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Unformatted text preview: Chapter 15: fl Sections 15.2, 15.3, and 15.5 Nonparametric Statistical Inference:
Comparing Two Populations i Outline . Sections 15.2, 15.3, and 15.5 will cover:  The Wilcoxon Rank Sum Test: Independent
Random Samples  The Sign Test: Paired Experiment . The Wilcoxon Signed—Rank Test: Paired
Experiment I Table used:
. Tables 1, 7, and 8 i Nonparametric Statistical Inference . Nonparametric statistical inference methods... . ...do not require the normality assumption to be
satisfied. . ...specify hypotheses in terms of population
distributions rather than population parameters. . ...use the ranks or signs of observations rather
than the actual observations themselves.  ...are usually more powerful than their parametric
counterparts when normality is not satisfied. 3 The Wilcoxon Rank Sum Test: $ Independent Random Samples  The Wilcoxon Rank Sum Test is used when we wish
to compare two populations and the following
conditions have been met: . We have two independent SRSs, one from each
population. . The normality assumption has been violated or
cannot be validated, and the sample sizes are
not large enough to apply the Central Limit
Theorem. i Step 1: State The Hypotheses H0: The distributions for Population 1 and
Population 2 are identical. versus one of the following alternative hypotheses: : The distributions for Population 1 and
Population 2 are different. (twotailed test) : The distribution for Population 1 lies to the
left of that for Population 2. (lefttailed test) : The distribution for Population 1 lies to the
right of that for Population 2. (righttailed tests) i Step 2: Compute Test Statistic T  In order to compute the test statistic, we must a. Let n1 denote the smaller of the two sample
sizes, and denote the population of the origin of this sample as Population 1. b. Rank all nl + n2 observations from smallest (1)
to largest (111+ n2). c. Compute T1 = the sum of the ranks of the first sample. This is the test statistic Tfor a left tailed
test, so if that is the test of interest, stop.
Otherwise, proceed to d. i Step 2: Compute Test Statistic T d. Compute Tf = n,(nl + n2 +1)— T,. This is the sum
of the ranks of the first sample if the assigned ranks
had been reversed from large to small. This is the
test statistic T for a right tailed test, so if that is the
test of interest, stop. Otherwise, proceed to e. e. The test statistic T for a twotailed test is the
minimum of TI and T1 NOTE: When ranking the observations, tied
observations are assigned the average of the ranks that the observations of differing values would have
received. 7 i Step 3: Critical Value  The critical value for the test statistic T computed
in Step 2 can be found in Table 7.  These critical values are given for onetailed tests
with a = 0.05, 0.025, 0.01, and 0.005.  NOTE: For a twotailed test, the values of a are
doubled.  The table gives the critical value c such that
P(T S c) S a . i Step 4: Decision . If our observed test statistic T is less than or equal
to the critical value c, reject H0 and accept HA. Exercise 1 . The weights of turtles caught in two different lakes
were measured to compare the effects of the two
lake environments on turtle growth. Lake Weight (in 02) At 05 = 005 do the data provide
1 14.1 15.2 13.9 14.5 14.7 sufficientevidence to 13'8 14'0 16'1 12'7 15'3 indicateadifference 2 12.2 13.0 141 136 124 in the distributionsof
11.9 12.5 13.8 turtle weights between the two
lakes? The Wilcoxon Rank Sum Test: i Large Sample Approximation . If our samples sizes are large enough (say at least
10), the Wilcoxon rank sum test statistic T1 will
follow an approximately normal distribution with 111(n1 +n2 +1) and 02 : I’lll’l2(l’ll +n2 +1) . 2 "' 12 T _ [UT approx ~ N(0, 1). #1' = . Thus, 2 =
0T
. We can use this as the basis for a test statistic, and carry out a hypothesis test in the usual way, using
either the pvalue or critical value method. 11 i Exercise 2  Suppose you wish to detect a shift in distribution 1
to the right of distribution 2 based on sample sizes
HI = 12 and n2 = 14. If T1 = 193, what do you
conclude? Use a = 0.05. i The Sign Test: Paired Experiment  The Sign Test is a nonparametric counterpart of
the paireddifference test in Section 10.5. It should
be used when any of the paireddifference
assumptions have been violated or cannot be
validated. These assumptions are:  Sample differences are at least approximately
normally distributed. . Equal variances across pairs. i The Sign Test: Paired Experiment . Recall that our data consists of ordered pairs (A, B),
and that we take the difference between these pairs to obtain A , B. When the two populations distributions are identical,
the probability that A exceeds B would bep = 0.5. LetX represent the number of times A exceeds B. Then under the assumption that the two population
distributions are identical, it follows that X~ B(n, 0.5), where n is the number of observed
ordered pairs. 14 i Step 1: State The Hypotheses H0: The population distributions are identical and
P(A exceeds B) =p = 0.5.
versus one of the following alternative hypotheses:
HA: The population distributions different and p i 0.5. HA: The population ofA measurements is shifted to
the right of the population of B measurements
andp > 0.5. HA: The population ofA measurements is shifted to
the left of the population of B measurements
and p < 0.5. i Step 2: Compute Test Statistic x  x is simply the number of times that A exceeds B.  In other words, x is the number of times the
difference A , B is positive. . NOTE: Delete tied pairs — that is, pairs with A = B
— from the data set. Reduce the sample size n
accordingly. i Step 3: pvalue . While it is possible to use a critical value method
here, it’s much easier to compute the pvalue. The
way in which this is done depends on the direction of HA, and on the assumption thatX~ B(n, 0.5):
HA : p>0.5:> p—value=P(X2x) HA :p<0.5:>p—value= P(XSx) 2P(X2x) if x>n/2
2P(XSx) if x<n/2 HAzp¢0.5:>p—value={ 17 i Step 4: Decision . We make a decision based on the pvalue in the
usual way. i Exercise 3 . An investment analyst wants to test whether
differences exist between the returns on two
mutual funds. Paired data of annualized rates of
return (in 0/o) for the two mutual funds during 16
randomly chosen months are given on the next
slide. Is there evidence at a = 0.10 to suggest whether
or not the distribution of returns are equal? Exercise 3 The Sign Test: i Large Sample Approximation  If our samples size is large enough (say at least
25), we can approximate our binomial random variable/test statistichith a normal distribution.
 Assuming H0 is true, this means approx X N N(O.5n,0.25n).
x — 0.5n WW ~ N 0,1).
0.5% ( . We can use this as the basis for a test statistic, and
carry out a hypothesis test in the usual way. 21 I Thus, 2 = i Exercise 4  A clinic is interested in comparing the popularity of
two doctors on staff, Dr. Pepper and Dr. Feelgood.
They select 30 patients at random to see both
doctors about the same problem and then ask
them to rate each doctor on a scale from 1 to 10. . Three of the patients rated the doctors the same.
Of the remaining patients, only 8 preferred Dr.
Pepper. At a = 0.05, is there evidence to suggest
that Dr. Pepper is less popular than Dr. Feelgood? 22 The Wilcoxon SignedRank Test: i Paired Experiment . The Wilcoxon SignedRank Test is another non
parametric counterpart of the paireddifference test
in Section 10.5. Like the Sign Test, it too should be
used when any of the usual paireddifference
assumptions have been violated or cannot be
validated. The Wilcoxon SignedRank Test:
Paired Experiment  Recall that our data consists of ordered pairs (X, Y),
and that we take the difference between these pairs
to obtain X— Y.  When the two populations distributions are identical,
we would expect half, or n/2, of these differences to
be negative and half to be positive.  We would also expect that positive and negative
difference of equal magnitude to occur with equal
probability. i Step 1: State The Hypotheses H0: The two population distributions are identical.
versus one of the following alternative hypotheses: H A: The two population distributions differ. HA: The Population 1 distribution is shifted to
the right (>) of the Population 2 distribution. HA: The Population 1 distribution is shifted to
the left (<) of the Population 2 distribution. Step 2: Compute Test Statistic T . In order to compute the test statistic, we must
a. Compute the differences X— onr each of the n
pairs. Tied pairs are eliminated and n is reduced. b. Rank the absolute values of the differences from
smallest (1) to largest. Average tied ranks. c. For the right tailed test, compute T’ = the sum of the (absolute value of the) ranks of the negative
differences. For the left tailed test, compute T = the sum of the ranks of the positive
differences. For the twotailed test, take the minimum of TT and T . i Step 3: Critical Value . The critical value for the test statistic T computed
in Step 2 can be found in Table 8.  These critical values are given for onetailed tests
with a = 0.05, 0.025, 0.01, and 0.005, or,
equivalently, for twotailed tests with a = 0.10,
0.05, 0.02, and 0.01.  The table gives the critical value T0 such that
P(T 3 To) 3 a . i Step 4: Decision . If our observed test statistic T is less than or equal
to the critical value To, reject H0 and accept HA. i Exercise 5 . Eight people were asked to perform a simple
puzzleassembly task under normal conditions and
under stressful conditions. During the stressful
time, mild shocks were given every 30 seconds.
The highest blood pressure readings were recorded
under both conditions, as given on the next slide.
Do the data present sufficient evidence to indicate
higher blood pressure readings under stressful
conditions? Use a = 0.05. i Exercise 5 Stressful 130
118
125
120
121
125
130
120 10 The Wilcoxon SignedRank Test:
Large Sample Approximation  If our samples sizes are large enough (say at least
25), the Wilcoxon signedrank test statistic T‘ will
follow an approximately normal distribution with x = ”W” and 0,2 w. 4 "* 24 T+ _ #T approx  ThUS,z= ~ N(0,1) UT
 We can use this as the basis for a test statistic, and carry out a hypothesis test in the usual way, using
either the pvalue or critical value method. 31 i Exercise 6 . Suppose you wish to detect a shift in distribution 1
to the right of distribution 2 based on a paired difference experiment consisting of n = 30 pairs.
 You are given T ‘= 249. Use a 20.05 . Appendix
Normal Probability Plots 11 1 Normal Probability Plots . We want to check our assumption that observed
values are coming from a normal distribution.  We could check this assumption in a rough way by
graphing a histogram or a stemandleaf plot of our
sample data and checking to see if its shape is
more or less “bell shaped”.  However, a more accurate/standard tool is a
normal probability plot. 1 Normal Probability Plots . To start, observed values are first ranked from
smallest to largest.  The ranked values are then plotted against their
observed cumulative frequency. . If the observations are coming from a normal
distribution, the plotted points will fall
approximately along a straight line.  The determination of how well the points fall along
a straight line is usually subjective. Normal Probability Plots 1 in MINITAB . Enter your data in one of the spreadsheet columns.
Go to the top of the screen and click on Graph,
then Probability Plot. Then, select Single. . Under Variables, simply indicate which column you
are graphing data from. The default name is C1.  Click Distribution and under the Distribution tab,
ensure “Normal” is selected. Under the Data
Display tab, you can uncheck the option to “Show
Confidence Interval”.  Click OK. Click OK again. 12 Normal Probability Plots 1 in MINITAB . In each of the three examples that follow, 500
observations were randomly generated (simulated)
from some distribution. . We will look at the corresponding histogram and
normal probability plot in each case. 1 Example 1: N(70,144) 50 p
O o)
O M
O Frequency _x
O j K—r—l
l I
100 110 38 l l Percentile lllllll l 13 1 Example 2: eXp(5) 90
80
70
60
50
40
30
20
10 O Frequency 2
'5
C
<1)
0
L
a.)
a. 1 Example 3: U(0,1) 30 N
O Frequency 14 3 Example 3: U(0,1) 2
‘E
(1)
U
L
d)
o. 15 ...
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 Winter '10
 LaniHaque
 Linear Algebra, Algebra

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