The reader should recall that the procedures discussed in Chapter 5 for testing the null hypothesis...

Question

The reader should recall that the procedures discussed in Chapter 5 for testing the null hypothesis that ?? = ??0are valid only if the population is approximately normal or if the sample is large. If n??~) = 0.5 andP (X??~)=P(X Chapter 9  Nonparametric Tests 9.1 Sign Test  The reader should recall that the procedures discussed in Chapter 5  for testing the null hypothesis that ? = ?0are valid only if the  population is approximately normal or if the sample is large. If n�̃�) ≤ 0.5 andP (X�̃�)=P(X �̃�0,  we reject ?0 in favor of ?1 only if the proportion of plus signs is  suffi ciently greater than 1/2, that is, when the value ? of our random  variable is small. Hence, if the computed P-value  ? =  ?(? ≥  ? ?ℎ?? ? = 1 /2) is less than ?, we reject ?0 in favor of ?1.  Finally, to test the hypothesis  ?0: �̃� = �̃�0, ?1: �̃� ≠ �̃�0,  we reject ?0 in favor of ?1when the proportion of plus signs is  significantly less than or greater than 1/2. This, of course, is  equivalent to x being sufficiently small or sufficiently large.  Therefore, if ?   ?/2 and the computed P-value  ? = 2?(? ≥  ? ?ℎ?? ? = 1 /2)  is less than or equal to α, we reject ?0 in favor of ?1. Example 9.1  The following data represent the number of hours that a rechargeable  hedge trimmer operates before a recharge is required:  1.5,2.2,0.9,1.3,2.0,1.6,1.8,1.5,2.0,1.2,1.7.  Use the sign test to test the hypothesis, at the 0.05 level of  significance, that this particular trimmer operates a median of  1.8hours before requiring a recharge.  Solution  1. ?0: �̃� = 1.8.  2. ?1: �̃� ≠ 1.8.  3. ? = 0 .05.   4. Test statistic: Binomial variable X with ? = 1 /2.   5. Computations: Replacing each value by the symbol “+” if it  exceeds 1.8 and by the symbol “−” if it is less than 1.8 and discarding  the one measurement that equals 1.8, we obtain the sequence  1.5 − , 2.2 + , 0.9 − , 1.3 − , 2.0 + , 1.6 − , 1.8 , 1.5 − , 2.0 + , 1.2 − , 1.7 −   for which n = 10, x = 3, and n/2 = 5. Therefore, the computed P-value  is   ? = 2?(? ≤  3 ?ℎ?? ? = 1 /2) = 2 ∑ ? (?; 10, 1 2 ) 3 ?=0 = 2 ∑ ( 10 ? ) ( 1 2 ) ? ( 1 2 ) 10−?3 ?=0 = 2 ∑ ( 10 ? ) ( 1 2 ) 103 ?=0 = ( 1 2 ) 10 (( 10 0 ) + ( 10 1 ) + ( 10 2 ) + ( 10 3 )) = 2 × 0.0000977(1 + 10 + 45 + 120) = 0.3438 > 0.05  6. Decision: Do not reject the null hypothesis and conclude that the  median operating time is not significantly different from 1.8 hours. We can also use the sign test to test the null hypothesis �̃�1 − �̃�2 = ?0for paired observations. Here we replace each diff erence, ??, with a  plus or minus sign depending on whether the adjusted difference,  ?? − ?0, is positive or negative. Throughout this section, we have  assumed that the populations are symmetric. However, even if  populations are skewed, we can carry out the same test procedure, but  the hypotheses refer to the population medians rather than the means. 9.2 Signed-Rank Test  The reader should note that the sign test utilizes only the plus and  minus signs of the differences between the observations and �̃�0 in the  one-sample case, or the plus and minus signs of the differences  between the pairs of observations in the paired-sample case; it does  not take into consideration the magnitudes of these differences. A test  utilizing both direction and magnitude, proposed in 1945 by Frank  Wilcoxon, is now commonly referred to as the Wilcoxon signed- rank test. The analyst can extract more information from the data in a  nonparametric fashion if it is reasonable to invoke an additional  restriction on the distribution from which the data were taken. The  Wilcoxon signed-rank test applies in the case of a symmetric  continuous distribution. Under this condition, we can test the null  hypothesis �̃� = �̃�0. We first subtract �̃�0 from each sample value,  discarding all differences equal to zero. The remaining differences are  then ranked without regard to sign. A rank of 1 is assigned to the  smallest absolute difference (i.e., without sign), a rank of 2 to the next  smallest, and so on. When the absolute value of two or more  differences is the same, assign to each the average of the ranks that  would have been assigned if the differences were distinguishable. For  example, if the fifth and sixth smallest differences are equal in  absolute value, each is assigned a rank of 5.5. If the hypothesis �̃� = �̃�0 is true, the total of the ranks corresponding to the positive  differences should nearly equal the total of the ranks corresponding to  the negative differences. Let us represent these totals by ?+and?−,  respectively. We designate the smaller of ?+and?−by w. In selecting  repeated samples, we would expect ?+and?−, and therefore w, to  vary. Thus, we may think of ?+,?− and ? as values of the  corresponding random variables ?+,?−andW. The null hypothesis  �̃� = �̃�0 can be rejected in favor of the alternative �̃�  �̃�0 can be  accepted only if ?+is large and ?−is small. For a two-sided  alternative, we may reject ?0 in favor of ?1if either ?+or?−, and  hence w, is sufficiently small. Therefore, no matter what the  alternative hypothesismay be, we reject the null hypothesis when the  value of the appropriate statistic ?+,?−, or W is sufficiently small. Two Samples with Paired Observations  To test the null hypothesis that we are sampling two continuous  symmetric populations with �̃�1 − �̃�2 for the paired-sample case, we  rank the differences of the paired observations without regard to sign  and proceed as in the single-sample case. The various test procedures  for both the single- and paired-sample cases are summarized in Table  9.2. Table 9.2: Signed-Rank Test  ?? ?? Compute  �̃� = �̃�0  { �̃�  �̃�0 �̃� ≠ �̃�0   ?+ ?− ?   �̃�1 = �̃�2  { �̃�1  �̃�2 �̃�1 ≠ �̃�2   ?+ ?− ?  It is not difficult to show that whenever n

The reader should recall that the procedures discussed in Chapter 5 for testing the null hypothesis that ?? = ??0are valid only if the population is approximately normal or if the sample is large. If...

Get Answer To This Question

Related Questions & Answers

Submit New Assignment