Freshman 15?For the matched pairs listed in Exercise,identify the following components used in the Wilcoxon signed-ranks test: a.?Differences?d b.?The ranks corresponding to the nonzero values of |d? | c.?The signed ranks d.?The sum of the positive ranks and the sum of the absolute values of the negative ranks e.?The value of?T f.?The critical value of?T?(assuming a 0.05 significance level in a test of no difference between September weights and April weights) Exercise Wilcoxon Signed-Ranks Test for Freshman 15?The following table lists some of the weights from Data Set 4 in Appendix B. Those weights were measured from college students in September and later in April of their freshman year. Assume that we plan to use the Wilcoxon signed-ranks test to test the claim of no difference between September weights and April weights. What requirements must be satisfied for this test? Is there any requirement that the populations must have a normal distribution or any other specific distribution? In what sense is this Wilcoxon signed-ranks test a “distribution-free test”?
Solution 2BSC Step 1 Now,we use the second approach to perform sign test. Then 2nd approach method suggest indicates to treat half of the 0s as positive and half as negative signs. The values are: The total subjects are 106 where 68 had temperatures below 98.6°F , 23 had temperatures above 98.6°F and 15 had temperatures equal to 98.6°F. Check the claim that the median is less than 98.6°F. Step 2 The Requirement for applying Sign Test is: 1. The given data is a Simple Random Sample. The null hypothesis is a population parameter is equal to value and it is denoted H . 0 The alternative hypothesis is what we might believe to be true or hope to prove H . 1 Then we have to test the hypothesis follow the step by step in below. Here the null hypothesis H an0 the alternative hypothesis H . 1 Then the null hypothesis H is 0 H : The median = 98.6°F that is m edian = 98.6 0 Against the alternative hypothesis H 1 H :1the median is less than 98.6°F it means median < 98.6 2. Now,we are using the Sign Test for second approach method for Median of Single population.Here positive sign is above temperature 98.6°F and negative temperature sign is below 9 8.6°F . Then,the equal number of size is equal to 98.6°F are 15 which is odd, we exclude one sign to divide the 0s equally. We know that half of signs as positive and the other half as negative of the 14 zero signs. So, 7 are negative signs and the other 7 are positive signs. Then we have to add 7 both positive and negative signs. Hence, positive signs = 23+7 = 30 and Negative signs = 68 + 7 = 75 From the Test statistic x as smaller of the two signs that is 30 positive signs. 3. Here we observe that the sample data do not contradict the Alternative HypothesisH . 1 the below temperatures = 98.6°F . Here, n is The total number of positive and negative signs combined is 105. Therefore n = 105 x is the number of times the less frequent signs occur is 30. Therefore x = 30 Here we have to find z test statistic. Then n > 25. Therefore we convert the Test Statistic x to z by using the correction for continuity. The z test statistic is given by ( x + 0.5 )2 ( ) z = 2 ( 30 + 0.5 ) 2 ) z = 105 2 105 ( 30.5 ) 2 ) z = 10247 z = 5.124 z = -4.29 Therefore the z test statistic is z = -4.29 We have to find Critical value. Critical value = - 1.645 Then z test statistic is z = -4.29 Here the evel of significance =0.05 Hence the critical value corresponding to the level of significance =0.05 . So, it is left tail in the given Table A-2. Then the table is given in below. Cu mul T ativ a e b Are l a e fro A m - the 2 LE FT . . . . . . . . . z .00 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 . . . . . . . . . 1 9 9 9 9 9 9 9 9 9 .94 . 52 4 4 4 4 5 5 5 5 5 6 6 7 8 9 0 1 2 3 4 3 4 4 5 5 5 5 5 5 . . . . . . . . . 1 9 9 9 9 9 9 9 9 9 .95 . 5 5 5 5 5 6 6 6 6 54 7 6 7 8 9 9 0 1 2 3 4 3 2 1 9 8 6 5 3 Here we are comparing the Test Statistic value and Critical value. Now, The Test Statistic value is smaller than the Critical value. Hence -4.29 is smaller than -1.645 Now,we cannot accept the null hypothesis. Here we reject the null hypothesis. Then it falls in the rejection region bounded by the Critical Value of –1.645.