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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...
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This document was uploaded on 05/15/2010.
 Spring '08

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