HW3_solutionR3

PH 1700 HW3 solutions 4.60 If a group of 200 IDDM 30- to 39-year-old women is followed, what is the probability that at least 2 will go blind over a 1-year period? 20pt (8 setting up the equation+8 plug in the right number+2 computation+2 the sentence) Expected answer We use the binomial distribution with n = 200, p=0.0074. We wish to compute Pr(X ≥2) = 1-Pr(X≤1)=1-[Pr(X=0)+Pr(X=1)] = 1 − [ ( 2000 ) 0.0074 0 0.9926 200 + ( 2001 ) 0.0074 1 0.9926 199 ] =0.436 (Alternative answer , as N=200 >100, p=0.0074 <0.01, we can use the Poisson distribution approximation. Must include the check assumption (if not, -1pt) μ=np=200*0.0074=1.48 1pt (out of 8 setting up the equation ) Pr(X ≥2) = 1-Pr(X≤1)=1-[Pr(X=0)+Pr(X=1)] = 1 −( e − 1.48 1.48 0 0 ! + e − 1.48 1.48 1 1 ! ) =0.4354) The probability that at least 2 will go blind over a 1-year period is 0.436. (0.4354) 4.61 What is the probability that a 30-year-old IDDM male patient will go blind over the next 10 years? 20pt (8 setting up the equation+8 plug in the right number+2 computation+2 the sentence) Method 1: If we calculate the probability that EXACTLY one male goes blind over the next 10 years: P ( X = k ) = e − μ μ k k! K=1, λ = 0.0067 µ=0.0067 * 10 = λ*t = 0.067 P ( X = 1 ) = e − 0.067 0.067 1 1 ! = 0.0627 Thus, the probability that exactly one 30-year old IDDM male patient will go blind over the next 10 years is 0.0627 Method 2: Pr(male go blind within the next 10 years) =1-Pr( male not blind over 10 years) =1- Pr Pr ( male not blind 1 year ) 10 =1- ( 1 − 0.0067 ) 10 =1-0.9350 =0.065

The probability that a 30-year-old IDDM male patient will go blind over the next 10 years is 0.065. 4.62 After how many years of follow-up would we expect the cumulative incidence of blindness to be 10% among 30-year-old IDDM females, if the incidence rate remains constant over time? 20pt Method 1 In Poisson formula, μ = λ * Δ t (6pt) Here we have the cumulative incidence μ =0.1. The annual incidence rate of female λ =0.0074. (6pt) So Δ t=0.1/0.0074=13.5 years (6pt) We expect the cumulative incidence of blindness to be 10% among 30-year-old IDDM females after 13.5 years. (2pt) Method 2 Similar like the method 2 in 4.61 5pt formula + 5pt plug in right number + 5pt right answer -3 it’s more accurate to use expectation value since it talks about expectation We expect the cumulative incidence of blindness to be 10% among 30-year-old IDDM females after 14.2 years. 2pt Additional problems: 1. Using the data from lead.csv answer the following questions: 1) Compute and interpret the 95% Confidence interval for the  variance  of Full-scale IQ (iqf). What does it tell us about the standard deviation of IQ scores? 20pt (4pt alpha (2pt each) + 4pt chi value (2pt each) +4pt computation (2pt each) +4 take the square root+4 interpretation)

In lead group 1, s1=15.34 Lower bound= ( 78 − 1 ) 15.34 2 χ 77 , 0.975 2 = 77 × 235.32 103.1581 =175.65 Upper bound= ( 78 − 1 ) 15.34 2 χ 77 , 0.025 2 = 77 × 235.32 54.6234 =331.72 SD = √ ❑ SD lower bound= √ ❑ =13.25 SD upper bound= √ ❑ =18.21 In lead group 2, s2=10.19 Lower bound= ( 24 − 1 ) 10.19 2 χ 23 , 0.975 2 = 23 × 103.84 38.0756 =62.72 Upper bound= ( 24 − 1 ) 10.19 2 χ 23 , 0.025 2 = 23 × 103.84 11.6886 =204.32 SD = √ ❑ SD lower bound= √ ❑ =7.92 SD upper bound= √ ❑ =14.29 In lead group 3, s3=14.29 Lower bound= ( 22 − 1 ) 14.29 2 χ 21 , 0.975 2 = 21 × 204.20 35.4789 =120.87 Upper bound= ( 22 − 1 ) 14.29 2 χ 21 , 0.025 2 = 21 × 204.20 10.2829 =417.03 SD = √ ❑ SD lower bound= √ ❑ =10.99 SD upper bound= √ ❑ =20.42

use the “ci” command in STATA For group 1, we are 95% confident that the true variance of iqf lies between (175.65, 331.72), and the true standard deviation of iqf lies between (13.25, 18.21). For group 2, we are 95% confident that the true variance of iqf lies between (7.92, 14.29), and the true standard deviation of iqf lies between (13.25, 18.21). For group 3, we are 95% confident that the true variance of iqf lies between (120.87, 417.03), and the true standard deviation of iqf lies between (10.99, 20.42). 2. Concerning the information used in 7.30, 7.31, and 7.32 answer the following questions 1) construct the appropriate 95% confidence interval for the proportion of deaths from lung cancer in the sample 10pt (2pt for check assumption of normal approximation and rejection+6pt for command+2pt 95% confident sentence)

With a sample size of 20, we have 21.42(45.14%) power to detect an effect size of at least 0.13 difference in the proportion of lung cancer deaths if it does exist. 4pt Or With a sample size of 20, and assuming the population rate is 0.12, we have 21.42(45.14%) power to detect an effect size of at least 0.13 difference in the proportion of lung cancer deaths if it does exist.