Think about the following game: A fair coin is tossed 10 times. Each time the to

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? As a statistics student, you are aware that the game Is a 10-tnal 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 8ayesian estimation approach. Journal of Applied Economerrics, 23(6), 843-860.) Someone is described as "risk neutral" if the maximum amount he or she is willing to pay to play is equal to $50, the game's expected value. Suppose in this 1, 536-pcrson sample, 34 people arc risk neutral. Let p denote the proportion of the adult population aged 22 to 55 who arc risk neutral and 1 = p, the proportion of the same population who are not risk neutral. Use the sample results to estimate the proportion p. The proportion p cap of adults in the sample who arc risk neutral is ______. The proportion 1 - p cap of adults in the sample who are not risk neutral is _____. You _____ conclude that the sampling distribution of p cap can be approximated by a normal distribution, because ______. The sampling distribution of p cap has a mean ______ and an estimated standard deviation of ______. Use the Distributions tool to develop a 95% confidence interval estimate of the proportion of adults aged 22 to 55 who are risk neutral. You can be 95% confident that the interval estimate LCL = _____ to UCL = ______ includes the population proportion p, the proportion of adults aged 22 to 55 who are risk neutral.

Explanation / Answer

p cap = proportion who are risk neutral = 0.022

1 - p cap = 0.9778

You can conclude that the sampling distribution of pcap can be approximated by a normal distribution because no. of observations is greater than 30.

Mean = 0.022

Standard deviaiton = sqrt( 0.22 * 0.9778/1536)=0.0118

for 95%, z=1.96

LCL = 0.022-0.0118*1.96=-0.0011

UCL = 0.022+0.0118*1.96=0.045

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