Hw6_soln - HW Set 6 Due 5/11 in quiz session HW16 hw_Z hw_AA hw_AB(lecture 18 7.4(a 7.11(a,b,c HW17 7.15 7.18 7.22(using messy eqn in lecture 19

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW Set 6: Due 5/11 in quiz session. HW16: hw_Z, hw_AA, hw_AB (lecture 18), 7.4(a), 7.11(a,b,c). HW17: 7.15, 7.18, 7.22 (using messy eqn in lecture 19), 7.23 (using simple eqn in lecture 19; typo in answer), 7.76 (using simple eqn in lect 19). HW18: 7.27 (There is a typo in the statement of the problem: ln(pi1,pi2) should be ln(pi1/ pi2), 7.28 (Hint: what value of pi gives the largest CI?), 7.33 (typo in answer.) HW19: 7.39, 7.40(a), 7.48 (use df=Welch's formula), 7.50 (use df=37), 7.53. In all of these, any boxplots and qqplots should be done by computer, and the rest by hand. Graded problems (and points): TBA. Hw_AA. (a) μ p = π = .80. σ p = n ) 1 ( π π- = 25 ) 80 . 1 ( 80 .- = .08 b (b) Since 20% do not favor the proposed changes (so π = .20), the mean & standard deviation of the sampling distribution of this proportion are μ p = π = .20 and σ p = n ) 1 ( π π- = 25 ) 20 . 1 ( 20 .- = .08. (c) For n = 1000 and π = .80, μ p = π = .80. σ p = n ) 1 ( π π- = 100 ) 80 . 1 ( 80 .- = .04. Notice that it was necessary to quadruple the sample size (from n=25 to n=100) in order to cut σ p in half (from σ p = .08 to σ p = .04). Hw_AB (a) x has a binomial distribution with n = 5 and π = .05. Writing P(.05-.01 ≤ p ≤ .05+.01) in terms of x, we find P(.05-.01 ≤ x/5 ≤ .05+.01) = P((.04)5 ≤ x ≤ (.06)5) = P(.2 ≤ x ≤ .3). Because x can only have integer values, there is no x between .2 and .3, so the probability of this event is 0. (b) For n= 25, P(.04 ≤ p ≤ .06) = P((.04)25 ≤ x ≤ (.06)25) = P(1 ≤ x ≤ 1.5) = P(x = 1) = 24 1 25 1 ) 95 (. ) 05 (. = 0.36498. (c) For n= 1005, P(.04 ≤ p ≤ .06) = P((.04)100 ≤ x ≤ (.06)100) = P(4 ≤ x ≤ 6). Using the normal approximation to the binomial, with μ = n π = 100(.05) = 5 and σ 2 = n π (1- π ) = 100(.05)(.95) = 4.75 and σ = 2.7195: P(4 ≤ x ≤ 6) ≈ P( 1795 . 2 5 6 7195 . 2 5 4- ≤ ≤- z ) = P(-.46 ≤ z ≤ .46) = .3544. Using the continuity correction makes a substantial difference in this problem because the interval from 4 to 6 contains the mean of the distribution (and hence, a large amount of the probability): P(4 ≤ x ≤ 6) ≈ P( 1795 . 2 5 5 . 6 7195 . 2 5 5 . 3- ≤ ≤- z ) = P(-.69 ≤ z ≤ .69) = .5098. 7.4 (a) Recall: ( 29 n p p π π σ π μ- = = 1 and Even though π is unknown, we can still set an upper bound on p σ . When p σ π , 5 . = is maximized. So, n p 2 1 = σ . In our case, 15811 . 10 = ⇒ = p n σ ( 29 ( 29 ( 29 4714 . 63 . 63 . 15811 . 10 . 15811 . 10 . 10 . 10 . So, = < <- = < <- = <- <- z P z P p P π 7.11 (a) Decreasing the confidence level from 95% to 90% will decrease the associated z value and therefore make the 90% interval narrower than the 95% interval. (Note: see the answer to Exercise 9 above) (b) The statement is not correct. Once a particular confidence interval has been created/ calculated, then the true mean is either in the interval or not. The 95% refers to the...
View Full Document

This document was uploaded on 05/15/2010.

Page1 / 9

Hw6_soln - HW Set 6 Due 5/11 in quiz session HW16 hw_Z hw_AA hw_AB(lecture 18 7.4(a 7.11(a,b,c HW17 7.15 7.18 7.22(using messy eqn in lecture 19

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online