Procedures for Handling Ties In the sign test procedure described in this section, we exclude ties (represented by 0 instead of a sign of + or ?). A second approach is to treat half of the 0s as positive signs and half as negative signs. (If the number of 0s is odd, exclude one so that they can be divided equally.) With a third approach, in two-tailed tests make half of the 0s positive and half negative; in one-tailed tests make all 0s either positive or negative, whichever supports the null hypothesis. Repeat Example 4 using the second and third approaches to handling ties. Do the different approaches lead to very different results?

Solution 21BB Step 1 In the second approach, to treat half of the zero’s as positive signs and the other half as negative signs. With a third approach in two tailed tests. Let us denote negative sign (-) for temperature below 98.6 F and positive sign (+) for temperature above 98.6 F, we have 15 zero’s we are excluding one zero in order to make it even, now we get 7 zero’s as positive and 7 zero’s as negative. Now, Positive signs = 30 Negative signs = 75 Now, the Test Statistic is the less frequent sign i.e., positive sign = 30 Therefore, x = 30 is the required value of test statistic. Hence for sign test the required sample size used is 105 i.e., n = 30 + 75 = 105 which is greater than 25 (n 25). Step 3 Hence the Test Statistic x = 20 can be converted to the Test Statistic as shown below (x + 0.5) ( ) z = n 2 2 (30 + 0.5) 2() z = 105 2 z = (30.5) (52.5) 5.12 On substitution we get, z = -1.645 is the required test statistic. From A-2 table, the critical values are z = -1.645 and z -1.645 for one tailed test(left tailed test) at 5% level of significance. From the above figure, we see that z = -4.29 does not fall within the critical region. Hence we reject the Null hypothesis and conclude that there is sufficient evidence to 0 0 claim that the the median body temperature is less than 98.6 F and it is not equal to 98.6 F.