Nonparametric and Parametric Test Questions Choose two of the following three questions and post your answer: 1. In your own words, what is a nonparametri

Nonparametric and Parametric Test Questions Choose two of the following three questions and post your answer:

1. In your own words, what is a nonparametric test? What is a parametric test (Section 13-2)?
2. In your own words, identify an advantage of using rank correlation instead of linear correlation (Section 13-6).
3. In your own words, what is the difference between a nonparametric test and a distribution-free test (Section 13-7)?

You may use the information in the textbook or cite a source.

Then answer your classmates’ posts and comment on their description. Elementary Statistics
Thirteenth Edition
Chapter 13
Nonparametric
Tests
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Nonparametric Tests
13-1 Basics of Nonparametric Tests
13-2 Sign Test
13-3 Wilcoxon Signed-Ranks Test for Matched Pairs
13-4 Wilcoxon Rank-Sum Test for Two Independent
Samples
13-5 Kruskal-Wallis Test for Three or More Samples
13-6 Rank Correlation
13-7 Runs Test for Randomness
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Key Concept
This section describes the Wilcoxon rank-sum test,
which uses ranks of values from two independent
samples to test the null hypothesis that the samples
are from populations having equal medians.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Caution
Don’t confuse the Wilcoxon rank-sum test for two
independent samples with the Wilcoxon signedranks test for matched pairs.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test
• Wilcoxon Rank-Sum Test
– The Wilcoxon rank-sum test is a nonparametric test that
uses ranks of sample data from two independent populations
to test this null hypothesis:
H0: Two independent samples come from populations with equal
medians.
The alternative hypothesis H1 can be any one of the following three
possibilities:
• The two populations have different medians.
• The first population has a median greater than the median of the
second population.
• The first population has a median less than the median of the
second population.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test: Objective
Use the Wilcoxon rank-sum test with samples from
two independent populations for the following null and
alternative hypotheses:
H0: The two samples come from populations with equal
medians.
H1: The median of the first population is different from (or
greater than, or less than) the median from the second
population.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test: Notation
n1 = size of Sample 1
n2 = size of Sample 2
R1 = sum of ranks for Sample 1
R2 = sum of ranks for Sample 2
R = same as R1 (sum of ranks for Sample 1)
µR = mean of the sample R values that is expected when
the two populations have equal medians
σR = standard deviation of the sample R values that is
expected with two populations having equal medians
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test:
Requirements
1. There are two independent simple random samples.
2. Each of the two samples has more than 10 values.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test: Test
Statistic
n1 = size of the sample from which the rank sum R is found
n2 = size of the other sample
R = sum of ranks of the sample with size n1
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test: P-Values
P-Values
P-values can be found from technology or by using the
z test statistic and Table A-2.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Wilcoxon Rank-Sum Test: Critical
Values
Critical Values
Critical values can be found in Table A-2 (because the test
statistic is based on the normal distribution).
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the Test Statistic (1 of 4)
Step 1: Temporarily combine the two samples into one
big sample, then replace each sample value with its
rank. (The lowest value gets a rank of 1, the next lowest
value gets a rank of 2, and so on. If values are tied,
assign to them the mean of the ranks involved in the tie.
See Section 13-1 for a description of ranks and the
procedure for handling ties.)
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the Test Statistic (2 of 4)
Step 1:
EXAMPLE: In the table, the ranks of
the 27 student course evaluations are
shown in parentheses. The rank of 1 is
assigned to the lowest sample value of
2.9, the rank of 2 is assigned to the
next lowest value of 3.1, and the rank
of 3 is assigned to the next lowest
value of 3.3. The next lowest value is
3.4, which occurs four times, so we
find the mean of the ranks of 4, 5, 6, 7,
which is 5.5, and we assign the rank of
5.5 to each of those four tied values.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Female Professor
Male Professor
4.3 (20.5)
4.5 (25.5)
4.3 (20.5)
3.7 (8)
4.4 (23.5)
4.2 (17.5)
4.0 (13.5)
3.9 (11)
3.4 (5.5)
3.1 (2)
4.7 (27)
4.0 (13.5)
2.9 (1)
3.8 (9.5)
4.0 (13.5)
3.4 (5.5)
4.3 (20.5)
3.4 (25.5)
3.4 (5.5)
3.8 (9.5)
3.4 (5.5)
4.3 (20.5)
3.3 (3)
4.4 (23.5)
Blank
4.1 (16)
Blank
4.2 (17.5)
Blank
4.0 (13.5)
n1 = 12
n2 = 15
R1 = 159.5
R2 = 218.5
Procedure for Finding the Value of
the Test Statistic (3 of 4)
Step 2:
Find the sum of the ranks for either one
of the two samples.
EXAMPLE: In the table, the sum of the
ranks from the first sample is 159.5.
R1 = 20.5 + 20.5 + 23.5 + … + 3 = 159.5
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the Test Statistic (4 of 4)
Step 3: Calculate the value of the z test statistic as
shown previously, where either sample can be used as
“Sample 1.” (If both sample sizes are greater than 10,
then the sampling distribution of R is approximately
normal with mean µR and standard deviation σR, and
the test statistic is as shown previously.)
EXAMPLE: Calculations of µR and σR and z are shown in
the example which follows.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (1 of 11)
The table lists course evaluation
ratings for courses taught by
female and male professors.
Use a 0.05 significance level to
test the claim that female and
male professors have the same
median course evaluation rating.
Female Professor
Male Professor
4.3 (20.5)
4.5 (25.5)
4.3 (20.5)
3.7 (8)
4.4 (23.5)
4.2 (17.5)
4.0 (13.5)
3.9 (11)
3.4 (5.5)
3.1 (2)
4.7 (27)
4.0 (13.5)
2.9 (1)
3.8 (9.5)
4.0 (13.5)
3.4 (5.5)
4.3 (20.5)
3.4 (25.5)
3.4 (5.5)
3.8 (9.5)
3.4 (5.5)
4.3 (20.5)
3.3 (3)
4.4 (23.5)
Blank
4.1 (16)
Blank
4.2 (17.5)
Blank
4.0 (13.5)
n1 = 12
n2 = 15
R1 = 159.5
R2 = 218.5
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (2 of 11)
Solution
Requirement Check (1) The
sample data are two independent
simple random samples. (2) The
sample sizes are 12 and 15, so
both sample sizes are greater than
10. The requirements are satisfied.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (3 of 11)
Solution
The null and alternative hypotheses are as follows:
H0: The median of the course evaluation ratings for courses
taught by female professors is equal to the median of
the course evaluation ratings for courses taught by male
professors.
H1: The median course evaluation rating for courses taught
by female professors is different from the median course
evaluation rating for courses taught by male professors.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (4 of 11)
Solution
Rank the combined list of all 27 course
ratings, beginning with a rank of 1
(assigned to the lowest value of 2.9).
The ranks corresponding to the
individual sample values are shown in
parentheses in the table. R denotes the
sum of the ranks for the sample we
choose as Sample 1. If we choose the
course evaluations of female professors
as Sample 1, we get
R = 20.5 + 20.5 + 23.5 + … + 3 = 159.5
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (5 of 11)
Solution
Because there are course
evaluation ratings for 12 female
professors, we have n1 = 12.
Also, n2 = 15 because there are
course evaluation ratings of 15
male professors. The values of
µR and σR and the test statistic z
can now be found as follows.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (6 of 11)
Solution
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (7 of 11)
Solution
The test is two-tailed because a large positive value of z
would indicate that disproportionately more higher ranks
are found in Sample 1, and a large negative value of z
would indicate that disproportionately more lower ranks are
found in Sample 1. In either case, we would have strong
evidence against the claim that the two samples come from
populations with equal medians.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (8 of 11)
Solution
The significance of the test statistic z can be treated as
in previous chapters. We are testing (with α = 0.05) the
hypothesis that the two populations have equal medians,
so we have a two-tailed test.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (9 of 11)
Solution
P-Value Using the unrounded z score, the P-value is 0.678,
so we fail to reject the null hypothesis that the populations
of female professors and male professors have the same
median course evaluation rating.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (10 of 11)
Solution
Critical Values If we use the critical values of z = ±1.96, we
see that the test statistic of z = −0.41 does not fall within the
critical region, so we fail to reject the null hypothesis that the
populations of female professors and male professors have
the same median course evaluation rating.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Course Evaluation Ratings
for Female and Male Professors (11 of 11)
Interpretation
There is not sufficient evidence to warrant rejection of the
claim that female professors and male professors have the
same median course evaluation rating. Based on the
available sample data, it appears that female professors and
male professors teach courses that are rated about the same.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Elementary Statistics
Thirteenth Edition
Chapter 13
Nonparametric
Tests
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Nonparametric Tests
13-1 Basics of Nonparametric Tests
13-2 Sign Test
13-3 Wilcoxon Signed-Ranks Test for Matched Pairs
13-4 Wilcoxon Rank-Sum Test for Two Independent
Samples
13-5 Kruskal-Wallis Test for Three or More Samples
13-6 Rank Correlation
13-7 Runs Test for Randomness
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Key Concept
This section describes the Kruskal-Wallis test, which
uses ranks of data from three or more independent
simple random samples to test the null hypothesis that
the samples come from populations with the same
median.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis test
• Kruskal-Wallis test
– The Kruskal-Wallis test (also called the H test)
is a nonparametric test that uses ranks of
combined simple random samples from three or
more independent populations to test the null
hypothesis that the populations have the same
median.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis Test: Objective
Use the Kruskal-Wallis test with simple random
samples from three or more independent populations
for the following null and alternative hypotheses:
H0: The samples come from populations with the same
median.
H1: The samples come from populations with medians that
are not all equal.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis Test: Notation
N = total number of observations in all samples
combined
k = number of different samples
R1 = sum of ranks for Sample 1
n1 = number of observations in Sample 1
For Sample 2, the sum of ranks is R2 and the number of
observations is n2, and similar notation is used for the
other samples.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis Test: Requirements
1. We have at least three independent simple random
samples.
2. Each sample has at least five observations.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis Test: Test Statistic
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis Test: P-Values
P-values are often provided by technology. By using
the test statistic H and the number of degrees of
freedom (k − 1), Table A-4 can be used to find a
range of values for the P-value.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Kruskal-Wallis Test: Critical Values
1. The test is right-tailed and critical values can be
found from technology or from the chi-square
distribution in Table A-4.
2. df = k − 1 (where df is the number of degrees of
freedom and k is the number of different samples)
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the H Test Statistic (1 of 4)
Step 1: Temporarily combine all samples into one big
sample and assign a rank to each sample value. (Sort
the values from lowest to highest, and in cases of ties,
assign to each observation the mean of the ranks
involved.)
EXAMPLE: In the table on the next slide, the numbers in
parentheses are the ranks of the combined data set. The
rank of 1 is assigned to the lowest value of 64, the rank of 2
is assigned to the next lowest value of 78, and so on. In the
case of ties, each of the tied values is assigned the mean of
the ranks involved in the tie.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the H Test Statistic (2 of 4)
Performance IQ Scores (Ranks in parentheses)
Low Blood Lead Level Medium Blood Lead Level High Blood Lead Level
80 (6.5)
78 (2)
93 (10)
90 (8.5)
97 (12.5)
100 (15.5)
107 (18.5)
107 (18.5)
97 (12.5)
85 (6.5)
80 (4)
79 (3)
100 (15.5)
90 (8.5)
97 (12.5)
97 (12.5)
83 (5)
Blank
101 (17)
Blank
Blank
64 (1)
Blank
Blank
n1 = 8
n2 = 6
n3 = 5
R1 = 86
R2 = 50.5
R3 = 53.5
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the H Test Statistic (3 of 4)
Step 2: For each sample, find the sum of the ranks and
find the sample size.
EXAMPLE: In the table, the sum of the ranks from the first sample
is 86, the sum of the ranks for the second sample is 50.5, and the
sum of the ranks for the third sample is 53.5.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Procedure for Finding the Value of
the H Test Statistic (4 of 4)
Step 3: Calculate H using the results of Step 2 and
the notation and test statistic given.
EXAMPLE: The test statistic is computed in the next
example.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (1 of 11)
The table on the next slide lists performance (nonverbal) IQ scores from samples of subjects with low
blood lead level, medium blood lead level, and high
blood lead level. Use a 0.05 significance level to test the
claim that the three samples of performance IQ scores
come from populations with medians that are all equal.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (2 of 11)
Low Blood Lead Level Medium Blood Lead Level High Blood Lead Level
80 (6.5)
78 (2)
93 (10)
90 (8.5)
97 (12.5)
100 (15.5)
107 (18.5)
107 (18.5)
97 (12.5)
85 (6.5)
80 (4)
79 (3)
100 (15.5)
90 (8.5)
97 (12.5)
97 (12.5)
83 (5)
Blank
101 (17)
Blank
Blank
64 (1)
Blank
Blank
n1 = 8
n2 = 6
n3 = 5
R1 = 86
R2 = 50.5
R3 = 53.5
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (3 of 11)
Solution
Requirement Check (1) Each of the three samples is a
simple random independent sample. (2) Each sample size
is at least 5. The requirements are satisfied.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (4 of 11)
Solution
The null and alternative hypotheses are as follows:
H0: The population of subjects with low lead exposure, the
population with medium lead exposure, and the population
with high lead exposure all have performance IQ scores with
the same median.
H1: The three populations of performance IQ scores have
three medians that are not all the same.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (5 of 11)
Solution
Test Statistic First combine all of the sample data and rank
them, then find the sum of the ranks for each category. In the
table, ranks are shown in parentheses next to the original
sample values. Next, find the sample size (n) and sum of
ranks (R) for each sample. Those values are shown at the
bottom of the table. Because the total number of observations
is 19, we have N = 19.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (6 of 11)
Solution
We can now evaluate the test statistic as follows:
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (7 of 11)
Solution
Because each sample has at least five observations, the
distribution of H is approximately a chi-square distribution
with k − 1 degrees of freedom. The number of samples is
k = 3, so we have 3 − 1 = 2 degrees of freedom.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (8 of 11)
Solution
P-Value With H = 0.694 and df = 2, Table A-4 shows that
the P-value is greater than 0.10. Using technology, we get
P-value = 0.707. Because the P-value is greater than the
significance level of 0.05, we fail to reject the null
hypothesis of equal population medians.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (9 of 11)
Solution
Critical Value Refer to Table A-4 to find the critical value
of 5.991, which corresponds to 2 degrees of freedom and
a 0.05 significance level (with an area of 0.05 in the right
tail). The figure on the next slide shows that the test
statistic H = 0.694 does not fall within the critical region
bounded by 5.991, so we fail to reject the null hypothesis
of equal population medians.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (10 of 11)
Solution
The figure shows the test
statistic of H = 0.694 and the
critical value of χ² = 5.991. The
test statistic does not fall in the
critical region, so we fail to
reject the null hypothesis of
equal medians.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Example: Effect of Lead on IQ
Scores (11 of 11)
Interpretation
There is not sufficient evidence to reject the claim that
the three samples of performance IQ scores come from
populations with medians that are all equal. The
population medians do not appear to be significantly
different.
Copyright © 2018, 2014, …
Purchase answer to see full
attachment