Question

5. A window that is manufactured for an automobile has five studs to attach it. A company that manufactures these windows performs “pull-out tests” to determine the force needed to pull a stud out of the window. Let X,, i = 1,2,3,4,5, equal the force required at position i and assume that the distribution of x, is N(4,0?). Observations 87 105 86 X, : 92 90 83 102 X,: 100 108 98 110 114 97 94 149 138 136 139 120 145 Xx: 143 X,: 147 144 160 149 152 131 134 Xs: 142 155 119 134 133 146 152 a. Test the null hypothesis H.:1 = H2 = Mz = HA = H, using seven independent observations at each position. Use a 1% significance level. (11 marks) b. Draw a figure illustrating your critical region. (2 marks) c. What is your conclusion? (2 marks)

Answer 1

> At an 2= 0.01 significance level, Ho is rejected if the compu- ted SSCT) /5-1) 4.02 = Fo.,(4,30) SS CE)/(35-5) The observed data along with certain sums are given in table - 1. For these data, SS (TO) = 556 174 I (4334)2 19500. 97. SS(T) $(6452 + 7272 +970? +10172 +9812) (4334) - 16672.11 SSCE) = 19500.97-16672.11 = 2828.86 35

Table 1 Pull-out test data 7 z xij observations 90 87 105 86 83 102 X,: 92 .645 59847 74 609 X2: 100 108 98 110 114 97 94 721 Xz: 143 149 138 136 139 120 145 970 134 936 144 160 149 152 131 134 1017 148 367 X4: 147 X5: 142 155 119 134 133 146 152 981 138415 4334 1 556174 Total Since the computed fis, 16672.11/4 = 44.20 2828.86/30 The null hypothesis is clearly rejected. Which is concluded below, square (MS) 4168.03 Treatment 4 44.20 94.30 2828.86 Total 19500.97 34 Mean DOF source Sum of squares (SS) 16672.11 27ΎΟΥ 30 To understand why Ho is rejected, the box & whisker diagram shown in figure below can be understood. X2 X3 XA X5 30 Too 12.0 140 160

The forces required to pull out studs in position 182 are similar and those in positions 3, 4 & 5 are quite similas but different from I & 2. An examination of the window would confirm that its the same case.

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