Homework1

Question 4: (Textbook problem 1.12) To collect data in an introductory statistics course, I gave the students a questionnaire. One question asked whether the student was a vegetarian. Of 25 students, 0 answered yes. They were not a random sample, but use these data to illustrate inference for a proportion. Let π denote the population proportion who would say yes. Consider H 0 : π = 0 . 50 and H a : π ̸ = 0 . 5 a. What happens when you conduct the Wald test, which uses the estimated standard error in the z test statistic? Sample proportion who answered yes π ¿ =0/25=0 Z test statistic= 0 − 0.5 √ 0 × ( 1 − 0 ) 25 =− ∞ Z test statistic does not give a specific test statistic because the estimated standard error is 0 and the test statistic − ∞ . But − ∞ is in the critical region, so there is enough evidence to reject the null hypothesis at any significance level. b. Find the 95% Wald confidence interval for π . Is it believable? 95% Wald confidence interval =0±1.96*0=(0,0) It does not contain any range, so it is not believable. c. Conduct the score test, which uses the null standard error in the z test statistic. Report and interpret the P-value. For the score test, Z test statistic= 0 − 0.5 √ 0.5 × ( 1 − 0.5 ) 25 = − 0.5 0.1 =− 5 P-value for the two sided test= 2*P(Z>|z|)=2*0=0 p-value is 0, there is enough evidence to reject the null hypothesis at the 5% significance level.

× Question 5: (Textbook problem 2.2) For diagnostic testing, let X = true status (1 = disease, 2 = no disease) and Y = diagnosis (1 = positive, 2 = negative). Let π 1 = P ( Y = 1 X = 1) and π 2 = P ( Y = 1 X = 2). Let γ denote the probability that a subject has the disease. a. Given that the diagnosis is positive, use Bayes’ Theorem to show that the probability a subject truly has the disease is P ( X = 1 | Y = 1) = π 1 γ /[ π 1 γ + π 2 (1− γ )]. π 1 =P(Y=1|X=1) π 2 =P(Y=1|X=2) γ =P(X=1) P(X=1|Y=1)= P ( X = 1 ,Y = 1 ) P ( Y = 1 ) = P ( X = 1 ,Y = 1 ) P ( X = 1 ,Y = 1 ) + P ( X = 2 ,Y = 1 ) = P ( X = 1 ) P ( Y = 1 ∨ X = 1 ) P ( X = 1 ) P ( Y = 1 | X = 1 ) + P ( X = 2 ) P ( Y = 1 ∨ x = 2 ) = γ π 1 γ π 1 +( 1 − γ ) π 2 b. For mammograms for detecting breast cancer, suppose γ = 0 . 01, sensitivity = π 1 = 0 . 86, and specificity = 1− π 2 = 0 . 88. Find the positive predictive value. The positive predictive value=P(X=1|Y=1)= γ π 1 γ π 1 +( 1 − γ ) π 2 = 0.01 ∗ 0.86 0.01 ∗ 0.86 + ( 1 − 0.01 ) ∗ 0.12 =0.0086/0.1274=0.0675 c. To better understand the answer in (b), find the joint probabilities for the 2 2 cross-classification of X and Y . Discuss their relative sizes in the two cells that refer to a positive test result. P(X=1,Y=1)=P(X=1)P(Y=1|X=1)=0.01*0.86=0.0086 P(X=1,Y=2)=P(X=1)P(Y=2|X=1)=0.01*(1-0.86)=0.0014 P(X=2,Y=1)=P(X=2)P(Y=1|X=2)=0.99*(1-0.88)=0.1188 P(X=2,Y=2)=P(X=2)P(Y=2|X=2)=0.99*0.88=0.8712

| | Question 8: The odds ratio can be defined as θ = P ( Y = 1 | X = 1)/( P ( Y = 2 | X = 1)) P ( Y = 1 | X = 2)/( P ( Y = 2 | X = 2)) In case control studies we are not able to estimate P ( Y = y X = x ) because the number of subjects that have each outcome level y is fixed by design. Instead we are able to estimate P ( X = x Y = y ). Show mathematically why this enables us to estimate odds ratios from case-control studies, i.e. show that the odds ratio can be written in terms of things we can estimate. P(Y=y|X=x) = P ( X = x | Y = y ) P ( Y = y ) P ( X = x ) ϴ= P ( Y = 1 | X = 1 ) / P ( Y = 2 ∨ X = 1 ) P ( Y = 1 | X = 2 ) / P ( Y = 2 ∨ X = 2 ) = P ( Y = 1 | X = 1 ) ∗ P ( Y = 2 ∨ X = 2 ) P ( Y = 1 | X = 2 ) ∗ P ( Y = 2 ∨ X = 1 ) = P ( X = 1 ∨ Y = 1 ) P ( Y = 1 ) P ( X = 1 ) · P ( X = 2 ∨ Y = 2 ) P ( Y = 2 ) P ( X = 2 ) P ( X = 2 ∨ Y = 1 ) P ( Y = 1 ) P ( X = 2 ) · P ( X = 1 ∨ Y = 2 ) P ( Y = 2 ) P ( X = 1 ) = P ( X = 1 ∨ Y = 1 ) P ( X = 2 ∨ Y = 2 ) P ( X = 2 ∨ Y = 1 ) P ( X = 1 ∨ Y = 2 ) Since we know the number of subjects that have exposed or not among cases and controls in this study design, we can estimate the odds ratio.