Chapter 9

Chapter 9 - Chapter 9. Inferences based on two samples...

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Unformatted text preview: Chapter 9. Inferences based on two samples Extend confidence interval and hypotheses testing for single mean, proportion, variance to two different means, proportions, variances.- comparison of two means and two proportions 1 2 1 2 ? ? p p μ μ- =- =- comparison of two variances 9.1 z-tests and confidence intervals for a difference between two population means Basic Assumptions: 1. 1 2 , , , m X X X L ~ rs from mean 1 μ and variance 2 1 σ 2. 1 2 , , , n Y Y Y L ~ rs from mean 2 μ and variance 2 2 σ 3. X and Y are independent of one another Prop) X Y- ~ 2 2 1 2 1 2 ( , ) m n σ σ μ μ- + 1 Case I : normal distribution with known variances X Y- ~ 2 2 1 2 1 2 ( , ) N m n σ σ μ μ- + 1 2 2 2 1 2 ( ) X Y m n μ μ σ σ--- + ~ ( 0, 1) N Test procedure: 1. Null Hypothesis: 1 2 : H μ μ- = ∆ 2. test statistic: 2 2 1 2 X Y z m n σ σ-- ∆ = + 3. Alternative Hypothesis Rejection Region P-value 1 2 : a H μ μ- ∆ z z α ≥ 1 ( ) z- Φ 1 2 : a H μ μ- < ∆ z z α ≤ - ( ) z Φ 1 2 : a H μ μ- ≠ ∆ / 2 z z α ≥ or / 2 z z α ≤ - 2[ 1 ( | | ) ] z- Φ 2 Ex 9.1) Two different treatments (treatment vs control) the difference of two averages strengths of steels 29. 8, 20 34. 7, 25 x m y n = = = = 1. Parameter of interest: 1 2 μ μ- =difference bet average strengths for the two types of steel 2. Null Hypothesis: 1 2 : H μ μ- = 3. Alternative Hypothesis: 1 2 : a H μ μ- ≠ 4. Test Statistic: 2 2 1 2 x y z m n σ σ- = + 5. Rejection Region: 0. 005 2. 58 z z ≥ = or 0. 005 2. 58 z z ≤ - = - 6. Compute Statistic: 29. 8 34. 7 3. 66 16. 0 25. 0 20 25 z- = = - + 7. Decide: -3.66<-2.58, we reject the null hypothesis. The data strongly suggest that the two average strengths are different....
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This note was uploaded on 01/06/2011 for the course STAT 511 taught by Professor Bud during the Fall '08 term at Purdue.

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Chapter 9 - Chapter 9. Inferences based on two samples...

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