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6.4 One and Two-Sample Inference for Variances. Example - Problem 26 – Page 435  D. Kim did some crude tensile strength testing on pieces of some nominally.

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Presentation on theme: "6.4 One and Two-Sample Inference for Variances. Example - Problem 26 – Page 435  D. Kim did some crude tensile strength testing on pieces of some nominally."— Presentation transcript:

1 6.4 One and Two-Sample Inference for Variances

2 Example - Problem 26 – Page 435  D. Kim did some crude tensile strength testing on pieces of some nominally 0.012 in. diameter wire of various lengths.  Perform a hypothesis test to determine if there is a significant difference between the mean tensile strengths between 25 cm and 30 cm lengths of nominal 0.012 in. diameter wire using a significance level of 0.05.  To determine if the equal variance approach can be applied, the following hypothesis test should be performed.

3 Test for Equal Variances  Hypotheses 

4 Minitab Output  Test for Equal Variances  F-Test (normal distribution)  Test Statistic: 0.487  P-Value : 0.364  Levene's Test (any continuous distribution)  Test Statistic: 2.710  P-Value : 0.122

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6 Additional Output  Descriptive Statistics: Strength by Length  Variable Length N Mean StDev  Strength 25 cm 8 4.4313 0.2314  30 cm 8 4.2880.331

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8 Two-Sample T-Test and CI: 25cm, 30cm Two-sample T for 25cm vs 30cm N Mean StDev SE Mean 25cm 8 4.431 0.231 0.082 30cm 8 4.288 0.331 0.12 Difference = mu 25cm - mu 30cm T-Test of difference = 0 (vs not =): T-Value = 1.01 P-Value = 0.334 DF = 12 Two-Sample T-Test and CI: 25cm, 30cm Two-sample T for 25cm vs 30cm N Mean StDev SE Mean 25cm 8 4.431 0.231 0.082 30cm 8 4.288 0.331 0.12 Difference = mu 25cm - mu 30cm T-Test of difference = 0 (vs not =): T-Value = 1.01 P-Value = 0.331 DF = 14 Both use Pooled StDev = 0.286


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