Test of Hypothesis

# Null hypothesis 2 alternave hypotheses can have three

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Unformatted text preview: but equal –  Case 3: The variances of both popula\$ons are unknown but unequal •  Test concerning the ra\$o of variances of two normal popula\$ons •  Tests concerning the parameter p of the binomial distribu\$on Dr. Hatem Elayat, AUC 16 Tests Concerning the Mean of a Normal Popula\$on Case 1: Known Variance Procedure H0 : µ = µ0 1.Null Hypothesis 2. Alterna\$ve Hypotheses (can have three possible alterna\$ve hypotheses a) H1 : µ > µ 0, b) H1 : µ < µ 0, c ) H1 : µ ≠ µ 0, 3. Test Sta\$s\$c Z = X − µ 0 4. Cri\$cal Region σ/ n Dr. H Reject if Z > Zα Reject if Z <atem Zlayat, AUC − Eα Reject if Z > Zα / 2 17 Example Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteris\$c. Speciﬁca\$ons require that the mean burning rate must be 50 cen\$meters per second. We know that the standard devia\$on of burning rate is σ = 2 cen\$meters per second. The experimenter decides to specify a type I error probability or signiﬁcance level of α = 0.05 and selects a random sample of n = 25 and obtains a sample average burning rate of x = 51.3 cen\$meters per second. What conclusions should be drawn? Dr. Hatem Elayat, AUC 18 Example (Contd.) Data: X = 51.3, n = 25, σ = 2, α = 0.05, Z 0.025 = 1.96 Procedure H 0 : µ = 50 1.Null Hypothesis 2. Alterna\$ve Hypotheses H1 : µ ≠ 50 3. Test Sta\$s\$c X − µ 0 51.3 − 50 Z= = = 3.25 σ/ n 2 / 25 4. Cri\$cal Region Reject H 0 if Z > Zα / 2 → 3.25 > 1.96 5. Conclusion: Since .25 > 1.96 , we reject H 0 : µ = 50 at the 0.05 3...
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