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Assignment 13.docx

# Assignment 13.docx - 5.15 a Using the Minitab output...

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5.15 a) Using the Minitab output, P-values = 0.001. Since P = 0.001 < 0.05, therefore, the null hypothesis H 0 : μ 1 μ 2 = 0 can be rejected at the 0.05 level of significance. b) It is a two-sided test because the alternative hypothesis is H 1 : μ 1 μ 2 0. c) Parameter of interest : We are interested in determining if μ 1 μ 2 = 2 . using μ 1 μ 2 . Null Hypothesis : H 0 : μ 1 μ 2 = 2 Alternative Hypothesis : H 1 : μ 1 μ 2 2 Test Statistic : t 0 = ´ X 1 ´ X 2 2 s p 1 n 1 + 1 n 2 standard deviation S p = ( n 1 1 ) S 1 2 + ( n 2 1 ) S 2 2 n 1 + n 2 2 Assume α = 0.05 Reject H 0 if : t 0 > t α 2 ,n 1 + n 2 2 t 0 <− t α 2 , n 1 + n 2 2 Computations : Using Minitab output, the estimated values for standard deviation is S p = 2.1277. Since ´ X 1 = 50.19, ´ X 2 = 52.52, n 1 = n 2 = 20.

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: t 0 = 50.19 52.52 2 2.1277 1 20 + 1 20 = -6.435 Conclusion : Using excel function = T.INV.2T (a, n 1 + n 2 2 ¿ .n 1 + n 2 2 = 38. . t 0.025,38 = 2.024. Since t 0 =− 6.435 <− 2.024 =− t 0.025,38 , we reject the null hypothesis H 0 : μ 1 μ 2 = 2 in favor of the Alternative hypothesis H 1 : μ 1 μ 2 2 at the 0.05 level of significance. Also using the P-value approach to perform the test, the decision rule which is H 0 if P < 0.05 . the null hypothesis will be rejected at the 0.05 level of significance. d) We obtained that the null hypothesis H 0 would be rejected at the 0.05 level of significance if the hypothesis had been H 0 : μ 1 μ 2 = 2 versus H 1 : μ 1 μ 2 2. if the hypotheses had been H 0 : μ 1 μ 2 = 2 versus H 1 : μ 1 μ 2 < 2, the T-value for the one-sided test would remain unchanged but the P-value would be that half of the two-sided P-value. Therefore, the null hypothesis H 0 would alsobe rejectedat the 0.05 levelof significance when testing H 0 : μ 1 μ 2 = 2 versus H 1 : μ 1 μ 2 < 2. e) UCL = ´ X 1 ´ X 2 + t a, n 1 + n 2 2 s p 1 n 1 + 1 n 2 = 50.19 -52.52 + 1.68 (2.1277) 1 20 + 1 20
= -1.1956 f) P = 2P ( T 38 > |-6.435|) = T.DIST.2T (6.435, 38) = 0.0000 5.17a ) Parameter of Interest: μ single μ double = 0 Null Hypothesis : H 0 : : μ single μ double = 0 Alternative Hypothesis : H 1 :: μ single μ double 0 Test Statistic : t 0 = ´ ´ X 1 ´ ´ X 2 2 sp 1 n 1 + 1 n 2 Assuming α = 0.05 Reject Computation: 66.385 45.278 8,26182 1 15 + 1 15 = 6.997 P = T DIST .2

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• Fall '17
• Dr Bheem

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