Non-parametric Hypothesis Testing

Transcription

1 Non-parametric Hypothesis Testing Procedures

2 Hypothesis Testing General Procedure for Hypothesis Tests 1. Identify the parameter of interest.. Formulate the null hypothesis, H Specify an appropriate alternative hypothesis, H 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. Sec 9-1 Hypothesis Testing

3 9-7 Testing for Goodness of Fit 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 O i be the observed frequency in the i-th class interval. Let E i be the expected frequency in the i-th class interval. The test statistic is X 0 k i 1 ( O i E E i i ) (9-16) Sec 9-7 Testing for Goodness of Fit 3

4 9-7 Testing for Goodness of Fit EXAMPLE 9-1 Printed Circuit Board Defects 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. Number of Observed Defects Frequency Sec 9-7 Testing for Goodness of Fit 4

5 Example Testing for Goodness of Fit 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, (3*0 + 15*1 + 9* + 4*3)/ From the Poisson distribution with parameter 0.75, we may compute p i, 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 p i as follows: p p p p P ( X P( X P( X P( X 0) 1) ) 3) e e 1 e ( 0.75 ) 0! ( 0.75 ) 1! ( 0.75 )! ( p + p + p ) Sec 9-7 Testing for Goodness of Fit 5

6 Example Testing for Goodness of Fit The expected frequencies are computed by multiplying the sample size n 60 times the probabilities p i. That is, E i np i. The expected frequencies follow: Expected Number of Defects Probability Frequency (or more) Sec 9-7 Testing for Goodness of Fit 6

7 9-7 Testing for Goodness of Fit Example 9-1 Since the expected frequency in the last cell is less than 3, we combine the last two cells: Number of Observed Expected Defects Frequency Frequency (or more) The chi-square test statistic in Equation 9-16 will have k p degree of freedom, because the mean of the Poisson distribution was estimated from the data. Sec 9-7 Testing for Goodness of Fit 7

8 Example Testing for Goodness of Fit 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.. Null hypothesis: H 0 : The form of the distribution of defects is Poisson. 3. Alternative hypothesis: H 1 : The form of the distribution of defects is not Poisson. 4. α Test statistic: The test statistic is χ 0 k i 1 ( o E ) i E i i Sec 9-7 Testing for Goodness of Fit 8

9 Example Testing for Goodness of Fit 6. Reject H 0 if: Reject H 0 if the P-value is less than Computations: χ 0 ( ) ( ) ( ) Conclusions: We find from Appendix Table III that χ 0.10,1.71 and χ 0.05, Because χ lies between these values, we conclude that the P-value is between 0.05 and 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 Sec 9-7 Testing for Goodness of Fit 9

10 Minitab

11 9-9 Nonparametric Procedures H : % µ % µ H : % µ < % µ X % µ i 0 H : % µ % µ H : % µ % µ Sec 9-9 Nonparametric Procedures 11

12 EXAMPLE 9-15 Propellant Shear Strength Sign Test Montgomery, Peck, and Vining (01) 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 0 randomly selected motors are shown in Table 9-5. Test the hypothesis that the median shear strength is 000 psi, using α Observation i Sec 9-9 Nonparametric Procedures Table 9-5 Propellant Shear Strength Data Shear Strength x i Differences x i Sign

13 EXAMPLE 9-15 Propellant Shear Strength Sign Test - Continued 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.. Null hypothesis: H : % µ p si 0 3. Alternative hypothesis: H : % µ p si 1 4. α Test statistic: The test statistic is the observed number of plus differences in Table 9-5, i.e., r Reject H 0 : If the P-value corresponding to r + 14 is less than or equal to α Computations : r + 14 is greater than n/ 0/ P-value : P P R 1 4 w h e n p r 1 4 r r ( ) ( ) 8. Conclusions: Since the P-value is greater than α 0.05 we cannot reject the null hypotheses that the median shear strength is 000 psi. 0 r Sec 9-9 Nonparametric Procedures 13

14 Rmin(R +,R -- ) Reject if R r*

15 Minitab

16 9-9 Nonparametric Procedures 9-9. The Wilcoxon Signed-Rank Test A test procedure that uses both direction (sign) and magnitude. Suppose that the hypotheses are H : µ µ H : µ µ Test procedure : Let X 1, X,...,X n be a random sample from continuous and symmetric distribution with mean (and Median) μ. Form the differences X µ. Rank the absolute differences X i µ 0 in ascending order, and give the ranks 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 H 0. i 0 For one-sided alternatives H 1: µ > µ 0 H : µ < µ 1 0 reject H 0 if W critical value reject H 0 if W + critical value Sec 9-9 Nonparametric Procedures 16

17 Sec 9-17

18 EXAMPLE 9-16 Propellant Shear Strength-Wilcoxon Signed-Rank Test 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 000 psi, using α 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.. Null hypothesis: H : µ p si 0 3. Alternative hypothesis: H : p si 1 4. α Test statistic: The test statistic is W min(w +, W ) 6. Reject H 0 if: W 5 (from Appendix Table IX). 7. Computations : The sum of the positive ranks is w + ( ) 150, and the sum of the absolute values of the negative ranks is w - ( ) 60. Sec 9-9 Nonparametric Procedures 18

19 EXAMPLE 9-16 Propellant Shear Strength-Wilcoxon Signed-Rank Test Observation i Differences x i Signed Rank W min(w +, W ) min(150, 60) Conclusions: Since W 60 is not 5 we fail to reject the null hypotheses that the mean or median shear strength is 000 psi. Sec 9-9 Nonparametric Procedures 19

20 Minitab Note: Minitab uses a modified procedure of the Wilcoxon Signed Rank test.

21 10-3 Wilcoxon Rank-Sum Test X 1 and X with means µ 1 and µ, but we are unwilling to assume that they are (approximately) normal. Want to test: Let X 11, X 1,...,X 1n1 and X 1, X,...,X n be two independent random samples of sizes n1 n 1

22 Procedure Arrange all n 1 + n observations in ascending order of 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. Let W 1 be the sum of the ranks in the smaller sample (1), and define W to be the sum of the ranks in the other sample. Then,

23 Procedure Get the critical value (w α ) from Appendex Table X Reject if either w 1 or w is less than w α For H, reject H 0 if w 1 w 1 :µ 1 < µ α for H, reject H0 if w w 1 :µ 1 > µ α 3

24 Table X 4

25 Table X 5

26 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 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-. Using α 0.05, we wish to test the hypothesis that the means of the two stress distributions are identical. 6

27 Example 1. Parameter of interest: The parameters of interest are the means of the two distributions of axial stress.. Null hypothesis: H 0 : µ 1 µ. 3. Alternative hypothesis: H : µ µ α Test statistic: We will use the Wilcoxon rank-sum test statistic in Equation

28 Example 6. Rejection criteria : If either w1 or w is less than or equal to w , we will reject H0: µ1 µ. 7. Computations: 8. Since both w 1 and w > 78 We cannot reject H 0. 8

29 Minitab Note: Another name for Wilcoxon Rank-Sum test is Mann-Whitney U-test