Midterm exam A solution

Midterm exam A - H 0 p 1 = p 2 vs H a p 1 ≠ p 2 138 1 = x 155 2 = x 65 250 200 155 138 ˆ = = p p Test statistics 55 1 1 1 ˆ 1 ˆ ˆ ˆ 2 1 2 1

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Statistics 217 – L06A Midterm Exam Solution 1. Use nonpooled t-interval since two population standard deviations are different Sample 1: Group 1 5985 . 7 1 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 1 2 1 = - + - + = n n s n n s n s n s γ Round it down to 7. df =7 t 0.025,7 = 2.3646 The 95% confidence interval is 2 2 2 1 2 1 2 / 2 1 n s n s t x x + ± - α =(26.22, 113.78) 2. Sample 1: Strawberry sample Two samples are paired H 0: μ 1 = μ 2 vs H a μ 1 ≠ μ 2 , 5 = d x s d = 19.748 test stat 62 . 0 = = n s x t d d critical value ±t 0.025, 5 = ±2.5706 Do not reject the null hypothesis since |t| < t 0.025, 5 3. Reject H 0 (a) Sample 1: boys aged 12 - 14
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Unformatted text preview: H 0: p 1 = p 2 vs H a p 1 ≠ p 2 138 1 = x 155 2 = x , , 65 . 250 200 155 138 ˆ = + + = p p Test statistics 55 . 1 ) 1 1 )( ˆ 1 ( ˆ ˆ ˆ 2 1 2 1 = +--= n n p p p p z p p Critical value: ± z 0.025 = ±1.96 Do not reject H since test statistic |z| < 1.96 4. (a) p-value = 0.0314 (b) (i) Reject H since p-value < 0.05 (ii) Do not reject H since p-value > 0.01 5. (a) 0.002 <p-value < 0.01 (b) Extremely trong evidence against H...
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This note was uploaded on 09/10/2009 for the course STAT 217 taught by Professor Dan during the Winter '07 term at University of Calgary.

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