Question

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? Define a success as a toss that lands on heads. Then the probability of a success is 0.5, and the expected number of successes after 10 tosses is 10(0.5) = 5. Since each success pays $10, the total amount you would expect to win 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 loving” if the maximum amount he or she is willing to pay to play is greater than $50, the game’s expected value. Suppose in this 1,536-person sample, 20 people are risk loving. Let p denote the proportion of the adult population aged 22 to 55 who are risk loving and 1 – p, the proportion of the same population who are not risk loving. Use the sample results to estimate the proportion p.

1. The proportion p̄ of adults in the sample who are risk loving is ____. 2. The proportion 1 – p̄ of adults in the sample who are not risk loving is _____. 3. You (can or cannot) conclude that the sampling distribution of p̄ can be approximated by a normal distribution, because _____. 4. The sampling distribution of p̄ has an estimated standard deviation of ____.

Answer options 1. 0.01302, 0.20101, 0.79899, 0.98698 2. 0.01302, 0.20101, 0.79899, 0.98698 3. can or cannot, 4. 0.00030, 0.1023, 0.01290, 0.00289

Answer 1

1. The proportion p̄ of adults in the sample who are risk loving is 20/1536 = 0.01302

2. The proportion 1 – p̄ of adults in the sample who are not risk loving is 1 - 0.01302 = 0.98698

3. You can conclude that the sampling distribution of p̄ can be approximated by a normal distribution, because

4. The sampling distribution of p̄ has an estimated standard deviation of