Test of Hypothesis

# Dr hatem elayat auc 29 example contd

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Unformatted text preview: 53, α = 0.05, Procedure H 0 : σ 2 = 0.01 1.Null Hypothesis 2. Alterna\$ve Hypotheses H1 : σ 2 > 0.01 2 χ 0.05,19 = 30.14 ( n − 1) S 2 (20 − 1)(0.0153) 3. Test Sta\$s\$c 2 χ= = = 29.07 2 σ0 0.01 4. Cri\$cal Region Dr. Hatem Elayat, AUC 27 Tests Concerning the Difference of Means of Two Normal Populations Case 1: The Variances of both populations are known Procedure 1.Null Hypothesis 2. Alterna\$ve H 0 : µ1 = µ 2 a) H1 : µ1 > µ 2, 3. Test Sta\$s\$c Z = 4. Cri\$cal Region Reject if Z > Zα b) H1 : µ1 < µ 2, c ) H1 : µ1 ≠ µ 2, X1 − X 2 2 σ 12 σ 2 + n1 n 2 Dr. DH Reject if Z r. <atem Zlayat, AUC − Eα Reject if Z > Zα / 2 28 Example A product developer is interested in reducing the drying \$me of a primer paint. Two formula\$ons of the paint are tested; formula\$on 1 is the standard chemistry, and formula\$on 2 has a new drying ingredient that should reduce the drying \$me. From experience, it is known that the standard devia\$on of drying \$me is 8 minutes, and this inherent variability should be un ­ aﬀected by the addi\$on of the new ingredient. Ten specimens are painted with formula\$on 1, and another 10 specimens are painted with formula\$on 2; the 20 specimens are painted in random order. The two sample average drying \$mes are X1 121 minutes X 2 = 112 minutes, respec\$vely. = What conclusions can the product developer draw about the Eﬀec\$veness of the new ingredient, using α = 0.05 ? Dr. Hatem Elayat, AUC 29 Example (Contd.) Data: X1 = 121, σ 1 = σ 2 = 8, X 2 = 112 n1 = n 2 = 10, Procedure 1.Null Hypothesis 2. Alterna\$ve 3. Test Sta\$s\$c 4. Cri\$cal Region α = 0.05, Z 0.05 = 1.645 H 0 : µ1 = µ 2 H1 : µ1 > µ 2, Z= X1 −...
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