Test2 Revision2_sol.pdf - Test2 Revision2 1 SOLUTIONS[7 marks 260 − 179.1(a P X > 260 = P Z > = P(Z > 2.04 39.7 = 1 − P Z < 2.04 = 1 −.9793

Test2 Revision2_sol.pdf - Test2 Revision2 1 SOLUTIONS[7...

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Test2 Revision2 SOLUTIONS 1. [7 marks] (a) [1] [1] [total = 2] (b) If P ( Z > z ) = .03, then z 1.88. [1] Hence X = μ + zσ = 179.1 + 1.88 × 39.7 = 253.74 mg/dL [1] [total = 2] (c) Since this probability is very small (< 1/600), and 152.7 is below the population mean, such a sample mean would be unusually small. [1, total = 3] 2. [8 marks] (a) 997 4 993 ~ = + = n , 54965 . 997 ) 2 546 ( ~ = + = p 90% CI or .524 to .576 [total = 4] (b) Couples using non-invasive sex selection methods to increase the chance of having a girl. [1] (c) [total = 3] 3. [3 marks] X = number of female babies in sample. X ~ Bin( 12, .9). P ( X ≥ 11) = 1 – P ( X ≤ 10) = 1 - .3410 = .6590 [total = 3] 4. [9 marks] (a) H 0 : p = .019 H 1 : p > .019 where p is the proportion of patients who report flu-like symptoms following treatment with 10 mg of atorvastatin. [1, total = 2] (b) 02781 . 863 24 ˆ = = p 89 . 1 00465 . 00881 . 863 981 . 019 . 019 . 02781 . ) 1 ( ˆ 0 0 0 = = × - = - - = n p p p p
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Unformatted text preview: Z [total = 3] (c) P-value = P ( Z &gt; 1.89) = 1 – P ( Z &lt; 1.89) = 1 - .9706 = .0294 [2] (d) Since P-value &lt; .05, we reject H . [2] 017 , 2 3 . 7 . 02 . 96 . 1 ) 1 ( 2 2 = × × =- ≥ p p e z n 0259 . 5497 . 01576 . 645 . 1 54965 . 997 45035 . 54965 . 645 . 1 54965 . ~ ) ~ 1 ( ~ ~ ± = × ± = × ± =-± = n p p k p 0207 . 9793 . 1 ) 04 . 2 ( 1 ) 04 . 2 ( 7 . 39 1 . 179 260 ) 260 ( =-= &lt;-= &gt; = -&gt; = &gt; Z P Z P Z P X P 1 1 1 1 1 1 2 1 1 1 1 0015 . ) 97 . 2 ( 20 7 . 39 1 . 179 7 . 152 ) 7 . 152 ( =-&lt; = -&lt; = &lt; Z P Z P X P 1 rounding doesn’t matter formal statement not required 1 ½ if 3 decimals, 0 if fewer than 3 decimals. exact = .0015 exact = .0208 exact = 253.77 1 exact = .0290 exact = 1.8957 1...
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