Question

hink 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 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.

The proportion p̄p̄ of adults in the sample who are risk averse is___________ . The proportion 1 – p̄p̄ of adults in the sample who are not risk averse is ________ .

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

The sampling distribution of p̄p̄ has an estimated standard deviation of_______ .

0123Standard Normalt Distribution

Select a Distribution

Use the Distributions tool to develop a 95% confidence interval estimate of the proportion of adults aged 22 to 55 who are risk averse.

You can be 95% confident that the interval estimate _______ to ________includes the population proportion p, the proportion of adults aged 22 to 55 who are risk averse.

Answer 1

The proportion p̄p̄ of adults in the sample who are risk averse is 1482/1536=0.96484

The proportion 1 – p̄p̄ of adults in the sample who are not risk averse is =1-0.96484 =0.03516

You can conclude that the sampling distribution of p̄p̄ can be approximated by a normal distribution, because . np̄>=5 and n(1-p̄) >=5

estimated standard deviation = se= √(p*(1-p)/n) =

0.00470

for 95 % CI value of z=		1.96
margin of error E=z*std error   =		0.0092
lower bound=p̂ -E                       =		0.9556
Upper bound=p̂ +E                     =		0.9741

You can be 95% confident that the interval estimate 0.956 to 0.974 includes the population proportion p, the proportion of adults aged 22 to 55 who are risk averse.