2. Organizing and Summarizing Data 3. Numerically Summarizing Data 4. Describing the Relation Betwe...

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answer guide
statistics
informed
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using data
7th edition
sullivan
9780138317409
solutions manual
7e

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Statistics Informed Decisions Using Data
7th Edition
by Michael Sullivan III

Complete Chapter Solutions Manual
are included (Ch 1 to 15)

** Immediate Download
** Swift Response
** All Chapters included

,The chapters are arranged in reverse order, starting from the
last. This structure helps you focus on the most recent and
advanced concepts first. You can also view a sample in this order
before purchasing, allowing you to preview the material for
better insight into the content.

Part 1 - Chapter 15 to 10 (Page 1 to Page 425)

Part 2 - Chapter 9 to 1 (Page 426 to 814)

Complete Chapters included (Chapters 1 to 15)

, Chapter 15
Nonparametric Statistics

Section 15.1 Section 15.2
1. Nonparametric statistics are inferential 1. run
procedures that are not based on parameters.
They do not require the population to follow a 2. n; n1; n2; r
specific type of distribution. Parametric
statistics test claims regarding parameters and 3. (a) n  14; nM  6; nF  8; r  6
often require that the distribution of the
population follow a specific distribution. (b) Since nM  20, nF  20, and   0.05,
Table XI should be used to find the
2. Nonparametric methods have fewer critical values. The table gives a lower
requirements but require larger sample sizes critical value of 3 and an upper critical
than equivalent parametric methods. The value of 12.
efficiency of a nonparametric test is the
proportion of the sample size required for the (c) Because the test statistic, r  6, is not
parametric test that would be required to less than or equal to the lower critical
achieve the same probability of a Type I error value, 3, and is not greater than or equal
with the equivalent nonparametric test. to the upper critical value, 12, do not
reject H0. The sequence of data is random.
3. The power of the test is the probability that the
null hypothesis is rejected when the
4. (a) n  15; nM  5; nF  10; r  4
alternative hypothesis is true.

4. Advantages of nonparametric procedures (b) Since nM  20, nF  20, and   0.05,
include the following: Table XI should be used to find the
critical values. The table gives a lower
1. Nonparametric tests have few critical value of 3 and an upper critical
requirements, so it is unlikely that they value of 12.
will be used improperly.
(c) Because the test statistic, r  4, is not
2. The computations are fairly easy. less than or equal to the lower critical
value, 3, and is not greater than or equal
3. The procedures can be used for count data to the upper critical value, 12, do not
or rank data. reject H0. The sequence of data is random.
5. Disadvantages of nonparametric procedures
5. (a) n  19; nY  9; nN  10; r  5
include the following:
1. They are often misused. That is, they are (b) Since nY  20, nN  20, and   0.05,
often used when more powerful Table XI should be used to find the
parametric methods are appropriate. critical values. The table gives a lower
2. Nonparametric methods are less powerful critical value of 5 and an upper critical
than parametric procedures. value of 16.
3. Nonparametric methods are less efficient (c) Because the test statistic, r  5, is less
than parametric procedures. than or equal to the lower critical value, 5,
reject H0. The sequence of data is not
6. It is appropriate to call nonparametric
random.
procedures distribution-free procedures, since
they do not require any particular underlying
distribution.

Copyright © 2025 Pearson Education, Inc.

, Section 15.2: Runs Test for Randomness 779

6. (a) n  21; nY  11; nN  10; r  10 11. n  20, nA  14, nR  6, and r  4. Since
nA  20, nR  20, and   0.05, Table XI
(b) Since nY  20, nN  20, and   0.05, should be used to find the critical values. The
Table XI should be used to find the table gives a lower critical value of 5 and an
critical values. The table gives a lower upper critical value of 14. Since r  5, reject
critical value of 6 and an upper critical H0. There is enough evidence at the   0.05
value of 17. level of significance to indicate that the way
the machine overfills and underfills does not
(c) Because the test statistic, r  10, is not
occur randomly.
less than or equal to the lower critical
value, 6, and is not greater than or equal 12. n  18, nA  15, nR  3, and r  3. Since
to the upper critical value, 17, do not
reject H0. The sequence of data is random. nA  20, nR  20, and   0.05, Table XI
should be used to find the critical values. The
7. n  20, nF  9, nC  11, and r  12. Since table gives a lower critical value of 3 and an
nF  20, nC  20, and   0.05, Table XI upper critical value of 8. Since r  3, reject
H0. There is enough evidence at the   0.05
should be used to find the critical values. The
table gives a lower critical value of 6 and an level of significance to indicate that the way
upper critical value of 16. Since 6  r  16, do the machine overfills and underfills does not
occur randomly.
not reject H0. There is not enough evidence at
the   0.05 level of significance to indicate 13. n  45, nP  25, nN  20, and r  21. Since
that Bumgarner’s pitches are not random.
nP  20, the standard normal distribution
8. n  15, nR  9, nP  6, and r  9. Since should be used.
nR  20, nP  20, and   0.05, Table XI 2(25)(20)
r   1  23.222
should be used to find the critical values. The 45
table gives a lower critical value of 4 and an
upper critical value of 13. Since 4  r  13, do 2(25)(20)  2(25)(20)  45
r   3.274
not reject H0. There is not enough evidence at 452 (45  1)
the   0.05 level of significance to indicate
that the sequence of plays is not random. 21  23.222
z0   0.68
3.274
9. n  14, nL  6, nO  8, and r  7. Since
The critical values at   0.05 are
nL  20, nO  20, and   0.05, Table XI
 z0.025  1.96 and z0.025  1.96. Since the
should be used to find the critical values. The
table gives a lower critical value of 3 and an test statistic, z0  0.68, is between 1.96
upper critical value of 12. Since 3  r  12, do and 1.96 (that is, does not lie within the
not reject H0. There is not enough evidence at critical region), do not reject H0. There is not
the   0.05 level of significance to indicate enough evidence at the   0.05 level of
that the arrival status is not random. significance to indicate that the stock price
fluctuations do not behave in a random
10. n  15, nL  5, nO  10, and r  6. Since manner. (Minitab: P-value  0.497  0.05.)
nL  20, nO  20, and   0.05, Table XI
should be used to find the critical values. The
table gives a lower critical value of 3 and an
upper critical value of 12. Since 3  r  12, do
not reject H0. There is not enough evidence at
the   0.05 level of significance to indicate
that the arrival status is not random.

Copyright © 2025 Pearson Education, Inc.

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