Homework1

STA4504 Homework 1, Spring 2021 Please turn in your own work, though you may discuss the problems with classmates, the TA, the Professor, the internet, etc. The most important thing is that you understand the problems and how they are solved as they will prepare you for future exams. Please turn in your work by Thursday, February 4th at midnight. Question 1: (Textbook problem 1.3) Each of 100 multiple-choice questions on an exam had four possible answers but one correct response. For each question, a student randomly selects one response as the answer. a. Specify the probability distribution of the student’s number of correct answers on the exam. Binomial distribution with n=100 and π =0.25 b. Based on the mean and standard deviation of that distribution, would it be surprising if the student made at least 50 correct responses? Explain your reasoning. μ = 100 × 0.25 = 25 σ = √ 100 × 0.25 × ( 1 − 0.25 ) = 4.33 Z test statistic=(50-25)/4.33=5.77 It would be surprising if the student made at least 50 correct response because 50 is 5.77 standard deviations above the mean. Question 2: (Textbook problem 1.6c) Genotypes AA, Aa, and aa occur with probabilities ( π 1 , π 2 , π 3 ). For n = 3 independent observations, the observed frequencies are ( y 1 .y 2 , y 3 ). Suppose ( π 1 , π 2 , π 3 ) = (0 . 25 , 0 . 5 , 0 . 25). What probability distribution does y 1 alone have? y 1 alone has the binomial distribution with sample size 3 and parameter of the distribution π 1 =0.25

Question 3: (Textbook problem 1.8) When the 2010 General Social Survey asked subjects in the US whether they would be willing to accept cuts in their standard of living to protect the environment, 486 of 1374 subjects said yes. a. Estimate the population proportion who would say yes. Construct and interpret a 99% confidence interval for this proportion. The sample proportion who would say yes=π π =486/1374=0.354 The standard error= √ 0.354 ( 1 − 0.354 ) 1374 = 0.0129 Z 0.005 =2.58 based on the standard normal distribution table 99% confidence interval=0.354±2.58*0.0129=(0.321,0.387) We are 99% confident that the population proportion who would say yes is between 0.321 and 0.387. b. Conduct a significance test to determine whether a majority or minority of the population would say yes. Report and interpret the P-value. Let H o : π=0.5 , H a : π ≠0.5 Sample proportion who would say yes=0.354 Test statis tic= 0.354 − 0.5 √ 0.5 × ( 1 − 0.5 ) 1374 = − 0.146 0.0135 =− 10.815 Based on the standard normal distribution table, the corresponding p-value for the test statistic -10.815 is <0.0001. Two-sided p-value is <0.0001. Under the null hypothesis, it is unlikely to obtain observed result or extreme. At the 1% significance level, there is enough evidence to reject the null hypothesis. There is strong evidence that the minority of the population would say yes.

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.