Suppose that insurance companies did a survey. They randomly surveyed 410 drivers and found that 310 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Construct a 95% confidence interval for the population proportion that claim to always buckle up. (i) State the confidence interval. (Round your answers to four decimal places.) (:lll:l) (ii) Sketch the graph. 3 = CL. 2 a Cl F (iii) Calculate the error bound. (Round your answer to four decimal places.)