Test of Hypothesis

# Test stasc t 2 2 sx n1 2 n1 1 sy n

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Unformatted text preview: X 2 − 0 σ σ + n1 n 2 2 1 2 2 = 121 − 112 8 2 10 Dr. Hatem Elayat, AUC + 8 2 = 2.52 10 30 Case 2: The Variances of Both Popula\$ons are Unknown but Equal Procedure 2 H 0 : µ 1 = µ 2 where σ 12 = σ 2 = σ 2 1.Null Hypothesis 2. Alterna\$ve Hypotheses a) H1 : µ1 > µ 2, b) H1 : µ1 < µ 2, c ) H1 : µ1 ≠ µ 2, 3. Test Sta\$s\$c X1 − X 2 t= 4. Cri\$cal Region S 1 + 1 p n1 n 2 where α /2 α tα ,n −1 Cri\$cal Region − tα ,n −1 Reject if t > tα ,n −1 Sp = Dr. H Reject if t <atem tEα AUC − layat, 2 2 ( n − 1) SX + ( m − 1) SY n+m−2 α /2 − tα / 2,n −1 tα / 2,n −1 Cri\$cal Region Reject if t > tα / 2,n −1 31 Example Two catalysts are being analyzed to determine how they aﬀect the mean yield of a chemical process. Speciﬁcally, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in Table 10 ­1. Is there any diﬀerence between the mean yields? Use = , and assume equal variances. α 0.05 Dr. Hatem Elayat, AUC 32 Data: Example (Contd.) X1 = 92.255, X 2 = 89.19, n1 = n 2 = 8, S1 = 2.39, S2 = 2.98,α = 0.05, t 0.025,14 = 2.145 Procedure 2 H 0 : µ 1 = µ 2 where σ 12 = σ 2 = σ 2 1.Null Hypothesis 2. Alterna\$ve Hypotheses H1 : µ1 ≠ µ 2, 3. Test Sta\$s\$c 4. Cri\$cal Region X1 − X 2 92.255 − 89.19 t= = = −0.35 1 1 11 Sp + 2.7 + n1 n 2 88 Dr. Hatem Elayat, AUC 33 Case 3: The Variances of Both Popula\$ons are Unknown but Unequal Procedure 1.Null Hypothesis 2. Alterna\$ve H 0 : µ1 = µ 2 a) H1 : µ1 > µ 2, b) H1 : µ1 < µ 2, c ) H1 : µ1 ≠ µ 2, ( SX / n1 + SY / n 2 ) X1 − X 2 ν = 3....
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## This document was uploaded on 11/09/2013.

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