HW4-SolR2

7.26 Suppose we identify 50 women 50 to 54 years of age who have both a mother and a sister with a history of breast cancer. Five of these women themselves have developed breast cancer at some time in their lives. If we assume that the expected prevalence rate of breast cancer in women whose mothers have had breast cancer is 4%, does having a sister with the disease add to the risk? Explain. 14pt (1 pt per hypothesis +3pt why exact binomial +1 closing sentence (interpretation/ answering the question) + 2pt recognize the significance<0.05+6pt computation) We wish to test the hypothesis H 0 : p=p 0 =0.04 H 1 : p>0.04 (One-sided test) Two-sided test acceptable We can use one-sided because the research questions asks if having a sister with the disease “adds” to the risk. Since np 0 q 0 = 50*0.04*(1-0.04) =1.92 <5, we cannot use the normal test. Instead, we must use the exact binomial test. P= ∑ k = 5 50 ( 50 k ) 0.04 k 0.96 50 − k =1- ∑ k = 0 4 ( 50 k ) 0.04 k 0.96 50 − k In stata, display 1-binomial(50,4,0.04) =1-0.9510 =0.049 <0.05 We can reject H 0 and say there is a significant additional risk associated with having a sister with breast cancer. The proportion of deaths due to lung cancer in males ages 15–64 in England and Wales during the period 1970– 1972 was 12%. Suppose that of 20 deaths that occur among male workers in this age group who have worked for at least 1 year in a chemical plant, 5 are due to lung cancer. We wish to determine whether there is a difference between the proportion of deaths from lung cancer in this plant and the proportion in the general population. 7.30 State the hypotheses to use in answering this question. 5pt(1pt what p is+ 2pt Hypothesis*2) Let p = true proportion of deaths are due to lung cancer among workers in the chemical plant. We wish to test the hypothesis: H 0 : p=p 0 =0.12 H 1 : p ≠ 0.12 7.31 Is a one-sided or two-sided test appropriate here? 3pt (one sided -3) A two-sided test is appropriate here since we have no prior information that the proportion of lung cancer deaths is higher or lower than in the general population. Also the research question is only compatible with a two sided hypothesis, focusing on “difference”.

7.32 Perform the hypothesis test, and report a p -value. 11pt(3pt check exact binomial +1 closing sentence (interpretation\answer question) + 2pt recognize the significance>0.05+5pt computation) Since np 0 q 0 = 20*0.12*(1-0.12) =2.1 <5, we cannot use the normal test. Instead, we must use the exact binomial test. ^ p = 5 20 = .25 > p 0 =0.12 p=2 × ∑ k = 5 20 ( 20 k ) 0.12 k 0.88 20 − k =2 × ¿ In stata, display 2*(1-binomial(20,4,0.12)) =0.165 >0.05 Thus, there is no significant excess or deficit in the proportion of deaths due to lung cancer in this plant. -0.5 if only no significant, no interpretation The mortality experience of 8146 male employees of a research, engineering, and metal- fabrication plant in Tonawanda, New York, was studied from 1946 to 1981 [2]. Potential workplace exposures included welding fumes, cutting oils, asbestos, organic solvents, and environmental ionizing radiation, as a result of waste disposal during the Manhattan Project of World War II. Comparisons were made for specific causes of death between mortality rates in workers and U.S. white-male mortality rates from 1950 to 1978. Suppose that 17 deaths from cirrhosis of the liver were observed among workers who were hired prior to 1946 and who had worked in the plant for 10 or more years, whereas 6.3 were expected based on U.S. white-male mortality rates. 7.48 What is the SMR for this group? 5pts (1 pt for in percent) SMR=(observed # deaths/ Expected # deaths) = (17/6.3)=2.7 4pts The SMR for this group is 270%. 7.49 Perform a significance test to assess whether there is an association between long duration of employment and mortality from cirrhosis of the liver in the group hired prior to 1946. Report a p -value. 15pt(4pt hypothesis+11pt same as 7.26) H 0 : μ = μ 0 H 1 : μ ≠ μ 0 Where μ 0 = 6.3 . Since the expected number of events(under the null, μ0) is 6.3 <10 , we must use exact methods. p = 2 × ∑ k = 17 ∞ e − 6.3 6.3 k k ! = 2 × ¿ in stata, display 2*(1-poisson(6.3,16)) =0.0006 <0.05

μ + t 1 − ❑ 2 , n − 1 × s √ ❑ = 388+2.0322*37/ √ ❑ = 400.71 Or use the stata command cii means 35 388 37 400 mcg is within the 95% CI (375.29, 400.71) We cannot reject H0. We don’t have evidence to say that the intake of women whose income is below the poverty level differ from the recommended amount. A.2) What is the sample size you would need to have 90% power to answer question 1? 10pt(5 command,+5 sentence) For a two-sided test, we have 90% power to detect a difference of 12 mcg/day assuming the standard deviation is 37 mcg with a sample size is 100. 1pt for each piece in codes for power -0.5 if sampsi

B. Consider the lighter smoking twins from the bone density study in the case study topic. Do the lighter smoking twins have a healthy bone density if we assume that a healthy bone mineral density of the femoral neck bone is 0.77g/cm 2 ? B.1)What test would we use to answer the research question? (Hint: remember to assess assumptions of the test you choose.) 6pts The variables we are interested in is fn1. Since we want to compare fn1 with 0.77g/cm 2 , a one- sample t-test is used. We need to check the assumption of the normality. 2pts 1 sample t test 1pt hist or box 2pts qq plot 1pt normally distributed Histogram Boxplot

Critical value method: . display invttail(40,0.025) 2pt correct critical value 2.0210754 Since t=-6.45< -2.02, we reject the null hypothesis. There is sufficient evidence to suggest that the mean femoral neck density among lighter smoking is different than 0.77. 1pt comparison+ 1pt interpretation + 1pt reject P value method: 2pt p value+1pt reject Since p<0.0001, we reject the null hypothesis. There is sufficient evidence to suggest that the mean femoral neck density among lighter smoking is different than 0.77. B.4) Compute and interpret the appropriate 99% confidence interval for the research question: Is the bone mineral density of the lighter smoking twin lower than heathy levels? (Hint: translate to hypotheses, and then construct the appropriate confidence interval) Let μ1= the average of the femoral neck density among lighter smoking twins H0: μ1= 0.77 Ha: μ1 < 0.77 1pt per hypothesis Since it’s a one-sided test, a one-sided 99% CI is used. In STATA, we calculate a two- sided 98% CI to find a one-sided 99% CI. Stata code: ttest fn1==0.77, level (98) 2pt CI We choose the lower one-sided 99% CI, which is (-∞, 0.704). Since the null value 0.77 is not within the one-sided 99% confidence interval (-∞, 0.704), we can reject the null hypothesis. There is sufficient evidence to suggest that the

average of the femoral neck density among lighter smoking twins is smaller than 0.77. 2pt interpretation A two-sided 99% CI is also accepted. Since the null value 0.77 is not within the two-sided 99% confidence interval (0.62, 0.71), we can reject the null hypothesis. There is sufficient evidence to suggest that the average of the femoral neck density among lighter smoking twins is smaller than 0.77. *Healthy bone density value computed from Jacqueline R. Center, Tuan V. Nguyen, Nick A. Pocock, John A. Eisman, Volumetric Bone Density at the Femoral Neck as a Common Measure of Hip Fracture Risk for Men and Women, The Journal of Clinical Endocrinology & Metabolism , Volume 89, Issue 6, 1 June 2004, Pages 2776–2782, https://doi.org/10.1210/jc.2003-030551