Download Document

Chapter 14: Elements of
Nonparametric Statistics
Weight
Person
Mrs. Smith
Mrs. Brown
Mrs. White
Mr. Collins
Mr. Gray
Ms. Collins
Mrs. Allen
Mrs. Noss
Ms. Wagner
Mr. Carroll
Mrs. Black
Before
146
175
150
190
220
157
136
146
128
187
172
After
142
178
147
187
212
160
135
138
132
187
171
Sign of Difference
After - Before









0

Chapter Goals
• Introduce the basic concepts of
nonparametric statistics, or distribution-free
techniques.
• Nonparametric statistics are versatile and
easy to use.
• Consider some of the most common tests
and applications.
14.1: Nonparametric Statistics
• Parametric methods: Assume the population
is at least approximately normal, or use the
central limit theorem.
• Nonparametric methods, or distribution-free
methods: Assume very little about the
population, subject to less confining
restrictions.
Nonparametric statistics have become popular:
1. Require few assumptions about the underlying population.
2. Generally easier to apply than their parametric
counterparts.
3. Relatively easy to understand.
4. Can be used in situations where the normality assumptions
cannot be made.
5. Generally only slightly less efficient than their parametric
counterparts.
Disadvantages?
14.2: Comparing Statistical Tests
• Four nonparametric tests presented in this
chapter. There are many others.
• Many nonparametric tests may be used as
well as certain parametric tests.
• Which statistical test is appropriate: the
parametric or nonparametric.
Which test is best?
1. When comparing two tests they must be equally qualified
for use; they must both be appropriate test procedures.
2. Each test has a set of assumptions that must be satisfied.
3. The best test: The test that is best able to control the risks
of error and at the same time keeps the sample size
reasonable.
4. A larger sample size usually means a higher cost.
The Risk of Error:
1. Type I Error: Controlled directly (set) by the level of
significance a.
2. P(type I error) = a,
P(type II error) = b
3. We try to control b.
4. The power of a statistical test = 1  b
The power of a test is the probability that we reject the null
hypothesis when it is false (a correct decision).
If two appropriate statistical tests have the same
significance level a, the one with the greater power is
better.
The sample size:
1. Set acceptable values for a and b. Determine the sample
size necessary to satisfy these values.
2. The statistical test that requires the smaller sample size is
better.
3. Efficiency: The ratio of the sample size of the best
parametric test to the sample size of the best nonparametric
test when compared under a fixed set of risk values.
Example: Efficiency rating for the sign test is
approximately 0.63. This means that a sample of size 63
with a parametric test will do the same job as a sample of
size 100 for the sign test.
To determine the choice of test:
1. Often forced to use a certain test because of the nature of
the data.
2. When there is a choice, consider three factors:
a. The power of the test.
b. The efficiency of the test.
c. The data (and the sample size).
Note: The following table shows a comparison of
nonparametric tests (presented in this chapter) with the
parametric tests presented earlier.
Comparison of Parametric and Nonparametric Tests:
Test
Situation
One mean
Parametric
Test
t test
(p. 773)
Two
t test
independent (p. 910)
means
Two
t test
dependent (p. 886)
means
Correlation Pearson's
(p. 1137)
Randomness
Nonparametric
Efficiency of
Test
Nonparametric Test
Sign test
0.63
(p. 1219)
U test
0.95
(p. 1243)
Sign test
(p. 1225)
0.63
Spearman test
(p. 1274)
Runs test
(p. 1260)
0.91
Not meaningful
14.3: The Sign Test
• Versatile, easy to apply, uses only plus and
minus signs.
• Three sign test applications: confidence
interval for a median, hypothesis test
concerning a median, hypothesis test
concerning the median difference (paired
difference) for two dependent samples.
Assumptions for inferences about the population median
using the sign test: The n random observations forming the
sample are selected independently and the population is
continuous in the vicinity of the median, M.
Procedure for using the sign test to obtain a confidence
interval for an unknown population median, M:
1. Arrange the data in ascending order (smallest to largest):
x1 (smallest), x2, x3, . . . , xn (largest)
2. Use Table 12, Appendix B to obtain the critical value, k
(the maximum allowable number of signs).
3. k indicates the number of positions to be dropped from
each end of the ordered data.
4. The remaining extreme values are the bounds for a 1  a
confidence interval.
Confidence Interval: xk+1 to xnk
Note: Based on the binomial distribution.
Example: Suppose 20 observations are selected at random and
are given in ascending order (x1, x2, x3, . . . , x20).
19 21 23 28 31 32 33 34 34 35
38 41 43 43 44 46 47 48 52 55
Find a 95% confidence interval for the population median.
Solution:
Table 12: n = 20, a = 0.05 k = 5
Drop the last 5 values on each end.
The confidence interval is bounded by x6 and x10.
The confidence interval: 32 to 44 (inclusive).
In general: xk1 to xnk is a 1  a confidence interval for M.
Single-Sample Hypothesis Test Procedure:
1. The sign test may be used when the null hypothesis
concerns the population median M.
2. The test may be either one- or two-tailed.
Example: A random sample of 88 tax payers was selected and
each was asked the amount of time spent preparing their
federal income tax return. Test the hypothesis “the median
time required to prepare a return is 8 hours” against the
alternative that the median is greater than 8 hours.
The data is summarized by:
Under 8: 37; Equal 8: 3; Over 8: 48
Use the sign test with a = 0.025.
Solution:
The data is converted to () and () signs according to
whether the data is more or less than 8.
A plus sign is assigned to each observation greater than 8.
A minus sign is assigned to each observation less than 8.
A zero is assigned to each observation equal to 8.
The sign test uses only the plus and minus signs.
The zeros are discarded.
Usable sample size = 88  3 = 85
n() = 48
n() = 37
n()  n() = n = 85
1. The Set-up:
a. Population parameter of concern: M, population median
time to prepare a federal income tax return.
b. The null and alternative hypothesis:
H0: M = 8
Ha: M > 8
2. The Hypothesis Test Criteria:
a. Assumptions: The 88 observations were randomly
selected and the variable time to prepare a return is
continuous.
b. Test statistic: x = the number of the less frequent sign = n()
c. Level of significance: a = 0.025
3. The Sample Evidence:
n = 85; x = n() = 37
4. The Probability Distribution (Classical Approach):
a. Critical value: The critical region is one-tail.
Table 12 is for two-tailed tests.
At the intersection of the column a = 0.05 (= 2  0.025)
and the row n = 85: k = 32.
The critical value: k = 32.
b. x is not is the critical region.
4. The Probability distribution (p-Value Approach):
a. The p-value: Using Table 12: P > (0.25/2) = 0.125
Using a computer: P = 0.1928
b. The p-value is larger than the level of significance, a.
5. The Results:
a. Decision: Do not reject H0.
b. Conclusion: At the 0.025 level of significance, there is
no evidence to suggest the median time required to
complete a federal income tax return is greater than 8
hours.
Two Sample Hypothesis Test Procedure:
1. The sign test may also be used in tests concerning the
median difference between paired data that result from two
dependent samples.
2. A common application: the use of before-and-after testing
to determine the effectiveness of some activity.
3. The signs of the differences are used to carry out the test.
Zeros are discarded.
Assumptions for inferences about median of paired
differences using sign test:
The paired data is selected independently and the variables
are ordinal or numerical.
Example: A new automobile engine additive (included during
an oil change) is designed to decrease wear and improve
engine performance by increasing gas mileage. Sixteen
randomly selected automobiles were selected and the beforeand-after miles per gallon were recorded. (The same driver
was used before and after the engine treatment.) Is there any
evidence to suggest the engine additive improves gas
mileage? Use a = 0.05.
Note: The claim being tested is that the additive improves gas
mileage. Form all the differences, After  Before. We will
only reject the null hypothesis if there are significantly more
plus signs.
Data:
Car
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Before
17.5
30.7
28.1
25.5
23.2
23.3
17.8
27.4
22.3
24.2
20.9
15.8
24.8
15.1
22.6
22.2
After
24.1
23.8
27.9
26.0
24.2
23.9
16.9
26.0
33.0
27.1
22.4
20.9
22.2
27.2
18.6
29.7
Sign
















1. The Set-up:
a. Population parameter of concern:
M, median gain in miles per gallon.
b. The null and alternative hypothesis:
H0: M = 0 (no mileage gain)
Ha: M > 0 (mileage gain)
2. The Hypothesis Test Criteria:
a. Assumptions: The automobiles were randomly selected
and the variables, miles per gallon before and after, are
both continuous.
b. Test statistic: The number of the less frequent sign.
In this example: x = n()
c. Level of significance: 0.05
3. The Sample Evidence:
n = 16; n() = 10; n() = 6
Observed value of the test statistic: x = n() = 6
4. The Probability Distribution (Classical Approach):
a. Critical Value: The critical region is one-tail.
Table 12 is for two-tailed tests.
At the intersection of the column a = 0.10 (= 2  0.05)
and the row n = 16: k = 4.
The critical value: k = 4.
b. x is not in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: Using Table 12: P > (0.25/2) = 0.125
Using a computer: P = 0.2272
b. The p-value is larger than the level of significance, a.
5. The Results:
a. Decision: Do not reject H0.
b. Conclusion: At the 0.05 level of significance, there is no
evidence to suggest the engine additive increases the
miles per gallon.
Normal Approximation:
1. The sign test may be carried our using a normal
approximation and the standard normal variable z.
2. The normal approximation is used if Table 12 does not
show the desired level of significance or if n is large.
Procedure:
1. x is the number of the less frequent sign or the most
frequent sign; consistent with the alternative hypothesis.
2. x is a binomial random variable with p = 0.5.
1 n
 x  np  n  
2 2
1 1 1
 x  npq  n   
n
2 2 2
3. x is a binomial random variable, but it does become
approximately normal for large n.
Problem: A binomial random variable is discrete and a
normal random variable is continuous.
Solution: Use the continuity correction: an adjustment in
the normal random variable so that the approximation is
more accurate.
Continuity Correction:
a. For the binomial random variable, the area of a
rectangular bar represents probability: width 1, from 1/2
unit below to 1/2 unit above the value of interest.
b. When z is used, make a 1/2 unit adjustment before
calculating the observed value of z.
c. x’ is the adjusted value for x.
If x > n/2 then x’ = x  (1/2)
If x < n/2 then x’ = x  (1/2)
Continuity Correction Illustration:
0
.
2
0
0
.
1
5
p(x)
0
.
1
0
0
.
0
5
0
.
0
0
6
.
0
6
.
5
7
.
0
7
.
5
8
.
0
P(x = 7) P(6.5  x  7.5)
discrete
continuous
1  a confidence interval for M:
Using the normal approximation (including the continuity
correction), the position numbers are:
1
1 1
(n)     z (a / 2)  n 
2
2 2

The interval is xL to xU where
n 1 z (a / 2)
L  
 n
2 2
2
and
n 1 z (a / 2)
U  
 n
2 2
2
Note: L should be rounded down and U should be rounded up
to be sure the level of confidence is at least 1  a.
Example: Estimate the population median with a 95%
confidence interval for a given data set with 55 observations:
x1, x2, x3, . . . , x54, x55.
Solution:
The position numbers are:
1
1 1
 1
1 1

(n)     z (a / 2)  n   (55)    1.96  55 
2
2 2
 2
2 2

 27.5  (0.50  7.27)
 27.5  7.77
L = 27.5  7.77 = 19.73; rounded down, L = 19.
U = 27.5  7.77 = 35.27; rounded up, U = 36.
Therefore: 95% confidence interval for M: x19 to x36.
Hypothesis test concerning M:
Using the standard normal distribution, z is computed using
the formula:
x'(n / 2)
z* 
n/2
Example: In a recent study children between the ages of 8 and
12 were reported to watch a median of 18 hours of television
per week. In order to test this claim, 105 children between 8
and 12 were selected at random and the number of hours of
television watched per week were recorded. A plus sign was
coded if the number of hours was greater than 18, a minus
sign if less than or equal to18: there were 71 plus signs and 34
minus signs. Use the normal approximation to the sign test to
determine if there is any evidence to suggest the median
number of hours watched is greater than 18. Use a = 0.05
Solution:
1. The Set-up:
a. Population parameter of concern: M, the median number
of hours of television watched per week.
b. The null and alternative hypothesis:
H0: M = 18 () (at least as may minus signs as plus signs)
Ha: M > 18 (fewer minus signs than plus signs)
2. The Hypothesis Test Criteria:
a. Assumptions: The random sample of 105 students was
independently surveyed and the variable, hours of television
watched per week, is continuous.
b. Test statistic: z*
c. Level of significance: a = 0.05
3. The Sample Evidence:
a. Sample information: n() = 71, n() = 34
n = 105 and x = 71
b. Calculate the value of the test statistic:
x'(n / 2) 70.5  (105 / 2)
z* 

n/2
105 / 2
70.5  52.5
18


 3.51
10.25 / 2
5.125
4. The Probability Distribution (Classical Approach):
a. Critical value: z(0.05) = 1.65
b. z* is in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = P(z* > 3.51) 0.0002
Using a computer: P  0.000224
b. The p-value is smaller than the level of significance, a.
5. The Results:
a. Decision: Reject H0.
b. Conclusion: At the 0.05 level of significance, there is
evidence to suggest the median number of hours of
television watched per week is greater than 18.
14.4: The Mann-Whitney U Test
• Nonparametric alternative for the t test for
the difference between two independent
means.
• Null hypothesis: the two sampled
populations are identical.
Assumptions for inferences about two populations using
the Mann-Whitney test:
The two independent random samples are independent within
each sample as well as between samples, and the random
variables are ordinal or numerical.
Note:
1. This test procedure is often applied in situations in which
the two samples are drawn from the same population of
subjects, but different treatments are used on each sample.
2. Test procedure described in the following example.
Example: A recent study claimed that adults who exercise
regularly tend to have lower pulse rates. To test this claim,
two independent random samples of adult males were
selected, one from those who exercise regularly (A), and one
from those who are more sedentary (B). The data is given
below. Is there any evidence to suggest that adults who
exercise regularly have lower pulse rates than those who do
not exercise regularly. Use a = 0.05.
Sample
A
B
63
83
71
75
61
63
66
69
Data
61 63
77 76
68
69
69
65
78
70
70
68
Solution:
1. The Set-up:
a. Population parameter of concern: The distribution of
pulse rates for each population of adult males.
b. The null and alternative hypothesis:
H0: Populations A and B have pulse rates with identical
distributions.
Ha: The two distributions are not the same.
2. The Hypothesis Test Criteria:
a. Assumptions: The two samples are independent, and the
random variable (pulse rate) is numerical.
b. Test statistic: Mann-Whitney U Statistic, described below.
c. Level of significance: a = 0.05
3. The Sample Evidence:
a. Sample information: Data given in the table above.
b. Calculate the value of the test statistic:
na = sample size from population A
nb = sample size from population B
Combine the two samples and order the data from
smallest to largest.
Assign each observation a rank number.
The smallest observation is assigned rank 1, the next
smallest is assigned rank 2, etc., up to the largest, which
is assigned rank na  nb.
For ties: assign each of the tied observations the mean
rank of those rank positions that they occupy.
The rankings:
Ranked
Data
61
61
63
63
63
65
66
68
68
69
Rank
1.5
1.5
4
4
4
6
7
8.5
8.5
11
Source
A
A
A
A
B
B
A
A
B
A
Ranked
Data
69
69
70
70
71
75
76
77
78
83
Rank
11
11
13.5
13.5
15
16
17
18
19
20
Source
B
B
A
B
A
B
B
B
A
B
To Compute the U Statistic:
1. Compute the sum of the ranks for each of the two samples:
Ra and Rb.
2. Compute the U score for each sample:
(nb )(nb  1)
U a  na  nb 
 Rb
2
(na )(na  1)
U b  na  nb 
 Ra
2
3. The test statistic, U*, is the smaller of Ua and Ub.
In this example:
Ra  1.5  1.5  4  4  7  8.5  11  13.5  15  19  85
Rb  4  6  8.5  11  11  13.5  16  17  18  20  125
(10)(10  1)
U a  (10)(10) 
 125  30
2
(10)(10  1)
U b  (10)(10) 
 85  70
2
U *  30
Background:
1. Suppose the two samples are very different.
Small ranks are associated with one sample, large ranks
with the other.
U* would tend to be small, and we would want to reject
the null hypothesis.
2. Suppose the two samples are very similar.
The ranks are evenly distributed between the two samples.
Ua and Ub tend to be about equal, U* tends to be larger.
Note: Ua  Ub = na  nb
Therefore: only need to consider the smaller U-value.
4. The Probability Distribution (Classical Approach):
a. Critical value: Use Table 13B, one-tailed, a = 0.05
Critical value is at the intersection of column n1 = 10
and row n2 = 10: 27
b. U* is not in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = P(U* 30, for n1 = 10 and n2 = 10)
Using Table 13: P > 0.05
Using a computer: P  0.0694
b. The p-value is not smaller than a.
5. The Results:
a. Decision: Do not reject H0.
b. Conclusion: At the 0.05 level of significance, there is no
evidence to suggest the two populations are different.
Normal Approximation:
If the sample sizes are large, then U is approximately normal
with
na  nb
U 
2
na  nb  (na  nb  1)
U 
12
The standard normal distribution may be used if both sample
sizes are greater than 10; the test statistic is
U  U
z* 
U
Example: The data below represents the number of hours two
different cellular phone batteries worked before a recharge
was necessary. Is there any evidence to suggest battery type
B lasts longer than battery type A. Use the Mann-Whitney
test with a =0.05.
Battery
A
44
44
42
53
41
47
B
42
44
49
38
49
45
51
45
49
48
40
43
44
35
52
47
41
37
49
53
55
Solution:
1. The Set-up:
a. Population parameter of concern: The distribution of
battery life for each brand.
b. The null and alternative hypothesis:
H0: The distributions for battery life are the same for
both brands.
Ha: The distributions are not the same.
2. The Hypothesis Test Criteria:
a. Assumptions: The two samples are independent and the
random variable, battery life, is continuous.
b. Test statistic: Mann-Whitney U statistic (normal
approximation).
c. Level of significance: a = 0.05
3. The Sample Evidence:
a. Sample information: Data given in the table above.
b. Calculate the value of the test statistic:
Rankings for battery life:
Ranked
Data
35
37
38
40
41
41
42
42
43
44
44
44
44
45
Rank
1
2
3
4
5.5
5.5
7.5
7.5
9
11.5
11.5
11.5
11.5
14.5
Source
B
B
A
B
A
B
A
A
B
A
A
A
B
A
Ranked
Data
45
47
47
48
49
49
49
49
51
52
53
53
55
Rank
14.5
16.5
16.5
18
20.5
20.5
20.5
20.5
23
24
25.5
25.5
27
Source
B
A
B
B
A
A
B
B
B
B
A
B
B
The sums:
RA  3  5.5  7.5  7.5  11.5  11.5  11.5  14.5  16.5
 20.5  20.5  25.5  155.5
RB  1  2  4  5.5  9  11.5  14.5  16.5  18  20.5  20.5
 23  24  25.5  27  222.5
The U scores:
(15)(15  1)
 222.5  77.5
2
(12)(12  1)
U B  (12)(15) 
 155.5  102.5
2
U *  77.5
U A  (12)(15) 
Determine the z statistic:
U 
n A  nB 15 12

 90
2
2
n A  nB  (n A  nB  1)
U 
12
15 12  (15  12  1)
(180)(28)


 420  20.49
12
12
z
U  U
U
77.5  90

 .6101
20.49
4. The Probability Distribution (Classical Approach):
a. Critical value: z(0.05) = 1.65
b. z* is not in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = P(z* < -.6101) = .2709
b. The p-value is not smaller than a.
5. The Results:
a. Decision: Do not reject H0.
b. Conclusion: At the 0.05 level of significance, there is no
evidence to suggest the battery life for brand B is longer
than the life for brand A.
14.5: The Runs Test
• Used to test the randomness of data (or lack
of randomness).
• Run: a sequence of data with a common
property.
• Test statistic, V: the number of runs
observed.
Example: A coin is tossed 15 times and a head (H) or a tail
(T) is recorded on each toss. The sequence of tosses was
T H T T T H H T T T H H T T T
The number of runs is V = 7.
T H TTT HH TTT HH TTT
Note:
1. No randomness: only two runs (all heads, then all tails, or
the other way around). Or H and T alternate.
2. n1 = number of data with property 1.
n2 = number of data with property 2.
n = n1 + n2 = sample size.
Assumptions for inferences about randomness using the
Runs test:
Each observation may be classified into one of two
categories.
Note:
1. A large number of runs, or a small number of runs, (more
or less than what we would expect by chance), suggests the
data is not random.
2. Another aspect of randomness: the ordering of
observations above or below the mean or median of the
sample.
Example: Consider the following sample and use the runs test
to determine if the sequence is random with respect to being
above or below the mean value.
24
24
24
27
25
23
30
31
28
24
19
25
29
15
26
26
23
22
33
18
24
27
20
15
32
28
26
35
30
32
25
25
17
Test the null hypothesis that this sequence is random.
Use a = 0.05.
Solution:
1. The Set-up:
a. Population parameter of concern: Randomness of the
values above or below the mean.
26
31
38
b. The null and the alternative hypothesis:
H0: The numbers in the sample form a random sequence
with respect to the two properties above and below the
mean value.
Ha: The sequence is not random.
2. The Hypothesis Test Criteria:
a. Assumptions: Each observation may be classified as
above or below the mean.
b. Test statistic: V, the number of observed runs.
c. Level of significance: a = 0.05
3. The Sample Evidence:
x  25.75
Compare each number in the original sample to the value
of the mean to obtain the following sequence of a’s
(above) and b’s (below).
b a a b a a a a a a b a b b a b b b
b b a a b a b b a b a b b b a a b a
na = 18,
nb = 18,
V = 20
If n1 and n2 are both less than or equal to 20, and a twotailed test with a = 0.05 is conducted, use Table 14,
Appendix B.
4. The Probability Distribution (Classical Approach):
a. Critical value: Two-tailed test, a = 0.05, Use Table 14.
Critical values at the intersection of column n1 = 18 and
row n2 = 18: 12 and 26.
b. V is not in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = 2  P(V  20, for na = 18 and nb = 18)
Using Table 14: P > 0.05
Using a computer: 0.7352
b. The p-value is not smaller than the level of significance, a.
5. The Results:
a. Decision: Do not reject H0.
b. Conclusion: At the 0.05 level of significance, there is no
evidence to reject the null hypothesis that the sequence
is random with respect to above and below the mean.
Normal Approximation:
1. If n1 and n2 are larger than 20, or if a is different from
0.05, a normal approximation may be used.
2. V is approximately normally distributed with
2n1  n2
(2n1  n2 )  (2n1  n2  n1  n2 )
V 
1
V 
n1  n2
(n1  n2 ) 2 (n1  n2  1)
3. Test statistic:
V  V
z
V
Example: The letters in the following sequence represent the
direction each car turned after exiting at a certain ramp on the
New Jersey Turnpike (L - left, R - right).
L L R R R R R R R L L L R R R R L L R R
R R L L L L L R R L L L R L R L L R R L
L R R R R R R L L L L L R R R R R R L L
Test the null hypothesis that the sequence is random with
regards to direction. Use a = 0.01.
Solution:
1. The Set-up:
a. Population parameter of concern: Randomness with
respect to direction turned after exiting the turnpike.
b. The null and alternative hypothesis:
H0: The sequence of directions (L and R) is random.
Ha: The sequence is not random.
2. The hypothesis Test Criteria:
a. Assumptions: Each observation may be classified an L or R.
b. Test statistic: V, the number of runs.
c. Level of significance: a = 0.01
3. The Sample Evidence:
Calculate the value of the test statistic:
From the table above: nL = 27, nR = 33,
Determine the z statistic:
V = 19
2n1  n2
2  27  33
V 
1 
 1  30.7
n1  n2
27  33
(2n1  n2 )  (2n1  n2  n1  n2 )
(2  27  33)(2  27  33  27  33)
V 

2
(n1  n2 ) (n1  n2  1)
(27  33) 2 (27  33  1)

z
(1782)(1722)
 14.4473  3.801
2
(60)  59
V  V
V
19  30.7

 3.078
3.801
4. The Probability Distribution (Classical Approach):
a. Critical values: z(0.005) = 2.58 and z(0.005) = 2.58
b. z* is in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = 2  P(z < 3.078) = 0.0021
b. The p-value is smaller that the level of significance, a.
5. The Results:
a. Decision: Reject H0.
b. Conclusion: At the 0.01 level of significance, there is
evidence to suggest the turning direction for cars exiting
the turnpike is not random.
14.6: Rank Correlation
• Charles Spearman developed the rank
correlation coefficient.
• A nonparametric alternative to the linear
correlation coefficient (Pearson’s product
moment, r).
Spearman rank correlation coefficient:
rS  1 
6 ( d i ) 2
n(n 2  1)
di = the difference in the paired rankings.
n = the number of pairs.
Note:
1. rS will range from 1 to 1.
2. rS used in a the same way as the linear correlation
coefficient r.
Calculation of rS:
1. Rank the x-values from smallest to largest: 1, 2, ... , n.
2. Rank the y-values from smallest to largest: 1, 2, ... , n.
3. Use the ranks instead of the actual numerical values in the
formula for r, the linear correlation coefficient.
4. If there are no ties, rS is equivalent to r.
5. rS is an easier procedure that uses the differences between
the ranks: di.
6. In practice, rS is used even when there are ties.
7. For ties: assign each of the tied observations the mean
rank of those rank positions that they occupy.
Assumptions for inferences about rank correlation:
The n ordered pairs of data form a random sample and the
variables are ordinal or numerical.
Null Hypothesis:
There is no correlation between the two rankings.
Alternative Hypothesis:
Two-tailed: There is correlation between rankings.
May be one-tailed if positive or negative correlation is
suspected.
Critical Region:
On the side(s) corresponding to the specific alternative.
Table 15, Appendix B: positive critical values only, add a
plus or minus sign to the value found in the table, as
appropriate.
Example: A researcher believes a certain toxic chemical
accumulates in body tissues with age and may eventually
cause heart disease. Twelve subjects were selected at
random. Their age and the chemical concentration (in parts
per million) in tissue samples is given in the table below. Is
there any evidence to suggest the chemical concentration in
tissue samples increases with age? Use a = 0.01.
Chemical
Chemical
Age, x Concentration, y Age, x Concentration, y
82
170
70
48
83
40
62
34
64
64
34
3
53
5
27
7
47
15
75
50
50
5
28
10
Solution:
1. The Set-up:
a. Population parameter of concern: Rank correlation
coefficient between age and chemical concentration, rS.
b. The null and alternative hypothesis:
H0: Age and chemical concentration are not related.
Ha: Older people tend to have higher chemical
concentrations in their tissues.
2. The Hypothesis Test Criteria:
a. Assumptions: The 12 ordered pairs of data form a
random sample; both variables are continuous.
b. Test statistic: Rank correlation coefficient, rS.
c. Level of significance: a = 0.01.
3. The Sample Evidence:
The ranks and differences:
Age
82
83
64
53
47
50
70
62
34
27
75
28
Age Rank
11
12
8
6
4
5
9
7
3
1
10
2
Chemical
Chem. Con.
Concentration
Rank
Difference (d i )
170
12
-1
40
8
4
64
11
-3
5
2.5
3.5
15
6
-2
5
2.5
2.5
48
9
0
34
7
0
3
1
2
7
4
-3
50
10
0
10
5
-3
(d i )2
1.00
16.00
9.00
12.25
4.00
6.25
0.00
0.00
4.00
9.00
0.00
9.00
70.5
Use the formula for rS:
6 ( d i ) 2
(6)(70.5)
rS  1 
 1
2
n(n  1)
(12)(12 2  1)
423
 1
 1  0.2465  0.7535
1716
4. The Probability Distribution (Classical Approach):
a. Critical value: The critical region is one-tailed.
Table 15 lists critical values for two-tailed tests.
The critical value is located at the intersection of the
a = 0.02 column (2  0.01) and the n = 12 row: 0.703
b. rS is in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = P(rS  0.7535, for n = 12)
Using Table 15: P < 0.005
Using a computer: P  0.0025
b. The p-value is smaller than the level of significance, a.
5. The Results:
a. Decision:Reject H0.
b. Conclusion: At the a = 0.01 level of significance, there
is evidence to suggest that older people tend to have
higher levels of chemical concentration in their tissues.
Normal Approximation:
1. As n gets large, rS approaches a normal distribution.
2. When n exceeds the values in Table 15, the following test
statistic may be used:
z
rS  0
 rS n  1
1/ n 1