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Transcript of Non-parametric Hypothesis Testing Procedures haalshraideh/Courses/IE347/Non... Non-parametric...

Microsoft PowerPoint - Non Parametric tests1. Identify the parameter of interest.
2. Formulate the null hypothesis, H0 .
3. Specify an appropriate alternative hypothesis, H1.
4. Choose a significance level, α.
5. Determine an appropriate test statistic.
6. State the rejection criteria for the statistic.
7. Compute necessary sample quantities for calculating the test statistic.
8. Draw appropriate conclusions.
2Sec 9-1 Hypothesis Testing
• The test is based on the Chi-square distribution.
• Assume there is a sample of size n from a population whose
probability distribution is unknown.
• Let Oi be the observed frequency in the i-th class interval.
• Let Ei be the expected frequency in the i-th class interval.• Let Ei be the expected frequency in the i-th class interval.
The test statistic is
∑ =
− =
Poisson Distribution
The number of defects in printed circuit boards is hypothesized to follow a
Poisson distribution. A random sample of n = 60 printed boards has been
collected, and the following number of defects observed.
4Sec 9-7 Testing for Goodness of Fit
Number of
9-7 Testing for Goodness of Fit Example 9-12
The mean of the assumed Poisson distribution in this example is unknown and
must be estimated from the sample data. The estimate of the mean number of
defects per board is the sample average, that is, (32*0 + 15*1 + 9*2 + 4*3)/60 =
0.75. From the Poisson distribution with parameter 0.75, we may compute pi, the
theoretical, hypothesized probability associated with the i-th class interval. Since
each class interval corresponds to a particular number of defects, we may find the
pi as follows:
( )
( )
( )
9-7 Testing for Goodness of Fit Example 9-12
The expected frequencies are computed by multiplying the sample size n = 60
times the probabilities pi. That is, Ei = npi. The expected frequencies follow:
Number of Defects Probability
1 0.354 21.24
2 0.133 7.98
9-7 Testing for Goodness of Fit Example 9-12
Since the expected frequency in the last cell is less than 3, we combine the
last two cells:
7Sec 9-7 Testing for Goodness of Fit
The chi-square test statistic in Equation 9-16 will have k− p − 1 = 3 −1 − 1 = 1
degree of freedom, because the mean of the Poisson distribution was
estimated from the data.
9-7 Testing for Goodness of Fit Example 9-12
The eight-step hypothesis-testing procedure may now be applied, using
α = 0.05, as follows:
1. Parameter of interest: The variable of interest is the form of the distribution
of defects in printed circuit boards.
2. Null hypothesis: H0: The form of the distribution of defects is Poisson.
8Sec 9-7 Testing for Goodness of Fit
3. Alternative hypothesis: H1: The form of the distribution of defects is not
Poisson.
( ) ∑ =
9-7 Testing for Goodness of Fit Example 9-12
6. Reject H0 if: Reject H0 if the P-value is less than 0.05.
7. Computations:
8. Conclusions: We find from Appendix Table III that and
. Because lies between these values, we conclude
that the P-value is between 0.05 and 0.10. Therefore, since the P-value
exceeds 0.05 we are unable to reject the null hypothesis that the distribution
of defects in printed circuit boards is Poisson. The exact P-value computed
from Minitab is 0.0864.
71.2 2 =χ 0.10,1
Minitab
0i X µ− %
0 0 1 0: :H Hµ µ µ µ= ≠% % % %
Montgomery, Peck, and Vining (2012) reported on a study in which a rocket motor is
formed by binding an igniter propellant and a sustainer propellant together inside a metal
housing. The shear strength of the bond between the two propellant types is an important
characteristic. The results of testing 20 randomly selected motors are shown in Table 9-5. Test the hypothesis that the median shear strength is 2000 psi, using α = 0.05.
EXAMPLE 9-15 Propellant Shear Strength Sign Test
Table 9-5 Propellant Shear Strength Data
Observation i
Shear Strength
The eight-step hypothesis-testing procedure is:
1. Parameter of Interest: The variable of interest is the median of the distribution
of propellant shear strength.
2. Null hypothesis:
3. Alternative hypothesis:
4. α= 0.05
5. Test statistic: The test statistic is the observed number of plus differences in
EXAMPLE 9-15 Propellant Shear Strength Sign Test - Continued
0 : 2 0 0 0 p s iH µ =%
1 : 2 0 0 0 p s iH µ ≠%
13
5. Test statistic: The test statistic is the observed number of plus differences in
Table 9-5, i.e., r+ = 14.
6. Reject H0 : If the P-value corresponding to r+ = 14 is less than or equal to
α= 0.05
7. Computations : r+ = 14 is greater than n/2 = 20/2 = 10.
P-value :
8. Conclusions: Since the P-value is greater than α= 0.05 we cannot reject
the null hypotheses that the median shear strength is 2000 psi.
( ) ( ) 2 0
2 0
1 4
2
0 . 1 1 5 3
r r
Minitab
• A test procedure that uses both direction (sign) and magnitude.
• Suppose that the hypotheses are
Test procedure : Let X1, X2,... ,Xn be a random sample from continuous and symmetric distribution with mean (and Median) μ. Form the differences .
• Rank the absolute differences in ascending order, and give the ranks
to the signs of their corresponding differences.
0i X µ−
0i X µ−
to the signs of their corresponding differences.
• Let W+ be the sum of the positive ranks and W– be the absolute value of the sum of the
negative ranks, and let W = min(W+, W−).
Critical values of W, can be found in Appendix Table IX.
• If the computed value is less than the critical value, we will reject H0 .
• For one-sided alternatives reject H0 if W– ≤ critical value
reject H0 if W+ ≤ critical value 1 0:H µ µ>
1 0:H µ µ<
Sec 9-9 Nonparametric Procedures
Sec 9- 17
Let’s illustrate the Wilcoxon signed rank test by applying it to the propellant shear strength
data from Table 9-5. Assume that the underlying distribution is a continuous symmetric
distribution. Test the hypothesis that the mean shear strength is 2000 psi, using α = 0.05.
The eight-step hypothesis-testing procedure is:
1. Parameter of Interest: The variable of interest is the mean or median of the
distribution of propellant shear strength.
2. Null hypothesis:
0 : 2 0 0 0 p siH µ =
18
5. Test statistic: The test statistic is W = min(W+, W−)
6. Reject H0 if: W ≤ 52 (from Appendix Table IX).
7. Computations : The sum of the positive ranks is
w+ = (1 + 2 + 3 + 4 + 5 + 6 + 11 + 13 + 15 + 16 + 17 + 18 + 19 + 20) = 150, and the
sum of the absolute values of the negative ranks is
w- = (7 + 8 + 9 + 10 + 12 + 14) = 60.
1 : 2 0 0 0 p s iH µ ≠
Sec 9-9 Nonparametric Procedures
Observation i
W = min(W+, W−) = min(150 , 60) = 60
7. Conclusions: Since W = 60 is not ≤ 52 we fail to reject the null hypotheses that the mean
or median shear strength is 2000 psi.
10 256.70 11
6 –291.70 –12
3 316.00 13
2 –321.85 –14
14 336.75 15
9 357.90 16
12 399.55 17
17 414.40 18
8 575.10 19
19 654.20 20
Minitab
Note: Minitab uses a modified procedure of the Wilcoxon Signed Rank test.
10-3 Wilcoxon Rank-Sum Test
• X1 and X2 with means µ1 and µ2, but we are unwilling to
assume that they are (approximately) normal.
• Want to test:• Want to test:
• Let X11, X12,...,X1n1 and X21, X22,...,X2n2 be two
independent random samples of sizes
21
magnitude and assign ranks to them.
– If two or more observations are tied (identical), use
the mean of the ranks that would have been assigned
if the observations differed.if the observations differed.
• Let W1 be the sum of the ranks in the smaller sample (1),
and define W2 to be the sum of the ranks in the other
sample. Then,
Table X
• Reject if either w or w is less than w• Reject if either w 1
or w 2
– For , reject H0 if w1≤ w α
– for , reject H0 if w2 ≤ w α
23
25
Example
• The mean axial stress in tensile members used in an aircraft structure is being
studied. Two alloys are being investigated. Alloy 1 is a traditional material, and
alloy 2 is a new aluminum-lithium alloy that is much lighter than the standard
material. Ten specimens of each alloy type are tested, and the axial stress is
measured. The sample data are assembled in Table 10-2. Using α= 0.05, we
wish to test the hypothesis that the means of the two stress distributions are
identical.identical.
26
Example
1. Parameter of interest: The parameters of interest are the means of the two
distributions of axial stress.
: µ 1
= µ 2
: µ 1 ≠ µ
: µ 1 ≠ µ
4. α=0.05
5. Test statistic: We will use the Wilcoxon rank-sum test statistic in Equation 10-
21.
27
Example 6. Rejection criteria :
If either w1 or w2 is less than or equal to
w 0.05
7. Computations:
and w 2
.