EX 27 - STAT 400 Examples for 11/10/2011 (1) Fall 2011 1....

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Examples for 11/10/2011 (1) Fall 2011 1. The overall standard deviation of the diameters of the ball bearings is σ = 0.005 mm. The overall mean diameter of the ball bearings must be 4.300 mm. A sample of 81 ball bearings had a sample mean diameter of 4.299 mm. Is there a reason to believe that the actual overall mean diameter of the ball bearings is not 4.300 mm? a) Perform the appropriate test using a 10% level of significance. Claim: μ 4.300 H 0 : μ = 4.300 vs. H 1 : μ 4.300 Test Statistic: σ is known Z = 81 005 0 300 4 299 4 X 0 μ . . . n - = - σ = – 1.80 . Rejection Region: 2 – tailed. Reject H 0 if Z < – z α / 2 or Z > z α / 2 α = 0.10 α 2 = 0.05. z 0.05 = 1.645 . Reject H 0 if Z < – 1.645 or Z > 1.645. Decision: The value of the test statistic does fall into the Rejection Region. Reject H 0 at α = 0.10 . OR P-value: p-value = P( Z 1.80 ) + P( Z 1.80 ) = 0.0359 + 0.0359 = 0.0718 . Decision: 0.0718 < 0.10. P-value < α . Reject H
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EX 27 - STAT 400 Examples for 11/10/2011 (1) Fall 2011 1....

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