Question

7. 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?

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 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-person sample, 34 people are risk neutral. Let p denote the proportion of the adult population aged 22 to 55 who are 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 p̄ of adults in the sample who are risk neutral is:

• 0.65989
• 0.97786
• 0.02214
• 0.34011

. The proportion 1 – p̄ of adults in the sample who are not risk neutral is:

• 0.97786
• 0.02214
• 0.34011
• 0.65989

conclude that the sampling distribution of p̄ can be approximated by a normal distribution, because:

• np̄p̄ ≥ 5 and n(1 – p̄p̄ ) ≥ 5
• np̄p̄ ≥ 2 and n(1 – p̄p̄ ) ≥ 2
• np̄p̄ < 2 and n(1 – p̄p̄ ) < 2
• np̄p̄ ≥ 4 and n(1 – p̄p̄ ) ≥ 4

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

• 0.01209
• 0.00375
• 0.02160
• 0.00060

Use the Distributions tool to develop a 90% confidence interval estimate of the proportion of adults aged 22 to 55 who are risk neutral.

You can be 90% confident that the interval estimate:

• 0.016
• 0.000
• 0.017
• 0.021

to:

• 0.058
• 0.023
• 0.028
• 0.027

includes the population proportion p, the proportion of adults aged 22 to 55 who are risk neutral.

Answer 1

1)

sample proportion p̂ =x/n=

0.02214

2)

The proportion 1 – p̄ of adults in the sample who are not risk neutral is =1-0.02214 =0.97786

3)

np̄≥ 5 and n(1 – p̄ ) ≥ 5

4)

std deviation se= √(p*(1-p)/n) =

0.00375

5)

for 90 % CI value of z=		1.64
margin of error E=z*std error   =		0.0062
lower bound=p̂ -E                       =		0.0160
Upper bound=p̂ +E                     =		0.0283

0.016 to 0.028