8. Interval estimation of a population proportion Think about the following game: A fair coin is tossed 10 times. Each time the toss results in heads, you receive $10; for tails, you get nothing. What is the maximum amount you would pay to play the game? As a statistics student, you are aware that the game is a 10-trial binomial experiment. A toss that lands on heads is defined as a success, and because the probability of a success is 0.5, the expected number of successes is 10(0.5) = 5. Since each success pays $10, the expected value of the game is 5($10) = $50. Suppose each person in a random sample of 1,536 adults between the ages of 22 and 55 is invited to play this game. Each person is asked the maximum amount they are willing to pay to play. (Data source: These data were adapted from Ben Mansour, Selima, Jouini, Elyes, Marin, Jean-Michel, Napp, Clotilde, & Robert, Christian. (2008). Are risk-averse agents more optimistic? A Bayesian estimation approach. Journal of Applied Econometrics, 23(6), 843-860.) Someone is described as "risk averse" if the maximum amount he or she is willing to pay to play is less than $50, the game's expected value. Suppose in this 1,536-person sample, 1,482 people are risk averse. Let p denote the proportion of the adult population aged 22 to 55 who are risk averse and 1 - p, the proportion of the same population who are not risk averse. Use the sample results to estimate the proportion p.You can be 95% condent that the interval estimate LCL = V to UCL = Y includes the population proportion pr the proportion of adults aged 22 to 55 who are risk averse. Use the Distributions tool to develop a 95% confidence interval estimate of the proportion of adults aged 22 to 55 who are risk averse