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ams572_notes_89

# ams572_notes_89 - AMS 572 Lecture Notes 8 Oct 8th 2010 Ch8...

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AMS 572 Lecture Notes 8 Oct. 8 th , 2010 Ch8 Inference on two population means (and two population variances). 1. The samples are paired paired sample t-test 2. The samples are independent independent sample t-test a) 2 2 1 2 σ σ =     pooled-variance t-test b) 2 2 1 2 σ σ unspooled-variance t-test (Note: to check if 2 2 1 2 σ σ = , use F-test) 2a.Inference on 2 population means, when both populations are normal. We have 2 independent samples, the population variances are unknown but equal ( 2 2 2 1 2 σ σ σ = = ) pooled-variance t-test. Data: 1 2 1 1 1 , , ( , ) iid n X X N μ σ K : 2 2 1 2 2 , , ~ ( , ) iid n Y Y N μ σ K Goal: Compare 1 μ and 2 μ 1) Point estimator: · 2 2 1 2 1 2 1 2 1 2 ~ ( , ) X Y N n n σ σ μ μ μ μ - = - - + 2 1 2 1 2 1 1 ( ,( ) ) N n n μ μ σ - + 2) Pivotal quantity:

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1 2 1 2 ( ) ( ) ~ (0,1) 1 1 X Y Z N n n μ μ σ - - - = + not the PQ, since we don’t know 2 σ Q 1 2 2 1 1 1 2 ( 1) ~ n n S χ σ - - , 2 2 2 2 2 1 2 ( 1) ~ n n S χ σ - - , and they are independent ( 2 1 S & 2 2 S are independent because these two samples are independent to each other) 1 2 2 2 2 1 1 2 2 2 2 2 ( 1) ( 1) ~ n n n S n S W χ σ σ + - - - = + Definition: 2 2 2 1 2 k W Z Z Z = + + + L , when . . . ~ (0,1) i i d i Z N , then 2 ~ k W χ . Definition:  t-distribution: ~ (0,1) Z N , 2 ~ k W χ , and Z & W are independent, then Z T W k = ~ k t 1 2 1 2 1 2 1 2 2 2 2 1 1 2 2 2 2 1 2 1 2 1 2 ( ) ( ) 1 1 ( ) ( ) ~ 1 1 ( 1) ( 1) 2 2 n n p X Y n n X Y Z T t W n S n S S n n n n n n μ μ σ μ μ σ σ + - - - - + - - - = = = - - + + + - + - . where 2 2 2 1 1 2 2 1 2 ( 1) ( 1) 2 p n S n S S n n - + - = + - is the pooled variance. This is the PQ of the inference on the parameter of interest 1 2 ( ) μ μ - 3) Confidence Interval for 1 2 ( ) μ μ -
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2, 2, 2 2 1 2 2, 2, 2 2 1 2 1 2 2, 2, 1 2 1 2 2 2 1 2 2, 1 2 2 1 ( ) ( ) ( ) 1 ( ) 1 1 1 1 1 1 1 ( ( ) ( ) ) 1 1 1 ( n n n n n n n n p p p n n n n p n n n n P t T t X Y P t t S n n P t S X Y t S n n n n P X Y t S X Y t n n α α α α α α α α μ μ α α μ μ α μ μ + - + - + - + - + - + - + - + - = - - - - - = - + - = - + - - - + - = - - + - - + 2, 1 2 2 1 1 ) p S n n α - + This is the 100(1 )% α - C.I for 1 2 ( ) μ μ - 4) Test: Test statistic: 0 T = 0 1 2 0 2 1 2 ( ) ~ 1 1 H n n p X Y c t S n n + - - - + a) 0 1 2 0 1 2 0 : : a H c H c μ μ μ μ - = - (The most common situation is 0 c =0 0 1 2 : H μ μ = )

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At the significance level α , we reject 0 H in favor of a H iff 1 2 0 2, n n T t α + - If 0 0 0 ( | ) P value P T t H α - = < , reject 0 H b) 0 1 2 0 1 2 0 : : a H c H c μ μ μ μ - = - < At significance level α , reject 0 H in favor of a H iff 1 2 0 2, n n T t α + - ≤ - If 0 0 0 ( | ) P value P T t H α - = < , reject 0 H c) 0 1 2 0 1 2 0 : : a H c H c μ μ μ μ - = - At α =0.05, reject 0 H in favor of a H
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