A random sample of 20 bofles results in a sample

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Unformatted text preview: = 0.83725, n = 15, S = 0.02456, α = 0.05, t 0.05,14 = 1.761 Procedure H 0 : µ = 0.82 1.Null Hypothesis 2. Alterna$ve Hypotheses H1 : µ > 0.82 3. Test Sta$s$c X − µ 0 0.83725 − 0.82 t= = = 2.72 S/ n 0.02456 / 15 4. Cri$cal Region Reject H 0 if t > tα ,n −1 → 2.72 > 1.761 5. Conclusion: Since .72 > 1.761 we reject H 0 : µ at the 0.05 = 0.82 2 , level of significance that the mean coefficient of res$tu$on exceeds 0.82. Dr. Hatem Elayat, AUC 23 Example (Contd.) Minitab Output Dr. Hatem Elayat, AUC 24 Test Concerning the Variance of a Normal Popula$on Procedure 2 H0 : σ 2 = σ 0 1.Null Hypothesis 2. Alterna$ve Hypotheses (can have three possible alterna$ve hypotheses a) H1 : σ > σ 2 2 0 3. Test Sta$s$c 4. Cri$cal Region 2 2 Reject if χ > χα ,n −1 b) H1 : σ < σ 2 ( n − 1) S 2 χ2 = 2 σ0 2 0 c ) H1 : σ ≠ σ 2 2 Reject if χ < χ1−α ,n −1 Dr. Hatem Elayat, AUC 2 2 0 2 χ 2 > χα / 2,n −1 Reject if 2 2 χ < χ25 α / 2,n −1 1− Example An automa$c filling machine is used to fill bofles with liquid detergent. A random sample of 20 bofles results in a sample variance of fill volume of S 2 = 0.0153 (fluid ounces)2. If the variance of fill volume exceeds 0.01 (fluid ounces)2, an unacceptable propor$on of bofles will be under filled or overfilled. Is there evidence in the sample data to suggest that the manufacturer has a problem with under filled or overfilled bofles? Use α = 0.05 , and assume that fill volume has a normal distribu$on. Dr. Hatem Elayat, AUC 26 Example (Contd.) Data: n = 20, S = 0.01...
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This document was uploaded on 11/09/2013.

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