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Inference on Single Mean

# This is called a s type i error type prej h0h0 iis

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Unformatted text preview: ed a s type I error. type P(Rej H0|H0 iis true) = α s 2) Fail to Reject H0 when Ha iis true at some s value. This is called a type II error. value. P(Fail to Rej H0|Ha iis true at some value) = β s to Rej G. Baker, Department of Statistics G. University of South Carolina; Slide 69 University 69 Avg Life of Light Bulb - Type I Error Life H0: µ < 2000 Ha: µ > 2000 Z Fail to reject H0. Assumes H0 is true. α = Probability that we will reject Ho when Ho is true. G. Baker, Department of Statistics G. University of South Carolina; Slide 70 University 70 Type I and Type II Errors H0: µ = 2000 β = Probability we will fail to reject Ho when Ha is true at µ = 2200 What if µ = 2200 α = Probability that we will reject Ho when Ho is true. G. Baker, Department of Statistics G. University of South Carolina; Slide 71 University 71 How can we control the size of β? How The value of α. Location of our point of interest. Location Sample size. Sample G. Baker, Department of Statistics G. University of South Carolina; Slide 72 University 72 Calculating β Calculating If µ = 2200, what is the probability of a type II 2200, error? error? Given: α = 0.05 and we are assuming µ = 2000. We will also assume we know σ = 2000. 216. 216. P( Z > 1.645) = 0.05 y − 2000 1.645 = → y = 2091 216 / 15 G. Baker, Department of Statistics G. University of South Carolina; Slide 73 University 73 Calculating β Calculating H0: µ = 2000 Fail to Reject Ho What if µ = 2200 2091 Reject Ho P( y < 2091 | µ = 2200) = β G. Baker, Department of Statistics G. University of South Carolina; Slide 74 University 74 Calculating β Calculating β = P( y < 2091 | µ = 2200) 2091 − 2200 β = P z < = P( z < −1.9544) = 0.0254 216 / 15 β = P(Fail to Reject H 0 | µ = 2200) = 0.0254 G. Baker, Department of Statistics G. University of South Carolina; Slide 75 University 75 α, β and Power α = P(Reject H0|µ = 2000) = 0.05 β = P(Fail to Rej H0| µ = 2200) = 0.0254 P(Fail Rej We say that the power of this test at of µ = 2200 is 1 – 0.0254 = 0.9746 2200 Power = 1 –β Power = P(Rej H0|µ is at some Ha level) P(Rej G. Baker, Department of Statistics G. University of South Carolina; Slide 76 University 76 Plastic Injection Molding A plastic injection molding process for a part that has a critical width dimension historically follows a normal distribution. historically A recent sample of n = 4 yielded a sample mean of 101.4 and sample standard deviation of 8. deviation Does this data support the statement: “The true average width is greater than The 95.”? 95. G. Baker, Department of Statistics G. University of South Carolina; Slide 77 University 77 Plastic Injection Molding Confidence Interval Approach 95% confidence interval on µ: y ± t df =3, 0.025 s n 8 101.4 ± 3.182 = 101.4 ± 12.728 4 (93.56,109.24) G. Baker, Department of Statistics G. University of South Carolina; Slide 78 University 78 Plastic Injection Molding Hypothesis Test Approach H0: α = 0.05 Ha: Test statistics is p-value = Conclusion: G. Baker, Department of Statistics G. University of South Carolina; Slide 79 University 79...
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