8.3.133: You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview? We need to find the sample size in this case which is given as: n = p*(1-p)*z2/E2 Here we have p = 0.5, 1-p = 0.5, E = 0.05 or 5%, z = 1.96 So n = (0.5*0.5*1.96*1.96)/(0.05*0.05) = 384.16 So the minimum students must be interviewed are 385. 8.3.119. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job. a. Define the random variables X and P in words. X = number of people who feel the president is doing an acceptable job P' = Proportion of people who feel the president is doing an acceptable job b. Which distribution should you use for this problem? Explain your choice. Since here the sample size is large and it is a binomial proportion so we will use z distribution for binomial response. c. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 90% confidence interval is given as Sample proportion Z(Critical Value)*SE Here we have Sample proportion = 0.61 Z (critical value at 10%) = 1.645 SE = sqrt(P'*(1-P')/n) = sqrt(0.61*0.39/1200) = 0.0141 Lower Limit = 0.61-0.0232 = 0.5868 Upper Limit = 0.61+0.0232 = 0.6332 90% confidence interval is (0.5868, 0.6332) Error bound = 1.645*0.014080128 = 0.0232